Naked Science Forum
On the Lighter Side => New Theories => Topic started by: Malamute Lover on 31/07/2020 17:46:57
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[Mod note: This topic was split from the "Is angular momentum frame dependent?" thread,
https://www.thenakedscientists.com/forum/index.php?topic=80177.msg611143 ]
Halc,
that's the answer I expected.
Mamalute Lover,
if you deny Bell's spaceship 'paradox' then you are going against Einstein and his original 1905 paper.
He starts his paper with a definition of simultaneity:
http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Einstein_1905_relativity.pdf
If you deny the simultaneity then you have nothing to back up your statements.
You cannot argue this is like this because the relativity says so. You denied the relativity.
Do you have your own math? Do you have your own hypothesis?
Jano
You failed to quote the important part of Einstein’s paper.
“So we see that we cannot attach any absolute signification to the concept of simultaneity, but that two events which, viewed from a system of co-ordinates, are simultaneous, can no longer be looked upon as simultaneous events when envisaged from a system which is in motion relatively to that system.”
Are the two space ships in motion relative to each other? In fact, yes.
Assume they start off with synchronized clocks and turn on their engines at an agreed time and begin to accelerate.
Spaceship 1 (in the lead) sees Spaceship 2 start up its engine a little bit later because it took a little time for the light from SP2 to reach SP1. As a result, as long as the engines are on SP1 will always see SP2 going slower, having started acceleration later. This implies that the distance between them is getting longer. A string connecting them should be under increasing tension.
However, SP2 sees SP1 start up its engine later than SP2 because of the lightspeed delay. That is, SP2 will always see SP1 going slower because SP2 started its engine first. The implies that the distance between them is getting shorter. A string connecting them should go increasingly slack.
The two spaceships are not in the same reference frame and their clocks are not synchronized. They will not agree on what they see.
Now what about that string?
SP1 thinks the string is under tension. That means that a force is being transmitted down the string at the speed of sound in the string, lightspeed at most. Assuming identical strength for the string all along its length, the string will not snap until the force wave hits the end of the string and cannot be passed on to the next segment.
SP2 thinks the string is slack, and will not snap at all. We might imagine SP1’s force wave meeting SP2’s slack somewhere in the middle, resulting in a string with neither slack nor tension. :D But that is assigning absolute reality to perceptions.
How can SP1 and SP2 agree on what is happening to the string? By turning off their engines at an agreed time by their originally synchronized clocks and getting back into an inertial frame of reference. SP1 will see SP2 turn off its engine a little late and catch up in speed, restoring the distance gap. SP2 will also see SP1 turn off its engines a little late and take up the slack. Once they are in a common inertial frame of reference again, they will discover that they are separated by the original length of the string, which suffered no harm.
A side note: SP1 thinks SP2 is going slower and therefore has a faster clock. SP2 thinks SP1 is going slower and has a faster clock. The perceived time difference between engine shutoffs will be the same on each side.
But what about Lorentz contraction on the string?
If SP1 and SP2 each look straight along the length of the string, they will be unable to judge its length. If each looks at the string from enough to the side to judge its length, they will observe the string to be curved because of the aforementioned perceived speed differences and the time it takes for the light to reach them from different points along the length of the string.
Once they are in a common inertial frame of reference, an outside observer in inertial motion relative to them will observe Lorentz contraction of the entire system, SP1 and SP2 and the string. No broken string, no slack string.
Lorentz contraction is relative. It is not real.
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Mamalute Lover,
as Halc said, the spaceships are stationary to each other in the Bell's spaceship paradox prior to the acceleration.
They are in the same inertial reference frame, their clocks can be synchronized.
This is the problem in physics.
Over a hundred years after the SR paper and the relativists do not agree on the Lorentz contraction.
I see your point, and I can show that the SR is reciprocal, the Lorentz contraction cannot be real and therefore self-contradictory.
Having said that the current 'more winning' view is that the Lorentz contraction is real, as per the Bell's spaceship paradox,
Jano
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Mamalute Lover,
as Halc said, the spaceships are stationary to each other in the Bell's spaceship paradox prior to the acceleration.
They are in the same inertial reference frame, their clocks can be synchronized.
As I demonstrated, the mere fact of being separated in space means that they are not synchronized. Each sees the other start the engine at a different time even though each turn on their engine when the alarm clock goes off. Each continues to see the other as being slower until after the engines are turned off when the second alarm goes off, which is observed to be at different times by each of them. The ‘catch up’ is observed by both sides and they again agree that they are motionless to each other. The clocks are not synchronized.
This is the problem in physics.
Over a hundred years after the SR paper and the relativists do not agree on the Lorentz contraction.
I see your point, and I can show that the SR is reciprocal, the Lorentz contraction cannot be real and therefore self-contradictory.
Having said that the current 'more winning' view is that the Lorentz contraction is real, as per the Bell's spaceship paradox,
Jano
Assuming synchronization of acceleration by some means so that the spaceships are always motionless with respect to each other, there is no Lorentz contraction and the string will not break. To an outside observer, the two ships and the string all contract together and the string does not break. Again the problem arises from inconsistent application of Lorentz transformation to only some of the components of the system but not to all of them. It is the space between the two ships that is contracted, not just the string. No tension is put on the string. To the two spaceships moving at the same speed, there is no contraction. To the outside observer, it is all contracted.
Lorentz contraction is relative, not real.
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Of course, the spokes experience Lorentz contraction. They are moving too.
They're moving sideways, so they get thinner but no shorter. Hence my saying that the wheel pictured in post 2 needed to be manufactured already spinning since the spokes are too short to reach the stationary rim circumference.
Because they are in motion, and the light from the outer portions takes longer to reach the observer in the center, they appear curved.
Yes, they appear curved from some points of view. That doesn't make them actually curved in any frame.
Yes, the spokes get shorter as the rim contracts.
This contradicts SR theory, which says contraction is only in the direction of motion, and spoke velocity along its length is always zero.
That wheel could not have ever been stationary. Spinning it would have broken it since the rim would contract but the spoke lengths would not. That's why it needs to be manufactured spinning. The picture in post 29 doesn't have that problem since there's no rim to contract, just blocks that contract, leaving real gaps between them. The spokes there stay the same length.
Since the rim is moving along with the blocks, it will experience exactly the same Lorentz contraction as the blocks. No increasing gap sizes.
Now you're confusing the two examples again. There's no rim in the example with the blocks. Discuss one at a time please. Be clear as to which scenario (2 or 29) you are referring.
But as we have seen, the radius does shrink because the spokes also experience Lorentz contraction and that is visible to the observer in the center.
Again a direct contradiction to SR theory. The spokes do not contract in length, only in width. Hence scenario 29 shows the blocks spinning at constant radius, necessarily forming gaps as they contract.
Sorry, no fixed radius when the wheel is spinning. [/quote]Well this seems to be the point about which we disagree. Perhaps you can defend this, or more in particular, explain how SR theory allows a spoke moving sideways to contract in length. Look at the picture in post 2. That wheel on the right is contracted only in the direction of motion (horizontal) Vertically (perpendicular to the velocity vector pointing left), the wheel is not contracted at all. But you seem to assert that it must be.
Different example. Don't mix them.
The increasing gap size points to the blocks shrinking but the rim not shrinking.
There is no scenario with both a rim and blocks. You again are mixing scenarios, and right after I say not to do that. Picture 2 has a rim and no blocks. Same wheel in both pictures, no change in proper angular speed. Picture 29 has blocks and no rim. Right side spinning, left side not.
Detached objects following a circular track requires an additional force to keep them on track. That is, there must be something comparable to a rim to exert that inward force. This changes nothing. Whatever is keeping the blocks on track will be subject to the same Lorentz contraction.
We're seemingly talking about scenaro 29 here. The spokes have zero velocity along their length, so are not contracted at all, and radius is held at a constant. Again, this seems to be the same point you're getting wrong multiple times per post.
To quote a line in the opening paragraph of the wiki article on length contraction:
"Length contraction is only in the direction in which the body is travelling."
You seem to insist that the spokes, moving perpendicular to their length, should contract in length. This is just wrong.
If you want something that holds the blocks on track but is not itself moving, you are introducing all sorts of complications.
The spokes work fine since they don't contract in length. If that stresses you out, then a roller-coaster track also works nicely.
For one thing, from the viewpoint of the blocks, the whatever it is outside will be the thing that is moving and will therefore be Lorentz contracted and have a smaller circumference/radius.
Yes it will, but only locally. That's an interesting effect, since the rider in the 2-meter car occupies 2 meters of track (which has meter markings on it), and is adjacent to the cars in front and behind. As the speed picks up, the gaps form. At .877c, you can fit a whole new car between each of them, and the rider cannot help but notice this. And as you point out, he locally (in the momentary inertial frame of his car) measures 4 meters of contracted track under his 2 meter car. But the inertial frame of each car is different, and if the Lorentz transform is applied while moving to each car in turn, it all adds up to there being gaps. There's no asymmetry violation going on here since rotation is absolute, and between the track and the cars riding on them, it is very clear which of the two is spinning.
For another, the force holding the blocks in place will result in the outside influence thing being set in motion and the blocks slowing down. (Newton #3)
Fail on basic mechanics now. Really? You reach for the action/reaction law? The reaction to the block at the top is balanced by the block at the bottom. There's your reaction force. What, do you suggest a spinning wheel in free-fall is suddenly going to accelerate to one side due to some misinterpretation of Newton's 3rd law?
The spinning disk (grey)
The grey does not appear to be spinning. They're fixed in position and not length contracted in either picture. They seem to serve as a series of measuring sticks to compare with the contracting red blocks. All the red stuff spins, as depicted by the red arrow. There are no spokes, so the grey blocks perhaps serve as the roller coaster track keeping the blocks in their path. OK, there's one 'spoke' that also consists of 4 blocks that don't quite reach the edge blocks, and that spins with the rest of the red stuff. Notice the red blocks in that bar are not contracted in length at all. It should be contracted in width, near the tips at least, but the artist apparently declined to attempt to depict that.
and the red boxes on it will all undergo Lorentz contraction along the direction of motion. This will reduce the circumference and therefore the radius. No broken strings.
It would if the grey boxes went around with the red stuff, but the picture (and the accompanying text) does not indicate that. It shows uncontracted grey boxes and identical radius in both pictures.
OK, the text talks about a disk, but a rigid disk cannot be spun up to relativistic speeds without it breaking. Not sure where the picture was found. It's just a stock imgur thing.
If you photograph the whole thing from a distant point. You will see something different from what an observer at the center sees.
Agree, but that's like saying a picture of a train from the middle of it is very different than a picture from well to the side.
But it will still exhibit Lorentz contraction since it is moving with respect to that distant observer.
Only in width, because its motion is all in that direction. 4th time this post you've made that mistake. My point was that it will appear straight from that distant location. It will also appear full length. I mentioned Coriolis effect, but I think the curvature observed by the PoV on the rotating spoke is better described as light aberration.
I am denying that gaps form because the circumference is contracting along with the blocks. There is no frame in which the gaps do form.[/quote]Five.
There's an awful lot of websites describing the gaps forming. Maybe you should correct them all.
The one in post 29 is just a picture taken out of context. It doesn't explicitly mention spokes or track or another means of keeping the radius fixed. I agree that a solid ring with no spokes or disk will just shrink its radius as it spins, but scenario 29 isn't about a solid ring, but rather some detached objects travelling about a fixed circuit. That's what the Ehrenfest scenario is all about. A spinning rim is a very different scenario than what Ehrenfest describes.
As I explained in a prior post, the clocks are out of sync when the engines start because each space ship sees a delay in the start of the other space ship’s engine.
Then all clocks are out of sync because each appear slower than the local one due to light-travel delays. You seem to be completely unaware of what it means for two clocks to be in sync in a given inertial frame. Clue: It doesn't mean that all clocks appear to read the same time from any given vantage point.
Really, is this the stance you're going to take? Denial that the ships can devise a way to depart simultaneously in the frame in which they're both stationary? It seems so:
It is not possible to start the engines at the same time.
:o
No cop outs going on. I understand the scenarios perfectly.
Pony up the numbers or shut up then. Assume, lacking intelligent clock operators, that the ships just happen by accident to set their clocks to zero and depart simultaneously in the frame in which they were stationary.
As I described in the previous post, even starting at the same prearranged time on previously synchronized clocks, each spaceship thinks the other one is going slower with opposite expectations.
Irrelevant since both has a schedule. Hit the gas at time zero and maintain say 1g of proper acceleration for amount of time T, then shut it off. No need to look at the other ship to follow that plan. Furthermore, the front ship trails a rigid rod behind it with meter markings that go right by the window of S2 following, so when S2 shuts off his engines, he as but to look out his window to see the distance between the two ships. OK, infinite rigidity is not plausible, so he'll wait for the bouncing to stop before taking the reading. It should stabilize somewhere because both ships having done the same thing, they'll be going the same speed in any frame after both have shut off the engines.
OK, I've gone that far, so some numbers.
Ships are initially 300,000 km apart (about a light second).
They accelerate at 1g proper acceleration, since we don't want to break our rod with insane g forces. Accelerate at 1g for 388 days, 19:49 hours (ship clock), which should get them up to a delta-V of 0.8c if I did that correctly. Once both ships have shut off their engines, they should both be stationary in the same new inertial frame, although your posts seem to deny even that. Feel free to give the actual values if you disagree.
My question is: Once the measuring rod stops its boinging around, what distance does S2 see outside his window? The rod is right there so he doesn't need to worry about speed of light to compute it. Of course if the two ships are not stationary relative to each other, then the numbers will be just flying by and there's no fixed separation, in which you just need to tell me the relative speed between the two.
Even if you manage to get them agreeing that they are both going at the same speed, the two spaceships and the string would all be in a common inertial reference frame and they would see no Lorentz contraction and no broken string.
So you assert the S2 guy will see 300,000 km outside his window then? Show me the math behind that assertion, because SR says he'll see 500,000 km mark on the rod.
As I pointed out in my prior post, an observer in a different inertial reference frame would see the entire complex of ships and strings equally Lorentz contracted and no broken string.
If it was a string, stretching a 300,000 km string to 500,000 km would break it. Please show the math behind your intact string, proving wrong all the websites that say otherwise.
If a string was tied inside a spaceship along the direction of travel, and the spaceship accelerated (slowly) to a high speed, would the string break?
Nope. You hopefully know that the ship and string contract together, but in that case the acceleration of the front of the ship is lower than the acceleration of the rear, and the front accelerates for a longer time (measured by a pair of clocks at each end of the ship), which is not the case with the two separate ships tied with a string.
Contraction is real, not just relative. Kryptid points this out.
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Are the two space ships in motion relative to each other? In fact, yes.
That answer is frame dependent. Both ships have identical proper acceleration. That much is fact.
In the inertial frame in which their clocks are synced, their speeds must therefore be identical at all times, contradicting your statement above.
In any other inertial frame, one ship starts accelerating before the other, and thus always has a greater velocity (assuming positive acceleration) than the other. So which ship has the greater velocity is frame dependent.
The string breaking is objective, so that fact is not frame dependent, and must happen even in a frame where the rear ship starts accelerating first and thus always has a greater velocity than the front one.
“at all times”? The basic lesson of Special Relativity is that there is no universal ‘now’. And ‘first’ is observer dependent. SP2 says it started first. SP1 says it started first. Different observers viewing the experiment from different angles will each have their own take on which one was first.
There is no string breaking. To an outside observer, the space containing the string and the ships has contracted. It is not material objects that contract independent of space. It is space itself. The string never experiences any tension.
Assume they start off with synchronized clocks and turn on their engines at an agreed time and begin to accelerate.
Spaceship 1 (in the lead) sees Spaceship 2 start up its engine a little bit later because it took a little time for the light from SP2 to reach SP1. As a result, as long as the engines are on SP1 will always see SP2 going slower, having started acceleration later.
What each ship sees when looking at the other is something from the past and is scant evidence for what the other ship is doing now. The observer must then complicate the situation by computing the time the light took, and compensating for that much additional acceleration.
There is no absolute ‘now’. If you do not understand that then you do not understand even the fundamentals of Relativity Theory. There is no Newtonian Absolute Space or Absolute Time.
This implies that the distance between them is getting longer.
Not in the initial inertial frame it isn't. But yes, in either ship's momentary inertial frame, that distance is increasing, as evidenced by the string breaking.
A string connecting them should be under increasing tension.
In the initial inertial frame, there was no relative motion so that is irrelevant.
As I demonstrated SP1 thinks it is pulling away from SP2 because it started its engine first, but SP2 thinks it is gaining on SP1 because it started its engine first. You cannot say whether the distance is increasing or decreasing. When they get back into a common inertial frame, the distance is unchanged. To the outside observer, all the distances are contracted. And the string is never under tension.
Lorentz contraction is not objective. It is not real, it is relative.
However, SP2 sees SP1 start up its engine later than SP2 because of the lightspeed delay. That is, SP2 will always see SP1 going slower because SP2 started its engine first. The implies that the distance between them is getting shorter. A string connecting them should go increasingly slack.
Again you equivocate what one sees with what is. This is faulty reasoning. As I said above, in an inertial frame where both ships initially have negative velocity, S2 really does start his engines first and thus always has a greater velocity than S1. The string nevertheless breaks.
Part of the problem is trying to envision this with an absurd non-stretchy string. Physics demands that the string is stretchy, so no matter what material it is made of, the string in front of S2 is going to go completely slack when S2 starts accelerating. That part of the string simply cannot notice that S1 has also begun moving. So it stretches and breaks after a while.
What is, is the 4D Minkowski spacetime. There is no absolute space. There is no absolute time. There is only spacetime. Observers see different aspects of it according to relative speeds.
Now what about that string?
SP1 thinks the string is under tension.
By same reasoning, this is objective fact. There is tension on the string, else the string cannot begin accelerating. The string must immediately physically stretch (or immediately break if it is infinitely non-stretchy), even if the other end is attached to nothing.
By the same reasoning, the distance is decreasing, which is what SP2 sees. The string should not break because of the slack at SP2’s end.
The Lorentz contraction assigned to the string cannot be given objective reality.
That means that a force is being transmitted down the string at the speed of sound in the string, lightspeed at most. Assuming identical strength for the string all along its length, the string will not snap until the force wave hits the end of the string and cannot be passed on to the next segment.
The string breaks because it ends? That can't be right. The end of the string relieves the stress, not adds to it. Yes, I tow a sufficiently long string at 1g of proper acceleration, and it must eventually break if it is long enough, even if it's tied to nothing. About a light-year is the maximum, although it still takes time to break it, so I have to maintain that 1g of acceleration. Yes, the faster the speed of sound in the string, the longer the string can be before it breaks on its own.
It is right. The energy is transmitted along the string until there is no longer any place to send it. Then the string must stretch and if the force is sufficient, it will break. That is an ideal string. In the real world, the internal composition of the string will not be totally consistent along its length and the force will not be transmitted cleanly. It may break at a weak spot. But we are talking ideal here.
In any case, if you pull on a string by walking away from it and I am holding the other end and walking as fast as you, the string is not going to break.
All this is a bit of a side-track to the main issue. Suppose our ships are point-ships and have only 10 meters between them. No need to get silly with the distance between the ships. 1g, 10 meters. The string breaks if it stretches more than 2%. When does it break? Who cares? Point is, it will break.
Why will it break? If you want to ignore the clock synchronization problem, the entire system will accelerate at 1g. From inside the system there is no Lorentz contraction, since relative speeds never change. No tension. To an external observer, the Lorentz contraction will be the same everywhere in the system. No tension.
How can SP1 and SP2 agree on what is happening to the string?
The tension from S1 will eventually move (at speed of sound) to S2 and begin to take up the slack. The stress is obviously greatest back at S1, so the string, with unform strength along its length, is probably going to eventually break right where it ties to S1 when the resistance from S2 pully back on the no-longer slack string moves at speed of sound back up to S1.
But from the SP2 viewpoint, there never is any tension because the distance is decreasing and the string is always slack. The two viewpoints are artifacts of perceived speed differences and naturally would not agree on the degree of Lorentz contraction, which is after all what we are taking about. Once they get back in a common inertial frame they will see that the distance between them is the same as at the start and that the unbroken string is still there.
By turning off their engines at an agreed time by their originally synchronized clocks and getting back into an inertial frame of reference. SP1 will see SP2 turn off its engine a little late and catch up in speed, restoring the distance gap. SP2 will also see SP1 turn off its engines a little late and take up the slack. Once they are in a common inertial frame of reference again, they will discover that they are separated by the original length of the string, which suffered no harm.
Ow, Ow! No, the gap between them in this new frame will be much greater, which is why the broken string between them is gone. Back it up with mathematics, because this assertion is a guess and totally wrong.
When they start off in a common inertial frame at a given distance and have both accelerated at a constant rate in the same direction for the same amount of time and stop accelerating, they obviously will again be in a common inertial frame with the same separation. Why does this need mathematics?
But since you insist.
A and B start off motionless (common inertial frame) at a separation of 10 meters.
A starts his engine and accelerates at 1g for 10 seconds then turns off his engine, having reached a speed of 98 m/s and a distance of 490 meters at the engine off point, which is 500 meters from where B originally was.
B starts his engine when A does, according to synchronized clocks, and also accelerates at 1g for 10 seconds in the same direction then turns off his engine. He is at 490 meters from his starting point. They are now at the same speed - 98 m/s – in the same common inertial frame.
At the moment when both engines are turned off, A is 500 meters from where B started and B is 490 meters from where B started. They are 10 meters apart. Not only that but they were 10 meters apart the whole time because at every point they had been accelerating at the same rate for the same time and therefore had the same speed and the same distance covered.
Why is this hard?
You are assuming the string will break and building everything else around that. You even want to introduce the assumption that the string will break as evidence that it will.
A side note: SP1 thinks SP2 is going slower and therefore has a faster clock. SP2 thinks SP1 is going slower and has a faster clock.
Gee whiz, and I thought that the perspective of any observer is the one where that observer is stationary and thus any other observer by definition cannot be moving slower. You're thinking in absolute terms, talking about both ships moving apparently fast, (no reference) instead of just 'moving relative to me' which always makes the other guy's clock slower.
I typed that wrong. I meant that going slower meant a slower clock and going faster meant a faster clock. 40 lashes with a wet noodle. :(
This is straight line motion, not lateral, as is the case with the example spaceships.
Going slower than the other guy means increasing distance and a longer time between ticks on the observed clocks because the distance the light has to travel is getting longer. Going slower than the other guy means seeing the other guys clock tick slower. Going faster than the other guy, decreasing distance, means seeing the other guy’s clock tick faster clock because the other guy’s clock ticks will be seen at shorter intervals because the distance the light has to travel is decreasing. This is not relativity but simply a consequence of light having a finite speed.
The perceived time difference between engine shutoffs will be the same on each side.
Is that so? In what frame, since you seem to be omitting a lot of frame references for one suggesting that contraction is relative and not real. Again, this is totally wrong. You're ignoring relativity of simultaneity here. S2 will witness the shutoff of S1's engine at a much earler time (on S2's clock) than the time that S1 sees S2 shut his engine off.
On the contrary I have been talking about all sort of different reference frames. SP1, SP2 and an external observer. And the string itself.
Because SP2 thinks it started its engine first and kept it on for a specified time, SP2 will stop its engine before seeing SP1’s engine stop. Because SP1 thinks it started its engine first and kept it on for a specified time, SP1 will stop its engine before seeing SP2 stop its engine. If you think otherwise, let’s see the math.
But what about Lorentz contraction on the string?
The string is dangling behind S2, having broken off S1 some time ago.
The string never broke. Lorentz contraction is relative, not absolute. From the viewpoint of the string, it never changed length. From the viewpoint of an outside observer, everything contracted including the distance between the spaceships. No tension, no break.
Different observers will see a different degree of contraction. If as in your example, the string can stretch 2% before breaking, which observer is correct about when it will break?
If SP1 and SP2 each look straight along the length of the string, they will be unable to judge its length. If each looks at the string from enough to the side to judge its length, they will observe the string to be curved because of the aforementioned perceived speed differences and the time it takes for the light to reach them from different points along the length of the string.
Why would it curve to one side or the other? Are you envisioning ships going side-by-side in parallel? The Bell's scenario is one behind the other, in which there is no reason for the string to curve at all in the y or z directions. If stretchy enough, it remains absolutely straight between the ships, and the enwise view of that is a dot, so nobody can really see it. If it is stretchy, then its length is pretty irrelevant.
As should have been obvious from all of my examples, and as I think I explicitly stated, the two spaceships are in line not parallel or anything like that.
Looking directly along the length, it is not possible to determine the length. You have to look at it from an angle. The light takes longer to get to your eye from the far end and does not appear to come from the same direction as from closer sections. Although it is totally unnoticeable at low speeds, it is there and can be seen in computer simulations of relativistic speeds.
Once they are in a common inertial frame of reference, an outside observer in inertial motion relative to them will observe Lorentz contraction of the entire system, SP1 and SP2 and the string. No broken string, no slack string.
Another ouch, but you seem determined to defend this:
Lorentz contraction is relative. It is not real.
Back up all the above with some numbers so I can apologize.
Lorentz contraction calculations.
Courtesy of https://keisan.casio.com/exec/system/1224059837
Relative speed: 0 km/s
Perceived length: 1 m
Relative speed: 1,000 km/s
Perceived length: 0.99999443673425 m
Relative speed: 10,000 km/s
Perceived length: 0.99944352013705 m
Relative speed: 100,000 km/s
Perceived length: 0.94272742316888 m
Relative speed: 200,000 km/s
Perceived length: 0.74494293578673 m
Relative speed: 250,000 km/s
Perceived length: 0.55190009509556 m
Relative speed: 275,000 km/s
Perceived length: 0.39819391646464 m
Imagine a coil spring connected between two walls. In your scenario, the spring will be trying to contract but cannot because it is attached to the walls and in effect being pulled apart by them.
At what flyby speed will the spring be stretched to structural deformity and no longer be able to spring back?
The answer is that the relative speed of a flyby observer has no bearing whatsoever on the spring getting deformed. The spring will not be affected by flyby speed. Lorentz contraction will apply not only to the spring but to the distance between the walls. The spring will never be stretched.
Lorentz contraction is relative. It is not locally real. Everything at the same relative speed will be seen to be contracted by the same factor, that factor dependent on the relative speed of the observer. But locally, there is no contraction.
And that is all for a while. I need to go traveling for a bit. I will continue on and off when I have time to spare.
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Are the two space ships in motion relative to each other? In fact, yes.
That answer is frame dependent. Both ships have identical proper acceleration. That much is fact.
In the inertial frame in which their clocks are synced, their speeds must therefore be identical at all times, contradicting your statement above.
“at all times”? The basic lesson of Special Relativity is that there is no universal ‘now’.
Read what I wrote. I specified a frame, not a 'universal time'. You're the one making statements that leave off the frame references. Want examples?
And ‘first’ is observer dependent. SP2 says it started first. SP1 says it started first. Different observers viewing the experiment from different angles will each have their own take on which one was first.
There is no string breaking. To an outside observer, the space containing the string and the ships has contracted. It is not material objects that contract independent of space. It is space itself. The string never experiences any tension.
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As I demonstrated SP1 thinks it is pulling away from SP2 because it started its engine first, but SP2 thinks it is gaining on SP1 because it started its engine first. You cannot say whether the distance is increasing or decreasing. When they get back into a common inertial frame, the distance is unchanged. To the outside observer, all the distances are contracted. And the string is never under tension.
Lorentz contraction is not objective. It is not real, it is relative.
All repetition of prior posts, rather than addressing my replies to them.
There is no absolute ‘now’.
I never said there was. I give frame references. See bold above.
What is, is the 4D Minkowski spacetime. There is no absolute space. There is no absolute time. There is only spacetime. Observers see different aspects of it according to relative speeds.
That's an indistinct way of saying that observers order events in Minkowski spacetime by selection of an inertial frame. Our spaceship guys have done that, so all events are nicely ordered in that frame, and thus are entirely capable of starting at the same time relative to that selected frame. Do you actually deny this? The distance between them never changes in that inertial frame since their motion is always identical in that frame. But the length contracted string between them no longer reaches that constant separation distance in that frame, and so the string breaks.
All this is a bit of a side-track to the main issue. Suppose our ships are point-ships and have only 10 meters between them. No need to get silly with the distance between the ships. 1g, 10 meters. The string breaks if it stretches more than 2%. When does it break? Who cares? Point is, it will break.
I changed the scenario to a more realistic distance of 1 light second and 1g for about a year. Kindly compute numbers for that scenario, or chose a different specific one.
Why will it break? If you want to ignore the clock synchronization problem
There's no sync problem. Einstein in his 1905 paper gives a very specific sync convention that allows the two to sync their clocks relative to their inertial frame.
the entire system will accelerate at 1g.
No it will not. The system of choice is the inertial frame, and that stays put, not accelerating at all. Each ship accelerates at a proper 1g, but each ship is a system unto itself and in the accelerating reference frame of either ship, the other ship is not accelerating at 1g. So there is no system containing both ships that is accelerating at 1g. The rest of your argument seems to depend on this faulty assertion that seems to totally ignore relativity of simultaneity.
But from the SP2 viewpoint, there never is any tension because the distance is decreasing and the string is always slack.
Relative to SP2's accelerated reference frame, the distance between the ships is always increasing, so your assertion otherwise is wrong. Just because the light from the SP1's departure hasn't reached SP2 yet doesn't mean it hasn't done it. They synced their clocks in the initial frame. They both depart simultaneously in that frame, regardless of their inability to immediately see that event. You still seem to equate simultaneity with what one sees, which isn't how it works. If I see Betelgeuse go supernova tomorrow, I will know that it happened 640 years ago in Earth's frame. I don't assume that because I don't see it do that today that it hasn't already happened.
My specifiic example put the two ships a light second apart. If SP2 departs at midnight and one second later sees SP1 depart, then he knows they both departed at the same time. He doesn't even have to look. Both know the plan. All he has to do is look at the markings on that rod outside his window.
By turning off their engines at an agreed time by their originally synchronized clocks and getting back into an inertial frame of reference. SP1 will see SP2 turn off its engine a little late and catch up in speed, restoring the distance gap. SP2 will also see SP1 turn off its engines a little late and take up the slack. Once they are in a common inertial frame of reference again, they will discover that they are separated by the original length of the string, which suffered no harm.
Ow, Ow! No, the gap between them in this new frame will be much greater, which is why the broken string between them is gone. Back it up with mathematics, because this assertion is a guess and totally wrong.
When they start off in a common inertial frame at a given distance and have both accelerated at a constant rate in the same direction for the same amount of time and stop accelerating, they obviously will again be in a common inertial frame with the same separation.
Same separation in the original frame, but not in the new frame.
Why does this need mathematics?
Because you've not done the Lorentz transformation from the old frame to the new one. In the original frame I can specify the start events of both ships (say x=0, t=0 for SP1, x=-300,000, t=0 for SP2). I can give the coordinates of the two events where the ships cease acceleration, which is still 300,000 km apart in that frame, but a Lorentz transformation is needed to compute the coordinates of those events in the new frame, and the separation will be 500,000 km after we do that. I can compute the exact coordinates of the events if you like.
You thinking that there's nothing to compute seems to be a big problem.
A and B start off motionless (common inertial frame) at a separation of 10 meters.
OK, back to the 10 meter example. I can deal with that.
A starts his engine and accelerates at 1g for 10 seconds then turns off his engine, having reached a speed of 98 m/s and a distance of 490 meters at the engine off point, which is 500 meters from where B originally was.
There's going to be some dang trivial length contraction at a mere 98 m/s. Maybe we should do the faster example so we don't need to compute these things to 20 digits of precision to see the strain on the string.
One other problem is that your're computing 1g of acceleration for 10 seconds measured by a clock in the original frame. That's OK, but the scenario was specified as 1g of proper acceleration for some fixed ship time. It makes little difference at 98 m/sec, but it makes a lot of difference after enough time. 1g of proper acceleration can be maintained indefinitely and is not frame dependent, but 1g of acceleration is frame dependent and can only be maintained for a bit under 1 year before infinite power is required. My example had it keep up proper acceleration for a bit over a year.
B starts his engine when A does, according to synchronized clocks
Agree, but this is something you've been denying can be done. Now you admit it can?
and also accelerates at 1g for 10 seconds in the same direction then turns off his engine. He is at 490 meters from his starting point. They are now at the same speed - 98 m/s – in the same common inertial frame.
But not 10 meters apart in the new frame. You've done no Lorentz transformation to compute the coordinates in that new frame.
At the moment when both engines are turned off, A is 500 meters from where B started and B is 490 meters from where B started.
This statement assume absolute time. In fact it is only true relative to the original frame. Relative to the new frame, relativity of simultaneity says the two events when the engines are shut off are not simultaneous, so there's no 'moment when both engines are turned off' in that frame. Or do you also deny relativity of simultaneity? Tell me where I'm wrong in any of this.
They are 10 meters apart. Not only that but they were 10 meters apart the whole time because at every point they had been accelerating at the same rate for the same time and therefore had the same speed and the same distance covered.
They are 10 meters apart relative to the original frame, but in that frame the string is contracted and can't reach the full 10 meters. It breaks. In the new frame the ships are more than 10 meters apart, so the 10 meter non-contracted string can't span that gap, and it breaks. Mind you, the string misses the target by some distance that needs said 20 or more digits of precision to compute. The 98 m/sec speed is unreasonably low to illustrate relativity.
Why is this hard?
Why'd you think it was that easy? You're using Newtonian physics to do the computations.
This is straight line motion, not lateral, as is the case with the example spaceships.
The wheel was not linear motion. The ship example is straight 1 dimensional, yes. The ships are moving one behind the other, not on parallel paths.
Going slower than the other guy means increasing distance and a longer time between ticks on the observed clocks
The distance between them is quite frame dependent, regardless of what they see when they look at each other's clocks. The clock comparison is better done once the acceleration quits and the two clocks are once again at least ticking at the same rate, if not in sync in the new inertial frame.
Both engines shut off at the same time on the local clock, and since those clocks are not in sync in the new frame, the engine shutoff events are not simultaneous in that frame.
Going slower than the other guy means seeing the other guys clock tick slower.
If the other guy appears to be 'ticking slower', how does that make me the one 'going slower'? I think this notation is confusing. I'm not having either guy looking at the other ship. If you like, once both ships have stopped accelerating, each can send a light pulse to a mirror on the other ship and time the round trip of the pulse. That's a complicated way to measure the distance between them instead of using the simple method of looking at the rod with the meter marks outside the window.
But as you said, I'm assuming an answer when computing my numbers. So ignore what anybody sees and simply compute the coordinates of the events in question and use those coordinates to determine the separation between the ships. No need to have anybody peer at the other and try to glean information he already has.
Because SP2 thinks it started its engine first
No frame reference, making this sound like an absolute statement. And both observers very much know that both ships departed simultaneously relative to the original frame. Try again.
The string never broke. Lorentz contraction is relative, not absolute.
Assertions without numbers. The only numbers you've shown use Newtonian physics. Use SR physics, not just bumper sticker quotes from SR.
From the viewpoint of the string, it never changed length.
True, but from the viewpoint of the string, the ships are increasing their separation distance. So the string breaks.
From the viewpoint of an outside observer, everything contracted including the distance between the spaceships.
You computed otherwise above. Relative to the outside observer stationary in the original frame, the ships were at respective spatial locations 0, 10 at the start, and 490, 500 at the end. No contraction at all. So you're contradicting yourself.
Different observers will see a different degree of contraction. If as in your example, the string can stretch 2% before breaking, which observer is correct about when it will break?
The string breaking is objective, and if all facts are known (like how much strain the string can take before it breaks), then any observer will agree on the event of the string breaking, despite them each computing a different contraction. For instance, the space ship guys will see no contraction at all, just the ships slowly separating.
Looking directly along the length, it is not possible to determine the length.
I listed 3 ways to do it above. Yes, just peering down the length of it isn't going to help. Getting out a calculator is probably the best way, and since we're not physically in those ships, it is our only option here.
Lorentz contraction calculations.
Courtesy of https://keisan.casio.com/exec/system/1224059837
...
Relative speed: 275,000 km/s
Perceived length: 0.39819391646464 m
Cute, but you've not computed the distance between the two ships moving at 98 m/sec in their own frame. Do that (to more digits than your tool can seemingly handle). That or use an example where we don't need so bloody many digits such as the case I did where the separation of the ships (in the frame in which both ships are stationary) is 500,000 km after the shut down. You can use the word 'perceived' if it makes you happier. SP1 perceives SP2 to be 500,000 km away, and thus perceives that the 300,000 km uncontracted string is insufficient to reach it.
Imagine a coil spring connected between two walls. In your scenario, the spring will be trying to contract but cannot because it is attached to the walls and in effect being pulled apart by them.
At what flyby speed will the spring be stretched to structural deformity and no longer be able to spring back?
Depends on the change in motion of the walls of course. You've not described that. It is completely specified in the spaceship example.
The answer is that the relative speed of a flyby observer has no bearing whatsoever on the spring getting deformed.
True, but we're no looking at parked spaceships in several different inertial frames. We're accelerating the ships on very specific trajectories/worldlines.
By relative, I mean that it is observer dependent, which is how Einstein meant it.. Different observers are seeing different things. Which one is real?
I agree that contraction is relative, in the way that Einstein meant it. But that doesn't make it not real. All observers (well, at least the ones that know their physics) will predict the string breaking and the gaps forming between the bumper cars on the track. Those objective effects are what I mean by length contraction being real and not just observer dependent.
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A note before I hit the road again.
The gaps between blocks are said to appear because the radius of the circle is taken to be constant. This is because there is no motion along the length of the radial spokes and therefore no contraction of the spokes. As a result, the rotating rim will experience length contraction but be unable to reduce the circumference. Gaps appear to account for the contraction. That is the contention presented.
It is my contention that the spokes will maintain their length but will curve due to differing time dilation along the length. The fixed length of the spokes but bent into a curve will result in a reduced radius and the rim holding the blocks will contract as expected. No gaps will appear.
First some comments about time dilation. The pilot of a rocket ship that has accelerated to 0.99 c will experience a time dilation factor of about 7. The clock on the ship will be running 7 times slower than before the acceleration. That is, watching landmarks go by, he sees himself as traveling at almost 7 c. (Either that or he thinks the distances have shrunk.) But after several years of sightseeing, he decelerates and gets back home again, and finds that his twin brother has aged much more than him. He realizes that acceleration really did slow his clock and that he was not traveling as fast as he thought. He was really traveling much slower.
Unlike time dilation due to different relative inertial speeds, time dilation resulting from acceleration is real and has real consequences. This is the resolution of the so-called Twin Paradox.
In the case of the spinning wheel, different portions of each spoke are traveling at different speeds. The time dilation factor increases as one travels up the spoke from the center. A clock carried up from the axle to a certain point and back down again will have recorded less time than a clock that stayed at the center. The further up the clock is taken, the greater the discrepancy when brought back.
When the wheel was spun up, the different parts experienced different accelerations, and/or different centripetal accelerations as the wheel turns. This is acceleration-based time dilation and it is therefore real. At each level, the clock is slower than at the center and the real speed is less than an observer at that level thinks it is.
Disregarding relativistic effects, the speed at each level would of course be higher than at lower levels, in proportion to the radial distance. In the time dilation scenario, the speed at each level is still higher than at lower levels but no longer in exact proportion to radial distance. It has lagged behind some, the displacement from the radial line being a function of the Lorentz time dilation. If at some level the speed were 0.33 c, the angular displacement from the radial line would be 1 - 1/1.06 of a circle, or about 20 degrees.
Since the angular displacement would vary at each level, the spoke would curve. As has been pointed out, the spoke experiences no speed along its length and is therefore not Lorentz contracted. But because it is curved, it does not reach as far out radially as when it was stationary. That is, it follows a smaller circle now. Furthermore, as it is attached to the rim, the rim must be contracted to fit into the smaller circle, exactly as required by Lorentz contraction due to its speed and in exact proportion to the same Lorentz factor that has caused the spoke to be displaced that much at that radial distance.
There are no gaps. Lorentz contraction is relative, not objective.
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As I explained in a prior post, the clocks are out of sync when the engines start because each space ship sees a delay in the start of the other space ship’s engine.
Then all clocks are out of sync because each appear slower than the local one due to light-travel delays. You seem to be completely unaware of what it means for two clocks to be in sync in a given inertial frame. Clue: It doesn't mean that all clocks appear to read the same time from any given vantage point.
Really, is this the stance you're going to take? Denial that the ships can devise a way to depart simultaneously in the frame in which they're both stationary? It seems so:
You are ignoring light speed being finite. The two engines start at the same reading on their local clocks. But because of the lightspeed, it is not possible for two observers who are not in the same location to each see the same time on his own clock as he sees on the other clock. In the context of this example, each will see the other’s engine after his own. Why is this so difficult to understand?
If you think there is a way for the two ships separated in space to devise a way for each to see the other’s engine start at the same time as his own engine, please describe it.
No cop outs going on. I understand the scenarios perfectly.
Pony up the numbers or shut up then. Assume, lacking intelligent clock operators, that the ships just happen by accident to set their clocks to zero and depart simultaneously in the frame in which they were stationary.
Being in the same reference frame is insufficient. They must be coincident in space, or each will always see the other’s engine start after his own due to lightspeed delay.
You want math? I will use some of your numbers.
Two spaceships 300,000 km apart. Call it 1 light second. They previously synchronized their clocks at the midway point and proceeded at identical accelerations, speeds and decelerations in opposite directions to their starting points. Both agree at what time they will start their engines using the synchronized clocks, what acceleration they will use and when they will shut their engines off, and in what exact direction they will head. For simplicity the acceleration they will use will be 10 m/s, a touch over 1 g. All intervening variables – extraneous gravitational fields etc. – are hereby declared null and void.
SP1 starts his engine at the specified time. One second later he sees SP2’s engine start, when the light arrives. Actually, it will be a touch over 1 second because SP1 will have moved a little in that 1 second. Since the distances are so great, I will not explicitly mention that again. Assume the extra little bit is added to the numbers presented.
At the 1 second mark on his clock, SP1 is doing 10 m/s. At the 2 second mark, SP1 will be doing 20 m/s. The light that will reach him from SP2 will be from when SP2 was doing 10 m/s. Based on doppler shift readings in the light received from SP2, SP2 is going 10 m/s slower than SP1. Angular measurements made by SP1 of the light from SP2 show that the distance from SP1 to SP2 is 10 meters greater than at the start.
At the 10 second mark on SP1’s clock, SP1 will be doing 100 m/s. Observation of doppler shift in the light from SP2 shows that SP2 is doing 10 m/s slower, due to that 1 second delay at the start. From SP1’s viewpoint, SP2 has been going 10 m/s slower for 10 seconds and the distance back to SP2 is now 100 meters. Angular measurement will confirm this.
From SP1’s viewpoint, the distance between them is growing.
Now back to SP2.
SP2 starts his engine at the specified time. One second later he sees SP1’s engine start, when the light arrives. Again, a touch under 1 second because SP2 will have moved a little in that 1 second.
At the 1 second mark on his clock, SP2 is doing 10 m/s. At the 2 second mark, SP2 will be doing 20 m/s. The light that will reach him from SP1 will be from when SP1 was doing 10 m/s. Based on doppler shift readings in the light received from SP1, SP1 is going 10 m/s slower than SP1. Angular measurements made by SP2 of the light from SP1 show that the distance from SP2 to SP1 is 10 meters less than at the start.
At the 10 second mark on SP2’s clock, SP2 will be doing 100 m/s. Observation of doppler shift in the light from SP1 shows that SP1 is doing 10 m/s slower, due to that 1 second delay at the start. From SP2’s viewpoint, SP1 has been going 10 m/s slower for 10 seconds and the distance forward to SP1 is now 100 meters. Angular measurement will confirm this.
From SP2’s viewpoint, the distance between them is getting shorter.
They shut off their engines at the 10 second mark on their respective clocks.
SP1 sees SP2’s engine shut off 1 second later than his own because of the initial time delay. During that 1 second, SP1 sees SP2 catch up at the rate of 100 m/s, restoring the original distance.
SP2 sees SP1’s engine shut off 1 second later than his own because of the initial time delay. During that 1 second, SP2 will see SP1 pull away at 100 m/s, restoring the original distance.
Each will end up at the same relative speed and at the same original distance, despite disagreeing on what happened in between. Acceleration, even identical acceleration in the same direction, throws a monkey wrench in observations when there is any spatial separation. The reason for this is that there is no universal time and no universal space.
As I described in the previous post, even starting at the same prearranged time on previously synchronized clocks, each spaceship thinks the other one is going slower with opposite expectations.
Irrelevant since both has a schedule. Hit the gas at time zero and maintain say 1g of proper acceleration for amount of time T, then shut it off. No need to look at the other ship to follow that plan. Furthermore, the front ship trails a rigid rod behind it with meter markings that go right by the window of S2 following, so when S2 shuts off his engines, he as but to look out his window to see the distance between the two ships. OK, infinite rigidity is not plausible, so he'll wait for the bouncing to stop before taking the reading. It should stabilize somewhere because both ships having done the same thing, they'll be going the same speed in any frame after both have shut off the engines.
OK, I've gone that far, so some numbers.
Ships are initially 300,000 km apart (about a light second).
They accelerate at 1g proper acceleration, since we don't want to break our rod with insane g forces. Accelerate at 1g for 388 days, 19:49 hours (ship clock), which should get them up to a delta-V of 0.8c if I did that correctly. Once both ships have shut off their engines, they should both be stationary in the same new inertial frame, although your posts seem to deny even that. Feel free to give the actual values if you disagree.
My posts have the starting and ending separation distances identical. My point was that spatial separation and acceleration, even if identical, leads to non-intuitive results. Only inertial reference frames and minimal distances adhere to intuition.
My question is: Once the measuring rod stops its boinging around, what distance does S2 see outside his window? The rod is right there so he doesn't need to worry about speed of light to compute it. Of course if the two ships are not stationary relative to each other, then the numbers will be just flying by and there's no fixed separation, in which you just need to tell me the relative speed between the two.
Even if you manage to get them agreeing that they are both going at the same speed, the two spaceships and the string would all be in a common inertial reference frame and they would see no Lorentz contraction and no broken string.
So you assert the S2 guy will see 300,000 km outside his window then? Show me the math behind that assertion, because SR says he'll see 500,000 km mark on the rod.
Show me the math where the rod, which had the same starting and ending speeds and the same acceleration along the way has a different length contraction from the two ships. Once again, it is space itself that gets contracted as seen by an external observer, not material objects within a fixed space. SR tells us that different inertial reference frames see Minkowski spacetime from different angles.
As I pointed out in my prior post, an observer in a different inertial reference frame would see the entire complex of ships and strings equally Lorentz contracted and no broken string.
If it was a string, stretching a 300,000 km string to 500,000 km would break it. Please show the math behind your intact string, proving wrong all the websites that say otherwise.
The websites all assume that material objects contract within a fixed space. In SR, it is space itself that contracts as seen by an external observer, since he is seeing Minkowski spacetime from a different angle.
If you have math that shows otherwise, present it.
If a string was tied inside a spaceship along the direction of travel, and the spaceship accelerated (slowly) to a high speed, would the string break?
Nope. You hopefully know that the ship and string contract together, but in that case the acceleration of the front of the ship is lower than the acceleration of the rear, and the front accelerates for a longer time (measured by a pair of clocks at each end of the ship), which is not the case with the two separate ships tied with a string.
The ship has thrusters all along the sides. And the ship is very skinny and made of ultra-rigid material. Acceleration is the same everywhere. If you want to introduce real world engineering factors, then the string can be very stretchy and very strong. The idealized math goes out the window. Can’t mix and match here.
Contraction is real, not just relative. Kryptid points this out.
And I have shown Kryptid to be wrong,
Contraction is relative. If it were real then passengers on an ultrafast ship would see themselves compressed in the direction of travel and be squashed. This is contrary to SR, which has time dilation and energy changes deriving from the same equations that make all inertial reference frames look the same inside.
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You seem to be completely unaware of what it means for two clocks to be in sync in a given inertial frame. Clue: It doesn't mean that all clocks appear to read the same time from any given vantage point.
Really, is this the stance you're going to take? Denial that the ships can devise a way to depart simultaneously in the frame in which they're both stationary? It seems so
You are ignoring light speed being finite.
I make no mention of light speed in that. The discussion was about both ships leaving simultaneously in the frame in which they were stationary. They use their synced clocks to do this, and there are all sorts of ways to sync a pair of stationary clocks, many of which don't involve light at all.
The two engines start at the same reading on their local clocks. But because of the lightspeed, it is not possible for two observers who are not in the same location to each see the same time on his own clock as he sees on the other clock. In the context of this example, each will see the other’s engine after his own. Why is this so difficult to understand?
I never said otherwise. I just said that what either sees out of the window is irrelevant so long as both ships leave on schedule.
You want math? I will use some of your numbers.
Two spaceships 300,000 km apart. Call it 1 light second. They previously synchronized their clocks at the midway point and proceeded at identical accelerations, speeds and decelerations in opposite directions to their starting points.
OK, that's one way to do it.
Both agree at what time they will start their engines using the synchronized clocks, what acceleration they will use and when they will shut their engines off, and in what exact direction they will head.
That direction is +x. This is a simplified 1D scenario.
For simplicity the acceleration they will use will be 10 m/s, a touch over 1 g. All intervening variables – extraneous gravitational fields etc. – are hereby declared null and void.
I will have to redo my numbers, since I based the proper acceleration time on the time needed to get to .8c at 9.8 m/s². Can do. You seem to not plan to use proper acceleration.
At the 1 second mark on [SP1's] clock, SP1 is doing 10 m/s. At the 2 second mark, SP1 will be doing 20 m/s. The light that will reach him from SP2 will be from when SP2 was doing 10 m/s. Based on doppler shift readings in the light received from SP2, SP2 is going 10 m/s slower than SP1. Angular measurements made by SP1 of the light from SP2 show that the distance from SP1 to SP2 is 10 meters greater than at the start.
What are 'angular measurements'?
I really don't care what light is doing. I care where the ships are, and the distance between them as measured by a string trailing behind the lead ship. Neither ship needs to look at the other since all we're interested in is their separation, not what they'll see.
The part in bold is wrong. After 2 seconds, SP1 has moved 20 meters, and sees SP1 as having moved 5 meters after only 1 second. That's a difference of 15 (Newton physics), not 10. Same issue below.
At the 10 second mark on SP1’s clock, SP1 will be doing 100 m/s. Observation of doppler shift in the light from SP2 shows that SP2 is doing 10 m/s slower, due to that 1 second delay at the start. From SP1’s viewpoint, SP2 has been going 10 m/s slower for 10 seconds and the distance back to SP2 is now 100 meters. Angular measurement will confirm this.
From SP1’s viewpoint, the distance between them is growing.
Now back to SP2.
SP2 starts his engine at the specified time. One second later he sees SP1’s engine start, when the light arrives. Again, a touch under 1 second because SP2 will have moved a little in that 1 second.
At the 1 second mark on his clock, SP2 is doing 10 m/s. At the 2 second mark, SP2 will be doing 20 m/s. The light that will reach him from SP1 will be from when SP1 was doing 10 m/s. Based on doppler shift readings in the light received from SP1, SP1 is going 10 m/s slower than SP1. Angular measurements made by SP2 of the light from SP1 show that the distance from SP2 to SP1 is 10 meters less than at the start.
At the 10 second mark on SP2’s clock, SP2 will be doing 100 m/s. Observation of doppler shift in the light from SP1 shows that SP1 is doing 10 m/s slower, due to that 1 second delay at the start. From SP2’s viewpoint, SP1 has been going 10 m/s slower for 10 seconds and the distance forward to SP1 is now 100 meters. Angular measurement will confirm this.
You've not invoked relativity theory at all. You're adding velocities the Newtonian way, and you're using the original rest frame to compute what each observer sees, not the frame of the observer. The inconsistencies will pile up once you move faster than a city bus.
They shut off their engines at the 10 second mark on their respective clocks.
Oh no they don't. This is a relativity exercise and any example that is to illustrate a point in relativity needs to get up to a speed far greater than 100 m/sec. Try again. I had them doing 1g for over a year.
Each will end up at the same relative speed and at the same original distance, despite disagreeing on what happened in between.
So says Newton. Einstein would disagree on the same-distance part of the story but you've made no attempt to compute that Einstein's way. They'll be stationary relative to each other, yes.
Acceleration, even identical acceleration in the same direction, throws a monkey wrench in observations when there is any spatial separation. The reason for this is that there is no universal time and no universal space.
But you've not worked out any of the monkey wrench part.
My posts have the starting and ending separation distances identical.
Their separation is frame dependent, so this statement is ambiguous at best, and otherwise just wrong. In the new inertial frame of the ships, the statement is wrong.
My point was that spatial separation and acceleration, even if identical, leads to non-intuitive results. Only inertial reference frames and minimal distances adhere to intuition.
Which is why I'm asking you to do the calculations and not just assert your intuitions.
Show me the math where the rod, which had the same starting and ending speeds and the same acceleration along the way has a different length contraction from the two ships.
It has identical length contraction as the ships. The lengths of the ships are unspecified, but if they were 5 mm at rest, they're 3 mm in length in a frame where they're moving at .8c. For the exercise, they're treated as point objects, else we'd have to specify which part of the ship is actually accelerating at 1g or 100g or whatever. It seems to be a needless complication, but if you insist, then the point at which the string is attached is the part that accelerates at the specified rate.
There are plenty of relativity deniers out there, but you don't assert it's wrong, you just seem to know almost nothing about it. Just a few bumper sticker slogans as I said.
The websites all assume that material objects contract within a fixed space.
1) Not an assumption. It follows directly from the empirical observation that light speed is a frame independent constant.
2) It is not specific to a material object. The distance between the sun and Betelgeuse is about 630 light years in the frame of the sun, but is about 90 light years in any frame where the sun is moving towards or away from Betelgeuse at .99c. This despite the lack of an object between the sun and Betelgeuse, but if there were a phone line strung between the two, that line would also be contracted relative to such a frame.
In SR, it is space itself that contracts as seen by an external observer, since he is seeing Minkowski spacetime from a different angle.
A different angle is a funny way of putting it, but rotating a coordinate system doesn't contract space, it just puts different abstract coordinates on unchanged objective events.
Yes, the difference is just a different choice of coordinate system. As you say, contraction is relative. But that doesn't make it not real since it is the real distance between a pair of worldlines relative to the selected orientation of coordinate system. The string example demonstrates this reality. You deny the string breaking, but that denial is due to a lack of understanding of the theory. No physicist who knows his stuff will suggest the string doesn't break in the scenario described. You can't back that up. You're bucking the accepted view, and claiming that you're not.
If a string was tied inside a spaceship along the direction of travel, and the spaceship accelerated (slowly) to a high speed, would the string break?
Nope. You hopefully know that the ship and string contract together, but in that case the acceleration of the front of the ship is lower than the acceleration of the rear, and the front accelerates for a longer time (measured by a pair of clocks at each end of the ship), which is not the case with the two separate ships tied with a string.
The ship has thrusters all along the sides. And the ship is very skinny and made of ultra-rigid material.
Very good. To reduce strain, thrust must be applied along the length the ship and the rod, not just at one point. No distortion that way. No need to worry about speed of sound.
Acceleration is the same everywhere.
It is not. Learn some physics about acceleration of rigid objects. I've some links:
https://physics.stackexchange.com/questions/172739/is-the-lay-explanation-of-the-equivalence-principle-wrong
https://physics.stackexchange.com/questions/112645/which-clock-is-the-fastest-inside-an-accelerating-body
Found one on quora as well, but quora seems bent on promoting some of the worst answers, so I'll not use them.
If you want to introduce real world engineering factors, then the string can be very stretchy and very strong. The idealized math goes out the window. Can’t mix and match here.
Let's stick with idealized math then. To satisfy the engineers, we (a) make it a rod attached only to SP1, and (b) outfit the rod with its own thrust as you describe so at no time is it under stress or strain. Of course those /thrusters would also have to be programmed with the plan so they start and stop at the proper time. I put SP2 close enough (at least in the 1g scenario) that such a thing can be done. In the other scenario, I put SP2 too far back to have a rod trailing being SP1 that reaches it, so we'll instead have to attach the rod to the front of SP2, allowing SP1 to take the measurement.
If it were real then passengers on an ultrafast ship would see themselves compressed in the direction of travel and be squashed.
That would contradict the principle of relativity. SR predicts no such thing. Passengers seeing such effects would falsify the first premise of SR, immediately invalidating the theory.
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You seem to be completely unaware of what it means for two clocks to be in sync in a given inertial frame. Clue: It doesn't mean that all clocks appear to read the same time from any given vantage point.
Really, is this the stance you're going to take? Denial that the ships can devise a way to depart simultaneously in the frame in which they're both stationary? It seems so
You are ignoring light speed being finite.
I make no mention of light speed in that. The discussion was about both ships leaving simultaneously in the frame in which they were stationary. They use their synced clocks to do this, and there are all sorts of ways to sync a pair of stationary clocks, many of which don't involve light at all.
I already provided a way of synchronizing the clocks
You need to take light speed into account. In the example, the two ship are 1 light second apart. They start their engines at a pre-established time on previously synchronized clocks and accelerate at matching rates in the same direction. Each will see the other start their engine one second later than they start their own engine. The first (lead) ship will always see light from the trailing ship one second later, that is one second slower, and will see the second ship fall behind. However, the second (trailing) ship will always see light from the first ship one second later and will see itself catching up with the lead ship.
Although I did not explicitly mention it, imagine that each ship is displaying a clock so that the other one can see it. Each will start off seeing the other as being one second behind. Since this (initially) matches the apparent increase or decrease of separation, each one can consider that as evidence that their clocks are not synchronized after all.
But it gets worse. Since SP1 sees itself pulling ahead of SP2, the delay in SP2’s clock is increasing as it takes longer and longer for the light from SP2’s clock to reach SP1. SP2’s clock is not just offset, it is slow. This is independent of relativity, it is simply a question of finite light speed. OTOH, SP2 sees itself as gaining on SP1 and sees SP1’s clock as running fast because SP2 is catching up with the light from SP1. SP1 thinking SP2 has a slow clock and SP2 thinking SP1 has a fast clock might seem to cancel. But they are heading in opposite directions and will not converge, until after both have stopped accelerating.
Spatial separation rules out giving meaning to synchronized clocks if acceleration is involved.
Another interesting effect happens because each sees the light from different parts of the string between them at different times, ranging from zero to 1 second. The light from the far end of the string will be delayed by 1 second (initially). The light from the nearest part is not delayed at all. Because of the delay and the apparent increasing gap, the first ship will see the string get longer but not break. And because of the apparent decreasing gap, the second ship will see the string get shorter but not go slack.
If you think you can provide some way whereby the two accelerating ships can each appear to the other to be at a constant speed and distance relative to each other even if they are, please provide it. What is each ship supposed to do to accomplish this?
The two engines start at the same reading on their local clocks. But because of the lightspeed, it is not possible for two observers who are not in the same location to each see the same time on his own clock as he sees on the other clock. In the context of this example, each will see the other’s engine after his own. Why is this so difficult to understand?
I never said otherwise. I just said that what either sees out of the window is irrelevant so long as both ships leave on schedule.
The point of my exercise was that at the end of the story, the two ships are found to be in a common inertial frame at their original separation and with an unbroken string between them despite their interpretation of what they are seeing.
You want math? I will use some of your numbers.
Two spaceships 300,000 km apart. Call it 1 light second. They previously synchronized their clocks at the midway point and proceeded at identical accelerations, speeds and decelerations in opposite directions to their starting points.
OK, that's one way to do it.
Actually describing the experimental setup is very important in thought experiments.
Both agree at what time they will start their engines using the synchronized clocks, what acceleration they will use and when they will shut their engines off, and in what exact direction they will head.
That direction is +x. This is a simplified 1D scenario.
For simplicity the acceleration they will use will be 10 m/s, a touch over 1 g. All intervening variables – extraneous gravitational fields etc. – are hereby declared null and void.
I will have to redo my numbers, since I based the proper acceleration time on the time needed to get to .8c at 9.8 m/s². Can do. You seem to not plan to use proper acceleration.
Each should experience identical proper acceleration regardless of how long they accelerate or what speed they reach relative to their original inertial frame. No problem with using proper acceleration. It can often make calculations simpler as long as you keep in mind what an outside observer in some specific reference frame will see.
At the 1 second mark on [SP1's] clock, SP1 is doing 10 m/s. At the 2 second mark, SP1 will be doing 20 m/s. The light that will reach him from SP2 will be from when SP2 was doing 10 m/s. Based on doppler shift readings in the light received from SP2, SP2 is going 10 m/s slower than SP1. Angular measurements made by SP1 of the light from SP2 show that the distance from SP1 to SP2 is 10 meters greater than at the start.
What are 'angular measurements'?
Measurement of the angle that the width of the other ship subtends. With a known width, this can be used to determine distance. Width is not affected by Lorentz contraction.
I really don't care what light is doing. I care where the ships are, and the distance between them as measured by a string trailing behind the lead ship. Neither ship needs to look at the other since all we're interested in is their separation, not what they'll see.
The part in bold is wrong. After 2 seconds, SP1 has moved 20 meters, and sees SP1 as having moved 5 meters after only 1 second. That's a difference of 15 (Newton physics), not 10. Same issue below.
Got it. I mixed myself up talking about doppler shift determined speed difference and typed that number instead of the correct difference in distance.
At the 10 second mark on SP1’s clock, SP1 will be doing 100 m/s. Observation of doppler shift in the light from SP2 shows that SP2 is doing 10 m/s slower, due to that 1 second delay at the start. From SP1’s viewpoint, SP2 has been going 10 m/s slower for 10 seconds and the distance back to SP2 is now 100 meters. Angular measurement will confirm this.
From SP1’s viewpoint, the distance between them is growing.
Now back to SP2.
SP2 starts his engine at the specified time. One second later he sees SP1’s engine start, when the light arrives. Again, a touch under 1 second because SP2 will have moved a little in that 1 second.
At the 1 second mark on his clock, SP2 is doing 10 m/s. At the 2 second mark, SP2 will be doing 20 m/s. The light that will reach him from SP1 will be from when SP1 was doing 10 m/s. Based on doppler shift readings in the light received from SP1, SP1 is going 10 m/s slower than SP1. Angular measurements made by SP2 of the light from SP1 show that the distance from SP2 to SP1 is 10 meters less than at the start.
At the 10 second mark on SP2’s clock, SP2 will be doing 100 m/s. Observation of doppler shift in the light from SP1 shows that SP1 is doing 10 m/s slower, due to that 1 second delay at the start. From SP2’s viewpoint, SP1 has been going 10 m/s slower for 10 seconds and the distance forward to SP1 is now 100 meters. Angular measurement will confirm this.
You've not invoked relativity theory at all. You're adding velocities the Newtonian way, and you're using the original rest frame to compute what each observer sees, not the frame of the observer. The inconsistencies will pile up once you move faster than a city bus.
That is right, I did not incorporate relativistic considerations. My point was that an apparent speed difference is not a necessarily a real difference if the clocks do not correspond.
So let’s see some relativity math that does not depend on illusory difference in distance.
They shut off their engines at the 10 second mark on their respective clocks.
Oh no they don't. This is a relativity exercise and any example that is to illustrate a point in relativity needs to get up to a speed far greater than 100 m/sec. Try again. I had them doing 1g for over a year.
Let’s see the math.
Each will end up at the same relative speed and at the same original distance, despite disagreeing on what happened in between.
So says Newton. Einstein would disagree on the same-distance part of the story but you've made no attempt to compute that Einstein's way. They'll be stationary relative to each other, yes.
What do you expect to be different? The distance? Show the math.
Acceleration, even identical acceleration in the same direction, throws a monkey wrench in observations when there is any spatial separation. The reason for this is that there is no universal time and no universal space.
But you've not worked out any of the monkey wrench part.
On the contrary, I showed that what appears to be the separation is not necessarily the actual separation. Monkey wrench time.
My posts have the starting and ending separation distances identical.
Their separation is frame dependent, so this statement is ambiguous at best, and otherwise just wrong. In the new inertial frame of the ships, the statement is wrong.
The separation as measured by the measuring rods they brought along is the same at the end as at the beginning. Why would that not be the case?
Since the acceleration rate and duration are the same for the two ships, why would there be any difference in the shape of the worldline path each follows? That is like saying that calculating the worldline path for one ship and then calculating it again will give a different result. I really have to see this math and how you get around the two being simply being Poincaré translations at every point.
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My point was that spatial separation and acceleration, even if identical, leads to non-intuitive results. Only inertial reference frames and minimal distances adhere to intuition.
Which is why I'm asking you to do the calculations and not just assert your intuitions.
SP1, string, SP2 all experience identical acceleration beginning at time zero on synchronized clocks.
Lorentz contraction factors at various speeds, as seen by an observer in a common inertial frame with SP1, string, SP2 at the start.
Speed: 0.0 c
Lorentz factor
SP1: 1
Str: 1
SP2: 1
Speed: 0.5 c
Lorentz factor
SP1: .866
Str: .866
SP2: .866
Speed: 0.8 c
Lorentz factor
SP1: .6
Str: .6
SP2: .6
Speed: 0.95 c
Lorentz factor
SP1: .31
Str: .31
SP2: .31
Now what about the Lorentz contraction as seen by an observer sitting on the string? It is 1 for all three components. Observer and all parts are motionless with respect to each other.
Now let’s see your math.
Show me the math where the rod, which had the same starting and ending speeds and the same acceleration along the way has a different length contraction from the two ships.
It has identical length contraction as the ships. The lengths of the ships are unspecified, but if they were 5 mm at rest, they're 3 mm in length in a frame where they're moving at .8c. For the exercise, they're treated as point objects, else we'd have to specify which part of the ship is actually accelerating at 1g or 100g or whatever. It seems to be a needless complication, but if you insist, then the point at which the string is attached is the part that accelerates at the specified rate.
There are plenty of relativity deniers out there, but you don't assert it's wrong, you just seem to know almost nothing about it. Just a few bumper sticker slogans as I said.
Total hogwash. I know Special Relativity inside and out, and General Relativity very well indeed as I have demonstrated repeatedly. If you think I do not, show me – I repeat SHOW ME – where I have been wrong. In detail. OTOH it is very clear that you have trouble with even the basics.
The points and string ensemble are 0.6 of L0 at a relative speed of 0.8 c and 0.866 of L0 at a relative speed of 0.5 c and – here is the big one – 1.0 of L0 at a relative speed of 0.0 c, which they are to each other and to the string. Which is objective? None of them. They are all relative. This is why it is called Relativity Theory.
The websites all assume that material objects contract within a fixed space.
1) Not an assumption. It follows directly from the empirical observation that light speed is a frame independent constant.
Which means that the distance from the front to the back of the spaceship will be the old familiar length to those on board but not the same to observers in different frames. And observers in different frames will disagree on the degree of contraction.
Lorentz contraction is relative, not objective.
2) It is not specific to a material object. The distance between the sun and Betelgeuse is about 630 light years in the frame of the sun, but is about 90 light years in any frame where the sun is moving towards or away from Betelgeuse at .99c. This despite the lack of an object between the sun and Betelgeuse, but {/if there were a phone line strung between the two, that line would also be contracted relative to such a frame.
Incorrect. If two objects are stationary relative to each other in any inertial frame of reference, they are stationary relative to each other in any other inertial frame of reference. Observers in different inertial reference frames can disagree about the separation but none will see a change in the separation.
Lorentz contraction is relative, not objective.
In SR, it is space itself that contracts as seen by an external observer, since he is seeing Minkowski spacetime from a different angle.
A different angle is a funny way of putting it, but rotating a coordinate system doesn't contract space, it just puts different abstract coordinates on unchanged objective events.
Yes, the difference is just a different choice of coordinate system. As you say, contraction is relative. But that doesn't make it not real since it is the real distance between a pair of worldlines relative to the selected orientation of coordinate system. The string example demonstrates this reality. You deny the string breaking, but that denial is due to a lack of understanding of the theory. No physicist who knows his stuff will suggest the string doesn't break in the scenario described. You can't back that up. You're bucking the accepted view, and claiming that you're not.
I see you do not understand Minkowski spacetime either. The time dimension is represented by imaginary numbers, the three spatial dimensions by real numbers. Observations made from different reference frames (polar coordinate systems) accounts for the various relativistic effects. But the spacetime interval, the distance between two points in spacetime, is always invariant no matter how one transforms coordinates. Changing the coordinate system cannot create different results for different observers, only different observations.
However, if you do understand Minkowski spacetime, perhaps you can explain why Poincaré symmetries may be non-Abelian.
I am backing my extensive understanding of Relativity Theory and the consequences of how it works. I am not saying that the string does not break because that is the accepted view. I am saying it because I understand relativity Theory. But it seems you have a problem with that. Unless you can demonstrate the math behind your claim.
At any moment, the ships and the string are all moving at the same speed in the same direction. Any observer will see the same degree of Lorentz contraction for all three. Why should the string break? Stop claiming it is a demonstrated fact or the opinion of some others. Prove it.
If a string was tied inside a spaceship along the direction of travel, and the spaceship accelerated (slowly) to a high speed, would the string break?
Nope. You hopefully know that the ship and string contract together, but in that case the acceleration of the front of the ship is lower than the acceleration of the rear, and the front accelerates for a longer time (measured by a pair of clocks at each end of the ship), which is not the case with the two separate ships tied with a string.
The ship has thrusters all along the sides. And the ship is very skinny and made of ultra-rigid material.
Very good. To reduce strain, thrust must be applied along the length the ship and the rod, not just at one point. No distortion that way. No need to worry about speed of sound.
OK. At any moment all three are moving at the same speed in the same direction. So why does the string break?
Acceleration is the same everywhere.
It is not. Learn some physics about acceleration of rigid objects. I've some links:
https://physics.stackexchange.com/questions/172739/is-the-lay-explanation-of-the-equivalence-principle-wrong
https://physics.stackexchange.com/questions/112645/which-clock-is-the-fastest-inside-an-accelerating-body
Found one on quora as well, but quora seems bent on promoting some of the worst answers, so I'll not use them.
We have already eliminated the difference in acceleration due to mechanical limitations, so I do not see how any of those answers are relevant. If you want to have the acceleration force travel at a finite speed from one end of the ship to the other due to the speed of sound in the material, then you have to let the acceleration force from the lead ship travel at a finite speed from one end of the string to the back end. The string does not know it is tied at the back end until the force reaches there and by that time, the second ship will have accelerated enough to give it some slack. In the general case, the string does not break. You can rig the parameters – very long ships with low speed of sound, very short string with high speed of sound – but with reasonable assumptions no breakage, especially if you consider that string can stretch some. Low speed of sound in material suitable for a spaceship sounds rather unlikely, especially since that would lead to significant compression as the acceleration kept up.
They made sure I knew something about engineering as well as physics. You should learn some instead of copying from websites.
In any case, what does relativity have to do with any of this?
If you want to introduce real world engineering factors, then the string can be very stretchy and very strong. The idealized math goes out the window. Can’t mix and match here.
Let's stick with idealized math then. To satisfy the engineers, we (a) make it a rod attached only to SP1, and (b) outfit the rod with its own thrust as you describe so at no time is it under stress or strain. Of course those thrusters would also have to be programmed with the plan so they start and stop at the proper time. I put SP2 close enough (at least in the 1g scenario) that such a thing can be done. In the other scenario, I put SP2 too far back to have a rod trailing being SP1 that reaches it, so we'll instead have to attach the rod to the front of SP2, allowing SP1 to take the measurement.
No need to have the rod connected only at one end. If everything accelerates at the same time (previously synchronized clocks) and in the same degree, length contraction will be the same everywhere.
However, if you want differentials in accelerations for mechanical reasons, then you have to let the rod at the back end not feel acceleration from being pulled immediately. Until the energy wave reaches the back end, the rod does not know it is rigidly attached there and by then the second ship has already pushed it, sending an energy wave forward, canceling the one from the front end. Continued energy waves from each end as acceleration continues will continue to cancel each other. The rod experiences no net stress and therefore no strain.
If it were real then passengers on an ultrafast ship would see themselves compressed in the direction of travel and be squashed.
That would contradict the principle of relativity. SR predicts no such thing. Passengers seeing such effects would falsify the first premise of SR, immediately invalidating the theory.
Exactly. Contraction is relative, not objective.
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Malamute Lover,
here are a couple of diagrams:
(https://i.imgur.com/XSs3QXa.png)
(https://i.imgur.com/sdxp5gT.png)
This is a uniform acceleration as per the textbook.
You can see the length contraction of the ruler.
This discussion would be better in the SR reciprocal thread though,
Jano
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Malamute Lover,
here are a couple of diagrams:
(https://i.imgur.com/XSs3QXa.png)
(https://i.imgur.com/sdxp5gT.png)
This is a uniform acceleration as per the textbook.
You can see the length contraction of the ruler.
This discussion would be better in the SR reciprocal thread though,
Jano
What did you think this means?
It is showing the worldlines of a uniformly accelerated observer and an observer who is at fixed coordinates.
Did the term ‘comoving’ confuse you? It just means that the observer is moving with the expansion of the universe but that (for an observer in an inertial reference frame) the coordinates remain the same and that for another observer not in the same reference frame, the coordinate system used for the first observer remains in effect. Basically, it means to ignore the expansion of the universe. It simplifies the math by keeping things local and not in reference to any noticeably red-shifted distant object.
As expected, the two worldlines are different because the situations are different. But the worldlines for spaceships SP1 and SP2, uniformly accelerating at the same rate, the worldline shapes would be identical.
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Malamute Lover,
here are a couple of diagrams:
(https://i.imgur.com/XSs3QXa.png)
(https://i.imgur.com/sdxp5gT.png)
This is a uniform acceleration as per the textbook.
You can see the length contraction of the ruler.
This discussion would be better in the SR reciprocal thread though,
Jano
What did you think this means?
It is showing the worldlines of a uniformly accelerated observer and an observer who is at fixed coordinates.
Did the term ‘comoving’ confuse you? It just means that the observer is moving with the expansion of the universe but that (for an observer in an inertial reference frame) the coordinates remain the same and that for another observer not in the same reference frame, the coordinate system used for the first observer remains in effect. Basically, it means to ignore the expansion of the universe. It simplifies the math by keeping things local and not in reference to any noticeably red-shifted distant object.
As expected, the two worldlines are different because the situations are different. But the worldlines for spaceships SP1 and SP2, uniformly accelerating at the same rate, the worldline shapes would be identical.
So why the string between the spaceships would not try to shrink like the ruler does?
The end-result will be a broken string.
That's the logical conclusion when the spaceships worldlines are going to be identical,
Jano
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I already provided a way of synchronizing the clocks
Good. No need to elaborate. I only was reacting to your assertion that it couldn't be done.
Although I did not explicitly mention it, imagine that each ship is displaying a clock so that the other one can see it.
Can do that. I just didn't find it useful for either of them to look at the other.
Each will start off seeing the other as being one second behind. Since this (initially) matches the apparent increase or decrease of separation, each one can consider that as evidence that their clocks are not synchronized after all.
Their motion is identical in the initial frame and their clocks are synced in that frame. Thus their clock must remain synced in that frame.
I agree that they will not observe that one second lag since they would need to be stationary in that frame to observe that.
If you think you can provide some way whereby the two accelerating ships can each appear to the other to be at a constant speed and distance relative to each other even if they are, please provide it. What is each ship supposed to do to accomplish this?
I never suggested any such thing. I said I don't care how they appear to each other. I care about what they're doing.
The point of my exercise was that at the end of the story, the two ships are found to be in a common inertial frame at their original separation and with an unbroken string between them despite their interpretation of what they are seeing.
That is your goal, but you need to perform invalid mathematics to get that story. SR predicts otherwise.
So let's see some relativity math that does not depend on illusory difference in distance.
As suspected, you have no clue how to go about it. Yes, let's see some. I'll do the brutal method since it illustrates the concepts you're trying to avoid by keeping the speed low.
Two ship 10 light days apart each accelerate at 100g proper acceleration for exactly 4 days as measured on ship clock. That's today's scenario.
SP1 initiates acceleration at event SP1a and shuts the engine off at SP1b. We've chosen SP1a event as the origin of both frames.
SP2 initiates acceleration at event SP2a and shuts the engine off at SP2b.
In the original frame:
SP1a is at x=0,t=0 at the start (or 0,0). SP2a is at (-10, 0) Units in (light days, days)
After end of acceleration,
SP1b coordinates are (2.5123, 4.9082) and
SP2b coordinates are (-7.4877, 4.9082). Both ships are moving at .8112c which has a Lorentz factor λ of 1.71008. The SP1 clock reads 4 at SP1b and the SP2 clock reads 4 at SP2b.
At time 4.9082, The rod is still accelerating at various rates. It is stationary relative to SP2 only in SP2's frame, so SP1 observes a rod moving outside his window at event SP1b.
In the post-acceleration frame:
Both ships are initially moving at -.8112c at come to a halt after 4 subjective days of acceleration.
The easy part. By symmetry, SP1 begins acceleration SP1a at (0, 0) (our chosen common event) and SP1b coordinates are (-2.5123, 4.9082). If you don't use proper acceleration, the symmetry is lost and it takes much more work to compute that.
For SP2 we need to do a Relativity of Simultaneity calculation to compute how far out of sync their clocks will be in this new frame.
ΔT = λ * separation distance (in original frame) * Δv = 1.71008 * 10 * .8112 = 13.8721
Therefore:
event SP2a coordinates are (-17.1008, 13.8721) t = 0 on SP2 clock, v = -.8112
event SP2b coordinates are (- 19.6131, 18.7803) t = 4 on SP2 clock, v = 0
and while we're at it, we can compute
where SP2 is at time zero: (-5.8477, 0) t=-8.112 on SP2 clock, v = -.8112
At time 18.78 in this frame, both ships become stationary. SP2 clock reads 4, and SP1 clock reads 17.8721, and only at this time does the rod outside his window stop moving. When it does, he reads 17.1008 on it (the difference between their x coordinates). If there was a 10 light-day string between them, it would have broken long ago since it cannot span the 17+ light-day gap.
There's the mathematics. You've shown only Newtonian mathematics, which has been falsified a century and a half ago.
If you see an error in the above numbers where they conflict with SR, point it out. If however you feel the numbers do not correspond to your naive beliefs, then keep it to yourself.
Notice that at no point did I bother to have either ship 'observe' the other. The only observation done was SP1 looking at the markings on the rod in his presence.
The separation as measured by the measuring rods they brought along is the same at the end as at the beginning. Why would that not be the case?
The above analysis shows it not to be the case. Any other value would contradict light speed being a frame independent constant. It is that assumption alone from which all the rules I used to compute the above numbers were derived.
Total hogwash. I know Special Relativity inside and out, and General Relativity very well indeed as I have demonstrated repeatedly.
The other readers of this thread have noted otherwise. You've demonstrated a total lack of ability to apply relativity theory. I've seen no calculations except a few charts of Lorentz factors.
If you think I do not, show me, I repeat SHOW ME where I have been wrong.
I did. You're using Newtonian mathematics (and choice of scenarios with non-relativistic speeds). That's wrong, and a copout respectively.
The ship has thrusters all along the sides. And the ship is very skinny and made of ultra-rigid material.
Very good. To reduce strain, thrust must be applied along the length the ship and the rod, not just at one point. No distortion that way. No need to worry about speed of sound.
OK. At any moment all three are moving at the same speed in the same direction.
That's not true. The rod in front of SP2 is always stationary with SP2 in the SP2 frame, but not in any other frame. SP1 likewise only has identical speed as SP2 in the orignal frame, and also after they've both ceased acceleration, but not otherwise. Your statement above assumes Newtonian mechanics, and shows a complete lack of awareness relativity of simultaneity.
So why does the string break?
See the numbers above. Your intuitions posted just above are completely naive.
No need to have the rod connected only at one end. If everything accelerates at the same time (previously synchronized clocks) and in the same degree, length contraction will be the same everywhere.
That's begging your conclusion, which is a serious fallacy.
Acceleration is absolute in that it is experienced by the accelerating observer. The magnitude of the acceleration is not absolute in that time dilation disassociates proper acceleration from that measured by an inertial frame observer.
The acceleration is experience by the accelerating observer and observered by everyone else. The magnitude of the proper acceleration is absolute. The magnitude of the acceleration is indeed frame dependent.
The traveling twin thinks he is continuing to accelerate at the same rate ? which he can feel - and therefore exceeds c
c is a speed. Acceleration is not. Different units. You?re equivocating the two here. It is meaningless to say that acceleration exceeds some speed.
This is why the pilot of the spaceship and the inertial frame observer disagree on how fast the ship is going.
They disagree because the pilot always says he is stationary. One cannot have a nonzero velocity relative to ones self.
As it turns out, the track is going to contract with the cars. How? Lense-Thirring frame dragging. (The names are Austrian and are pronounced LENsuh TEERing).
Congrats, you managed to find yet another way to add obfuscation to this topic. This is an SR example, so you can assume negligible mass of the moving parts. Nobody considers the mass of the space ship in the twins scenario, despite the fact that that mass will affect the dilation of the ship clock. It also is assumed to be negligible.
If you do not know the rules, do not try to play the game.
...
I know SR cold
Says the guy who cannot compute the coordinates of some spaceships except using Newtonian physics.
Acceleration puts different observers in different reference frames. Time dilation is observer dependent.
It makes them stationary in different reference frames. It doesn?t put them in different reference frames since you can?t enter or exit a reference frame under SR.
Bringing two clocks into a common inertial frame will allow comparison, with the slower clock the one that experienced more acceleration.
Not necessarily true. I can sync a pair of clocks, separate them and have clock A undergo 10x the acceleration of clock B, and yet clock B be the slower of the two when they?re compared in each other?s presence. It?s the temporal length of their respective worldlines that counts, not the amount of acceleration.
Do you have any understanding of the subject at all or was this another ad hom attempt that misfired?
If I point out actual mistakes being made, it?s not an ad-hom. I?m not (often) directing any argument against you personally. I?m directing my arguments against all the things you?re saying.
Einstein was not believed by everyone for some time. There are relativity deniers today, to use your phrase.
If you?re not a relativity denier, then why have you not shown any relativistic mathematics when computing the coordinates of the ships in the string example? That exercise is pretty trivial, and yet all I get is low speed Newtonian math from you, which indicates a fear to expose the inconsistencies in your view. Stop it with these repetitive assertions and give workable numbers.
Choose a different scenario preferrably, but not one from a website where somebody else has done it. Notice that I don?t use the same one twice. You must use a scenario with relativistic dilation. Ideally, the situation should become evident with a couple digits of precision, but I did mine with around 5 digits. A speed of 100 m/sec would require more digits than my calculator has. It demonstrates nothing.
I?ve put up numbers. You?ve not found any inconsistency in them except disagreements with your personal vision, which I do not accept as evidence that I did it wrong. If it?s wrong, then somewhere there will be a pair of events that don?t relate properly with fixed light speed. I can?t do that with yours because you?ve given no relativistic example to work with. You apparently don?t know your physics at all because you continue to decline to do this.
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Acceleration is absolute in that it is experienced by the accelerating observer. The magnitude of the acceleration is not absolute in that time dilation disassociates proper acceleration from that measured by an inertial frame observer.
The acceleration is experience by the accelerating observer and observered by everyone else. The magnitude of the proper acceleration is absolute. The magnitude of the acceleration is indeed frame dependent.
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A worldline has a 4-velocity.
A change of this 4-velocity -> 4-acceleration has to be the same for all observers.
If it is not then we do not have an invariance of the spacetime interval, we do not have proper local physics.
We know nothing.
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Although I did not explicitly mention it, imagine that each ship is displaying a clock so that the other one can see it.
Can do that. I just didn’t find it useful for either of them to look at the other.
Each will start off seeing the other as being one second behind. Since this (initially) matches the apparent increase or decrease of separation, each one can consider that as evidence that their clocks are not synchronized after all.
Their motion is identical in the initial frame and their clocks are synced in that frame. Thus their clock must remain synced in that frame.
I agree that they will not observe that one second lag since they would need to be stationary in that frame to observe that.
Spatial separation rules out giving meaning to synchronized clocks if acceleration is involved.
Oopsie. They must stay synced in the original frame if their acceleration is identical, which it is. Per RoS, any clocks synced in that frame are not synced in another, so indeed they’ll be out of sync in either ship frame.
My point was that the one second lag in the onset of acceleration, due to the distance between them, led to observations on each side that contradicted the other side. SP1 was convinced by all evidence that SP2 was increasingly lagging behind and that SP2’s clock was running slow. Not just one second behind, but increasingly behind. SP2 was convinced by all evidence that SP1 was going too slow and SP2 was gaining on SP1 and that SP1’s clock was increasingly behind. Not until they got back into a common inertial reference frame was the illusion broken. The distance between the ships is the same as at the start and the string is unbroken.
When acceleration is involved , different observers separated by space can see different things that are not necessarily the case. And this is not even in relativistic territory.
If you think you can provide some way whereby the two accelerating ships can each appear to the other to be at a constant speed and distance relative to each other even if they are, please provide it. What is each ship supposed to do to accomplish this?
I never suggested any such thing. I said I don’t care how they appear to each other. I care about what they’re doing.
I had a rod with distance marks affixed to the front of SP2 so all SP1 has to do is look out his window at the rod in his presence to know the separation in SP2’s frame. No dependence on light speed. The rod, being attached to SP2, is always stationary in SP2’s frame. I put it on SP2 because it might break if we put it on SP1.
As we will see, the rod is not always stationary in SP2’s frame.
The point of my exercise was that at the end of the story, the two ships are found to be in a common inertial frame at their original separation and with an unbroken string between them despite their interpretation of what they are seeing.
That is your goal, but you need to perform invalid mathematics to get that story. SR predicts otherwise.
My math was completely valid. The point was to show that measurements made at a distance during acceleration, even common acceleration are not necessarily reflective of the measurements made in common inertial frames.
You want math? I will use some of your numbers.
Two spaceships 300,000 km apart. Call it 1 light second.
Each should experience identical proper acceleration regardless of how long they accelerate or what speed they reach relative to their original inertial frame.
OK, this is the first you’ve stated the use of proper acceleration. Your mathematics in prior posts used constant acceleration, not constant proper acceleration. They yield different numbers. Proper is easier to use due to the frame symmetry of it all, but it works either way if you don’t mind the more complex arithmetic.
We have both been talking about proper acceleration all along as is very obvious from the context and the discussions. Nit picking again. Acceleration of the two ships would not have the same values for several observers in varying frames but for each observer it would still be the same for each of the ships independent of externally measured acceleration value.
If after coming back into a common inertial frame, they each emailed a copy of their acceleration record to each other, would either one see a difference between the other’s record and his own?
If you think there will be a difference in the acceleration records because of relativistic speeds, show me the math.
You've not invoked relativity theory at all. You're adding velocities the Newtonian way, and you're using the original rest frame to compute what each observer sees, not the frame of the observer. The inconsistencies will pile up once you move faster than a city bus.
That is right, I did not incorporate relativistic considerations. My point was that an apparent speed difference is not a necessarily a real difference if the clocks do not correspond.
I’ve not considered appearances at all. I want the coordinates of the two events where the ships turn off their engines relative to the ship frame. Newtonian physics does not answer that. Motion does not involve time or length dilation. It assumes an absolute frame, something at least you claim to deny, but here you are using only absolute physics, and only at low speeds to boot. The theory had been falsified 150 years ago.
No, I am demonstrating that to properly assess what is happening one must consider that the two spaceships are separated in space and that the speed of light is finite. This has ramifications as we will shortly see.
So let’s see some relativity math that does not depend on illusory difference in distance.
As suspected, you have no clue how to go about it. Yes, let’s see some. I’ll do the brutal method since it illustrates the concepts you’re trying to avoid by keeping the speed low.
Two ship 10 light days apart each accelerate at 100g proper acceleration for exactly 4 days as measured on ship clock. That’s today's scenario.
SP1 initiates acceleration at event SP1a and shuts the engine off at SP1b. We’ve chosen SP1a event as the origin of both frames.
SP2 initiates acceleration at event SP2a and shuts the engine off at SP2b.
In the original frame:
SP1a is at x=0,t=0 at the start (or 0,0). SP2a is at (-10, 0) Units in (light days, days)
After end of acceleration,
SP1b coordinates are (2.5123, 4.9082) and
SP2b coordinates are (-7.4877, 4.9082). Both ships are moving at .8112c which has a Lorentz factor λ of 1.71008. The SP1 clock reads 4 at SP1b and the SP2 clock reads 4 at SP2b.
At time 4.9082, The rod is still accelerating at various rates. It is stationary relative to SP2 only in SP2’s frame, so SP1 observes a rod moving outside his window at event SP1b.[/color
In the post-acceleration frame:
Both ships are initially moving at -.8112c at come to a halt after 4 subjective days of acceleration.
The easy part. By symmetry, SP1 begins acceleration SP1a at (0, 0) (our chosen common event) and SP1b coordinates are (-2.5123, 4.9082). If you don’t use proper acceleration, the symmetry is lost and it takes much more work to compute that.
For SP2 we need to do a Relativity of Simultaneity calculation to compute how far out of sync their clocks will be in this new frame.
ΔT = λ * separation distance (in original frame) * Δv = 1.71008 * 10 * .8112 = 13.8721
Therefore:
event SP2a coordinates are (-17.1008, 13.8721) t = 0 on SP2 clock, v = -.8112
event SP2b coordinates are (- 19.6131, 18.7803) t = 4 on SP2 clock, v = 0
and while we’re at it, we can compute
where SP2 is at time zero: (-5.8477, 0) t=-8.112 on SP2 clock, v = -.8112
At time 18.78 in this frame, both ships become stationary. SP2 clock reads 4, and SP1 clock reads 17.8721, and only at this time does the rod outside his window stop moving. When it does, he reads 17.1008 on it (the difference between their x coordinates). If there was a 10 light-day string between them, it would have broken long ago since it cannot span the 17+ light-day gap.
There’s the mathematics. You’ve shown only Newtonian mathematics, which has been falsified a century and a half ago.
If you see an error in the above numbers where they conflict with SR, point it out. If however you feel the numbers do not correspond to your naive beliefs, then keep it to yourself.
Notice that at no point did I bother to have either ship ‘observe’ the other. The only observation done was SP1 looking at the markings on the rod in his presence.
The separation as measured by the measuring rods they brought along is the same at the end as at the beginning. Why would that not be the case?
The above analysis shows it not to be the case. Any other value would contradict light speed being a frame independent constant. It is that assumption alone from which all the rules I used to compute the above numbers were derived.
For clarity, the rod is attached to SP2 who is pushing it and it is not attached to SP as per your comments above.
In that case, you are wrong. You are assuming that the end of the rod will stay stationary outside SP1’s window until he stops accelerating. It will not. It will be left behind.
The acceleration force cannot move through the rod at greater than the speed of sound in the material. 100 G acceleration would probably deform most materials, compromising its length. So let us assume that the rod is as strong as it needs to be and that the speed of sound in the material is the speed of light.
In that case, SP1 will be leaving the rod behind when it starts accelerating because the accelerating force from SP2 has not yet reached the front end of the rod. It will not reach the front end of the rod for ten days as counted by an observer at x=0 and in the same inertial frame as SP1 started from. Will the rod get compressed by the speed differentials? No, it is as stiff as possible and can transmit acceleration force at lightspeed. The rod will get progressively Lorentz contracted from the SP2 end forward as seen by an inertial observer.
When SP2 stops accelerating, all points on the rod will, over time, stop receiving acceleration forces and achieve a common metric everywhere, and will be the original proper length as measured by the measuring rods that SP1 and SP2 brought along.
SP1 saw SP2 start accelerating ten days after himself. He will also see SP2 stop accelerating ten days after himself. At the end of the ten days, he will look out the window and see that the rod he left behind ten days earlier has now caught up with him and is where it started. Everything is symmetric and everything will end up in the same relative positions, although all equally Lorentz contracted as per an inertial observer.
You assumed that SP1a and SP2a were simultaneous and that the rod would start moving at SP1a. A basic lesson of Relativity Theory, which you should learn someday, is that at any non-trivial distance, the idea of even approximate simultaneity has no meaning.
If you are going to do thought experiments, get your thoughts together before starting the math.
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Malamute Lover,
here are a couple of diagrams:
(https://i.imgur.com/XSs3QXa.png)
(https://i.imgur.com/sdxp5gT.png)
This is a uniform acceleration as per the textbook.
You can see the length contraction of the ruler.
This discussion would be better in the SR reciprocal thread though,
Jano
What did you think this means?
It is showing the worldlines of a uniformly accelerated observer and an observer who is at fixed coordinates.
Did the term ‘comoving’ confuse you? It just means that the observer is moving with the expansion of the universe but that (for an observer in an inertial reference frame) the coordinates remain the same and that for another observer not in the same reference frame, the coordinate system used for the first observer remains in effect. Basically, it means to ignore the expansion of the universe. It simplifies the math by keeping things local and not in reference to any noticeably red-shifted distant object.
As expected, the two worldlines are different because the situations are different. But the worldlines for spaceships SP1 and SP2, uniformly accelerating at the same rate, the worldline shapes would be identical.
So why the string between the spaceships would not try to shrink like the ruler does?
The end-result will be a broken string.
That's the logical conclusion when the spaceships worldlines are going to be identical,
Jano
The graph shows an accelerating observer and a non-accelerating observer. What does this have to do with the subject, which is two observers in a common inertial reference frame accelerating at the same rate in the same direction beginning at an agreed on moment on synchronized clocks?
In that scenario, the entire ensemble will not be aware of any contraction. An external observer in an inertial frame would observe increasing Lorentz contraction on the entire ensemble. Are you still imagining that there is an Absolute Space in which the shrinkage is happening? There isn't. Contraction is relative, not objective.
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My point was that the one second lag in the onset of acceleration, due to the distance between them
They leave simultaneously, so there is no lag in the onset of acceleration. If you mean a lag in what the see, then say that.
If their a light second apart and their acceleration is low enough, there might be an approximate 1 second lag in their seeing the departure of the other, but you’ve not determined that since you’ve only been using Newtonian mathematics which yields wrong numbers.
Look at my scenario. They’re 10 light days apart, but if SP1 never shut off his engines, he’d never see SP2 depart. Newtonian mathematics actually would probably agree with that since after 3.5 days SP1 would be moving faster than light and would be able to see nothing behind him at all.
SP1 was convinced by all evidence that SP2 was increasingly lagging behind and that SP2’s clock was running slow. Not just one second behind, but increasingly behind. SP2 was convinced by all evidence that SP1 was going too slow and SP2 was gaining on SP1 and that SP1’s clock was increasingly behind. Not until they got back into a common inertial reference frame was the illusion broken. The distance between the ships is the same as at the start and the string is unbroken.
Again, you used the wrong mathematics, so you’ve not shown this. Not saying it’s wrong, only that you’ve just not shown any of the above stuff. Some of it is true, and some not. You’re guessing because you’ve not done the math, and your assertions carry no weight.
As we will see, the rod is not always stationary in SP2’s frame.
I defined it to be, so it is, unless it produces a contradiction. Yes, as discussed above, it needs its own thrust to be able to do this. I can’t trail it behind SP1 because I get exactly such a contradiction. So I attached it to SP2.
The point of my exercise was that at the end of the story, the two ships are found to be in a common inertial frame at their original separation and with an unbroken string between them despite their interprettion of what they are seeing.
That is your goal, but you need to perform invalid mathematics to get that story. SR predicts otherwise.
My math was completely valid.
You used Newtonian math, falsified 150 years ago.
If after coming back into a common inertial frame, they each emailed a copy of their acceleration record to each other, would either one see a difference between the other’s record and his own?
Whether they used constant acceleration or constant proper acceleration, the scenario says they must do the same thing, so yes, by that statement, their acceleration records will agree precisely, else at least one of them violated the flight plan. We’re not positing cheating or deceit here.
I’m assuming you’re using constant proper acceleration then. Your mathematics posted earlier did not use it, which is why I’ve been careful to have it specified explicitly.
I’ll do the brutal method since it illustrates the concepts you’re trying to avoid by keeping the speed low.
Two ship 10 light days apart each accelerate at 100g proper acceleration for exactly 4 days as measured on ship clock. That’s today's scenario.
SP1 initiates acceleration at event SP1a and shuts the engine off at SP1b. We’ve chosen SP1a event as the origin of both frames.
SP2 initiates acceleration at event SP2a and shuts the engine off at SP2b.
In the original frame:
SP1a is at x=0,t=0 at the start (or 0,0). SP2a is at (-10, 0) Units in (light days, days)
After end of acceleration,
SP1b coordinates are (2.5123, 4.9082) and
SP2b coordinates are (-7.4877, 4.9082). Both ships are moving at .8112c which has a Lorentz factor λ of 1.71008. The SP1 clock reads 4 at SP1b and the SP2 clock reads 4 at SP2b.
At time 4.9082, The rod is still accelerating at various rates. It is stationary relative to SP2 only in SP2’s frame, so SP1 observes a rod moving outside his window at event SP1b.[/color
In the post-acceleration frame:
Both ships are initially moving at -.8112c at come to a halt after 4 subjective days of acceleration.
The easy part. By symmetry, SP1 begins acceleration SP1a at (0, 0) (our chosen common event) and SP1b coordinates are (-2.5123, 4.9082). If you don’t use proper acceleration, the symmetry is lost and it takes much more work to compute that.
For SP2 we need to do a Relativity of Simultaneity calculation to compute how far out of sync their clocks will be in this new frame.
ΔT = λ * separation distance (in original frame) * Δv = 1.71008 * 10 * .8112 = 13.8721
Therefore:
event SP2a coordinates are (-17.1008, 13.8721) t = 0 on SP2 clock, v = -.8112
event SP2b coordinates are (- 19.6131, 18.7803) t = 4 on SP2 clock, v = 0
and while we’re at it, we can compute
where SP2 is at time zero: (-5.8477, 0) t=-8.112 on SP2 clock, v = -.8112
At time 18.78 in this frame, both ships become stationary. SP2 clock reads 4, and SP1 clock reads 17.8721, and only at this time does the rod outside his window stop moving. When it does, he reads 17.1008 on it (the difference between their x coordinates). If there was a 10 light-day string between them, it would have broken long ago since it cannot span the 17+ light-day gap.
If you see an error in the above numbers where they conflict with SR, point it out. If however you feel the numbers do not correspond to your naive beliefs, then keep it to yourself.
Notice that at no point did I bother to have either ship ‘observe’ the other. The only observation done was SP1 looking at the markings on the rod in his presence.
For clarity, the rod is attached to SP2 who is pushing it and it is not attached to SP as per your comments above.
Kind of. I said it is always stationary in the inertial frame in which SP2 is stationary at any given moment in time. To do that, it has its own propulsion programmed with the flight plan, so SP2 exerts no force on it. To say it ‘pushes’ it is misleading. It does not accelerate at 100g along its length except at the point where it attaches to SP2. Likewise the SP2 ship only accelerates at 100g at that attachment point and not elsewhere.
In that case, you are wrong. You are assuming that the end of the rod will stay stationary outside SP1’s window until he stops accelerating. It will not.
Please don’t put words in my mouth. Critique what I said, and not what you think I’m assuming.
I’m assuming no such thing. I did not state any such thing. I said it wouldn’t stop moving outside his window until time 17.8721 on the SP1 clock. See the comment near that figure above.
In fact, the rod will be seen to be moving to the rear at a non-constant rate, eventually settling on 17.1 light days. I didn’t compute the intermediate states, only the end ones that matter.
Note that SP1’s clock is .9021 days slow in the 2nd reference frame. It was running slow while he was moving in it, and only runs at full speed once he comes to a stop. So I’m careful to distinguish between event coordinates in that frame, and what the clocks read on the respective ships.
The acceleration force cannot move through the rod at greater than the speed of sound in the material.
Indeed, which is why we both agreed to use idealized math for the rod. This necessitates giving it its own thrust and knowledge of the flight plan to avoid the speed of sound problem. No stress, no deformation. It’s a mathematical experiment, not an engineering one.
So let us assume that the rod is as strong as it needs to be and that the speed of sound in the material is the speed of light.
No. We agreed to idealized math, so no force or deformation to propagate at light speed. The rod stays stationary in the SP2 frame by definition.
SP1 saw SP2 start accelerating ten days after himself.
I did not state that. It’s wrong. Critique what I said, without injecting your ideas of what is going on.
He will also see SP2 stop accelerating ten days after himself.
I didn’t say that either. Not hard to compute. SP1 stops when his clock says 4. SP2 stops at 17.8721 according to SP1 clock, and it takes 17.1008 light days for the light to get to SP1, so he sees that event at time 34.9729, which is almost 31 days after he stopped his acceleration.
Critique what I said, without injecting your ideas of what is going on.
At the end of the ten days, he will look out the window and see that the rod he left behind ten days earlier has now caught up with him and is where it started.
I didn’t say that.
If you think this, fine, but without numbers, you’re just blowing steam. You’re making uniformed assertions, which is pathetic at this point. You’ve not found any inconsistency with what I’ve said, and are only comparing it to your unquantified guesswork. I expected my scenario to conflict with yours, so that makes my scenario meet my expectations spectacularly.
Everything is symmetric and everything will end up in the same relative positions, although all equally Lorentz contracted as per an inertial observer.
No numbers. I can hear the steam blowing.
You assumed that SP1a and SP2a were simultaneous and that the rod would start moving at SP1a.
The statement lacks a frame reference, deliberately it seems based on your comment below. That makes you a troll now. I was suspect, but didn’t want to put it out there without this sort of blatant move.
A basic lesson of Relativity Theory, which you should learn someday, is that at any non-trivial distance, the idea of even approximate simultaneity has no meaning.
A frame is specified, and you know it. Look at the very first line in the maroon print.
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Acceleration is absolute in that it is experienced by the accelerating observer. The magnitude of the acceleration is not absolute in that time dilation disassociates proper acceleration from that measured by an inertial frame observer.
The acceleration is experience by the accelerating observer and observered by everyone else. The magnitude of the proper acceleration is absolute. The magnitude of the acceleration is indeed frame dependent.
Since there is disagreement among multiple observers about the magnitude of the acceleration, saying that the magnitude of the proper acceleration is absolute gives the wrong impression, implying that what the accelerating observer sees and predicts will necessarily be correct for everyone. Acceleration is real, having a real effect on spacetime, which gets bent by the energy being expended, But since it is spacetime that is bent, and acceleration is rate of change over time different observers will disagree about the rate of acceleration.
The traveling twin thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds c
c is a speed. Acceleration is not. Different units. You’re equivocating the two here. It is meaningless to say that acceleration exceeds some speed.
Did you actually read what you wrote? Or what I wrote?
The traveling twin thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds c
The driver thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds the speed limit.
Where is the problem?
No ‘equivocation’ here. Or equating as the case may be. :D
This is why the pilot of the spaceship and the inertial frame observer disagree on how fast the ship is going.
They disagree because the pilot always says he is stationary. One cannot have a nonzero velocity relative to ones self.
The pilot feels acceleration and sees the landmarks go by faster and faster. He knows his speed is changing and that he is not stationary. Again, did you actually read what you wrote?
In the roller coaster thing, the cars are being held against centripetal force by an outside track which is not rotating with them.
As the cars accelerate, they push the track in the opposite direction. (Newton’s Third Law.) That is the acceleration will not be as much as expected, half of the energy going into the track.
No energy goes into the track. We’re assuming lack of friction here. No work is being done by or on the track. It just guides the cars. The cars are not thrusting against the track, so the track remains stationary.
If there is no friction, then the cars could not accelerate in the first place. The wheels would just slip. No reaction, no action. If the cars have always been racing around at the same speed, then the spacing, if any, is what it has always been. No changing contraction to worry about.
What is holding them in a circular orbit rather than flying out in a straight line and crashing into the track?
Wheels
The track is holding them. If the track disappeared, they would fly off in straight lines despite having wheels.
As it turns out, the track is going to contract with the cars. How? Lense-Thirring frame dragging. (The names are Austrian and are pronounced LENsuh TEERing).
Congrats, you managed to find yet another way to add obfuscation to this topic. This is an SR example, so you can assume negligible mass of the moving parts. Nobody considers the mass of the space ship in the twins scenario, despite the fact that that mass will affect the dilation of the ship clock. It also is assumed to be negligible.
The twins paradox has no answer in SR because acceleration is involved. SR cannot deal with acceleration. There is centripetal acceleration even with a constant speed. SR cannot handle this problem. You need GR and that is what GR says will happen. If you insist on staying in SR, then you have just created another fake twins paradox with no way to answer it.
The mass of the spaceship and the mass of the Earth for the stay at home twin will in fact alter how much the clocks will disagree by small amounts. But the key factor is that the traveling twin underwent sufficient acceleration to make those factors insignificant and this is what made him end up standing next to a twin who looks like his grandfather.
Going around a track involves centripetal acceleration. The faster those cars go, the more centripetal acceleration and the more obvious GR effects will be. No obfuscation in applying the proper tools for the situation. Substantial acceleration? This is not an SR problem anymore.
If you do not know the rules, do not try to play the game.
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I know SR cold
Says the guy who cannot compute the coordinates of some spaceships except using Newtonian physics.
I was demonstrating the principle that synchronizing clocks and separating them negates simultaneity. Assuming a non-existent simultaneity is what confused you. Clear concepts first, a realistic experimental setup taking those concepts into account, then plug and chug the math. Your belief that the SP2 rod would start moving at time SP1a showed that you did not understand this concept.
Since the situations are symmetric at the beginning and in the acceleration applied according to synchronized clocks points directly at the end results having the same appearance to the participants. Clear concepts first. No need for math in this.
Acceleration puts different observers in different reference frames. Time dilation is observer dependent.
It makes them stationary in different reference frames. It doesn’t put them in different reference frames since you can’t enter or exit a reference frame under SR.
Acceleration takes us out of SR. Acceleration changes reference frames. You need GR if there is any asymmetry of situations. SR snapshots could be used in the two spaceships scenario since the situations are symmetric. But you have to pay attention to the actual physical circumstances, like the fact that the rod attached to SP2 will not yet move at time SP1a.
Bringing two clocks into a common inertial frame will allow comparison, with the slower clock the one that experienced more acceleration.
Not necessarily true. I can sync a pair of clocks, separate them and have clock A undergo 10x the acceleration of clock B, and yet clock B be the slower of the two when they’re compared in each other’s presence. It’s the temporal length of their respective worldlines that counts, not the amount of acceleration.
Acceleration changes the length of the time component of a worldline. Two observers moving relative to each other will always see the other’s clock as running slower. But bring them into a common inertial frame and there will be a difference in their elapsed times from a synchronization event in the past according to their acceleration histories, with the one having experienced the most acceleration having less elapsed time. If you think you can set up a situation where that is not the case, show me. Don’t need to do complicated calculations. Keep it simple.
To an observer on the surface (and therefore to the mechanical properties of the cylinder) the same number of revolutions a minute are taking place
Now you’ve contradicted yourself again. We have a stationary marker by which a rotation can be measured. If their clocks are not running at the same pace, they necessarily measure a different time for one revolution.
The circle has contracted because the geometry has gone non-Euclidean. To an external observer, the circle is smaller and faster, but the clock is slower. An observer on the circle counts the same revolutions per minute.
You only need SR to understand that part. What is your problem?
Nope. The reduced circumference circle is spinning faster because tangential speed is still the same but the increased time dilation makes an observer on the circumference think a revolution takes as long as if the circumference were still longer.
You’re failing simple grade school arithmetic now. Angular rate is measured in say RPM which isn’t a function of radius or linear speed of any of the parts. If a rotation takes 12 seconds as measured by the central clock, then regardless of the radius and rim speed, that object is rotating at 5 RPM. If the clock at the rim runs slower and measures 10 seconds, then the object rotates at 6 RPM. It is only a function of the measured time to do one revolution, so if the two measured times are different, the measured number of revolutions a minute must be different.
Your concepts are mistaken. You must look at the details to get them right.
To begin with, tangential speed and radius and RPM are all related. Your claim that they are not related is wrong. If you know the values of two of them, you know the third. That is also true in non-Euclidean geometry as long as the details of the curvature between circle and center are consistent all around the circle. You just need to use the proper formulas consistent with the curvature.
We should note that the central observer will always see the clock on the circle as running slower, and vice versa, because they are moving relative to each other. However, the circle has shrunken and the tangential speed is now greater. As a result the two clocks will each be seen as running slower than before by the opposite observer. But the observer on the circle never looks at the center so we need only deal with increased time dilation on the circle as viewed by the central observer when the circle shrank.
The clock on the circle is running slower than before as viewed from the center. Previously it took 12 seconds according to the central clock for a marked point on the circle to pass a bright red line pointing out from the center, a line that is stationary to the central observer. But the circle has shrunk and sped up. Now the central clock says it takes only 10 seconds.
Imagine the circle has evenly spaced markings on it. Viewed from the center it previously took 1.2 seconds for these markings to pass that stationary red line. Now it takes only 1 second as viewed from the center since the circle has shrunk and is moving faster. Previously 5 RPM, now 6 RPM. Tangential speed is 20% faster.
However, the observer on the circle has had his clock slowed as compared to the central clock. The ticks do not come as often as seen from the center. How much slower are those circle ticks coming? Since the shrinking of the circle and consequent speed up was due to Lorentz contraction, the time dilation will also be in accordance with the Lorentz factor. Tangential speed is now 20% faster what it was as viewed from the center, so the perceived clock rate will be 1/1.2 = 0.83 as fast on the circle clock as compared to the central clock by the central observer.
Viewed from the center, the time for each marking to pass the stationary red line is 1 second but the clock on the circle is running at 0.83 speed. However the observer on the circle is unaware of this. He thinks the ticks on his clock are coming as fast as ever. He sees the markings pass the red line every 1/.083 = 1.2 seconds same as before the circle shrunk. He is unaware that the circle has shrunk. He still thinks it is going at 5 RPM and the mechanical stresses experienced on the circle per unit of time are the same as before.
Accelerated observers feel acceleration but are unaware of length contraction or time dilation that other observers may notice. Relative. Not objective.
And “a pilot always experiences being stopped”? No, the pilot experiences being accelerated to 0.99 c from his original reference frame. If he decelerates and finds himself in a matching inertial frame that is actually accelerated from the starting frame, is he stopped?
A pilot is an observer observing from a position of rest in his current reference frame. That means he experiences his clock running at normal rate, and experiences his proper mass, not something 7x his prior mass.
What you seem to be referring to is proper velocity, which is indeed 7c. Proper velocity is the result of integration of past acceleration, and there’s no limit to that, which is why, with a fast enough ship, I can get to the far side of the galaxy before I die.
Stopped is a relative term. There is no absolute space so there is no absolute stopped. The pilot can say he is stopped if he does not feel acceleration and does not observe any motion in other objects that he considers stopped.
Length contraction, time dilation, and mass-energy increase are not observed in the accelerated reference frame. They are relative to other reference frames, not objective.
I am reminded of Poul Andersen’s novel Tau Zero from years back. A ship with a gigantic fuel supply and a hyper-efficient engine suffers a problem. The engine won’t turn off. :D
If you were able to see the GPS clock it would be going at the same rate as your own clock because it was intentionally set to run slower.
Looking at a clock known to be designed to run slower and seeing it run the same rate as my normal clock is a direct observation of objective time dilation between those clocks.
Sorry, no. Not objective. Another observer in a different reference frame will observe a different a time dilation factor between you and the GPS satellite. Even if you exclude other observers from consideration, the GPS satellite will consider you time dilated since you are moving relative to it and are deeper in a gravity well and your clock is running slower. The same absolute value of time dilation but opposite signs. Not objective which would require identical values measured from either end. Time dilation is relative, not objective
Acceleration is in the domain of General Relativity.
Gravity actually
Gravity and acceleration are dealt with by the same mathematical framework. Mathematically gravity is acceleration and vice versa. SR cannot handle acceleration. GR can.
Einstein was not believed by everyone for some time. There are relativity deniers today, to use your phrase.
If you’re not a relativity denier, then why have you not shown any relativistic mathematics when computing the coordinates of the ships in the string example? That exercise is pretty trivial, and yet all I get is low speed Newtonian math from you, which indicates a fear to expose the inconsistencies in your view. Stop it with these repetitive assertions and give workable numbers.
Choose a different scenario preferably, but not one from a website where somebody else has done it. Notice that I don’t use the same one twice. You must use a scenario with relativistic dilation. Ideally, the situation should become evident with a couple digits of precision, but I did mine with around 5 digits. A speed of 100 m/sec would require more digits than my calculator has. It demonstrates nothing.
I’ve put up numbers. You’ve not found any inconsistency in them except disagreements with your personal vision, which I do not accept as evidence that I did it wrong. If it’s wrong, then somewhere there will be a pair of events that don’t relate properly with fixed light speed. I can’t do that with yours because you’ve given no relativistic example to work with. You apparently don’t know your physics at all because you continue to decline to do this.
I have shown the inconsistency with your concepts. Your math is wrong because it does not reflect the real world. You assume simultaneity at a distance which is invalid even in SR. If I showed math that was based on the correct concepts you would just say it was wrong because you do not understand the concepts. Get those straight first.
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...
The graph shows an accelerating observer and a non-accelerating observer. What does this have to do with the subject, which is two observers in a common inertial reference frame accelerating at the same rate in the same direction beginning at an agreed on moment on synchronized clocks?
In that scenario, the entire ensemble will not be aware of any contraction. An external observer in an inertial frame would observe increasing Lorentz contraction on the entire ensemble. Are you still imagining that there is an Absolute Space in which the shrinkage is happening? There isn't. Contraction is relative, not objective.
Malamute Lover,
let us talk some concrete analysis.
L0 - length of the spaceship in its rest frame.
5L0 - distance between the spaceships.
Let us choose the left frame as our origin.
(https://i.imgur.com/UYYtCix.png)
The right spaceships has to get much further away into the future, in the original rest frame, in order to maintain the simultaneity line of the left spaceship frame.
(https://i.imgur.com/xY6D6Zg.png)
Do you trust the Minkowski space-time diagrams?
Does this make sense now?
Jano
Edit:
This is the Special Relativity acceleration 101 (One-o-One) in transition towards general frames.
What do you know, this is the post #101 in this thread. :)
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The traveling twin thinks he is continuing to accelerate at the same rate which he can feel - and therefore exceeds c
The driver thinks he is continuing to accelerate at the same rate which he can feel - and therefore exceeds the speed limit.
Where is the problem?
No problem with that. Proper velocity has no upper limit and can exceed c, as I explained above.
What is holding them in a circular orbit rather than flying out in a straight line and crashing into the track?
Wheels
The track is holding them. If the track disappeared, they would fly off in straight lines despite having wheels.
That they would. Same thing if the spokes disappeared in the spoke example.
The twins paradox has no answer in SR because acceleration is involved. SR cannot deal with acceleration. There is centripetal acceleration even with a constant speed. SR cannot handle this problem. You need GR and that is what GR says will happen. If you insist on staying in SR, then you have just created another fake twins paradox with no way to answer it.
You're welcome to invoke GR if that's the only way you can figure it out. SR contains Lorentz transformations between frames, and that's all I used in my example.
Since the situations are symmetric at the beginning and in the acceleration applied according to synchronized clocks points directly at the end results having the same appearance to the participants. Clear concepts first. No need for math in this.
Until you actually do the math as I did and it shows otherwise. Do the trivial math then. I gave coordinates of the four important events (SP1a/b and SP2a/b). If I gave incorrect coordinates, then correct them. Because with my numbers, a 10-LD rod length-contracted down to size 5.8477 isn't going to bridge the gap between SP1b and SP2b which are still 10 LD apart. My string has broken. You have to have some kind of different numbers to have the string not break, but you've shown nothing but Newtonian math, examples with near stationary motion, plus a lot of assertions that don't add up.
like the fact that the rod attached to SP2 will not yet move at time SP1a.
Sp1a is an event, not a time. The 10 LD mark on the rod happens to be present at that event, and it indeed begins acceleration at that event, but since it has zero velocity at the moment it begins to accelerate, I'll have to agree that it does not yet move at that event.
I am reminded of Poul Andersen's novel Tau Zero from years back. A ship with a gigantic fuel supply and a hyper-efficient engine suffers a problem. The engine won't turn off.
And nobody thinks to just put the ship into a slow turn so the speed never gets stupid.
Not objective. Another observer in a different reference frame will observe a different a time dilation factor between you and the GPS satellite.
Oopsie. Math shows otherwise.
I have shown the inconsistency with your concepts.
You've shown my numbers to be inconsistent with your assertions is all. Without showing light speed to be not c somewhere, or to show that events seen in one frame differ from the observations made from the same events in the other frame, or that 1+2 for one observer is different than 1+2 for the other, you've not demonstrated any inconsistency at all. You've just make empty assertions with no numbers to back them. And that's all you'll continue to do, because if you post 'corrected' numbers for my scenario, I'll show exactly those inconsistencies.
Use my scenario: 100g for 4 days. You say there's no need for mathematics, so it should be trivial for you. Show the numbers in both frames, as I did. But you can't.
Sloppy.
Ooh, I hit a nerve.
You are comparing the 4.9082 elapsed time that an unaccelerated observer would see (unaccelerated with respect to the original frame of SP1 and SP2) with the 4 elapsed time that both SP1 and SP2 see.
The coordinate system of M is what your stationary observer in M would compute, yes. That coordinate system is unaffected by what one pair of ships happen to be doing.
The unaccelerated observer sees 4.9082 for both.
He can work out that time for both. He?s probably not present at either event, so what he sees of those events comes some time later. I expended no effort computing what anybody sees except things happening in their presence. But yes, if there was a series of unaccelerated observers with clocks synced in M, that just happen to be present at SP1b and SP2b, their respective clocks would read 4.9082 at those events.
The accelerated observers (SP1 and SP2) both see 4. The rod is stopped relative to both SP1 and SP2 in all frames of reference.
Sloppy.
I said no such thing, and if you look at my numbers in M, the rod is very much still moving relative to SP1 at event SP1b.
The rod is stopped by definition in frame N when SP2 stops in that frame. Outside SP1?s window, event eW thus happens immediately in N, but takes over 3 weeks to happen in M. Again, your statement suggests a complete lack of awareness of relativity of simultaneity, which for somebody claiming to know even the most basic fundamentals of relativity, is not so much sloppy as plain uneducated. If two separated events (SP2b and eW) happen simultaneously in one frame, they?re not going to be simultaneous in the other. There?s exceptions to this, but not in a 1D example.
In the post-acceleration frame:
Both ships are initially moving at -.8112c at come to a halt after 4 subjective days of acceleration.
The easy part. By symmetry, SP1 begins acceleration SP1a at (0, 0) (our chosen common event) and SP1b coordinates are (-2.5123, 4.9082). If you don?t use proper acceleration, the symmetry is lost and it takes much more work to compute that.
If you use proper acceleration the elapsed time for both is 4.
The elapsed ship proper time is 4. Elapsed time time in frame N is 4.9082. I explicitly said these are N coordinates (or rather the post-acceleration frame which I?m now calling N).
If you do not use proper acceleration, that is, if you use an unaccelerated frame then the elapsed time for both is 4.9082.
Frame N is an inertial reference frame, as is M. Neither are accelerated frames. The use of proper acceleration or not affects the flight plan and the symmetry of the mathematics, but has no effect on what has been defined as an inertial reference frame.
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Sloppy.
Ooh, I hit a nerve.
Merely an accurate observation as will be shown in detail later.
In the original frame [M]:
SP1a is at x=0,t=0 at the start (or 0,0). SP2a is at (-10, 0) Units in (light days, days)
After end of acceleration,
SP1b coordinates are (2.5123, 4.9082) and
SP2b coordinates are (-7.4877, 4.9082). Both ships are moving at .8112c which has a Lorentz factor λ of 1.71008. The SP1 clock reads 4 at SP1b and the SP2 clock reads 4 at SP2b.
At time 4.9082, The rod is still accelerating at various rates. It is stationary relative to SP2 only in SP2’s frame, so SP1 observes a rod moving outside his window at event SP1b.
For brevity, I will call the two frames M and N, conveniently M (maroon) for the original frame, and N (navy) for the frame of either ship after the end of acceleration. This corresponds to the colors I used for the text.
If both clocks read 4 and they are initially synchronized and undergo identical acceleration histories, then both engines have stopped. Why should the rod still be accelerating?
Your question again displays a lack of awareness of relativity of simultaneity. The event of the halting of the 17.1 LD mark on the rod (which I shall call event eW for where SP1 sees the rod stop outside his Window) is simultaneous with distant event SP2b in N, so it cannot be simultaneous with event SP2b in M. In fact, the coordinates of eW in M is (21.7559, 28.6306), so it actually takes 23.7224 days to stop in frame M after SP2 shuts down. If the rod is longer than this, extending well past SP1, then those parts have not yet stopped at that time in M.
I chose my numbers (high acceleration, separation greater than distance to Rindler horizon) to highlight such effects.
You are comparing the 4.9082 elapsed time that an unaccelerated observer would see (unaccelerated with respect to the original frame of SP1 and SP2) with the 4 elapsed time that both SP1 and SP2 see.
The coordinate system of M is what your stationary observer in M would see, yes. That coordinate system is unaffected by what one pair of ships happen to be doing.[
The unaccelerated observer sees 4.9082 for both.
He can work out that time for both. He’s probably not present at either event, so what he sees of those events comes some time later. I expended no effort computing what anybody sees except things happening in their presence. But yes, if there was a series of unaccelerated observers with clocks synced in M, that just happen to be present at SP1b and SP2b, their respective clocks would read 4.9082 at those events.
The accelerated observers (SP1 and SP2) both see 4. The rod is stopped relative to both SP1 and SP2 in all frames of reference.
Sloppy.
I said no such thing, and if you look at my numbers in M, the rod is very much still moving relative to SP1 at event SP1b.
The rod is stopped by definition in frame N when SP2 stops in that frame. Outside SP1’s window, event eW thus happens immediately in N, but takes over 3 weeks to happen in M. Again, your statement suggests a complete lack of awareness of relativity of simultaneity, which for somebody claiming to know even the most basic fundamentals of relativity, is not so much sloppy as plain uneducated. If two separated events (SP2b and eW) happen simultaneously in one frame, they’re not going to be simultaneous in the other. There’s exceptions to this, but not in a 1D example.
In the post-acceleration frame:
Both ships are initially moving at -.8112c at come to a halt after 4 subjective days of acceleration.
The easy part. By symmetry, SP1 begins acceleration SP1a at (0, 0) (our chosen common event) and SP1b coordinates are (-2.5123, 4.9082). If you don’t use proper acceleration, the symmetry is lost and it takes much more work to compute that.
If you use proper acceleration the elapsed time for both is 4.
The elapsed ship proper time is 4. Elapsed time time in frame N is 4.9082. I explicitly said these are N coordinates (or rather the post-acceleration frame which I’m now calling N).
If you do not use proper acceleration, that is, if you use an unaccelerated frame then the elapsed time for both is 4.9082.
Frame N is an inertial reference frame, as is M. Neither are accelerated frames. The use of proper acceleration or not affects the flight plan and the symmetry of the mathematics, but has no effect on what has been defined as an inertial reference frame.
If the flight plan is constant acceleration relative to M, then proper acceleration increases over time resulting in a lower average speed in the M frame than the same phase in the N frame where most of the acceleration (reduction of speed) is at the end, resulting in a higher average speed. The symmetry is lost and the elapsed time of accelration in frame M would be different than in frame N. I did not care to attempt to compute that.
Also, 100g of acceleration for 4 days from a stop relative to some frame is impossible. It involves infinite proper acceleration before 4 days.
For SP2 we need to do a Relativity of Simultaneity calculation to compute how far out of sync their clocks will be in this new frame.
The clocks are not out of sync at all. You are mixing two different reference frames. That does not work.
Your baseless assertion, not mine. You’ve shown no self-inconsistency with my numbers, only asserting inconsistency with your fictional ones where there is no RoS, which is of consequence to me only if I can see your alternate version of all the numbers.
The numbers done the right way.
SP1a is at x=0,t=0 at the start (or 0,0). SP2a is at (-10, 0) Units in (light days, days)
After end of acceleration,
Because the inertial observer mentioned below does not see proper acceleration, the spatial (light days) coordinates are wrong but I have no intention of calculating that since it is not relevant.
SP1b coordinates are (2.5123, 4.9082) relative to an unaccelerated observer in the original inertial frame of SP1 and SP2
SP2b coordinates are (-7.4877, 4.9082) relative to an unaccelerated observer in the original inertial frame of SP1 and SP2
Both ships are moving at .8112c which has a Lorentz factor λ of 1.71008 relative to an unaccelerated observer in the original inertial frame of SP1 and SP2
But since both ships have been accelerating at the same rate for the same time in the same direction beginning at an agreed time on synchronized clocks, they are in the same inertial frame and there is no difference in their λ. Neither is there any difference in their proper separation from the start, although that unaccelerated observer sees the whole ensemble contracted.
The SP1 clock reads 4 at SP1b and the SP2 clock reads 4 at SP2b.
At time 4.9082 relative to an unaccelerated observer in the original inertial frame of SP1 and SP2, which is time 4 according to synchronized clocks on the two ships, both ships and the rod are stationary relative to each other. Since both ships have stopped accelerating at the same time, they are in a common inertial frame.
Relativity of Simultaneity refers to the difference in apparent time separation between two events as viewed by two observers not in the same inertial frame. A light is turned on at point A and at Point B at Midnight UTC (synchronized clocks). Observer Alice is on a point midway on a straight line between A and B. Alice sees the light from both A and B at the same time. Bob is on a fast motorcycle on that straight line heading toward A and passes Alice at Midnight UTC on Alice’s watch. Bob will see the light from A earlier than the light from B.
A and B send each other information about the time they turned the lights on and when they saw the light from the other. Since they have synchronized clocks, A and B will then agree that they turned their lights on at the same time, even though each saw the other’s light go on later than his own. If they agree on the time delay value, they know they are in a common inertial frame and can even calculate the proper separation.
The Lorentz formula for Relativity of Simultaneity allows quantifying the degree of disagreement between two observers about the relative timing of two events, generalized to different locations in space for the two observers. It does not represent anything real, as can be seen in the above example. SP1 and SP2 observations of each other disagree about who started first but at the end when they send each other copies of their logs they agree everything was really symmetrical and simultaneous according to their synchronized clocks.
It appears that you do not understand Relativity of Simultaneity either. It is not applicable here.
I see that you have given up on the first picture you presented that attempted to show changing real lengths after I explained how it was wrong. And I see that you have again presented the second picture but this time without the explanatory text that, if you bothered to read and understand it, would have shown you that you were wrong again. Here it is, with the text.
(https://i.imgur.com/sdxp5gT.png)
The horizontal lines show the length of the ruler as it passes by the inertial observer at the origin, with the changes due to changing angles of observation. The horizontal distance (call it W) between where the dashed lines intersect the edges of the colored area correspond to the length of the ruler as seen by the observers at the ends of the ruler. That is the rectangles mentioned in the text formed by the points of intersection, those points being opposite corners of the rectangle. Note that the dashed lines change angles, being steeper where the colored area becomes narrower as seen by the inertial observer. Note especially that because of the changing angles, W is constant. The observers at the end of the ruler never see any change in the length of the ruler. SP1 and SP2 never see any change in their separation and the position of the rod.
What you have is a handful of buzzwords that you do not understand, some online calculating tools that you do not know how to properly apply, and some pretty pictures that do not show what you think they do. The truth is that you do not understand the basic concepts of Relativity Theory.
Give it up. You are wrong.
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The traveling twin thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds c
The driver thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds the speed limit.
Where is the problem?
No problem with that. Proper velocity has no upper limit and can exceed c, as I explained above.
Here is what you said.
The traveling twin thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds c
c is a speed. Acceleration is not. Different units. You’re equivocating the two here. It is meaningless to say that acceleration exceeds some speed.
As I showed, I was not equating speed and acceleration. You were wrong about what you said but rather than admit such a thing, you switch to another subject.
What is holding them in a circular orbit rather than flying out in a straight line and crashing into the track?
Wheels
The track is holding them. If the track disappeared, they would fly off in straight lines despite having wheels.
That they would. Same thing if the spokes disappeared in the spoke example.
The subject was that the track would necessarily move in a direction opposite to the cars courtesy of Newton #3. But you skip that part because you cannot deal with the consequences.
The twins paradox has no answer in SR because acceleration is involved. SR cannot deal with acceleration. There is centripetal acceleration even with a constant speed. SR cannot handle this problem. You need GR and that is what GR says will happen. If you insist on staying in SR, then you have just created another fake twins paradox with no way to answer it.
You’re welcome to invoke GR if that’s the only way you can figure it out. SR contains Lorentz transformations between frames, and that’s all I used in my example.
I did invoke GR and demonstrated that everything, cars and track, are going to contract. No gaps. SR cannot handle ongoing acceleration and that is what you have when circular motion.
Since the situations are symmetric at the beginning and in the acceleration applied according to synchronized clocks points directly at the end results having the same appearance to the participants. Clear concepts first. No need for math in this.
Until you actually do the math as I did and it shows otherwise. Do the trivial math then. I gave coordinates of the four important events (SP1a/b and SP2a/b). If I gave incorrect coordinates, then correct them. Because with my numbers, a 10-LD rod length-contracted down to size 5.8477 isn’t going to bridge the gap between SP1b and SP2b which are still 10 LD apart. My string has broken. You have to have some kind of different numbers to have the string not break, but you’ve shown nothing but Newtonian math, examples with near stationary motion, plus a lot of assertions that don’t add up.
I have addressed your math earlier and shown just how wrong it is. The contraction of the entire ensemble, ships and rod, is only as seen by the inertial observer. To the ships traveling at the same speeds because of identical acceleration histories, contraction does not exist because there is no relative motion.
like the fact that the rod attached to SP2 will not yet move at time SP1a.
Sp1a is an event, not a time. The 10 LD mark on the rod happens to be present at that event, and it indeed begins acceleration at that event, but since it has zero velocity at the moment it begins to accelerate, I’ll have to agree that it does not yet move at that event.
SP1a is the time on SP1’s clock when SP1 acceleration begins. SP2a is the time on SP2’s clock when SP2 acceleration begins. Observing each other’s clocks will not give consistent results because of light speed delays. This is the case even when in common inertial frames. In inertial frames. each will see the other’s clock as behind but ticking at the same rate. Acceleration makes this more complicated leading to strange and contradictory illusions as previously described. Only when the acceleration has ended can they see that the elapsed time on each clock is the same, even though each will still thinks the other’s clock is behind but running at the right rate. Comparing clocks must be done with a shovel of salt.
Bringing two clocks into a common inertial frame will allow comparison, with the slower clock the one that experienced more acceleration.
Not necessarily true. I can sync a pair of clocks, separate them and have clock A undergo 10x the acceleration of clock B, and yet clock B be the slower of the two when they’re compared in each other’s presence. It’s the temporal length of their respective worldlines that counts, not the amount of acceleration.
… with the one having experienced the most acceleration having less elapsed time. If you think you can set up a situation where that is not the case, show me. Don’t need to do complicated calculations. Keep it simple.
Clock A accelerates at 1g for however long it takes it to get to the nearest star and back. Clock B meanwhile gets put into a cyclotron on Earth and is subjected to 10g of continuous acceleration the whole time Clock A is gone. Clock A will have logged less time when they reunite, despite having accelerated far less. I didn’t do the numbers, so I can’t say how much. Can do if you think it matters. Clock B will age negligibly slower than one not in the cyclotron, depending on the radius of the thing.
Wrong. Clock B will have gone much faster and been time dilated much more than Clock A and will be way behind Clock A. It will be obvious which one had experienced more acceleration. Where they went or did not go is irrelevant as long as they end up side by side in a common inertial frame. What makes you think otherwise?
I am reminded of Poul Andersen’s novel Tau Zero from years back. A ship with a gigantic fuel supply and a hyper-efficient engine suffers a problem. The engine won’t turn off.
And nobody thinks to just put the ship into a slow turn so the speed never gets stupid.
And why will that matter? They will still be accelerating until all that fuel runs out. Comparing this and your response above, I get the impression that you think that straight line distance covered has something to do with time dilation. The ultra-high speed particles in accelerators are seriously time dilated as can be seen by the decay times of the unstable byproducts being much longer than usual.
Not objective. Another observer in a different reference frame will observe a different time dilation factor between you and the GPS satellite.
Oopsie. Math shows otherwise.
You said that it was knowledge of the clock rate adjustment on a GPS satellite that allowed you to know the time dilation, which does not really prove anything. You could have been told wrong.
If you were able to see an unadjusted clock riding on the GPS satellite you could then compare the tick rate of that clock with your own and know the time dilation difference per day, although dividing that up into the parts due to relative speed and to gravitational differences would require other information.
An observer in geosynchronous orbit (no motion relative to you to keep it simple) with an even faster clock than the unadjusted GPS clock and faster still than your clock would not see the same difference between you and GPS because its clock is ticking faster than either of the other two. The scale has changed and a different value will result. Moving that other satellite clock up and down will change the tick rate faster and slower. The values of the daily time dilation difference will vary because the scale is changing.
If you think otherwise, show the math.
Even if you exclude other observers from consideration, the GPS satellite will consider you time dilated since you are moving relative to it and are deeper in a gravity well and your clock is running slower.
Relative to the ISS, my clock runs objectively faster, despite my continuous speed relative to it, and despite my greater gravity well depth. I’m only slower relative to GPS because the gravity difference is enough to outweigh the motion difference. GPS satellites don’t move all that fast.
True. ISS has an average altitude of 322 km and a speed of about 28,000 km/h. GPS satellites are way up at 20,200 km and 14,000 km/h. That works out to a very convenient two orbits a day, simplifying ongoing recalibration of position.
Acceleration is in the domain of General Relativity.
Gravity actually
Gravity and acceleration are dealt with by the same mathematical framework. Mathematically gravity is acceleration and vice versa. SR cannot handle acceleration. GR can.
I have shown the inconsistency with your concepts.
You’ve shown my numbers to be inconsistent with your assertions is all. Without showing light speed to be not c somewhere, or to show that events seen in one frame differ from the observations made from the same events in the other frame, or that 1+2 for one observer is different than 1+2 for the other, you’ve not demonstrated any inconsistency at all. You’ve just make empty assertions with no numbers to back them. And that’s all you’ll continue to do, because if you post ‘corrected’ numbers for my scenario, I’ll show exactly those inconsistencies.
Use my scenario: 100g for 4 days. You say there’s no need for mathematics, so it should be trivial for you. Show the numbers in both frames, as I did. But you can’t.
No, you showed mixed numbers, partly from the accelerated frame and partly from an inertial frame and assigned different numbers to SP1 and SP2 despite identical acceleration histories. You do not understand Relativity Theory at all.
I have posted corrected numbers. Let’s see what mistakes you make this time.
Be back when I can,
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The numbers done the right way.
All the numbers are the same as mine, if only the frame M coordinates. You offer no frame N coordinates at all.
If you are correcting something, I expected different numbers. It appears you just copied my stuff, which contradicts your assertion that I had not done them correctly.
We each say the other is wrong, so it is a he-said/she-said sort of situation. But numbers don’t lie. If my numbers are wrong, you need to produce different ones. Producing the same ones means you are in agreement with my assessment of the situation. My numbers show the string breaking as there is over 17 light-days of proper separation between the ships after acceleration. You’ve not shown otherwise since you don’t compute the proper distance between the ships at all.
SP1a is at x=0,t=0 at the start (or 0,0). SP2a is at (-10, 0) Units in (light days, days)
After end of acceleration,
SP1b coordinates are (2.5123, 4.9082) relative to an unaccelerated observer in the original inertial frame pf SP1 and SP2
SP2b coordinates are (-7.4877, 4.9082) relative to an unaccelerated observer in the original inertial frame of SP1 and SP2
Both ships are moving at .8112c which has a Lorentz factor λ of 1.71008 relative to an unaccelerated observer in the original inertial frame of SP1 and SP2
All in agreement, and seemingly copied from my post, except for some of the rearangement of the wording.
But since both ships have been accelerating at the same rate for the same time in the same direction beginning at an agreed time on synchronized clocks, they are in the same inertial frame and there is no difference in their λ.
Agree. They both are moving at the same speed, and thus are stationary relative to each other after both have finished accelerating.
Neither is there any difference in their proper separation from the start, although that unaccelerated observer sees the whole ensemble contracted.
Words, which lie. Show numbers, which don’t. Without that, all I can do is assert that you’re wrong, and my numbers show it, and your lack of different numbers lends zero support to your assertion. I see no computation of their proper separation after acceleration.
The SP1 clock reads 4 at SP1b and the SP2 clock reads 4 at SP2b.
Yes, kind of by definition, since that’s exactly the flight plan.
At time 4.9082 relative to an unaccelerated observer in the original inertial frame of SP1 and SP2,
That took a whole line to say “At 4.9082 in M”. That’s why I gave the frames names, so I didn’t have to type all that each time.
which is time 4 according to synchronized clocks on the two ships, both ships and the rod are stationary relative to each other.
They’re both moving at the same speed in M at that time in M. That’s different than saying they’re stationary relative to each other, because you have to use the ship frame to state that they’re stationary at that time. I go so far as to agree that relative to the ships at event SP2b, both ships and the rod are stationary relative to each other, but I cannot agree with your statement above. Again, without numbers in N, your assertion is just words. I have no coordinates in N to work with, so I cannot show you the contradictions that result from your assertions.
Since both ships have stopped accelerating at the same time, they are in a common inertial frame
They’re ‘in’ all the frames, but each is stationary in only one. I think you meant that, but being clear is nice. It’s because they both accelerate to the same speed that makes them stationary in the same frame after acceleration. When they do it or how long they take to do it makes no difference, so long as net acceleration from M is .8112c. We’re more controlled than that of course, but them being stationary relative to each other has nothing to do with when they stopped accelerating, especially since the ‘when’ part is completely frame dependent.
Relativity of Simultaneity refers to the difference in apparent time separation between two events as viewed by two observers not in the same inertial frame.
It has to do with the frame, and not the observers, and again, all observers are in all frames. But yes, something like how you word it.
A light is turned on at point A and at Point B at Midnight UTC (synchronized clocks). Observer Alice is on a point midway on a straight line between A and B. Alice sees the light from both A and B at the same time.
No frame reference. I presume the clocks are stationary in the frame in which they are in sync, and Alice is also stationary in this same frame. Just making sure.
Bob is on a fast motorcycle on that straight line heading toward A and passes Alice at Midnight UTC on Alice’s watch. Bob will see the light from A earlier than the light from B. A and B send each other information about the time they turned the lights on and when they saw the light from the other. Since they have synchronized clocks, A and B will then agree that they turned their lights on at the same time, even though each saw the other’s light go on later than his own.
If they agree on the time delay value, they know they are in a common inertial frame and can even calculate the proper separation.
They already know they were stationary in the one frame, so the subsequent timings just confirm again that those clocks are indeed synced in that frame, yes.
You didn’t mention what Bob concludes since he doesn’t know where in his frame the light pulses originated, and therefore seems unable to assess which one flashed first. The one might have been before the other, or just closer. So what’s the purpose of Bob here? A stationary observer off to one side of Alice would similarly observe one pulse before the other. RoS seems not to have been well demonstrated with this example.
You didn’t label any events, but there are the two events where the light pulses were emitted, and various observation events. RoS says that the two emission events, if simultaneous in Alice’s frame are not simultaneous in Bob’s frame, even if that wasn’t demonstrated by your example.
I see that you have given up on the first picture you presented that attempted to show changing real lengths after I explained how it was wrong.
I don’t even remember posting a picture. What post is this? Have you now switched to replies to Jaaanosik without indication?
His picture shows a continuously accelerating rigid object (a ruler apparently), much like SP2 at 100g and the first ~5 light days of rod in front of it. It shows some lines of proper simultaneity for the object, but doesn’t bother with putting units on anything. The picture seems accurate enough.
I haven’t much been paying attention to what Jaaanosik has been trying to convey with this picture, but he used it in post 85 to illustrate length contraction, which it does.
Notice that SP2 would be the left end of the ruler (@) accelerating at 100g, and the right end of the ruler (@') corresponds approximately to the 5 LD mark on SP2’s rod. It is not accelerating at 100g as evidenced by the different hyperbolic curves of the two lines. So the two worldlines, accelerating differently, converge in the original frame. SP1 is not like that since it follows the same curve as the left side of the ruler, so there is no contraction of the distance between them, but the rod in front of SP2 does contract, so 10 LD is insufficient to reach SP1 once they’re moving.
… and therefore exceeds c
As I showed, I was not equating speed and acceleration. You were wrong about what you said but rather than admit such a thing, you switch to another subject.
I was commenting on the bit I quote just above. You didn’t say what exceeded c. Turns out you were talking about proper velocity, but apparently without knowing the term, and once that was clarified, I agreed that yes, that can exceed c.
The subject was that the track would necessarily move in a direction opposite to the cars courtesy of Newton #3. But you skip that part because you cannot deal with the consequences.
It would not since there is not just the one car, but a series of them equally spaced. The vector sum of the forces on the track is zero and it stays put. Surely you know this, and are just trying to nitpick. I was started to back off the troll assessment when I thought you were putting out numbers, but no, you just copied mine and put out comments like this one above.
I have addressed your math earlier and shown just how wrong it is.
You quoted a subset of the exact same numbers, either showing how much you agree with what I’ve said, or you are unable to do the math so you just copied them. Numbers don’t lie.
Then you added a bunch of words unbacked by different numbers.
The contraction of the entire ensemble, ships and rod, is only as seen by the inertial observer.
Wrong. It is contracted only relative to a frame in which it is moving. SP2 is inertial after it shuts down, and yet it is not then contracted in the SP2 frame (N), so that contradicts your statement above. The statement also implies that any accelerating observer will not see SP2 contracted, which is wrong as well. Some observer could be accelerating the other way for instance. SP2 will be contracted relative to him.
To the ships traveling at the same speeds because of identical acceleration histories, contraction does not exist because there is no relative motion.
Agree if you add ‘Relative’ in front of that statement. For instance, I claim no length contraction of the rod in frame N once SP2 has stopped accelerating. Look at the numbers. Don’t take my word for it. If the rod were contracted, a length 17.1 rod would not span from N coordinates -19.6131 to -2.5123 where SP1 is parked. It needs its full uncontracted length to reach between them. Numbers don’t lie.
Sp1a is an event, not a time.
SP1a is the time on SP1’s clock when SP1 acceleration begins.
OK, so you don’t know what an event is. An event is an objective point in spacetime, a frame-invariant fact. The location and time of an event is relation with a coordinate system, typically an inertial frame.
Anyway, the coordinates of SP1a is (0, 0) in both frames since we’ve chosen that event as our common origin. That’s two values. A time is one value. SP1a occurs at time 0 in both frame, but time zero does not define an event since SP2a also occurs at time 0, at least in frame M, and that is a different event.
Clock A accelerates at 1g for however long it takes it to get to the nearest star and back. Clock B meanwhile gets put into a cyclotron on Earth and is subjected to 10g of continuous acceleration the whole time Clock A is gone. Clock A will have logged less time when they reunite, despite having accelerated far less. I didn’t do the numbers, so I can’t say how much. Can do if you think it matters. Clock B will age negligibly slower than one not in the cyclotron, depending on the radius of the thing.
Wrong.
Thank you. I’d question my statement if you agreed with me. You’re getting predictable.
Clock B will have gone much faster and been time dilated much more than Clock A and will be way behind Clock A. It will be obvious which one had experienced more acceleration. Where they went or did not go is irrelevant as long as they end up side by side in a common inertial frame. What makes you think otherwise?
Want numbers? You first this time. Pick a real distance to a star and say a cyclotron with a 20m radius, which seems reasonable. Otherwise, just more words from you. You seem to have the mistaken impression that the magnitude of time dilation is a function of the magnitude of acceleration, so I picked a counterexample as requested.
No, you showed mixed numbers, partly from the accelerated frame and partly from an inertial frame and assigned different numbers to SP1 and SP2 despite identical acceleration histories.
They’re not mixed. They’re carefully separated in discussions relative to each frame, just like Einstein talks about the frame of the train and the frame of the platform, but with no statement mixing the two frames.
All events are consistent between the two frames. That shows I did it correctly.
If thinking in terms of observers helps you, just consider the N stuff to be relative to some inertial observer stationary the whole time in frame N, watching the ships go from -.8112c to a halt over the course of 4.9082 days each. All the N coordinates are relative to that observer. So if my numbers are wrong relative to that unaccelerated observer (who is at location 0 the whole time), then give your own numbers. I never once referenced an accelerated reference frame, so I used nothing but inertial SR mathematics to arrive at my numbers. Feel free to do it your own way.
I have posted corrected numbers.
You mean correct numbers, not corrected, since they were my numbers.
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The observers at the end of the ruler never see any change in the length of the ruler. SP1 and SP2 never see any change in their separation and the position of the rod.
This is not so straight forward.
Are you suggesting a rod between the two spaceships?
If yes, then you need to understand/explain the post #101.
Do you see where is the simultaneity line?
(https://i.imgur.com/xY6D6Zg.png)
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SP1 and SP2 never see any change in their separation and the position of the rod.
This is not so straight forward.
Are you suggesting a rod between the two spaceships?
Yes, the rod was first suggested in post 3 and defined exactly in post 8 where I affixed it to the front of the rear ship rather than having it trail behind the lead ship, which leads to it breaking even if not attached to anything.
So our scenario since post 14 has been a pair of negligible length ships (shorter than about 10 light minutes, the precision of the calculations) initially separated by 10 light days, which accelerate at 100g for 4 days ship time. The rod in front of the rear ship is 20 light days long, so it extends past the front ship, who therefore at any time can determine the separation between the two by just reading the markings on it. No need to send time-consuming light signals between the ships.
If yes, then you need to understand/explain the post #101.
Do you see where is the simultaneity line?
Why did you post the same picture twice in that post?
It is sort of appropriate, except it depicts continuous acceleration, whereas our scenario has them stop accelerating after 4 subjective days. M-L claims that the proper distance between the ships will somehow revert to the original 10 light-day separation after they stop accelerating. Your picture shows continuous acceleration, and thus perhaps doesn't clearly show it. It also lacks numbers on the ruler, making it difficult to assess the separation in a frame not depicted.
The picture BTW seems entirely accurate to me, even if it doesn't depict our exact scenario.
Based on the most recent post, M-L seems in total denial of the existence of the frame in which both ships are subsequently stopped, as he refuses to compute any numbers for it. It would of course show that length contraction is real, so I suppose that's why he feels the need to take that stance.
I can think of three possibilities why he would do that:
1) He actually has as little clue about relativity as he presents himself to have. The only relativistic math I've seen in any post is a chart of speed-to-Lorentz-factors. Never any math applied to a specific scenario except very slow speed examples using Newtonian mathematics.
2) He has a pathological fear of being shown wrong, and refuses to show math that would illustrate it.
I personally love being shown to be wrong, because I'm here to learn, not to be correct.
My numbers could very well have mistakes in them. I did them quickly without double-check. They need interval checks for instance, the first thing I'd do with somebody else's numbers.
3) He knows his relativity quite well, but is a troll getting his jollies from yanking our collective chains.
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SP1 and SP2 never see any change in their separation and the position of the rod.
This is not so straight forward.
Are you suggesting a rod between the two spaceships?
Yes, the rod was first suggested in post 3 and defined exactly in post 8 where I affixed it to the front of the rear ship rather than having it trail behind the lead ship, which leads to it breaking even if not attached to anything.
So our scenario since post 14 has been a pair of negligible length ships (shorter than about 10 light minutes, the precision of the calculations) initially separated by 10 light days, which accelerate at 100g for 4 days ship time. The rod in front of the rear ship is 20 light days long, so it extends past the front ship, who therefore at any time can determine the separation between the two by just reading the markings on it. No need to send time-consuming light signals between the ships.
If yes, then you need to understand/explain the post #101.
Do you see where is the simultaneity line?
Why did you post the same picture twice in that post?
It is sort of appropriate, except it depicts continuous acceleration, whereas our scenario has them stop accelerating after 4 subjective days. M-L claims that the proper distance between the ships will somehow revert to the original 10 light-day separation after they stop accelerating. Your picture shows continuous acceleration, and thus perhaps doesn't clearly show it. It also lacks numbers on the ruler, making it difficult to assess the separation in a frame not depicted.
The picture BTW seems entirely accurate to me, even if it doesn't depict our exact scenario.
Based on the most recent post, M-L seems in total denial of the existence of the frame in which both ships are subsequently stopped, as he refuses to compute any numbers for it. It would of course show that length contraction is real, so I suppose that's why he feels the need to take that stance.
I can think of three possibilities why he would do that:
1) He actually has as little clue about relativity as he presents himself to have. The only relativistic math I've seen in any post is a chart of speed-to-Lorentz-factors. Never any math applied to a specific scenario except very slow speed examples using Newtonian mathematics.
2) He has a pathological fear of being shown wrong, and refuses to show math that would illustrate it.
I personally love being shown to be wrong, because I'm here to learn, not to be correct.
My numbers could very well have mistakes in them. I did them quickly without double-check. They need interval checks for instance, the first thing I'd do with somebody else's numbers.
3) He knows his relativity quite well, but is a troll getting his jollies from yanking our collective chains.
Halc,
I see your point about acceleration and stopping.
This might be one way how to look at it.
Having said that, this is not a uniform acceleration between the two spaceships as per the textbook definition.
It is a correct uniform acceleration/deceleration per individual spaceships though.
Jano
(https://i.imgur.com/i0iXbzI.png)
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Halc,
I see your point about acceleration and stopping.
No, not stop moving, just cease accelerating. The ships continue at the same inertial speed off to the right, not slowing back to the original frame.
You picture also shows one rigid object accelerating up and down to a stop, not two separate objects accelerating identically.
There's seemingly strain being put on the object in your picture. If it was brittle (not bendy), then it would perhaps break given what being done to it. Hard to tell since the speed never gets high enough to make it obvious.
Having said that, this is not a uniform acceleration between the two spaceships as per the textbook definition.
Not sure what you consider uniform. Our ships accelerate at a constant proper acceleration of 100g for 4 days ship time. That's identical acceleration, not necessarily uniform, depending on your definition of uniform. If it means the object is always stationary along its length in its own frame, then yes, it illustrates that sort of motion.
Your prior picture I think showed uniform acceleration, where the left side of the ruler accelerates a higher g force than the right side. Those two endpoints are not representative of the motion of our two ships.
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SP1a [0,0]
SP1 proper acceleration is 100g
SP1 duration of proper acceleration is 4 days
SP1 final proper speed is 1.130 c
SP1 proper acceleration has been continuous, so average proper speed is 0.565 c
SP1 proper distance covered is 2.261 LD
SP1b proper [2.261,4]
SP2a [-10,0]
SP2 proper acceleration is 100g
SP2 duration of proper acceleration is 4 days
SP2 final proper speed is 1.130 c
SP2 proper acceleration has been continuous, so average proper speed is 0.565 c
SP2 proper distance covered is 2.261 LD
SP2b proper [-7.739,4]
2.261 -(-7.739) = 10 LD
Same proper separation between SP1 and SP2 at end of acceleration as at start of acceleration.
The rod has been defined as having constant acceleration along its length. Attached thrusters or whatever.
SP1 end [0,0]
SP1 end proper acceleration is 100g
SP1 end duration of proper acceleration is 4 days
SP1 end final proper speed is 1.130 c
SP1 end proper acceleration has been continuous, so average proper speed is 0.565 c
SP1 end proper distance covered is 2.261 LD
SP1 end at SP1b proper [2.261,4]
SP2 end [0,-10]
SP2 end proper acceleration is 100g
SP2 end duration of proper acceleration is 4 days
SP2 end final proper speed is 1.130 c
SP2 end proper acceleration has been continuous, so average proper speed is 0.565 c
SP2 end proper distance covered is 2.261 LD
SP2 end at SP2b proper [-7.739,4]
2.261 -(-7.739) = 10 LD
Same proper separation between SP1 end of rod and SP2 end of rod at end of acceleration as at start of acceleration. The rod is the same length at the beginning and at the end.
Let us define SP1x as Time = 2 on the SP1 clock
SP1 proper acceleration is 100g
SP1 duration of proper acceleration is 2 days
SP1 proper speed is 0.565 c
SP1 proper acceleration has been continuous, so average proper speed is 0.2825 c
SP1 at SP1x proper distance covered is 0.565 LD
SP1 at SP1x proper [0.565,2]
Let us define SP2x as Time = 2 on the SP2 clock
SP2 proper acceleration is 100g
SP2 duration of proper acceleration is 2 days
SP2 proper speed is 0.565 c
SP2 proper acceleration has been continuous, so average proper speed is 0.2825 c
SP2 at SP2x proper distance covered is 0.565 LD
SP2 at SP2x proper [-9.435,2]
Now the rod
SP1 end proper acceleration is 100g
SP1 end duration of proper acceleration is 2 days
SP1 end final proper speed is 0.565 c
SP1 end proper acceleration has been continuous, so average proper speed is 0.2825 c
SP1 end at SP1x proper distance covered is 0.565 LD
SP1 end at SP1x proper [0.565,2]
The difference between SP1’s proper distance covered and the SP1 end of the rod say SP1x is 0.565 - 0.565 = 0
The SP1 end of the rod is motionless outside SP1’s window for any SP1 time you choose.
Since we have already seen that the final proper separation is the same as the starting separation, the rod cannot be moving after SP1b.
Does the proper length of the rod ever change?
Let Rh be the point on the rod 5 LD forward of SP2 and 5 LD aft of SP1 at t=0 before any acceleration begins. That is, at the mid-point of the rod. Coordinates are [-5,0]
How long is the rod at the 2 day mark?
Rh proper acceleration is 100g
Rh duration of proper acceleration is 2 days
Rh proper speed is 0.565 c
Rh proper acceleration has been continuous, so average proper speed is 0.2825 c
Rh at 2 days proper distance covered is 0.565 LD
Rh at 2 days proper [-5.435,2]
Distance to SP1 at 2 day mark is 0.565 - (-4.435) = 5 LD
Distance to SP2 at 2 day mark is -4.435 - (-9.435) = 5 LD
The proper length of the rod does not change at any point in time.
As I have been saying all along:
Since both ships have been accelerating at the same rate for the same time in the same direction beginning at an agreed time on synchronized clocks, they are in the same inertial frame and there will be no difference in proper lengths.
If you had your head on straight about basic Relativity concepts this would have been obvious from the start. But you don’t. You blithely mix up proper values and external observer values and think you are saying something meaningful. You’re not.
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Halc,
I see your point about acceleration and stopping.
No, not stop moving, just cease accelerating. The ships continue at the same inertial speed off to the right, not slowing back to the original frame.
You picture also shows one rigid object accelerating up and down to a stop, not two separate objects accelerating identically.
There's seemingly strain being put on the object in your picture. If it was brittle (not bendy), then it would perhaps break given what being done to it. Hard to tell since the speed never gets high enough to make it obvious.
Having said that, this is not a uniform acceleration between the two spaceships as per the textbook definition.
Not sure what you consider uniform. Our ships accelerate at a constant proper acceleration of 100g for 4 days ship time. That's identical acceleration, not necessarily uniform, depending on your definition of uniform. If it means the object is always stationary along its length in its own frame, then yes, it illustrates that sort of motion.
Your prior picture I think showed uniform acceleration, where the left side of the ruler accelerates a higher g force than the right side. Those two endpoints are not representative of the motion of our two ships.
Then this is the better diagram.
(https://i.imgur.com/xY6D6Zg.png)
the proper distance between the ships will somehow revert to the original 10 light-day separation after they stop accelerating.
If both spaceship pause acceleration then the distance between the spaceships will not revert to the original 10 light-day interval according to the SR.
Jano
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Great. An attempt to obfuscate by specifying everything in proper terms instead of coordinate terms.
Let’s see if the numbers hold up...
SP1a [0,0]
SP1 proper acceleration is 100g
SP1 duration of proper acceleration is 4 days
SP1 final proper speed is 1.130 c
SP1 proper acceleration has been continuous, so average proper speed is 0.565 c
SP1 proper distance covered is 2.261 LD
SP1b proper [2.261,4]
Still using Newtonian mathematics I see. This is wrong. The actual proper speed is 1.387c
Now me just saying that is words, so let me let the numbers speak.
You apparently got the proper speed the Newtonian way by multiplying 980m/sec2 by 345600 seconds which indeed yields a proper speed (dx/dt’) of 1.13c. This corresponds to a coordinate speed (dx/dt) of about .749c, not .8112c. Doing it this way leads to contradictions:
Suppose we accelerated for 8 days instead of 4. Using your linear method, that would double the proper speed to 2.26c, but that contradicts the relativistic computation of proper speed:
.749c coordinate speed doubled using relativistic addition is .9596c, which has a λ of 3.5558 yielding a proper speed of 3.4123c which contradicts the 2.26c using your method. Proper speeds don’t add in a linear way like you’re doing. You can’t use Newtonian method of just multiplying proper time by proper acceleration.
The ‘proper distance’ covered (as you’re using the term) is 2.5123 as you had in your prior post, but I see you’ve changed it to 2.261 now.
SP2a [-10,0]
SP2 proper acceleration is 100g
SP2 duration of proper acceleration is 4 days
SP2 final proper speed is 1.130 c
SP2 proper acceleration has been continuous, so average proper speed is 0.565 c
SP2 proper distance covered is 2.261 LD
SP2b proper [-7.739,4]
2.261 -(-7.739) = 10 LD
OK, that much is the same I computed. In frame M (your figures are all relative to frame M), the ships are always 10 LD apart.
You’re suggesting that a moving 10 LD rod can bridge the 10 LD gap between (your) coordinates -7.739 and 2.261 in frame M. That cannot be unless the rod is not length contracted, so you seem to be in denial of relativistic length contraction. That’s at least one contradiction I can point out without any numbers in N.
Same proper separation between SP1 and SP2 at end of acceleration as at start of acceleration.
You’re using ‘proper separation’ incorrectly. Yes, the proper separation between the grid markers stationary in frame M is 10 in frame M, but proper separation only applies to mutually stationary things, so that’s only the proper distance between the M-coordinate grid markers, not the proper separation between the ships because they’re now moving in the M frame which is the only frame you’ve referenced in this entire post. The proper separation of the ships hasn’t been computed. Proper separation of the ships is by definition the distance between the ships in the frame in which they’re both stationary, which is frame N. You’ve performed no computations in frame N. so you’ve not computed their proper separation.
The rod has been defined as having constant acceleration along its length.
It most certainly has not. The rod has been defined to be stationary in the frame of SP2 at all times.
Asserting it to have constant acceleration along its length is just words. Back it with mathematics in the frame of the rod, and you’ll see the contradiction that results from this assertion. Jaaasonik’s picture he keeps posting over and over illustrates the situation exactly, showing a rigid rod accelerating with constant proper length. It accelerates at a different rate along its length. The picture illustrates that nicely, but you seem to be in denial of any illustrations taken from accepted texts, preferring instead to make up your own facts.
SP1 end [0,0]
What does this mean? That SP1 has gone nowhere in 4 days? This whole section lacks a frame reference, so it’s not immediately clear what you’re trying to convey here.
SP1 end proper acceleration is 100g
SP1 end duration of proper acceleration is 4 days
SP1 end final proper speed is 1.130 c
SP1 end proper acceleration has been continuous, so average proper speed is 0.565 c
SP1 end proper distance covered is 2.261 LD
SP1 end at SP1b proper [2.261,4]
SP2 end [0,-10]
SP2 end proper acceleration is 100g
SP2 end duration of proper acceleration is 4 days
SP2 end final proper speed is 1.130 c
SP2 end proper acceleration has been continuous, so average proper speed is 0.565 c
SP2 end proper distance covered is 2.261 LD
SP2 end at SP2b proper [-7.739,4]
2.261 -(-7.739) = 10 LD
This just seems to be a repeat of the numbers above, but with the word ‘end’ stuck in at various places. It’s all in frame M, so indeed, you seem to be asserting that between start and end, neither ship has changed coordinates, which means it hasn’t moved. I don’t think you mean that, but it is totally unclear what you’re trying to convey with this repeat of the same numbers. They’re all still mostly proper numbers relative to frame M, same as the first time.
Same proper separation between SP1 end of rod and SP2 end of rod at end of acceleration as at start of acceleration. The rod is the same length at the beginning and at the end.
Repeating the numbers and repeating a wrong conclusion doesn’t make it right the second time. You’re still not computing proper distance correctly. Look up the definition of it.
Let us define SP1x as Time = 2 on the SP1 clock
An event presumably?
SP1 proper acceleration is 100g
SP1 duration of proper acceleration is 2 days
SP1 proper speed is 0.565 c
SP1 proper acceleration has been continuous, so average proper speed is 0.2825 c
SP1 at SP1x proper distance covered is 0.565 LD
SP1 at SP1x proper [0.565,2]
Let us define SP2x as Time = 2 on the SP2 clock
SP2 proper acceleration is 100g
SP2 duration of proper acceleration is 2 days
SP2 proper speed is 0.565 c
SP2 proper acceleration has been continuous, so average proper speed is 0.2825 c
SP2 at SP2x proper distance covered is 0.565 LD
SP2 at SP2x proper [-9.435,2]
This is all Newtonian math, which has been shown above to lead to contradictions. The separation between them in M is still 10 of course, even with these wrong numbers. But that’s the proper distance of the grid markers in M, not the proper distance between the ships since A) the ships are not stationary, and B) no frame has been demonstrated to exist where both ships are simultaneously moving at this half-way speed.
Now the rod
SP1 end proper acceleration is 100g
Again I have no idea how this statement is distinct from “SP1 proper acceleration is 100g”. I don’t know what ‘end’ means inserted in like that in all these statements.
SP1 end duration of proper acceleration is 2 days
SP1 end final proper speed is 0.565 c
SP1 end proper acceleration has been continuous, so average proper speed is 0.2825 c
SP1 end at SP1x proper distance covered is 0.565 LD
SP1 end at SP1x proper [0.565,2]
The difference between SP1’s proper distance covered and the SP1 end of the rod say SP1x is 0.565 - 0.565 = 0
You’re assuming the end of the rod is accelerating at 100g, which it cannot be. Compute the location of the rod in frame N and this assumption will lead to contradictions. You perhaps know this, because you avoid doing such computations like the plague. If your assertions were correct, the numbers would show them to be since they’d be self consistent. Actually, I just think you’re not up to the math since all I’ve ever seen you use is Newtonian physics.
The SP1 end of the rod is motionless outside SP1’s window for any SP1 time you choose.
My numbers in N (with which no inconsistency has been demonstrated) show otherwise. Your complete lack of numbers relative to N lend zero support to these mere words. This seems to the main point of contention, so you really need to back this one up. How far out of sync are the ship clocks in frame in which both ships are now stationary?
Let Rh be the point on the rod 5 LD forward of SP2 and 5 LD aft of SP1 at t=0 before any acceleration begins. That is, at the mid-point of the rod. Coordinates are [-5,0]
Ooh, actual coordinates of an event. Things are improving.
How long is the rod at the 2 day mark?
Frame dependent question.
Rh proper acceleration is 100g
You’ve not established that with any numbers in the frame of the rod. So your argument falls apart here.
The ‘ruler’ in Jaaasonik’s picture shows a rod about exactly that length of about 5 LD at 100g. That means the left edge of the yellow region is SP2 and the right edge is Rh, which according to the textbook from which that illustration was taken, has significantly lower acceleration. You seem to have been denying the textbooks when it comes to SR, so that leaves your numbers to back your assertions. Still waiting for the N numbers.
The proper length of the rod does not change at any point in time.
By definition, yes. An actual statement with which I agree. That sort of wrecks your batting average, no?
Since both ships have been accelerating at the same rate for the same time in the same direction beginning at an agreed time on synchronized clocks, they are in the same inertial frame and there will be no difference in proper lengths.
But there will be a change in their proper separation, and the rod which did not change its proper length will not be able to span this new proper separation. The proper length of anything cannot change unless the object is deformed. It would then not be rigid, and we’ve defined all our objects to be rigid.
If both spaceship pause acceleration then the distance between the spaceships will not revert to the original 10 light-day interval according to the SR.
The distance between them is frame dependent.
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The numbers done the right way.
All the numbers are the same as mine, if only the frame M coordinates. You offer no frame N coordinates at all.
If you are correcting something, I expected different numbers. It appears you just copied my stuff, which contradicts your assertion that I had not done them correctly.
We each say the other is wrong, so it is a he-said/she-said sort of situation. But numbers don’t lie. If my numbers are wrong, you need to produce different ones. Producing the same ones means you are in agreement with my assessment of the situation. My numbers show the string breaking as there is over 17 light-days of proper separation between the ships after acceleration. You’ve not shown otherwise since you don’t compute the proper distance between the ships at all.
You asked for proper numbers, as seen by SP1 and SP2. As can be clearly seen from those numbers, at all times the proper lengths and location of SP1, the rod and SP2 never change. The rod stops when SP1 stops. The proper separation never changes. Your numbers are wrong because you mix proper numbers and external observer numbers. It is clear that you have no knowledge of how Relativity Theory works.
SP1a is at x=0,t=0 at the start (or 0,0). SP2a is at (-10, 0) Units in (light days, days)
After end of acceleration,
SP1b coordinates are (2.5123, 4.9082) relative to an unaccelerated observer in the original inertial frame pf SP1 and SP2
SP2b coordinates are (-7.4877, 4.9082) relative to an unaccelerated observer in the original inertial frame of SP1 and SP2
Both ships are moving at .8112c which has a Lorentz factor λ of 1.71008 relative to an unaccelerated observer in the original inertial frame of SP1 and SP2
All in agreement, and seemingly copied from my post, except for some of the rearangement of the wording.
You are still missing the point. The λ of SP1, the rod and SP2 are identical as seen by an unaccelerated observer. From their own viewpoints, all lengths are the same. And their relative λ values are zero. You apply the λ only to SP2. That is wrong. You cannot mix, you can only match.
But since both ships have been accelerating at the same rate for the same time in the same direction beginning at an agreed time on synchronized clocks, they are in the same inertial frame and there is no difference in their λ.
Agree. They both are moving at the same speed, and thus are stationary relative to each other after both have finished accelerating.
And as I have shown in a recent post, within their own common reference frame, no separations have changed. Relative to an unaccelerated observer, all separations have changed to the same degree. The rod fits between SP1 and SP2 exactly as before.
Neither is there any difference in their proper separation from the start, although that unaccelerated observer sees the whole ensemble contracted.
Words, which lie. Show numbers, which don’t. Without that, all I can do is assert that you’re wrong, and my numbers show it, and your lack of different numbers lends zero support to your assertion. I see no computation of their proper separation after acceleration.[/quote]
I have provided lots of numbers about proper separation. It is identical at all times.
The SP1 clock reads 4 at SP1b and the SP2 clock reads 4 at SP2b.
Yes, kind of by definition, since that’s exactly the flight plan.
At time 4.9082 relative to an unaccelerated observer in the original inertial frame of SP1 and SP2,
That took a whole line to say “At 4.9082 in M”. That’s why I gave the frames names, so I didn’t have to type all that each time.
I was emphasizing that SP1, rod and SP2 are all in one reference frame and the observer is in another, a point you are having a problem with. BTW did you know that SP1, rod, SP2 and M were all originally accelerated from frame Z before t=0? What separation change between SP1 and SP2 does observer Z see? After all he sees them as going even faster with an even bigger λ. You are insisting on a change in proper separation as seen by SP1 and SP2. So who is right, M or Z?
which is time 4 according to synchronized clocks on the two ships, both ships and the rod are stationary relative to each other.
They’re both moving at the same speed in M at that time in M. That’s different than saying they’re stationary relative to each other, because you have to use the ship frame to state that they’re stationary at that time. I go so far as to agree that relative to the ships at event SP2b, both ships and the rod are stationary relative to each other, but I cannot agree with your statement above. Again, without numbers in N, your assertion is just words. I have no coordinates in N to work with, so I cannot show you the contradictions that result from your assertions.
I have provided numbers. The fact that you needed numbers to understand the result shows that you are seriously lacking in comprehension of the basic concepts of Relativity Theory.
Since both ships have stopped accelerating at the same time, they are in a common inertial frame
They’re ‘in’ all the frames, but each is stationary in only one. I think you meant that, but being clear is nice. It’s because they both accelerate to the same speed that makes them stationary in the same frame after acceleration. When they do it or how long they take to do it makes no difference, so long as net acceleration from M is .8112c. We’re more controlled than that of course, but them being stationary relative to each other has nothing to do with when they stopped accelerating, especially since the ‘when’ part is completely frame dependent.
They are stationary in the common inertial frame that they (SP1, rod, SP2) are all in, as I have demonstrated in the recent proper numbers post. They are not moving relative to each other at any time. Proper separation and proper length of the rod never change.
Relativity of Simultaneity refers to the difference in apparent time separation between two events as viewed by two observers not in the same inertial frame.
It has to do with the frame, and not the observers, and again, all observers are in all frames. But yes, something like how you word it.
Wrong. An observer is in the frame he is in. He is not in any other frame. If he were in all frames, then he would have all length contractions, and all time dilations and all mass-energy values simultaneously. Not to mention that he would be traveling in all directions at all speeds.
The frame an observer is in is simply the one and only proper frame he is in where he will measure the speed of light to be the textbook value. He might be observed from all sorts of other frames but those are relative, not objective. This is why it is called Relativity Theory.
You don’t even talk the talk, but you want to claim you walk the walk.
A light is turned on at point A and at Point B at Midnight UTC (synchronized clocks). Observer Alice is on a point midway on a straight line between A and B. Alice sees the light from both A and B at the same time.
No frame reference. I presume the clocks are stationary in the frame in which they are in sync, and Alice is also stationary in this same frame. Just making sure.
That Alice is in the midpoint and see the lights at the same time tells us that they are all in a common inertial frame. No need to specify. Seems that you do not understand that point. Oh wait! You don’t! You think that the proper length of the rod will contract. Or stretch. Which one was it again?
Bob is on a fast motorcycle on that straight line heading toward A and passes Alice at Midnight UTC on Alice’s watch. Bob will see the light from A earlier than the light from B. A and B send each other information about the time they turned the lights on and when they saw the light from the other. Since they have synchronized clocks, A and B will then agree that they turned their lights on at the same time, even though each saw the other’s light go on later than his own.
If they agree on the time delay value, they know they are in a common inertial frame and can even calculate the proper separation.
They already know they were stationary in the one frame, so the subsequent timings just confirm again that those clocks are indeed synced in that frame, yes.
No, A and B did not know for sure that they were in the same inertial frame. Alice knew because she was in the midpoint because repeated readings in each direction with her laser range finder always had the same value. The lightkeepers at A and B did not know until they compared notes.
You didn’t mention what Bob concludes since he doesn’t know where in his frame the light pulses originated, and therefore seems unable to assess which one flashed first. The one might have been before the other, or just closer. So what’s the purpose of Bob here? A stationary observer off to one side of Alice would similarly observe one pulse before the other. RoS seems not to have been well demonstrated with this example.
Bob sees the light flashes at different times. He has no idea of how far off they are or if they are synchronized or moving (only Alice knows before the fact). He only knows that he saw them at different times. His observation is that they were not simultaneous.
You didn’t label any events, but there are the two events where the light pulses were emitted, and various observation events. RoS says that the two emission events, if simultaneous in Alice’s frame are not simultaneous in Bob’s frame, even if that wasn’t demonstrated by your example.
A flashed at Midnight UTC. B flashed at Midnight UTC. Alice observed both flashes simultaneously at some time after Midnight UTC, the exact time difference being irrelevant. Bob passed Alice at Midnight UTC. His exact speed is irrelevant as long as it is non-zero with respect to Alice and that it is in the direction of A.
If you need exact numbers and math to understand this, then you do not understand the principle behind Relativity of Simultaneity and do not know how to apply it properly.
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The numbers done the right way.
All the numbers are the same as mine, if only the frame M coordinates. You offer no frame N coordinates at all.
If you are correcting something, I expected different numbers. It appears you just copied my stuff, which contradicts your assertion that I had not done them correctly.
We each say the other is wrong, so it is a he-said/she-said sort of situation. But numbers don’t lie. If my numbers are wrong, you need to produce different ones. Producing the same ones means you are in agreement with my assessment of the situation. My numbers show the string breaking as there is over 17 light-days of proper separation between the ships after acceleration. You’ve not shown otherwise since you don’t compute the proper distance between the ships at all.
You asked for proper numbers, as seen by SP1 and SP2. As can be clearly seen from those numbers, at all times the proper lengths and location of SP1, the rod and SP2 never change. The rod stops when SP1 stops. The proper separation never changes. Your numbers are wrong because you mix proper numbers and external observer numbers. It is clear that you have no knowledge of how Relativity Theory works.
SP1a is at x=0,t=0 at the start (or 0,0). SP2a is at (-10, 0) Units in (light days, days)
After end of acceleration,
SP1b coordinates are (2.5123, 4.9082) relative to an unaccelerated observer in the original inertial frame pf SP1 and SP2
SP2b coordinates are (-7.4877, 4.9082) relative to an unaccelerated observer in the original inertial frame of SP1 and SP2
Both ships are moving at .8112c which has a Lorentz factor λ of 1.71008 relative to an unaccelerated observer in the original inertial frame of SP1 and SP2
All in agreement, and seemingly copied from my post, except for some of the rearangement of the wording.
You are still missing the point. The λ of SP1, the rod and SP2 are identical as seen by an unaccelerated observer. From their own viewpoints, all lengths are the same. And their relative λ values are zero. You apply the λ only to SP2. That is wrong. You cannot mix, you can only match.
But since both ships have been accelerating at the same rate for the same time in the same direction beginning at an agreed time on synchronized clocks, they are in the same inertial frame and there is no difference in their λ.
Agree. They both are moving at the same speed, and thus are stationary relative to each other after both have finished accelerating.
And as I have shown in a recent post, within their own common reference frame, no separations have changed. Relative to an unaccelerated observer, all separations have changed to the same degree. The rod fits between SP1 and SP2 exactly as before.
Neither is there any difference in their proper separation from the start, although that unaccelerated observer sees the whole ensemble contracted.
Words, which lie. Show numbers, which don’t. Without that, all I can do is assert that you’re wrong, and my numbers show it, and your lack of different numbers lends zero support to your assertion. I see no computation of their proper separation after acceleration.
I have provided lots of numbers about proper separation. It is identical at all times.
The SP1 clock reads 4 at SP1b and the SP2 clock reads 4 at SP2b.
Yes, kind of by definition, since that’s exactly the flight plan.
At time 4.9082 relative to an unaccelerated observer in the original inertial frame of SP1 and SP2,
That took a whole line to say “At 4.9082 in M”. That’s why I gave the frames names, so I didn’t have to type all that each time.
I was emphasizing that SP1, rod and SP2 are all in one reference frame and the observer is in another, a point you are having a problem with. BTW did you know that SP1, rod, SP2 and M were all originally accelerated from frame Z before t=0? What separation change between SP1 and SP2 does observer Z see? After all he sees them as going even faster with an even bigger λ. You are insisting on a change in proper separation as seen by SP1 and SP2. So who is right, M or Z?
which is time 4 according to synchronized clocks on the two ships, both ships and the rod are stationary relative to each other.
They’re both moving at the same speed in M at that time in M. That’s different than saying they’re stationary relative to each other, because you have to use the ship frame to state that they’re stationary at that time. I go so far as to agree that relative to the ships at event SP2b, both ships and the rod are stationary relative to each other, but I cannot agree with your statement above. Again, without numbers in N, your assertion is just words. I have no coordinates in N to work with, so I cannot show you the contradictions that result from your assertions.
I have provided numbers. The fact that you needed numbers to understand the result shows that you are seriously lacking in comprehension of the basic concepts of Relativity Theory.
Since both ships have stopped accelerating at the same time, they are in a common inertial frame
They’re ‘in’ all the frames, but each is stationary in only one. I think you meant that, but being clear is nice. It’s because they both accelerate to the same speed that makes them stationary in the same frame after acceleration. When they do it or how long they take to do it makes no difference, so long as net acceleration from M is .8112c. We’re more controlled than that of course, but them being stationary relative to each other has nothing to do with when they stopped accelerating, especially since the ‘when’ part is completely frame dependent.
They are stationary in the common inertial frame that they (SP1, rod, SP2) are all in, as I have demonstrated in the recent proper numbers post. They are not moving relative to each other at any time. Proper separation and proper length of the rod never change.
Relativity of Simultaneity refers to the difference in apparent time separation between two events as viewed by two observers not in the same inertial frame.
It has to do with the frame, and not the observers, and again, all observers are in all frames. But yes, something like how you word it.
Wrong. An observer is in the frame he is in. He is not in any other frame. If he were in all frames, then he would have all length contractions, and all time dilations and all mass-energy values simultaneously. Not to mention that he would be traveling in all directions at all speeds.
The frame an observer is in is simply the one and only proper frame he is in where he will measure the speed of light to be the textbook value. He might be observed from all sorts of other frames but those are relative, not objective. This is why it is called Relativity Theory.
You don’t even talk the talk, but you want to claim you walk the walk.
A light is turned on at point A and at Point B at Midnight UTC (synchronized clocks). Observer Alice is on a point midway on a straight line between A and B. Alice sees the light from both A and B at the same time.
No frame reference. I presume the clocks are stationary in the frame in which they are in sync, and Alice is also stationary in this same frame. Just making sure.
That Alice is in the midpoint and see the lights at the same time tells us that they are all in a common inertial frame. No need to specify. Seems that you do not understand that point. Oh wait! You don’t! You think that the proper length of the rod will contract. Or stretch. Which one was it again?
Bob is on a fast motorcycle on that straight line heading toward A and passes Alice at Midnight UTC on Alice’s watch. Bob will see the light from A earlier than the light from B. A and B send each other information about the time they turned the lights on and when they saw the light from the other. Since they have synchronized clocks, A and B will then agree that they turned their lights on at the same time, even though each saw the other’s light go on later than his own.
If they agree on the time delay value, they know they are in a common inertial frame and can even calculate the proper separation.
They already know they were stationary in the one frame, so the subsequent timings just confirm again that those clocks are indeed synced in that frame, yes.
No, A and B did not know for sure that they were in the same inertial frame. Alice knew because she was in the midpoint because repeated readings in each direction with her laser range finder always had the same value. The lightkeepers at A and B did not know until they compared notes.
You didn’t mention what Bob concludes since he doesn’t know where in his frame the light pulses originated, and therefore seems unable to assess which one flashed first. The one might have been before the other, or just closer. So what’s the purpose of Bob here? A stationary observer off to one side of Alice would similarly observe one pulse before the other. RoS seems not to have been well demonstrated with this example.
Bob sees the light flashes at different times. He has no idea of how far off they are or if they are synchronized or moving (only Alice knows before the fact). He only knows that he saw them at different times. His observation is that they were not simultaneous.
You didn’t label any events, but there are the two events where the light pulses were emitted, and various observation events. RoS says that the two emission events, if simultaneous in Alice’s frame are not simultaneous in Bob’s frame, even if that wasn’t demonstrated by your example.
A flashed at Midnight UTC. B flashed at Midnight UTC. Alice observed both flashes simultaneously at some time after Midnight UTC, the exact time difference being irrelevant. Bob passed Alice at Midnight UTC. His exact speed is irrelevant as long as it is non-zero with respect to Alice and that it is in the direction of A.
If you need exact numbers and math to understand this, then you do not understand the principle behind Relativity of Simultaneity and do not know how to apply it properly.
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I see that you have given up on the first picture you presented that attempted to show changing real lengths after I explained how it was wrong.
I don’t even remember posting a picture. What post is this? Have you now switched to replies to Jaaanosik without indication?
OOPS! You are right. I was preparing on reply and got distracted by some RLBS and when I went back, I started a different reply but got the two mixed up. It was Jano who posted the pictures. Apoplectic Apologies. :(
If you want to see the other picture Jano posted here it is.
https://i.imgur.com/XSs3QXa.png
It relates a fixed observer and a moving observer.
His picture shows a continuously accelerating rigid object (a ruler apparently), much like SP2 at 100g and the first ~5 light days of rod in front of it. It shows some lines of proper simultaneity for the object, but doesn’t bother with putting units on anything. The picture seems accurate enough.
I haven’t much been paying attention to what Jaaanosik has been trying to convey with this picture, but he used it in post 85 to illustrate length contraction, which it does.
Notice that SP2 would be the left end of the ruler (@) accelerating at 100g, and the right end of the ruler (@') corresponds approximately to the 5 LD mark on SP2’s rod. It is not accelerating at 100g as evidenced by the different hyperbolic curves of the two lines. So the two worldlines, accelerating differently, converge in the original frame. SP1 is not like that since it follows the same curve as the left side of the ruler, so there is no contraction of the distance between them, but the rod in front of SP2 does contract, so 10 LD is insufficient to reach SP1 once they’re moving.
Consider the ruler to be the rod with observers at each end (SP1 and SP2). The length of the ruler relative to the observer at the origin (0) changes as the ruler accelerates uniformly (constant rate). But the length of the ruler as determined by the observers at each of the ruler ends is determined by the width of a rectangle whose opposite corners are where the dashed lines intercept the colored area. That is constant. The observers on SP1 and SP2 never see any change in the length despite ongoing acceleration. It is always 10 LD. It is the observer at the origin who sees changing length contraction as the relative speeds differ.
BTW can anyone explain why the shape of the colored area first grows then shrinks as time goes on? Time is the vertical axis, increasing from bottom to top. The answer is somewhat subtle.
… and therefore exceeds c
As I showed, I was not equating speed and acceleration. You were wrong about what you said but rather than admit such a thing, you switch to another subject.
I was commenting on the bit I quote just above. You didn’t say what exceeded c. Turns out you were talking about proper velocity, but apparently without knowing the term, and once that was clarified, I agreed that yes, that can exceed c.
Here is what I said.
The traveling twin thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds c.
The traveling twin feels the continued acceleration and thinks he exceeds c. The reference frame is obvious – the traveling twin. And it was obviously proper acceleration and speed since no other frame was mentioned. This was all obvious but you nitpick because you are having trouble with your arguments and need to distract from that.
The subject was that the track would necessarily move in a direction opposite to the cars courtesy of Newton #3. But you skip that part because you cannot deal with the consequences.
It would not since there is not just the one car, but a series of them equally spaced. The vector sum of the forces on the track is zero and it stays put. Surely you know this, and are just trying to nitpick. I was started to back off the troll assessment when I thought you were putting out numbers, but no, you just copied mine and put out comments like this one above.
Completely and totally wrong. Each car is pushing the track in the opposite direction from its motion. Adding more cars does not change that.
Imagine a car on a gravel road that is not getting much traction. It is throwing gravel opposite to its direction of motion. Put that gravel road on the surface of a flat circular track. The track is supported on ball bearings. Put a bunch of cars on the track. Do they always throw gravel opposite to the direction of their individual motions? Yes. Are they all imparting energy to the gravel opposite to the direction they are moving? Yes. If they are all traveling counterclockwise are they not all throwing gravel clockwise? Yes.
Now instead of gravel, substitute macadam. Are the cars all imparting energy into the macadam opposite to their individual directions of motion? Yes. The track is supported on ball bearings, so it is going to move in the opposite direction from the cars. If the cars are going counterclockwise, the track is going to move clockwise. How fast it is going to go will depend on the relative mass of the track and the cars because angular momentum is conserved. And Newton #3 is validated.
Trying to apply math without first understanding the concepts behind it is going to give wrong results because the math will be misapplied.
I have addressed your math earlier and shown just how wrong it is.
You quoted a subset of the exact same numbers, either showing how much you agree with what I’ve said, or you are unable to do the math so you just copied them. Numbers don’t lie.
Then you added a bunch of words unbacked by different numbers.
I have provided lots of numbers about proper acceleration, length etc. Deal with it.
The contraction of the entire ensemble, ships and rod, is only as seen by the inertial observer.
Wrong. It is contracted only relative to a frame in which it is moving. SP2 is inertial after it shuts down, and yet it is not then contracted in the SP2 frame (N), so that contradicts your statement above. The statement also implies that any accelerating observer will not see SP2 contracted, which is wrong as well. Some observer could be accelerating the other way for instance. SP2 will be contracted relative to him.
WRONG. Contraction is relative to observers in other frames with the degree of contraction related to speed differences in accordance with the Lorentz formula. An observer never sees contraction in his own environment because the speed difference is zero. The contraction of the entire ensemble will only be seen by the observer who never accelerated with respect to SP1,rod,SP2. Since is said the[/i[ inertial observer, there is no doubt about who was meant, the same one we have been discussing all along. Different observers at different relative speeds will see different contraction factors. SP1,rod,SP2 will not be contracted with respect to themselves because there is never any speed difference among them. Nitpicking again to avoid dealing with the actual arguments.
To the ships traveling at the same speeds because of identical acceleration histories, contraction does not exist because there is no relative motion.
Agree if you add ‘Relative’ in front of that statement. For instance, I claim no length contraction of the rod in frame N once SP2 has stopped accelerating. Look at the numbers. Don’t take my word for it. If the rod were contracted, a length 17.1 rod would not span from N coordinates -19.6131 to -2.5123 where SP1 is parked. It needs its full uncontracted length to reach between them. Numbers don’t lie.
Sp1a is an event, not a time.
SP1a is the time on SP1’s clock when SP1 acceleration begins.
OK, so you don’t know what an event is. An event is an objective point in spacetime, a frame-invariant fact. The location and time of an event is relation with a coordinate system, typically an inertial frame.
SP1a time is obviously the time at event SP1a. But since you want to be ultraprecise…
There are no objective points in spacetime. SP1 and SP2 can be inertial with respect to each other but could be moving with respect to another inertial observer. There is no Newtonian Absolute Space and no Newtonian Absolute Time. There are only coordinates of convenience relative to some observer. An absolute distance in spacetime between events can be calculated in SR, although the general opinion among relativity types is that even this is simply a handy tool for calculation purposes and has no physical significance. In GR it is not as simple as the complex Pythagorean Theorem used in SR (where time is in i units. The possibility of ongoing acceleration screws the pooch. And in the real world there is always a gravitational gradient.
Clock A accelerates at 1g for however long it takes it to get to the nearest star and back. Clock B meanwhile gets put into a cyclotron on Earth and is subjected to 10g of continuous acceleration the whole time Clock A is gone. Clock A will have logged less time when they reunite, despite having accelerated far less. I didn’t do the numbers, so I can’t say how much. Can do if you think it matters. Clock B will age negligibly slower than one not in the cyclotron, depending on the radius of the thing.
Wrong.
Thank you. I’d question my statement if you agreed with me. You’re getting predictable.
Clock B will have gone much faster and been time dilated much more than Clock A and will be way behind Clock A. It will be obvious which one had experienced more acceleration. Where they went or did not go is irrelevant as long as they end up side by side in a common inertial frame. What makes you think otherwise?
Want numbers? You first this time. Pick a real distance to a star and say a cyclotron with a 20m radius, which seems reasonable. Otherwise, just more words from you. You seem to have the mistaken impression that the magnitude of time dilation is a function of the magnitude of acceleration, so I picked a counterexample as requested.
Clocks can only be compared when they are nearby in a common inertial frame. To get to a given point and back at a constant g factor means accelerating to the midpoint then decelerating, then accelerating again in the other direction to the midpoint then decelerating again ending up in the place and inertial frame where the trip started. Constant acceleration for half the trip and constant deceleration for half the trip.
The acceleration will slow the onboard clock relative to an inertial observer (in a common inertial frame with Earth) as the relative speed between them increases. The deceleration will speed up the clock relative to that inertial observer as the relative speed between them decreases eventually bringing the relative clock rate back to that of the original inertial frame.
In the following scenarios, all accelerations are proper, as felt by the accelerating entity, and unless specified otherwise all times are as per an unaccelerated clock on Earth.
Let us have a clock accelerate at 1 g for 1 year then decelerate at 1 g for 1 year . The clock will then reverse direction, accelerate at 1 g for 1 year then decelerate at 1 g for 1 year.
To make the situation symmetrical, let us have a clock in a cyclotron accelerate at 10 g for 1 year, decelerate at 10 g for 1 year, accelerate at 10 g for 1 year and decelerate at 10 g for 1 year.
What elapsed time do each of the traveling clocks show as compared to the Earth clock?
The formula for this is:
t = c/a*ASINH(a*T/c)
Where:
t is proper time of the accelerated clock
c is lightspeed
a is proper acceleration
T is unaccelerated clock time (Earth time)
The Earth clock elapsed time is 4 years.
The 1 g acceleration clock elapsed time is 3.50 years
The 10 g acceleration clock elapsed time is 1.17 years
The 10 g acceleration clock shows the least elapsed time indicating that it experienced the most proper acceleration.
No, you showed mixed numbers, partly from the accelerated frame and partly from an inertial frame and assigned different numbers to SP1 and SP2 despite identical acceleration histories.
They’re not mixed. They’re carefully separated in discussions relative to each frame, just like Einstein talks about the frame of the train and the frame of the platform, but with no statement mixing the two frames.
All events are consistent between the two frames. That shows I did it correctly.
If thinking in terms of observers helps you, just consider the N stuff to be relative to some inertial observer stationary the whole time in frame N, watching the ships go from -.8112c to a halt over the course of 4.9082 days each. All the N coordinates are relative to that observer. So if my numbers are wrong relative to that unaccelerated observer (who is at location 0 the whole time), then give your own numbers. I never once referenced an accelerated reference frame, so I used nothing but inertial SR mathematics to arrive at my numbers. Feel free to do it your own way.
Nope. You applied Lorentz time dilation to SP2 but not to SP1. Mixed numbers.
I have posted corrected numbers.
You mean correct numbers, not corrected, since they were my numbers.
No. I corrected your erroneous mismatch in time values for SP1 and SP2. You were mixing numbers from different reference frames. And once more, look at the proper distance numbers I posted showing that all distances are the same not just at the beginning and the end but during acceleration.
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Malamute Lover,
The acceleration at the front of the spaceship or rod is different from the acceleration at the back.
As per the textbook, FYI,
Jano
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These post are getting too long, so I will in general not respond to comments not backed by the language of science, which is mathematics. If you assert that my numbers are wrong, either show a self-inconsistency with them (as I have done for yours), or show corrected number, which you have declined. Showing an inconsistency with your assertions carries no weight with me since I find your assertions to be consistently wrong.
You asked for proper numbers, as seen by SP1 and SP2.
I asked for corrected numbers, since you asserted mine to be wrong. You corrected not a one of them, so by the language that doesn’t lie, I assume you are in full agreement with them.
As can be clearly seen from those numbers, at all times the proper lengths and location of SP1, the rod and SP2 never change.
SP1 accelerated at 100g for 4 days and its proper location never changed? I did see that at one point and pointed out the absurdity of that.
Anyway, of the new numbers you put out, I showed them to be self-inconsistent.
You are still missing the point. The λ of SP1, the rod and SP2 are identical as seen by an unaccelerated observer.
My numbers show otherwise. You lack of corrected numbers (or demonstration of self-inconsistency of my numbers) shows that you are in fact in agreement with this. Numbers don’t lie.
I have provided lots of numbers about proper separation. It is identical at all times.
You provided a small paragraph of what are claimed to be proper numbers and copied them (except for x being offset by -10 for SP2) seven times. The numbers have been shown to be self-inconsistent (same mistake in each). You used Newtonian math to generate all those numbers. This seems to be the limit of your ability to produce numbers, since that’s all I’ve ever seen from you if you don’t count the numbers copied unchanged from my post. You’re full of Einstein slogans, but you seem totally incapable of putting any of it into practice, always reaching for Newton’s mathematics instead.
I was emphasizing that SP1, rod and SP2 are all in one reference frame and the observer is in another
They are all in every reference frame. One cannot exit a reference frame.
You are insisting on a change in proper separation as seen by SP1 and SP2. So who is right, M or Z?
Proper separation isn’t something ‘as seen by’ anybody. It is objective. My numbers show the proper separation between the ships to be 17.1 after they both cease accelerating. Your lack of computing their proper separation shows you are in full agreement. The only thing you computed was the proper distance between the frame-M grid marker at SP1 and the frame-M grid marker at SP2, which is indeed always distance 10 at any time in M (and only in M). Since the ships are not stationary in that frame, you’ve not computed their proper separation. I have.
I have provided numbers.
You did. I showed them to be self-inconsistent.
They are stationary in the common inertial frame that they (SP1, rod, SP2) are all in
The ships are under acceleration at times, so they are not always stationary in that common frame N. You’ve not provided numbers as to when those ships become stationary in that frame, so I assume you agree with mine. My numbers show their proper separation to be 17, so you agree with that as well.
An observer is in the frame he is in. He is not in any other frame.
Then it would be wrong to say Fred is moving at 20 m/sec relative to me since he’s not in my frame at all. He must have blinked out of existence. This seems to be easily falsified by the simple observation of watching cars (with observers) go by me as I stand by the street.
If he were in all frames, then he would have all length contractions
His length contraction relative to a given frame F is a function of his velocity in F, not his existence in F. Relative to a different frame G, yes, he’d have a different length contraction in that frame, but he’d be in both those frames or he’d not have a length at all.
Not to mention that he would be traveling in all directions at all speeds.
He is indeed traveling at one velocity in frame F and another in frame G. Not at all speeds. He can’t be moving faster than light in any inertial frame.
If you want to see the other picture Jano posted here it is.
(https://i.imgur.com/XSs3QXa.png)
It relates a fixed observer and a moving observer.
I like that one since it includes some inertial coordinates for the grid (‘frame M’), and clock readings for the two moving observers. It doesn’t entirely mirror our scenario since the red observer is accelerating at a far higher rate than is the orange observer. They are defined to be fixed with respect to each other, which means they maintain a constant proper separation (of 1.5 in this case). Our two ships accelerate at the same rate, and so do not maintain a constant proper separation. They are not comoving with each other like the observers in the picture, despite your insistence otherwise.
Consider the ruler to be the rod with observers at each end (SP1 and SP2).
No. Our ships are separate accelerating objects. If they stay fixed relative to each other (as the picture above depicts), then SP1 would be accelerating far less than SP2.
If you want to assert otherwise, then pony up the number so I can tear them apart.
BTW can anyone explain why the shape of the colored area first grows then shrinks as time goes on?
Yes. The graph is depicted in some inertial frame, and the accelerating ruler becomes stationary at only one time in that frame, and is moving in all others. So only when it is stationary in that frame does the ruler project its full proper length onto that frame.
There is no Newtonian Absolute Space and no Newtonian Absolute Time. There are only coordinates of convenience relative to some observer.
And yet you refuse to compute the coordinates of convenience relative to the ships stationary in frame N.
About the cyclotron:
Clock A accelerates at 1g for however long it takes it to get to the nearest star and back. Clock B meanwhile gets put into a cyclotron on Earth and is subjected to 10g of continuous acceleration the whole time Clock A is gone. Clock A will have logged less time when they reunite, despite having accelerated far less.
Wrong. Clock B will have gone much faster and been time dilated much more than Clock A and will be way behind Clock A.
I agree that if B goes much faster, then it will be more dilated. You’ve not computed how fast it’s going, so your words are totally empty. You asserted earlier it was about acceleration, not about speed:
with the slower clock the one that experienced more acceleration.
My example demonstrates that it is not in fact a function of acceleration.
It will be obvious which one had experienced more acceleration.
Yes. I set it up so B experiences 10x the magnitude of acceleration as A.
Where they went or did not go is irrelevant as long as they end up side by side in a common inertial frame.
Again correct. It is not a function of where they go.
Want numbers? You first this time. Pick a real distance to a star and say a cyclotron with a 20m radius, which seems reasonable. Otherwise, just more words from you. You seem to have the mistaken impression that the magnitude of time dilation is a function of the magnitude of acceleration, so I picked a counterexample as requested.
Clocks can only be compared when they are nearby in a common inertial frame.
They all come back and get compared before and after the exercise.
To get to a given point and back at a constant g factor means accelerating to the midpoint then decelerating, then accelerating again in the other direction to the midpoint then decelerating again ending up in the place and inertial frame where the trip started. Constant acceleration for half the trip and constant deceleration for half the trip.
Right. I can just hear the actual math not happening.
Deceleration is the wrong term. If it accelerates at 1 m/sec for a year in the direction of x, then for the next 2 years it accelerates at -1 m/sec in the direction of x. The magnitude is always 1 and only the direction changes. I’m just saying that deceleration is the same as acceleration, but with the opposite sign.
The acceleration will slow the onboard clock relative to an inertial observer (in a common inertial frame with Earth) as the relative speed between them increases.
We can put a third clock C (control) for this. It sits right next to the cyclotron but isn’t in there like the B clock is. We can compare all three when we’re done.
The deceleration will speed up the clock relative to that inertial observer as the relative speed between them decreases eventually bringing the relative clock rate back to that of the original inertial frame.
In the following scenarios, all accelerations are proper, as felt by the accelerating entity, and unless specified otherwise all times are as per an unaccelerated clock on Earth.
Let us have a clock accelerate at 1 g for 1 year then decelerate at 1 g for 1 year . The clock will then reverse direction, accelerate at 1 g for 1 year then decelerate at 1 g for 1 year.
To make the situation symmetrical, let us have a clock in a cyclotron accelerate at 10 g for 1 year
The cyclotron clock maintains continuous magnitude of acceleration until the clock from the star gets back. If the clocks measure different times, then one is going to stop before the other. If that’s 4 years as measured by clock A or C (you didn’t specify, but I kind of assumed A), then it’s something else for B since we both agree that B will read a different time than A. I say B logs more time, and you say less (due to its greater continuous acceleration).
Let us have a clock in a cyclotron accelerate at 10 g for 1 year and decelerate at 10 g for 1 year.
There is no ‘deceleration’ in a cyclotron. The g force is continuous the whole time, but with the vector direction changing as it spins. The speed is a function of the radius, not a function of time. The ISS is accelerating continuously at nearly 1g, and yet it never gains speed.
What elapsed time do each of the traveling clocks show as compared to the Earth clock?
The Earth clock elapsed time is 4 years.
That answers the question of what clock measures the 4 years. I guessed wrong.
The 1 g acceleration clock elapsed time is 3.50 years
Impressive. I got that as well.
The 10 g acceleration clock elapsed time is 1.17 years
Sorry, but this assumes a straight out-and back path, not a cyclotron. Speed at a constant acceleration is a function of radius v, not of time. V = √(ar). If the speed in the cyclotron kept going up and up like you’re computing, the g force would also keep going up, and I said it was a constant 10g. Remember, acceleration is a vector quantity and is thus a change in velocity (dv/dt), not a change in speed.
Nope. You applied Lorentz time dilation to SP2 but not to SP1. Mixed numbers.
Show me where I did that. I asked you to point out a self-inconsistency with my numbers.
I corrected your erroneous mismatch in time values for SP1 and SP2. You were mixing numbers from different reference frames.
They were never mixed. I carefully separated all the numbers with colors, each color headed with what frame was used.
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...
The graph shows an accelerating observer and a non-accelerating observer. What does this have to do with the subject, which is two observers in a common inertial reference frame accelerating at the same rate in the same direction beginning at an agreed on moment on synchronized clocks?
In that scenario, the entire ensemble will not be aware of any contraction. An external observer in an inertial frame would observe increasing Lorentz contraction on the entire ensemble. Are you still imagining that there is an Absolute Space in which the shrinkage is happening? There isn't. Contraction is relative, not objective.
Malamute Lover,
let us talk some concrete analysis.
L0 - length of the spaceship in its rest frame.
5L0 - distance between the spaceships.
Let us choose the left frame as our origin.
(https://i.imgur.com/UYYtCix.png)
The right spaceships has to get much further away into the future, in the original rest frame, in order to maintain the simultaneity line of the left spaceship frame.
(https://i.imgur.com/xY6D6Zg.png)
Do you trust the Minkowski space-time diagrams?
Does this make sense now?
Jano
Edit:
This is the Special Relativity acceleration 101 (One-o-One) in transition towards general frames.
What do you know, this is the post #101 in this thread. :)
Malamute Lover,
The acceleration at the front of the spaceship or rod is different from the acceleration at the back.
As per the textbook, FYI,
Jano
Read the text that goes with the picture
(https://i.imgur.com/sdxp5gT.png)
The horizontal lines show the perceived length of the ruler as it accelerates relative to the inertial observer located on the Y axis, (the vertical time axis) as observed by that inertial observer. The changes in the colored areas, as observed by that inertial observer, are due to changing relative speeds. Notice that the length of the ruler and the width of the colored area match when the ruler matches speed with the inertial observer on the X axis.
The horizontal distance (call it W) between where the dashed lines intersect the edges of the colored area correspond to the length of the ruler as seen by the observers at the ends of the ruler. That is, the rectangles mentioned in the text formed by the points of intersection, those points being opposite corners of the rectangle. Note that the dashed lines change angles, being steeper where the colored area becomes narrower as seen by the inertial observer. Note especially that because of the changing angles, W is constant. Again, W is the width of the rectangle whose opposite corners are where the slanted dashed lines intersect the colored areas. The observers at the end of the ruler never see any change in the length of the ruler. SP1 and SP2 never see any change in their separation and the position of the rod.
A point of interest about this graph is that the ruler is accelerating from a speed lower than the inertial observer relative to the ruler’s starting point (not on the graph). The change in the width of the colored area from smaller to larger to smaller again is because the relative speed of the inertial observer and the ruler grows lesser until they match speeds on the X axis, and then the relative speed increases as the ruler continues to accelerate.
The observers on the ruler (the observers on SP1 and SP2) would see the length of the inertial observer initially contracted because of the speed difference, then the inertial observer would be seen to be less and less contracted as the speed difference reduced, with no contraction as they matched speeds. Then as the ruler continued to accelerate and the speed difference grew, the observers on the ruler would again see the inertial observer contract.
You cannot pull a picture out of a textbook on Relativity Theory and expect to understand what it means just by looking at it.
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The observers at the end of the ruler never see any change in the length of the ruler. SP1 and SP2 never see any change in their separation and the position of the rod.
This is not so straight forward.
Are you suggesting a rod between the two spaceships?
If yes, then you need to understand/explain the post #101.
Do you see where is the simultaneity line?
(https://i.imgur.com/xY6D6Zg.png)
The simultaneity line concept does not apply here. It relates to disagreements about the simultaneity of different events to observers in different frames, which includes different distances to the two events even if there is no relative speed difference. It has nothing to do with whether events are simultaneous according to observers in the same reference frame with synchronized clocks. Events that take place at the same time on synchronized clocks that have not been affected in different ways can be said to be simultaneous regardless of what different observers may see.
As I described long ago, two observers at a significant distance from each who start in the same direction at the same proper acceleration at an agreed time on synchronized clocks and who stop accelerating at an agreed time on those synchronized clocks will see odd and contradictory things along the way. But after they stop accelerating and wait for the light images from each other to stop changing, they will find themselves at the same proper distance from each other.
Relativity of Simultaneity is irrelevant.
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SP1 and SP2 never see any change in their separation and the position of the rod.
This is not so straight forward.
Are you suggesting a rod between the two spaceships?
Yes, the rod was first suggested in post 3 and defined exactly in post 8 where I affixed it to the front of the rear ship rather than having it trail behind the lead ship, which leads to it breaking even if not attached to anything.
So our scenario since post 14 has been a pair of negligible length ships (shorter than about 10 light minutes, the precision of the calculations) initially separated by 10 light days, which accelerate at 100g for 4 days ship time. The rod in front of the rear ship is 20 light days long, so it extends past the front ship, who therefore at any time can determine the separation between the two by just reading the markings on it. No need to send time-consuming light signals between the ships.
If yes, then you need to understand/explain the post #101.
Do you see where is the simultaneity line?
Why did you post the same picture twice in that post?
It is sort of appropriate, except it depicts continuous acceleration, whereas our scenario has them stop accelerating after 4 subjective days. M-L claims that the proper distance between the ships will somehow revert to the original 10 light-day separation after they stop accelerating. Your picture shows continuous acceleration, and thus perhaps doesn't clearly show it. It also lacks numbers on the ruler, making it difficult to assess the separation in a frame not depicted.
The picture BTW seems entirely accurate to me, even if it doesn't depict our exact scenario.
Based on the most recent post, M-L seems in total denial of the existence of the frame in which both ships are subsequently stopped, as he refuses to compute any numbers for it. It would of course show that length contraction is real, so I suppose that's why he feels the need to take that stance.
I can think of three possibilities why he would do that:
1) He actually has as little clue about relativity as he presents himself to have. The only relativistic math I've seen in any post is a chart of speed-to-Lorentz-factors. Never any math applied to a specific scenario except very slow speed examples using Newtonian mathematics.
2) He has a pathological fear of being shown wrong, and refuses to show math that would illustrate it.
I personally love being shown to be wrong, because I'm here to learn, not to be correct.
My numbers could very well have mistakes in them. I did them quickly without double-check. They need interval checks for instance, the first thing I'd do with somebody else's numbers.
3) He knows his relativity quite well, but is a troll getting his jollies from yanking our collective chains.
Halc,
I see your point about acceleration and stopping.
This might be one way how to look at it.
Having said that, this is not a uniform acceleration between the two spaceships as per the textbook definition.
It is a correct uniform acceleration/deceleration per individual spaceships though.
Jano
(https://i.imgur.com/i0iXbzI.png)
I have explained what the pictures mean in an earlier post. The width of the colored area is the width as perceived by an inertial observer, not as perceived by the observers on the ruler. Since the ruler is accelerating, the perceived width will change over time. The observers on the ruler will see the same thing happen to the inertial observer since the relativity of length is related to speed difference, there being no such thing as absolute speed.
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Halc,
I see your point about acceleration and stopping.
No, not stop moving, just cease accelerating. The ships continue at the same inertial speed off to the right, not slowing back to the original frame.
Correct. Good point to bring out. There is such thing as ‘stopped’ except relative to a specified frame (as you referenced). Since Absolute Space does not exist, any observer is always in motion relative to some other observer and each will see the other through the lens of Relativity.
You picture also shows one rigid object accelerating up and down to a stop, not two separate objects accelerating identically.
There's seemingly strain being put on the object in your picture. If it was brittle (not bendy), then it would perhaps break given what being done to it. Hard to tell since the speed never gets high enough to make it obvious
The picture shows the apparent length as perceived by an observer at a different and changing relative speed. The observers on the ruler see the same thing happen to the inertial observer. Different observers in different frames, inertial or accelerating will see different things. Who is right about who will bend or break and to what degree? The local frame is right. In most circumstances, inertial or uniform acceleration, the local frame is indistinguishable from a Newtonian frame. (The significant exception is when acceleration differential is so great (like near a black hole) that the idea of ‘local’ is not so obvious.)
Having said that, this is not a uniform acceleration between the two spaceships as per the textbook definition.
Not sure what you consider uniform. Our ships accelerate at a constant proper acceleration of 100g for 4 days ship time. That's identical acceleration, not necessarily uniform, depending on your definition of uniform. If it means the object is always stationary along its length in its own frame, then yes, it illustrates that sort of motion.
The standard definition of uniform acceleration in Relativity Theory is proper acceleration at a constant felt rate such as what an onboard recording accelerometer would show.
Your prior picture I think showed uniform acceleration, where the left side of the ruler accelerates a higher g force than the right side. Those two endpoints are not representative of the motion of our two ships.
The picture represents the viewpoint of an inertial observer, not what the observers in the accelerating frame see. It represents perceived length contraction. Another inertial observer who is in motion with respect to the first inertial observer will see a different contraction. Also, since several observers at different relative speeds will have their time axes angled differently w.r.t. each other, the perceived instantaneous slope of the edges of the colored area is not meaningful in any absolute sense.
Some comments about terms. The textbook uses the term ‘uniformly accelerated motion’. This refers to the motion resulting from speed that changes at a uniform rate over time. All terms proper. One might refer to ‘uniform acceleration’ which is the proper acceleration rate involved. But if you see ‘accelerated frame’, check the context. It may mean the proper acceleration is still ongoing at the time of interest, or that the proper acceleration has ended at the time of interest and the frame is being compared to some other frame. In SR it typical means the latter since SR does not deal with ongoing acceleration, only the resulting changes at specified times on someone’s clock.
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Halc,
I see your point about acceleration and stopping.
No, not stop moving, just cease accelerating. The ships continue at the same inertial speed off to the right, not slowing back to the original frame.
You picture also shows one rigid object accelerating up and down to a stop, not two separate objects accelerating identically.
There's seemingly strain being put on the object in your picture. If it was brittle (not bendy), then it would perhaps break given what being done to it. Hard to tell since the speed never gets high enough to make it obvious.
Having said that, this is not a uniform acceleration between the two spaceships as per the textbook definition.
Not sure what you consider uniform. Our ships accelerate at a constant proper acceleration of 100g for 4 days ship time. That's identical acceleration, not necessarily uniform, depending on your definition of uniform. If it means the object is always stationary along its length in its own frame, then yes, it illustrates that sort of motion.
Your prior picture I think showed uniform acceleration, where the left side of the ruler accelerates a higher g force than the right side. Those two endpoints are not representative of the motion of our two ships.
Then this is the better diagram.
(https://i.imgur.com/xY6D6Zg.png)
the proper distance between the ships will somehow revert to the original 10 light-day separation after they stop accelerating.
If both spaceship pause acceleration then the distance between the spaceships will not revert to the original 10 light-day interval according to the SR.
Jano
As previously discussed, the picture shows the perception of an inertial observer w.r.t. an accelerating frame. It does not indicate what the participants in the acceleration will see.
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Great. An attempt to obfuscate by specifying everything in proper terms instead of coordinate terms.
Let’s see if the numbers hold up...
SP1a [0,0]
SP1 proper acceleration is 100g
SP1 duration of proper acceleration is 4 days
SP1 final proper speed is 1.130 c
SP1 proper acceleration has been continuous, so average proper speed is 0.565 c
SP1 proper distance covered is 2.261 LD
SP1b proper [2.261,4]
Still using Newtonian mathematics I see. This is wrong. The actual proper speed is 1.387c
Now me just saying that is words, so let me let the numbers speak.
You apparently got the proper speed the Newtonian way by multiplying 980m/sec2 by 345600 seconds which indeed yields a proper speed (dx/dt’) of 1.13c. This corresponds to a coordinate speed (dx/dt) of about .749c, not .8112c. Doing it this way leads to contradictions:
Suppose we accelerated for 8 days instead of 4. Using your linear method, that would double the proper speed to 2.26c, but that contradicts the relativistic computation of proper speed:
.749c coordinate speed doubled using relativistic addition is .9596c, which has a λ of 3.5558 yielding a proper speed of 3.4123c which contradicts the 2.26c using your method. Proper speeds don’t add in a linear way like you’re doing. You can’t use Newtonian method of just multiplying proper time by proper acceleration.
For proper separation and proper speed that is exactly what you do. The Principle of Relativity states that the laws of physics have the same form in all arbitrary frames of reference. Uniform proper acceleration for a specified proper time will yield the expected proper speed and the expected proper distance covered. The motions of SP1 and SP2 are identical. Their initial proper separation and their final proper separation will be the same. If you cannot understand that they you do not understand the foundation of Relativity Theory and if you do not understand that you know nothing at all about Relativity Theory.
You insist on applying a nonsensical relativity correction to proper values but never stop to think that this relativity correction of yours depends on which observer you use as a base. And that has no influence whatsoever on what SP1 and SP2 will see.
The ‘proper distance’ covered (as you’re using the term) is 2.5123 as you had in your prior post, but I see you’ve changed it to 2.261 now.
SP2a [-10,0]
SP2 proper acceleration is 100g
SP2 duration of proper acceleration is 4 days
SP2 final proper speed is 1.130 c
SP2 proper acceleration has been continuous, so average proper speed is 0.565 c
SP2 proper distance covered is 2.261 LD
SP2b proper [-7.739,4]
2.261 -(-7.739) = 10 LD
OK, that much is the same I computed. In frame M (your figures are all relative to frame M), the ships are always 10 LD apart.
You’re suggesting that a moving 10 LD rod can bridge the 10 LD gap between (your) coordinates -7.739 and 2.261 in frame M. That cannot be unless the rod is not length contracted, so you seem to be in denial of relativistic length contraction. That’s at least one contradiction I can point out without any numbers in N.
There is no such thing as ‘relativistic computation of proper speed’. Saying that shows an utter ignorance of Relativity Theory. Proper speed is what an observer will calculate from proper acceleration history as confirmed by passage of landmarks. Constant proper acceleration does not require any relativistic adjustment in proper speed.
In the first post I copied your numbers but this time I calculated it myself, avoiding your unjustified ‘relativistic correction’ which does not exist in proper measurements.
In an inertial or uniformly accelerating frame (the definition of both including ‘proper’) local physics is Newtonian.
The entire frame in which SP1 and SP2 have accelerated identically has been contracted as viewed by an inertial observer, not as viewed by SP1 and SP2 because they are traveling at the same sapped and will not see any contraction. Different inertial observers will see different contraction factors. SP1 and SP2 will see all of those other observers as contracted because observation of Lorentz contraction is related to relative speed. You are still thinking in terms of there being some kind of absolute space underneath everything in which things really contract. There is not. If you do not understand that then you know nothing about Relativity Theory. But I think that has been amply demonstrated already.
Buzzwords and online calculators are useless unless you know to apply them. And that requires understanding fundamental concepts, an area in which you are seriously lacking.
Same proper separation between SP1 and SP2 at end of acceleration as at start of acceleration.
You’re using ‘proper separation’ incorrectly. Yes, the proper separation between the grid markers stationary in frame M is 10 in frame M, but proper separation only applies to mutually stationary things, so that’s only the proper distance between the M-coordinate grid markers, not the proper separation between the ships because they’re now moving in the M frame which is the only frame you’ve referenced in this entire post. The proper separation of the ships hasn’t been computed. Proper separation of the ships is by definition the distance between the ships in the frame in which they’re both stationary, which is frame N. You’ve performed no computations in frame N. so you’ve not computed their proper separation.
No, I am using proper separation correctly. The term proper refers to what is experienced, measured etc. by the participants. You cannot adjust actual proper separation by what some other observer sees, What about observers O, P, Q, R, S, T, U, V, W, X, Y and Z all of whom are inertial but have different speeds relative to each other and to SP1 and SP2? They will all want different contraction factors. Proper separation is what SP1 and SP2 will both see and once the light images catch up and the proper separation will be what is was at the start. What any other observer sees is irrelevant to that because different observers will see different things. That’s why it’s called Relativity Theory.
You have got buzzwords and online calculating tools but you have no idea how to apply them correctly because you do not understand the basic concepts underneath them.
The rod has been defined as having constant acceleration along its length.
It most certainly has not. The rod has been defined to be stationary in the frame of SP2 at all times.
Asserting it to have constant acceleration along its length is just words. Back it with mathematics in the frame of the rod, and you’ll see the contradiction that results from this assertion. Jaaasonik’s picture he keeps posting over and over illustrates the situation exactly, showing a rigid rod accelerating with constant proper length. It accelerates at a different rate along its length. The picture illustrates that nicely, but you seem to be in denial of any illustrations taken from accepted texts, preferring instead to make up your own facts.
In post #8 (new thread numbering) you agreed to add thrusters to the rod to avoid the speed of sound problem
…outfit the rod with its own thrust as you describe so at no time is it under stress or strain. Of course those thrusters would also have to be programmed with the plan so they start and stop at the proper time.
Later the rod got attached to SP2 instead of SP1 but still the same problem. If there are no thrusters on the rod and it is attached to SP2, then the force waves from the acceleration will proceed up the rod as the speed of sound in the material. If we say that the speed of sound is the maximum possible, the speed of light, then it will take ten days for the force waves to reach the front end. SP1 will be long gone. When the acceleration stops and the last of the waves reaches the front end, the acceleration of the rod will cease, the rod will be its original proper length and the front end will come to a halt relative to SP1 exactly where it started, outside SP1’s window.
Jaaasonik’s picture does not mean what you and he thought it means. It is about what an inertial observer sees. You need to do the rectangle analysis as the (omitted) explanatory text explains to discover that the observers on the ruler see. Once you do that you see that they never see any contraction. The width of the rectangle is always the same.
You cannot grab a picture from page 409 in a textbook on advanced physics, ignore the explanation, and think you understand what the picture is saying.
SP1 end [0,0]
What does this mean? That SP1 has gone nowhere in 4 days? This whole section lacks a frame reference, so it’s not immediately clear what you’re trying to convey here.
SP1 end proper acceleration is 100g
SP1 end duration of proper acceleration is 4 days
SP1 end final proper speed is 1.130 c
SP1 end proper acceleration has been continuous, so average proper speed is 0.565 c
SP1 end proper distance covered is 2.261 LD
SP1 end at SP1b proper [2.261,4]
SP2 end [0,-10]
SP2 end proper acceleration is 100g
SP2 end duration of proper acceleration is 4 days
SP2 end final proper speed is 1.130 c
SP2 end proper acceleration has been continuous, so average proper speed is 0.565 c
SP2 end proper distance covered is 2.261 LD
SP2 end at SP2b proper [-7.739,4]
2.261 -(-7.739) = 10 LD
This just seems to be a repeat of the numbers above, but with the word ‘end’ stuck in at various places. It’s all in frame M, so indeed, you seem to be asserting that between start and end, neither ship has changed coordinates, which means it hasn’t moved. I don’t think you mean that, but it is totally unclear what you’re trying to convey with this repeat of the same numbers. They’re all still mostly proper numbers relative to frame M, same as the first time.
You omitted the ‘Now the rod’ prefix to the section. If you had included that part it would have been obvious that SP1 end meant. SP1 end of the rod. But this way you can ignore the math.
Of course it is proper numbers relative to frame M. It is proper separation that we are talking about. It does not matter what frame N, O, P, Q, R, S, T, U, V, W, X, Y or Z see. It matters what SP1 and SP2 will see when acceleration stops and the light images catch up. There is no relativistic correction to proper measurements, only differing measurements made by observers in different frames. And since different observers will see different things it is clear that only frame M counts when making proper measurements.
Same proper separation between SP1 end of rod and SP2 end of rod at end of acceleration as at start of acceleration. The rod is the same length at the beginning and at the end.
Repeating the numbers and repeating a wrong conclusion doesn’t make it right the second time. You’re still not computing proper distance correctly. Look up the definition of it.
Proper distance is the same concept as proper length.
Proper length L0 is the distance between two points measured by an observer who is at rest relative to both of the points.
https://courses.lumenlearning.com/physics/chapter/28-3-length-contraction/
Now where is your definition that proper length/distance involves relativistic corrections based on an arbitrary observer not in the same frame. Put up or shut up.
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Let us define SP1x as Time = 2 on the SP1 clock
An event presumably?
No, a time, a clock reading. The clocks are all synchronized. Location of the clock does not matter.
SP1 proper acceleration is 100g
SP1 duration of proper acceleration is 2 days
SP1 proper speed is 0.565 c
SP1 proper acceleration has been continuous, so average proper speed is 0.2825 c
SP1 at SP1x proper distance covered is 0.565 LD
SP1 at SP1x proper [0.565,2]
Let us define SP2x as Time = 2 on the SP2 clock
SP2 proper acceleration is 100g
SP2 duration of proper acceleration is 2 days
SP2 proper speed is 0.565 c
SP2 proper acceleration has been continuous, so average proper speed is 0.2825 c
SP2 at SP2x proper distance covered is 0.565 LD
SP2 at SP2x proper [-9.435,2]
This is all Newtonian math, which has been shown above to lead to contradictions. The separation between them in M is still 10 of course, even with these wrong numbers. But that’s the proper distance of the grid markers in M, not the proper distance between the ships since A) the ships are not stationary, and B) no frame has been demonstrated to exist where both ships are simultaneously moving at this half-way speed.
Proper measurements within a frame of reference in which there is no relative motion are Newtonian. Identical proper acceleration in the same direction for the same proper duration means no relative motion. The frame is of course the common frame of observers not in motion relative to each other. There is no such thing as absolute motion. And there is no relativistic correction of proper measurements. If you do not understand that, you do not understand Relativity. But we know that already.
Now the rod
SP1 end proper acceleration is 100g
Again I have no idea how this statement is distinct from “SP1 proper acceleration is 100g”. I don’t know what ‘end’ means inserted in like that in all these statements.
SP1 end is the end of the rod outside SP1’s window. This time you did include the explanatory prefix but ignored it anyway.
SP1 end duration of proper acceleration is 2 days
SP1 end final proper speed is 0.565 c
SP1 end proper acceleration has been continuous, so average proper speed is 0.2825 c
SP1 end at SP1x proper distance covered is 0.565 LD
SP1 end at SP1x proper [0.565,2]
The difference between SP1’s proper distance covered and the SP1 end of the rod say SP1x is 0.565 - 0.565 = 0
You’re assuming the end of the rod is accelerating at 100g, which it cannot be. Compute the location of the rod in frame N and this assumption will lead to contradictions. You perhaps know this, because you avoid doing such computations like the plague. If your assertions were correct, the numbers would show them to be since they’d be self consistent. Actually, I just think you’re not up to the math since all I’ve ever seen you use is Newtonian physics.
Frame N is irrelevant as are frames O, P, Q, R, S, T, U, V, W, X, Y and Z all of which would provide different correction values. And all of which would be length contracted and time dilated relative to M because they have different relative speeds. There is no relativistic correction of proper distance. You got some buzzwords and online calculators but have no idea how to correctly apply them because you do not understand the concepts.
The SP1 end of the rod is motionless outside SP1’s window for any SP1 time you choose.
My numbers in N (with which no inconsistency has been demonstrated) show otherwise. Your complete lack of numbers relative to N lend zero support to these mere words. This seems to the main point of contention, so you really need to back this one up. How far out of sync are the ship clocks in frame in which both ships are now stationary?
Frame N is irrelevant as are frames O, P, Q, R, S, T, U, V, W, X, Y and Z all of which are moving at different relative speeds and would provide different correction values. The clocks on SP1 and SP2 all agree on the same time. Since they were all initially synchronized and experienced identical proper accelerations for identical proper times, why would they not be? Move the SP1 clock back to the midpoint between SP1 and SP2 and move the SP2 clock back to that same midpoint in the same identical way and they will agree. And as my math has showed, that midpoint will be the midpoint marker of the rod.
Let Rh be the point on the rod 5 LD forward of SP2 and 5 LD aft of SP1 at t=0 before any acceleration begins. That is, at the mid-point of the rod. Coordinates are [-5,0]
Ooh, actual coordinates of an event. Things are improving.
But sadly you are not.
How long is the rod at the 2 day mark?
Frame dependent question.
Evasion. SP1 and SP2 and the entire rod with thrusters have undergone identical proper acceleration in the same direction for the same proper time beginning at the same time on synchronized clocks. It is all the same reference frame. You want it somehow not to be but you cannot demonstrate why.
Rh proper acceleration is 100g
You’ve not established that with any numbers in the frame of the rod. So your argument falls apart here.
It is part of the previously agreed idea of thrusters all along the length of the rod.
The ‘ruler’ in Jaaasonik’s picture shows a rod about exactly that length of about 5 LD at 100g. That means the left edge of the yellow region is SP2 and the right edge is Rh, which according to the textbook from which that illustration was taken, has significantly lower acceleration. You seem to have been denying the textbooks when it comes to SR, so that leaves your numbers to back your assertions. Still waiting for the N numbers.
I am not denying the textbook. I am [i[understanding[/i] what it means. Which you are not. The graph shows the length contraction of the uniformly accelerating ruler perceived by the inertial observer as the relative speed between them changes. To get the length of the ruler as perceived by the two observers on the ruler, you have to construct the rectangles using the points where the slanted dashed lines as opposite corners. Do that and you will see that the width of the rectangle never changes. The length of the ruler, as perceived by observers on each end, never changes even though the ruler is accelerating. The use of rectangles in this manner is a standard technique in graphic representations of situations in Relativity Theory. But you would not know anything about that.
The proper length of the rod does not change at any point in time.
By definition, yes. An actual statement with which I agree. That sort of wrecks your batting average, no?
Then you are agreeing with my argument that the entire ensemble is in a common inertial frame and the proper distance relationships never change throughout. Kind of wrecks your argument, doesn’t it.
Since both ships have been accelerating at the same rate for the same time in the same direction beginning at an agreed time on synchronized clocks, they are in the same inertial frame and there will be no difference in proper lengths.
But there will be a change in their proper separation, and the rod which did not change its proper length will not be able to span this new proper separation. The proper length of anything cannot change unless the object is deformed. It would then not be rigid, and we’ve defined all our objects to be rigid.
The perceived contraction of the entire ensemble will be the same for any other given observer. There will be no change in separation. Different observers at different relative speeds will see different contraction factors but none of them will see any gaps open up.
Oh wait! The difference in contraction between the front half of SP1 and the back half of SP1 has caused the spaceship to break apart! How much are the two halves separated? Depends on which observer you ask.
Obviously nonsense. But it is really no different from your claim of a gap appearing between SP1 and the rod because somebody looked at them. Contraction does not happen in an absolute space. There is no absolute space. Contraction is what observers in other reference frames see. Observers in frame M will see the observer in frame N contracted. Who is right? Properly speaking they are both wrong. Neither reference frame will experience proper. No gap.
If both spaceship pause acceleration then the distance between the spaceships will not revert to the original 10 light-day interval according to the SR.
The distance between them is frame dependent.
And in their common frame of reference which they maintained during the entire experiment it is still a proper 10 light days.
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Due to serious personal developments I will not be participating in this forum until further notice.
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Edit:
This is the Special Relativity acceleration 101 (One-o-One) in transition towards general frames.
What do you know, this is the post #101 in this thread. :)
Sorry Jano. I split the topic and it isn't post 101 anymore :(
The observers at the end of the ruler never see any change in the length of the ruler. SP1 and SP2 never see any change in their separation and the position of the rod.
Proof by bold I see.
The former statement is true because the front of the ruler experiences less acceleration than the rear, as can be plainly seen in the diagram you posted.
SP1 on the other hand accelerates at the same g-force as SP2 (the rear of the ruler), so it pulls away from the end of the ruler that is accelerating at a lower pace.
Your numbers posted reflect this, and thus contradict what you’re asserting in bold here. SP1 and SP2 cover the same proper distance in their respective 4 subjective seconds.
Look at the other picture with the number to see how far the end of the ruler gets in 4 subjective time units, which is even less. The left end of the ruler (red line) after 1.5 time units has moved from proper location 0 (you seem to only be able to think in such terms) to location 1.4, but the right end of the ruler (orange line) has moved from 1.5 to just short of 2, or about a third as far after two seconds on its clock. The clocks are not staying in sync with each other, not even in the inertial frame. But they do stay in sync (in the original frame) with SP1 and SP2.
The standard definition of uniform acceleration in Relativity Theory is proper acceleration at a constant felt rate such as what an onboard recording accelerometer would show.
I will accept that. Then the ruler (or the bar in front of SP2) everywhere experiences uniform acceleration, but not identical acceleration since it is different everywhere along its length. SP1 cannot be comoving with the 10-LD mark on the rod then since SP1’s acceleration (100g) does not match the ~35g acceleration of the 10-LD mark on the rod.
SP1 final proper speed is 1.130 c
Still using Newtonian mathematics I see. This is wrong. The actual proper speed is 1.387c
Now me just saying that is words, so let me let the numbers speak.
You apparently got the proper speed the Newtonian way by multiplying 980m/sec2 by 345600 seconds which indeed yields a proper speed (dx/dt’) of 1.13c. This corresponds to a coordinate speed (dx/dt) of about .749c, not .8112c. Doing it this way leads to contradictions:
Suppose we accelerated for 8 days instead of 4. Using your linear method, that would double the proper speed to 2.26c, but that contradicts the relativistic computation of proper speed:
.749c coordinate speed doubled using relativistic addition is .9596c, which has a λ of 3.5558 yielding a proper speed of 3.4123c which contradicts the 2.26c using your method. Proper speeds don’t add in a linear way like you’re doing. You can’t use Newtonian method of just multiplying proper time by proper acceleration.
For proper separation and proper speed that is exactly what you do.
I notice that you did not point out any error in the mathematics used to demonstrate your proper speed calculation to be faulty. Therefore I presume that you agree that your numbers are self-inconsistent. Math doesn’t lie.
The Principle of Relativity states that the laws of physics have the same form in all arbitrary frames of reference. Uniform proper acceleration for a specified proper time will yield the expected proper speed and the expected proper distance covered.
This contradicts the pictures you and Jano keep posting that shows the front of the rod not covering as much proper distance as does SP2 to which it is attached. And yet you assert that the front of the rod somehow will be where SP1 is when it stops accelerating. This is you contradicting yourself.
Their initial proper separation and their final proper separation will be the same.
The proper distance covered by both will be the same. The proper separation between the two will not. You seem completely unaware of the difference in meaning between the two terms. I’ve said this before, and you’ll ignore me again because you know better.
I am using proper separation correctly. The term proper refers to what is experienced, measured etc. by the participants.
Indeed, and the proper separation thus means the separation as seen by comoving observers, which means the separation in their frame, and not in another. But you’re still measuring the separation by the frame of those grid markers, not by the frame of the observers.
There is no such thing as ‘relativistic computation of proper speed’.
I actually performed such a computation above where I worked out the contradiction in your numbers. Actual speed is dx’/dt’, and proper speed is dx/dt’, so all you need to do to get the relativistic proper speed is multiply actual speed by dx/dx’ which is λ.
Saying that shows an utter ignorance of Relativity Theory. Proper speed is what an observer will calculate from proper acceleration history as confirmed by passage of landmarks.
Yes, it is, but not by shoving that history through a Newtonian linear formula. That formula assumes no time dilation as speed increases.
Constant proper acceleration does not require any relativistic adjustment in proper speed.
Then show the error in my numbers instead of just asserting that you must be correct because you assume it must be that simple.
The rod has been defined as having constant acceleration along its length.
It most certainly has not. The rod has been defined to be stationary in the frame of SP2 at all times.
In post #8 (new thread numbering) you agreed to add thrusters to the rod to avoid the speed of sound problem
Which is how we keep it absolutely stationary relative to SP2 at all times in the frame of SP2.
If there are no thrusters on the rod and it is attached to SP2, then the force waves from the acceleration will proceed up the rod as the speed of sound in the material.
Which is why we didn’t posit a lack of thrusters. It allows us to use ideal math. The ruler in the pictures being posted are considered ideal rigid objects for simplicity. We’re sticking with that.
You omitted the ‘Now the rod’ prefix to the section. If you had included that part it would have been obvious that SP1 end meant. SP1 end of the rod. But this way you can ignore the math.
Got it. That was unclear because of course I don’t consider SP1 to be at the end of the rod.
SP1 end proper acceleration is 100g
SP1 end duration of proper acceleration is 4 days
SP1 end final proper speed is 1.130 c
SP1 end proper acceleration has been continuous, so average proper speed is 0.565 c
SP1 end proper distance covered is 2.261 LD
SP1 end at SP1b proper [2.261,4]
SP2 end [0,-10]
SP2 end proper acceleration is 100g
SP2 end duration of proper acceleration is 4 days
SP2 end final proper speed is 1.130 c
SP2 end proper acceleration has been continuous, so average proper speed is 0.565 c
SP2 end proper distance covered is 2.261 LD
SP2 end at SP2b proper [-7.739,4]
2.261 -(-7.739) = 10 LD
You don’t give any indication of duration in frame M or actual speed in frame M, so I’ll use the numbers you gave before, which is time 4.9082 days and .8112c. The exact time and speed doesn’t really matter so long as its the same for both SP1 and SP2 in frame M.
I cannot convert your proper speed to coordinate speed because you’re using Newtonian speed, which of course is the same as coordinate speed in Newtonian mechanics (things can move faster than light given those physics). But I’ll use your proper distance of 2.261 LD.
SP2 end is attached to SP2, so of course those numbers will be the same.
So I’m going to put some stationary observer here and there in frame M with clocks synced in that frame. Each never moves. You refuse to tell me how far out of sync the ship clocks are in their proper frame, so I cannot talk about what time each pilot sees on the other clock, so I’m forced to use stationary observers. Your total inability to produce any figures in frame N is a serious indication that you don’t know what’s going on.
There are 1000 grid markers every light day, all stationary in the original frame M. You report each ship’s proper speed, meaning you are having them count these markers going by. So each has counted 2261 markers after a subjective 4 seconds.
My first observers are at markers -7739 and 2261. They are present at the point where each ship stops accelerating as it flies by at .8112c. The time on their clocks is 4.908, and they can see the time of 4 on the ship/rod clock as it passes by. The guy at the 2261 mark does nothing except note the location of the event where both rod and SP1 pass by together.
The guy at -7739 (SP2b) sends a light signal towards the opposite end of the rod at time 4.9082. The third observer is at marker 45227. This marker is 52.966 light days away from the first observer, and it thus takes about 53 days for the light to get there at time 57.874 on that observer’s clock. Just at that exact time the right end of the rod goes by him. Verification of that: The far end of the rod was at marker 2261 at time 4.908. Marker 45227 is 42.966 LD from that, and the rod end is moving at .8112c, so it covers 42.966 LD in 52.966 days, the exact same time as the light signal to get to the same spot.
The signal is reflected back. There is a 4th observer at marker 39706, 5.521 LD distant from the 3rd guy with the mirror, so it takes 5.521 days to get there at time 63.395. Meanwhile, SP2 moving at .8112c takes 58.487 days to get to the same observer at the same time as the refleted light signal.
Total time (as measured by the stationary observers) is 58.487 to do the round trip. Now the ship clocks, moving inertially since the acceleration stopped, are moving at .8112c in that frame, so the time dilation factor is 1.71008, so SP2 clock will have advanced not 58.487 days, but only 34.201 days while waiting for the light to reach the other end of the rod and back.
This is a contradiction with your asserted length of 10 LD for the rod. It should have taken 20 days (ship time) for light to traverse a 10-LD rod (stationary relative to the ship taking the measurement) to SP1 where you assert the other end of the rod is, but it took 34.2 days, indicating a proper separation of exactly 17.1008 LD between the ships.
This again shows your numbers to be self-contradictory. Either that or I made a mistake above, in which case you’ll be happy to point it out.
If I got the coordinate speed wrong, you can tell me which one is correct. It’s easy enough to change the above figures for a different speed. The same contradiction will result, but 17.1 is unique to the 0.8112 speed. Talk to the numbers, not to the text.