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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.pdfIf 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

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

Of course, the spokes experience Lorentz contraction. They are moving too.

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, the spokes get shorter as the rim contracts.

Since the rim is moving along with the blocks, it will experience exactly the same Lorentz contraction as the blocks. No increasing gap sizes.

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.

Quote from: HalcDifferent example. Don't mix them.The increasing gap size points to the blocks shrinking but the rim not shrinking.

Different example. Don't mix them.

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.

If you want something that holds the blocks on track but is not itself moving, you are introducing all sorts of complications.

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.

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)

The spinning disk (grey)

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.

If you photograph the whole thing from a distant point. You will see something different from what an observer at the center sees.

But it will still exhibit Lorentz contraction since it is moving with respect to that distant observer.

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.

It is not possible to start the engines at the same time.

No cop outs going on. I understand the scenarios perfectly.

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.

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.

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 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?

Quote from: Malamute Lover on 31/07/2020 17:46:57Are 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.

Are the two space ships in motion relative to each other? In fact, yes.

QuoteAssume 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.

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.

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.

QuoteHowever, 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.

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.

QuoteNow 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.

Now what about that string?SP1 thinks the string is under tension.

QuoteThat 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.

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.

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.

QuoteHow 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.

How can SP1 and SP2 agree on what is happening to the string?

QuoteBy 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.

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.

QuoteA 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.

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.

QuoteThe 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.

The perceived time difference between engine shutoffs will be the same on each side.

QuoteBut what about Lorentz contraction on the string?The string is dangling behind S2, having broken off S1 some time ago.

But what about Lorentz contraction on the string?

QuoteIf 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.

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.

QuoteOnce 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:QuoteLorentz contraction is relative. It is not real.Back up all the above with some numbers so I can apologize.

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.

Quote from: HalcQuote from: MalamuteAre 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’.

Quote from: MalamuteAre 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.

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....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.

There is no absolute ‘now’.

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.

Why will it break? If you want to ignore the clock synchronization problem

the entire system will accelerate at 1g.

But from the SP2 viewpoint, there never is any tension because the distance is decreasing and the string is always slack.

Quote from: HalcQuoteBy 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?

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?

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

Going slower than the other guy means seeing the other guys clock tick slower.

Because SP2 thinks it started its engine first

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.

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?

Looking directly along the length, it is not possible to determine the length.

Lorentz contraction calculations.Courtesy of https://keisan.casio.com/exec/system/1224059837...Relative speed: 275,000 km/sPerceived 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.

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?

Quote from: Malamute Lover on 31/07/2020 23:19:26As 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:

QuoteNo 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.

QuoteAs 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.QuoteEven 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.

QuoteAs 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.

QuoteIf 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.

Quote from: Halc on 01/08/2020 02:27:04You 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 soYou are ignoring light speed being finite.

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

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?

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.

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.

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.

They shut off their engines at the 10 second mark on their respective clocks.

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.

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.

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.

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.

Quote from: HalcQuoteIf 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.

If it were real then passengers on an ultrafast ship would see themselves compressed in the direction of travel and be squashed.

Quote from: Malamute Lover on 02/08/2020 04:14:06Quote from: Halc on 01/08/2020 02:27:04You 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 soYou 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.

QuoteThe 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.

QuoteYou 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.

QuoteBoth 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.QuoteFor 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.

QuoteAt 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.

QuoteAt 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.

QuoteThey 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.

QuoteEach 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.

QuoteAcceleration, 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.

QuoteMy 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.

QuoteMy 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.

QuoteShow 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.

QuoteThe 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.

QuoteIn 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.

QuoteQuote from: HalcQuoteIf 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.

QuoteAcceleration 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-wronghttps://physics.stackexchange.com/questions/112645/which-clock-is-the-fastest-inside-an-accelerating-bodyFound one on quora as well, but quora seems bent on promoting some of the worst answers, so I'll not use them.

QuoteIf 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.

QuoteIf 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.

Malamute Lover,here are a couple of diagrams: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

Quote from: Jaaanosik on 04/08/2020 20:04:00Quote from: Jaaanosik on 04/08/2020 20:04:00Malamute Lover,here are a couple of diagrams: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,JanoWhat 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.

I already provided a way of synchronizing the clocks

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.

Spatial separation rules out giving meaning to synchronized clocks if acceleration is involved.

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 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.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.

Quote from: HalcWhat are 'angular measurements'?Measurement of the angle that the width of the other ship subtends.

What are 'angular measurements'?

Quote from: HalcYou'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.

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.

So let’s see some relativity math that does not depend on illusory difference in distance.

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?

SP1, string, SP2 all experience identical acceleration beginning at time zero on synchronized clocks.

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.

Quote from: Halc on 03/08/2020 04:36:302) 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.

I see you do not understand Minkowski spacetime either. The time dimension is represented by imaginary numbers, the three spatial dimensions by real numbers.

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.

Quote from: HalcQuoteThe 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.

QuoteThe 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.

The ship has thrusters all along the sides. And the ship is very skinny and made of ultra-rigid material.

So why does the string break?

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

They made sure I knew something about engineering as well as physics. You should learn some instead of copying from websites.

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.

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 traveling twin thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds c

This is why the pilot of the spaceship and the inertial frame observer disagree on how fast the ship is going.

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.

What is holding them in a circular orbit rather than flying out in a straight line and crashing into the track?

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).

If you do not know the rules, do not try to play the game....I know SR cold

Acceleration puts different observers in different reference frames. Time dilation is observer dependent.

Bringing two clocks into a common inertial frame will allow comparison, with the slower clock the one that experienced more acceleration.

Quote from: HalcQuoteTo an observer on the surface (and therefore to the mechanical properties of the cylinder) the same number of revolutions a minute are taking placeNow 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.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.

QuoteTo an observer on the surface (and therefore to the mechanical properties of the cylinder) the same number of revolutions a minute are taking placeNow 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.

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

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?

Do you have any understanding of the subject at all or was this another ad hom attempt that misfired?

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.

Acceleration is in the domain of General Relativity.

Einstein was not believed by everyone for some time. There are relativity deniers today, to use your phrase.

Quote from: Malamute Lover on 06/08/2020 00:52:24Acceleration 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....

Quote from: Malamute Lover on 04/08/2020 19:24:09Although 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.QuoteEach 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. QuoteSpatial 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.

QuoteIf 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.

QuoteThe 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.

QuoteYou 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.

QuoteQuote from: HalcYou'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.

QuoteSo 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) andSP2b 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 = 13.8721Therefore:event SP2a coordinates are (-17.1008, 13.8721) t = 0 on SP2 clock, v = -.8112event SP2b coordinates are (- 19.6131, 18.7803) t = 4 on SP2 clock, v = 0and while we’re at it, we can computewhere SP2 is at time zero: (-5.8477, 0) t=-8.112 on SP2 clock, v = -.8112At 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.QuoteThe 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.

Quote from: Malamute Lover on 05/08/2020 01:01:26Quote from: Jaaanosik on 04/08/2020 20:04:00Quote from: Jaaanosik on 04/08/2020 20:04:00Malamute Lover,here are a couple of diagrams: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,JanoWhat 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