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New Theories / Re: Split: Bell's paradox: Does the string break?
« on: 13/08/2020 07:48:18 »
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QuoteLet us define SP1x as Time = 2 on the SP1 clockAn event presumably?
No, a time, a clock reading. The clocks are all synchronized. Location of the clock does not matter.QuoteSP1 proper acceleration is 100gThis 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.
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]
QuoteNow the rodAgain 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 proper acceleration is 100g
QuoteSP1 end duration of proper acceleration is 2 daysYou’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.
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
QuoteThe 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?
QuoteLet 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.
QuoteHow long is the rod at the 2 day mark?Frame dependent question.
QuoteRh proper acceleration is 100gYou’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.
QuoteThe 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?
QuoteSince 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.
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]Still using Newtonian mathematics I see. This is wrong. The actual proper speed is 1.387c
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]
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.QuoteSP2a [-10,0]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.
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
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.
QuoteSame 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.
QuoteThe 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.
…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.
QuoteSP1 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.QuoteSP1 end proper acceleration is 100gThis 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.
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
QuoteSame 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 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/
Then this is the better diagram.Halc,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.
I see your point about acceleration and stopping.
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.QuoteHaving 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.Quotethe 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
Halc,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.
I see your point about acceleration and stopping.
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
QuoteHaving 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.
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.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?
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.QuoteIf yes, then you need to understand/explain the post #101.Why did you post the same picture twice in that post?
Do you see where is the simultaneity line?
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
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?
...
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.
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.
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
QuoteI 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.
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.… and therefore exceeds cAs 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.
QuoteThe 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.
QuoteI 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.
QuoteThe 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.
QuoteTo 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.QuoteOK, 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 is an event, not a time.SP1a is the time on SP1’s clock when SP1 acceleration begins.
QuoteThank you. I’d question my statement if you agreed with me. You’re getting predictable.Quote from: HalcClock 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.QuoteClock 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.
QuoteNo, 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.
QuoteI have posted corrected numbers.You mean correct numbers, not corrected, since they were my numbers.
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.
QuoteSP1a is at x=0,t=0 at the start (or 0,0). SP2a is at (-10, 0) Units in (light days, days)All in agreement, and seemingly copied from my post, except for some of the rearangement of the wording.
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
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.QuoteBut 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.
QuoteNeither 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.
QuoteThe 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.QuoteAt 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.
Quotewhich 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.
QuoteSince both ships have stopped accelerating at the same time, they are in a common inertial frameThey’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.
QuoteRelativity 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.
QuoteA 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.
QuoteBob 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.
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.
QuoteSP1a is at x=0,t=0 at the start (or 0,0). SP2a is at (-10, 0) Units in (light days, days)All in agreement, and seemingly copied from my post, except for some of the rearangement of the wording.
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
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.QuoteBut 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.[/quote]
QuoteThe 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.QuoteAt 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.
Quotewhich 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.
QuoteSince both ships have stopped accelerating at the same time, they are in a common inertial frameThey’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.
QuoteRelativity 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.
QuoteA 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.
QuoteBob 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.
The traveling twin thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds cNo problem with that. Proper velocity has no upper limit and can exceed c, as I explained above.
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?
QuoteThe traveling twin thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds cc is a speed. Acceleration is not. Different units. You’re equivocating the two here. It is meaningless to say that acceleration exceeds some speed.
QuoteThat they would. Same thing if the spokes disappeared in the spoke example.Quote from: HalcThe track is holding them. If the track disappeared, they would fly off in straight lines despite having wheels.QuoteWhat is holding them in a circular orbit rather than flying out in a straight line and crashing into the track?Wheels
QuoteThe 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.
QuoteSince 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.
Quotelike 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.
Quote from: HalcClock 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.Quote… 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.QuoteBringing 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.
QuoteI 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.
QuoteNot 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.
QuoteEven 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.
QuoteYou’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.QuoteQuoteQuoteAcceleration 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.
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.
QuoteIn the original frame [M]: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.
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.QuoteIf 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.QuoteYou 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.[QuoteThe 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.QuoteThe accelerated observers (SP1 and SP2) both see 4. The rod is stopped relative to both SP1 and SP2 in all frames of reference.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.
Sloppy.
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.QuoteThe 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).QuoteIn the post-acceleration frame:If you use proper acceleration the elapsed time for both is 4.
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.QuoteIf 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.QuoteYour 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.QuoteFor 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.
Based on Newton formula I have found that the total mass should be 75 M sun mass.
Both possess beliefs that I find unconvincing.The latter has abundant falsification tests. The latter, like any valid quantum interpretation, has none.
The former harder to prove erroneous than the latter.
I think that rates the flat Earthers on a different level of crazy than the MWI people.
FYI, Everett's interpretation does not posit parallel universes, despite that popular spin on it by the pop media.Sean Carroll is a researcher approaching quantum physics from the 'many worlds' viewpoint.Of all the quantum interpretations, I do believe Everett's hypothesis hold the record for fewest assumptions:
- He happily admits that the underlying hypothesis is unprovable
- But suggests that it requires fewer assumptions than some other interpretations of quantum physics
"All isolated systems evolve according to the Schrodinger equation"
That's it. No more. No positing of parallel universes.
I'm not an MWI guy, but I respect the simplicity of that.
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.
QuoteThe traveling twin thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds cc is a speed. Acceleration is not. Different units. You’re equivocating the two here. It is meaningless to say that acceleration exceeds some speed.
QuoteThis 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.
QuoteIn the roller coaster thing, the cars are being held against centripetal force by an outside track which is not rotating with them.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.
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.
QuoteWhat is holding them in a circular orbit rather than flying out in a straight line and crashing into the track?Wheels
QuoteAs 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.
QuoteIf you do not know the rules, do not try to play the game.Says the guy who cannot compute the coordinates of some spaceships except using Newtonian physics.
...
I know SR cold
QuoteAcceleration 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.
QuoteBringing 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.
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.
QuoteNope. 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.
QuoteAnd “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.
QuoteIf 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.
QuoteAcceleration is in the domain of General Relativity.Gravity actually
QuoteEinstein 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.
So why the string between the spaceships would not try to shrink like the ruler does?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
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.
The end-result will be a broken string.
That's the logical conclusion when the spaceships worldlines are going to be identical,
Jano
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.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.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.
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.
QuoteI’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.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.
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) 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.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.
QuoteTo 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. Both observers agree on what one rotation is, but if their clocks are not running at the same pace, they necessarily measure a different time for one revolution.
QuoteAn observer inside with a faster clock will be surprised to not see the cylinder shatter just as he is surprised when his Twin comes back much younger than him.The twin apparently doesn’t know his physics then, because if he did, there would be no surprise. The cylinder doesn’t shatter because it was always spinning.
At this point you go into a bend about gravity and GR, which seems a diversion from the more simple SR topic that you need to master first.
One comment though:QuoteIt will be crushed around the circumference. Where does the energy come from to crush it?It doesn’t take energy to crush something. It seemingly takes force, which means a strong but brittle object can be crushed by expenditure of arbitrarily small energy. The less brittle it is, the more that energy goes into strain and not into failure, so it takes more. I’m assuming insanely brittle and strong materials for our objects else they’d not be able to withstand the centripetal stresses being put on them. Anything else would just fly apart.
QuoteWhether and how much contraction an observer will see depends on the observer.The cases I enumerate above are observed by anybody. They were chosen for that purpose. The measuring rod between ships is also a real consequence.
QuoteAn observer on the spinning wheel sees nothing different because contraction is only visible from another reference frame.In all three cases, the observer on the wheel very much sees differences, which are pointed out in the cases above. You seem to agree that one ring fits through the other, something to which all observers agree. You don’t seem to have any fake physics that lets you deny the bumper-car-track thing, nor do you seem to deny the non-Euclidean dimensions of the ‘cylinder’. OK, the non-Euclidean dimensions are frame dependent. A stationary observer will measure normal dimensions, but the inability of the object to change its angular speed is an objective observation.
QuoteIt’s not an ad-hom. It’s sloppy because you’re describing ‘what the pilot experiences’ and a pilot always experiences being stopped. You’re referencing the pilot frame and also the original frame, which makes it confusing. That’s sloppy.Quote from: HalcI did provide a reference frame. The frame in which the ship is moving at .99 c is the one before the acceleration, which I explicitly mentioned right there: “accelerated to 0.99 c”. No sloppiness. We are not going ad hom., are we?QuoteFirst 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.No frame reference, so that statement is ambiguous. It will be running 7 times slower relative to the frame in which the ship is moving at .99c. Not saying you're wrong, just sloppy. It's running at normal rate relative to the ship of course.
Secondly, the bolded statement is wrong since no observer can experience time dilation. I can look at the GPS clocks and objectively notice I’m running slower than them, but I still don’t experience that dilation.
QuoteYou have to bring two observers into the same reference frame to judge which one is right, that is, which one underwent acceleration.Acceleration is absolute (at least in Minkowski spacetime). An accelerometer works inside a box. All observers will agree if something has accelerated. That’s twice you’ve made this mistake.
QuoteHis observation is that he is traveling faster than c. Known landmarks are whizzing by at 7 c. Why can’t he conclude that Einstein was wrong?He is free to propose a different theory, but none has been found so far. So are you a relativity denier then? It seems to be your goal here. You resist it at every step of the way.
Such deniers are dime a dozen on sites like this, but then don't go telling me that your stories conform to an established theory and mine don't. I've pointed out several self contradictions with your assertions.
Einstein didn’t just suggest that light speed yielded the same value in any frame. That because quite apparent by all the attempts to measure the difference as was predicted by the prevailing view of the time. So he can’t conclude Einstein was wrong, he’d have to conclude that all the decades of light speed measurement were wrong. Einstein didn’t perform any of those measurements.