1. Mar 3, 2016

JVNY

Here is a linear version of the Ehrenfest paradox with the goal of understanding the observations of someone in motion in the scenario, then solicit your views on whether the calculations are correct and whether one can extend it to circular motion.

Consider a one dimensional train of proper length 100 traveling at 0.6c relative to a straight track. Gamma for the train relative to the ground is 1.25. The train's ground length is 80. The train is shunted onto a rectangular track whose short sides are so small as can be ignored, and whose long sides (side A and side B) each have 40 ground length. In the ground frame, the entire train fits in the rectangle with its front touching its rear (the train's ground length is 80, and the sum of sides A and B is 80).

At any given time in the ground frame half of the train is moving to the right along side A at 0.6c, and half is moving to the left along side B at 0.6c. The train goes continuously around the rectangle. The train's proper length is not altered by going around the track -- it still has 100 proper length because each half of the train occupies 40 ground length of track traveling at 0.6c, so each half of the train still has 50 proper length. The relative speed of the sections of the train on the opposite sides of the rectangle is (by Einstein velocity addition) 0.8823529411764707, for which gamma equals 2.125.

Although half of the train is on each side of the track in the ground frame, the same should not be true in the frame of an observer on the train because of the relativity of simultaneity.

Any observer on the train is traveling inertially (except when he reaches the end of a side and turns around). In his inertial frame, both sides of the track are length contracted to length 32. Therefore the portion of the train's proper length that is in his frame (that is on the same section of track as he simultaneously in that frame) is 32. In his reference frame, the remaining 68 of the train's proper length is on the other side of the track. The 68 remaining proper length divided by 2.125 (the appropriate gamma) is 32. Therefore the total train length in his frame is 64.

Therefore the train fits into the rectangular track in the ground frame because the 100 proper length is contracted by gamma=1.25 to a ground length of 80. The train fits into the rectangular track in the frame of an observer on the train because the track's length is contracted to 64 in his frame, and the train's length in his reference frame is also 64 (proper length divided by gamma squared).

So one could say that the perimeter of the rectangle is 80 in the ground frame and 64 in the train observer's frame. But if the train observer walks around the train he would measure it to have proper length of 100 (it has not deformed in the length dimension), and the train is at all times within the rectangle, so he might also say that the perimeter of the rectangle is 100 for an observer on the train.

2. Mar 4, 2016

A.T.

Why not just use circular motion, which doesn't have the discontinuities at the corners and allows to define a rest frame for the entire train?

3. Mar 4, 2016

4. Mar 4, 2016

JVNY

This is a further development of the discussion in the circular barn/pole thread. Many say that it is not possible to have a single rest frame for the circumference of a rotating disk because none of the points shares a common simultaneity surface with any other. So I have designed the linear example in order to have inertial frames.

In the broader context of the Ehrenfest Paradox, Planck pointed out that we should consider (a) what disk-riding observers will measure as compared to (b) what stationary observers will measure. Wiki at https://en.wikipedia.org/wiki/Ehrenfest_paradox. Some argue that a disk rider will measure the circumference to be length contracted compared to the measurement of a nonrotating observer who is at rest with respect to the path of the circumference in the ground frame because a rotating observer following that path will measure it to be length contracted. See, e.g., http://abacus.bates.edu/~msemon/SemonMalinWortel.pdf [Broken], whose authors consider the path that a moving observer takes along an n-gon and show that each side of the n-gon is length contracted for the observer, so the perimeter of the n-gon must be contracted by gamma for that observer as compared to the measurement of the observer at rest with respect to the n-gon. On the other hand, some like Einstein say that the disk riding observer will measure the circumference to be greater than that measured by the nonrotating observer.

This linear model seems to show how both views can be understood. It shows that for an observer on the train the perimeter is contracted relative to the ground frame (64 rather than 80), and the length of the train is also contracted (64 rather than 80); yet it also shows that in another sense for an observer on the train the length of the train is still 100, therefore the perimeter must be 100 because the entire train is within the perimeter at all times.

Last edited by a moderator: May 7, 2017
5. Mar 4, 2016

Staff: Mentor

Yes, but there's still not one such frame in which all parts of the train are at rest. So trying to analyze what is going on as if there is will lead to errors.

This simple-looking statement is hiding a lot of complexity that you are not taking into account.

Here's a suggestion: instead of waving your hands and reasoning using imprecise ordinary language, try using precise math. Write down the coordinates of key events in the rest frame of the track, describing the train at a particular instant of time in that frame--say the instant where the front end of the train is just completing one circuit around the rectangle, and the rear end of the train is just starting one, so the two ends are just touching. Then Lorentz transform those event coordinates into the two other frames of interest--the rest frame of the right-moving portion of the train, and the rest frame of the left-moving portion of the train--and see what you find.

6. Mar 4, 2016

JVNY

Start with image 1 of the track, including the rectangular section with the two parallel long sides A and B, and the short sides exaggerated in length for clarity. Consider for purposes of the example that the two sides A and B are essentially next to each other. The left sides of A and B are at x=0 in the ground frame, and the right sides are at x=40.

Next, image 2 is the train when the front of the train ("F") reaches the right end of side A. Wristwatches ("W") of each observer on the train are synchronized in the train frame. F’s wristwatch reads 0 in the ground frame. The wristwatch of the observer at the center of the train (“C”) reads 30 in the ground frame. The wristwatch of the observer at the rear of the train (“R”) reads 60 in the ground frame.

Finally, image 3 illustrates the train after F and R have traveled to the left end of the rectangle and C has reached the right end. This also illustrates a fourth observer, M, who is on side A at x=25.6 in the ground frame. The length of the segment of the train RM measured in the ground frame is 25.6, and as measured in the inertial frame of that segment is 32 (25.6 * gamma of 1.25).

All of the train observers on the segment RM have agreed to photograph the tracks at their own W=113.33. They are in an inertial frame and their clocks are synchronized in their own frame. So their photos will capture the tracks at the same time in RM time.

In the ground frame, R’s camera flashes first and captures a photo of R at x=0 on side A, and of F at x=0 on side B. Ground time t=24 later, M reaches x=40, its wristwatch reads 113.33 in the ground frame, and it takes a photo of itself there, which also captures the section of the train just ahead of M as having just turned to be on side B. In between these two events, each observer between R and M took a flash photo in sequence, moving forward sequentially in the ground frame from R to M, and capturing themselves on side A and the entirety of train segment MF on side B. Segment MF is 68 of proper train length, and it is entirely on side B, length contracted for observers on RM to length 32.

This is no different from simply having a train of proper length 32 moving to the right at 0.6c in the ground frame, and a train of proper length 68 moving to the left at 0.6c. In the reference frame of the 32 train, the 68 train is length contracted to 32, and the ends of both trains align at the same time in the 32 train’s inertial frame.

7. Mar 4, 2016

Staff: Mentor

I would make it clear that this is a derived result, not an assumption. The motion is clearly not Born rigid, so you cannot assume it. That it works out this way is interesting and actually puzzles me.

8. Mar 4, 2016

JVNY

Yes, it is a derived result. It is an extension of the prior thread of the circular barn/pole linked in Ibix's response 3 above. The group agreed that if a one spatial dimensional train is shunted into a circular track then its proper length is unchanged as it becomes a circle and travels continuously around the circular track. The group concluded that if it has extension in the direction in which it bends (say its width if it bends sideways as it enters the circular track) then it would deform along that dimension as it bends. But if it has just one spatial dimensional, having only length, then the group agreed that it would not deform along its length.

The current linear example is even more extreme in that the one dimensional train effectively doubles back on itself, but it should still be the case that it keeps its proper length as it goes around the perimeter of the rectangular track, just as it does when it goes around the circular track.

9. Mar 4, 2016

A.T.

If you want the train to move inertially, then lay out the tracks across a closed cylindrical universe. Using co-moving rulers, you will again find different proper lengths for the track and the train which fills the track, just like on the circular track.

But when you start worrying about clocks and global simultaneity in both frames, you will still encounter the same issues as on the circular track, despite the track and train being inertial here. Which suggest that these issues are not caused by the non-inertiallity of the circular motion.

Last edited: Mar 4, 2016
10. Mar 4, 2016

jartsa

JVNY said: "Therefore the total train length in his [passenger's] frame is 64"

So the real proper length of the train is 64.

(edit: well maybe that's not exactly a real proper length, but rather as real proper length as these circumstances allow)

If a passenger walks through the whole train he will say that he had to walk 100 length units, this is some kind of "proper length" according to an accelerated observer.

Last edited: Mar 4, 2016
11. Mar 4, 2016

A.T.

That's the physically relevant kind of proper length, that tells you how many passengers you can fit into the train.

12. Mar 4, 2016

JVNY

Agreed. So my goal is more limited, which is to be able to understand and demonstrate what might be true about the two different theories of why the rotating disk is non-Euclidean (circumference larger v. shorter than 2 pi r).

The circular barn/pole discussion supports Einstein's conclusion that the circumference is greater than the ground circumference. But it is also true that an observer moving along the circular path would measure the path to be shorter than the ground circumference, just as here an observer on the train will measure the total distance he travels along the rectangle to be shorter (64) than a ground observer measures it (80). And the train is entirely within the rectangle at all times, which suggests that an observer on the train should measure its length to be 64, which some theorists claim is the case. The example explains how that can be: 32 of the proper length is also 32 long in the observer's frame, and the other 68 of proper length is 32 long in the observer's frame because of the greater length contraction that the high closing speed causes. The 64 length in the moving observer's frame is not proper length. But it is the length of the train in the observer's frame if you apply SR to the straight line travel of segment RM.

It is interesting that in this case the total length in the observer's frame is the sum of an uncontracted segment and a contracted segment. And that the result is symmetrical -- an observer on side B will measure her segment's length to be 32 long (the same as its proper length), and she will measure the segment on side A as being 32 long (also 68 proper length contracted to 32). So this example is similar to other SR examples of inertial motion in which each observer in relative motion symmetrically measures the other as length contracted.

13. Mar 4, 2016

Staff: Mentor

The portions of it that JVNY is considering are, since he's only considering the inertial segments of motion and ignoring what happens at the turnarounds at each end. Inertial motion is Born rigid in the trivial sense that there's nothing to adjust for rigidity since there's no proper acceleration.

14. Mar 5, 2016

Staff: Mentor

Once again, you're ignoring a number of issues lurking beneath this apparently simple and innocuous statement. Let's actually assign coordinates to events and do some math.

We'll call the time illustrated in your Figure 2 (when the front end of the train is just reaching the turnaround at the right end of the shunt for the first time) in the ground frame time $t = 0$. And we'll call the spatial location of the left end of the shunt (where the middle of the train is at this instant in the ground frame) $x = 0$ in the ground frame. Then we have three events illustrated in Figure 2:

Event O: The middle of the train at the left end of the shunt, at $(x, t) = (0, 0)$. This event is the spacetime origin.

Event F1: The front of train at the right end of the shunt, at $(x, t) = (40, 0)$.

Event R1: The rear of the train, still well to the left of the shunt, at $(x, t) = (-40, 0)$.

The coordinates of these events in the right-moving train frame (in which the train in its entirety is at rest at all these events) are:

Event O: $(x', t') = (0, 0)$.

Event F1: $(x', t') = (50, -30)$.

Event R1: $(x', t') = (-50, 30)$.

Note that these $t'$ values are offset by 30 from the "W" times in your Figure 2. This is for convenience.

We can already see that, if we are talking about "how things are" in the right-moving train frame, we have to adjust for the relativity of simultaneity; the above three events happen at different times in this frame. First, let's take the easy adjustment: at time $t' = -30$ in this frame, the length of the train is 100--because the center and rear of the train are at rest in this frame between $t' = -30$ and the times of the events above in this frame, so their $x'$ coordinates don't change. That means that at $t' = -30$ in the train frame, the center of the train is at $x' = 0$ and the rear of the train is at $x' = -50$, so the train's length is 100. So far, so good.

Now let's look at the way things are at time $t' = 0$ in the right-moving train frame. The front of the train has been moving to the left, in this frame, for 30 units of time. Its speed is $(0.6 + 0.6) / (1 + 0.6 * 0.6) \approx 0.88$, so in 30 units of time it will move $30 * 0.88 = 26.5$ units of distance. That means that, at time $t' = 0$ in this frame, the front of the train will be at $x' = 23.5$. Furthermore, the right end of the shunt, which was co-located with the front of the train at $t' = -30$, has also moved to the left at 0.6c, so at time $t' = 0$, it will have moved $30 * 0.6 = 18$ units of distance and will be at $x' = 32$. So the train at time $t' = 0$ consists of two segments: a segment from $x' = -50$ (the rear end) to $x' = 32$ (where the right end of the shunt is) that is at rest, and a segment from $x' = 32$ to $x' = 23.5$, which is moving to the left at 0.88c. The first segment is 82 units long, and the second has a length contracted length of 8.5, which equates to a proper length of 18 (the gamma factor is 2.125), so the train's total proper length is still 100.

Finally, let's look at the way things are at $t' = 30$ in this frame. Here the front of the train has moved another 26.5 units of distance, so it is at $x' = -3$. The right end of the shunt has moved another 18 units of distance, so it is at $x' = 14$. The train is thus divided into two segments again, the first from $x' = -50$ to $x' = 14$, which is at rest, and the second from $x' = 14$ to $x' = -3$, which is moving to the left at 0.88c. The first segment is 64 units long, and the second has a length contracted length of 17, which equates to a proper length of 36, for a total proper length of 100, once again. Notice that at this point in the right-moving train frame, the center of the train still has not reached the right end of the shunt, so we are still at an earlier time than your Figure 3, which we discuss next.

Now let's switch to your Figure 3 and label the key events there and give their coordinates in the ground frame. All of these events happen at time $t = 40 / 0.6 = 66.67$, since that is how long it takes for the front end of the train, at 0.6c, to cover the distance 40 from the left end of the shunt to the right end. So the events and their coordinates are:

Event R2: The rear of the train is at the left end of the shunt, at $(x, t) = (0, 66.67)$.

Event C2: The center of the train is at the right end of the shunt, at $(x, t) = (40, 66.67)$.

Event F2: The front of the train is at the left end of the shunt, at $(x, t) = (0, 66.67)$. Note that this event has the same coordinates as event R2, since the front and rear ends of the train are co-located at this event.

The coordinates of these events in the right-moving train frame (in which only the rear half of the train is at rest in the period between the previous set of events and these, in the ground frame) are:

Event R2/F2: $(x', t') = (- 50, 83.33)$.

Event C2: $(x', t') = (0, 53.33)$

Similar to what we did above, we can figure out where the rear and front of the train are, in this frame, at the time of event C2, and where the center is at the time of event R2/F2. At time $t' = 53.33$, 23.33 time units have elapsed since event R1 above, so the front of the train has moved an additional $23.33 * 0.88 = 20.5$ units of distance, and is at $x' = -23.5$, and the right end of the shunt has moved an additional $23.33 * 0.6 = 14$ units of distance and is at $x' = 0$. Of course the latter is exactly what we expect since the center of the train is at the right end of the shunt at this time in this frame. So the train consists of two segments, one from $x' = -50$ to $x' = 0$, and one from $x' = 0$ to $x' = -23.6$. The latter segment's proper length is $23.6 * 2.125 = 50$, so the train still has a total proper length of 50 at this time.

Now let's look at event R2/F2. Here an additional 30 units of time have passed in this frame since event C2, so the front of the train has moved 26.5 units of distance and is at $x' = -50$, as expected; and the center of the train has also moved 26.5 units of distance to the left and is at $x' = -26.5$. The right end of the shunt has moved 18 units of distance from event C2 and is at $x' = -18$. So the train consists of two segments, one from $x' = -50$ to $x' = -18$ at rest, and one from $x' = -18$ to $x' = -50$ which is length contracted by a factor of 2.125 and so has a proper length of $32 * 2.125 = 68$. So the train's total proper length is, once more, 100. And from this point on in this frame, the front and rear ends of the train are touching, so the train continues to form a "loop" with two segments having lengths as just given, and its proper length remains 100.

We can do a similar analysis in the left-moving train frame--the frame in which the portion of the train moving to the left in the ground frame is at rest. Here the Lorentz transformation is the inverse of the one we did above, so the event coordinates are:

Event O: $(x'', t'') = (0, 0)$.

Event F1: $(x', t') = (50, 30)$.

Event R1: $(x', t') = (-50, -30)$.

Event R2/F2: $(x'', t'') = (50, 83.33)$.

Event C2: $(x'', t'') = (100, 113.33)$.

Up until event F1 in this frame, the entire train is moving to the right at 0.88c (gamma factor 2.125). So we can evaluate the total length contracted length of the train in this frame at time $t'' = 30$, since it will be the same at any time up to then (the entire train will be length contracted by this factor). At this time, the rear end of the train will have moved for 60 units of time at 0.88c, or about 53 units of distance, so it will be at $x'' = 3$. That means the length contracted length of the train is 47 units, for a proper length of 47 * 2.125 = 100.

The next time of interest is $t'' = 83.33$, the time of event R2/F2. The rear end of the train will have moved a distance 53.33 * 0.88 or 47, so it will be at $x'' = 50$, as expected--co-located with the front of the train. But now we need to know where the right end of the shunt is at this time in this frame. It was co-located with the front of the train at event F1, and it moves to the right at 0.6c, so it will have moved a distance $0.6 * 53.33 = 32$ and be at $x'' = 82$. That means the train consists of two segments, one of length 32 at rest, and one of length 32 that is length contracted by a factor 2.125, so again the train has a total proper length of 100. And, once again, from this point on in this frame, the front and rear ends of the train touch and it continues to form two segments of the same lengths.

(Note, btw, that event C2, where the center of the train reaches the right end of the shunt, happens later, at time $t'' = 113.33$, which means that at the time of event R2/F2, the center of the train is still moving to the right. This is to be expected since the length of the segment of the train that's at rest is only 32 in this frame.)

So, when the math is done properly, it is evident that the proper length of the train is always 100. There is no justification for any other number.

15. Mar 5, 2016

A.T.

If you instead consider a track across a whole closed cylindrical universe, you don't have any corners.

16. Mar 5, 2016

Staff: Mentor

Yes, that would be a good "pure" idealized case.

17. Mar 5, 2016

JVNY

PeterDonis, I think that we actually agree except perhaps on language.

You write:
I wrote:
So we agree that the train's proper length is 100.

You wrote:
I wrote:
So we both agree that the two segments are 32 each, thus the sum of the segments is 64. As to this sum, I wrote:
Of course jartsa disagrees, writing:
For me, it is clear that the proper length remains 100. That was the result of the circular barn/pole discussion. But Planck suggests that we determine what disk-riding observers will measure, or in this case what a train riding observer will measure. It is clear that an observer on the train who is stationary on the train will measure the perimeter of the rectangle to be 64. He has to, because the perimeter is 80 in the ground frame, and the stationary train observer will measure the perimeter to be length contracted to 80%, or 64.

But what does it mean for a train riding observer to measure the length of the train? One way is to remain stationary in his position on the train. In that case, one segment of the train will be moving with respect to him, so he must measure the train to be shorter than its proper length. In this case, he measures the train to be 64 long, consisting of (1) a segment that is uncontracted and 32 long in his frame, and (2) a segment that is length contracted and is 32 long in his frame. This is important because it is the same length as his measurement of the length of the perimeter in his frame. A train that has 64 length in his inertial frame fits entirely within a perimeter that has 64 length in his frame.

A different way to measure the length of the train is to walk around the train with a measuring rod or tape and measure front to back. In this case he will find that the train is 100 long. This is the proper length of the train.

So, one can say that the length of the train (its proper length) is longer than the length of the perimeter in the ground frame because a walking train rider measures the length to be 100. Or, one can say that the length of the train (in the stationary rider's frame) is shorter than the length of the perimeter in the ground frame, because a stationary train rider measures the train's length in his inertial frame to be 64.

These are just two different measurements, one in an inertial frame and one that is not in any frame at all, because there is no single inertial frame in which the entire train is at rest.

18. Mar 5, 2016

Staff: Mentor

There is no way for the train riding observer to measure the length of the train "all at once". Even if he uses light beams bouncing off various parts of the train and measures the round-trip times by his clock, he still has to face the fact that the train as a whole does not have a single state of motion; different parts are moving at different velocities, and the velocity of one particular part changes during the time the light is making its round trip.

In many simpler scenarios, this issue does not arise, because all parts of the object being measured have the same state of motion during the entire period of the measurement. So one can view measurements made over time on different parts of the object as amounting to measuring the object "all at once", since any measurement can be translated in time without changing its results. But that is simply not true in the present scenario. And that means that some questions we are used to asking, like "what is the length of the train?" or "what is the perimeter of the rectangle as measured by the observer on the train?" simply do not have unique answers.

19. Mar 5, 2016

JVNY

On this we disagree. The segment RM is simply the platform in Einstein's example, and segment NF is the train. If two bolts of lightning strike simultaneously in the inertial RM frame at RM time 113.33 between sides A and B at points R and M, then the one on the left will scorch R and F, and the one on the right will scorch M and N. Then riders on the RM segment can determine that the entirety of segment NF was aligned with RM at the same time in the RM inertial frame. The entire train was on the two tracks, in two parallel segments, with the ends of each segment aligned at the same time (in the RM frame). It is no different from a 68 proper length train passing a 32 proper length platform at approx. 0.88c with lightning bolts striking both ends of the platform simultaneously in the platform frame. Observers on the platform can determine that the entirety of the train was between the two ends of the platform at the same time in the platform frame, that the train had length 32 in the platform frame, that the combined length of the train and the platform was 64 in the platform frame.

Similarly, if an infinite number of bolts of lightning strike from R to M simultaneously in the RM frame at RM time 113.33 between sides A and B, they will scorch every part of the segment RM and every part of the segment NF, and again riders on the segment RM can determine that the entirety of the train was on the two tracks, in two parallel segments at the same time in the RM frame.

So one correctly measured length of the train as measured in the RM inertial frame is 64.

On this I agree. 64 is one answer for the length of the train, and a correct one, but it is not unique. The train has a different proper length (100). And it will have yet a different length for an observer who is in relative motion to all of (a) the ground, (b) segment RM, and (c) segment NR.

20. Mar 5, 2016

Staff: Mentor

What is "this", exactly? It looks like you are describing basically what I was describing in the latter part of post #14; your two lightning strikes would be at the event I labeled as R2/F2, and at the event where the right end of the shunt is at that time in the right-moving train frame, which I didn't label, but which is at $(x'', t'') = (82, 83.33)$ in the coordinates I used in that post.

21. Mar 5, 2016

JVNY

By "this" I mean your statement that "some questions we are used to asking, like 'what is the length of the train?' . . . do not have unique answers." I agree that there is not a unique answer to the question "what is the length of the train?" There are many correct answers.

One correct answer is that its length is 100 (its proper length).

A second correct answer is that the train's length is 80 (its length in the reference frame of a ground observer relative to whom the train is moving at 0.6c, regardless of whether the train is straight as in image 2 above or doubled back on itself as in image 3 above).

A third correct answer is that the train's length is 64 (its length in the inertial reference frame of the segment RM as measured at RM time 113.33 in image 3 above). Yes, this is your event R2/F2. I took the second quotation of your writings in my post #17 directly from your paragraph describing event R2/F2. I will quote it again, including the introductory sentence of that paragraph:

The at-rest segment RM's length of 32 is uncontracted in its own frame. Adding the other segment's length contracted 32 length yields a total train length of 64 in the frame of RM. Thus your own calculations show that 64 is a correct answer for the length of the train.

A fourth correct answer is that the train's length is 60 (to an observer in relative motion of 0.8c with the train when the train is straight as in image 2 above).

Each of these answers is correct. None is uniquely correct.

22. Mar 5, 2016

Staff: Mentor

In other words, you agree with the "this" that you were referring to. What confused me was that you said

23. Mar 5, 2016

JVNY

Right. I disagreed with the statement "There is no way for the train riding observer to measure the length of the train 'all at once.' Just as Einstein measures the train passing the platform all at once (platform time), an observer on the segment RM can measure the length of the segment NF 'all at once' (in RM time) using for example lightning bolts as I described.

Here is another way to think about the issue, although it is quite late for me and so I can only sketch it out. What is the length of the train for an observer in an inertial reference frame R that is moving to the right at 0.6c relative to the ground when event O occurs? What is the length of the train in R when event R2/F2 occurs? What is the length of the train in R shortly after event R2/F2 occurs? How does the train move in R after event R2/F2 occurs? I can't draw it immediately, but essentially the worldlines of all of the points on the train from R to F should be straight upwards until event O occurs. Then, F should travel left at approx. 0.88c, then thereafter all points behind F should sequentially travel left at approx. 0.88c while all points in the segment RM continue to travel straight up, until event R2/F2 occurs. Then, segment RM and segment NF should be parallel and of equal length in R (32), thus a total length in R of 64 (although segment RM will also have proper length 32, whereas segment NF will have proper length 68).

After that it becomes more interesting, I think. I think that the two parallel segments then move leftward at 0.6c in R. Each keeps its 32 length in R; the segment of the train on side A continues to have 32 proper length, and the segment on side B continues to have 68 proper length. The train continues to go around the perimeter, so different points of the train are on side A or side B at any given time. While they are on side A the parts of the train are moving relative to the ground at 0.6c to the right, just like R. So they are at rest in R. But the rectangular section of track continues to move left in R at 0.6c.

24. Mar 5, 2016

Staff: Mentor

The lightning bolt strikes themselves do not measure any length at all; they are single events, points in spacetime, with no length. A measurement of length that uses the light travel time from the two bolt strikes to some central location does not measure the length "all at once" because it takes time for the light to travel.

Your idea about trying to draw the worldlines of all the points on the train in spacetime is a good one, and I think it would help to clarify the discussion.

25. Mar 6, 2016

JVNY

Start with Einstein's train and platform example. A train passes a platform. Simultaneously in the platform frame, lightning bolts strike the left side of the platform (L) and the right side of the platform (R). Platform observers know that the strikes were simultaneous in the platform frame because the light from the two bolts reach an observer at the center of the platform (LR/2) simultaneously. Einstein assumes that the distance between the frame and the platform is so small that it can be ignored. A lightning bolt that strikes the platform also strikes the train at the same place in the platform frame. The bolts scorch points L and R, and they also scorch the front of the train (TF) and the rear of the train (TR). Now, after the strikes occur observers can confirm that the bolts struck and scorched L and R simultaneously while L and R were are rest with respect to the platform, that the light from the bolts reached LR/2 simultaneously while LR/2 was at rest with respect to the platform, and that the bolts struck and scorched TF and TR. Therefore, the observers can confirm that the entirety of the train from TR to TF has the same length in the platform frame as the platform does. Thus if the train has 68 proper length, the platform has 32 proper length, and the train is moving at approx. 0.88c to the left, the scorch marks will confirm that the train has length 32 in the platform frame (proper length contracted by gamma = 2.125). In addition (which Einstein's example does not describe), it is only relevant that L and R be at rest with respect to the platform frame when the bolts strike. If the strike causes R to accelerate, say to start to move to the right in an alternate world line marked as a dashed line (Ralt), that does not in any way affect the conclusion that the train has 32 length in the platform frame.