Does Relativity Affect How We See Simultaneous Events on a Moving Train?

[stevmg]

****No****

 

[stevmg]

Sports Fans -

I am a doctor (MD) so give me a break here.

Blood doesn't move that fast and we cannot prove Fizeau's experiment with it...

 

[stevmg]

To JesseM

Reread #25 and you were clearer than Albert was in that you specifically identified both the observer (on the desert island) and the ship's separate time frames and the behaviour of light in each frame from the identical flash. Got it. See clearly why the observer on the desert island would not see the flashes as simultaneous and wouldn't know it. Now, I have to work with the second part of the experiment and understand that applicability to the problem.

Well, actually, that proves that the flashes are simulataneous in one frame and not another even though these flashes are physically the same flash for both frames.

Don't worry, I didn't do brain surgery when I was in practice.

 

[JesseM]

I’m having difficulty following this, especially with the bolded segments. Imo, the only time we can use/say that the train’s frame is at rest is for the very short moment both strikes hit the train

What do you mean? By definition, the inertial rest frame of an object moving inertially is the frame where it's always at rest, that's just what "rest frame" means, in both Newtonian physics and relativity. In the train's rest frame, each part of the train remains at a fixed position coordinate while the ground moves at constant velocity past it.

(later shown to be simultaneous, as confirmed by the stationary observer situated precisely at an equidistant location from the train’s back and front when the strikes hit), and that would be what you refer to as the single point in space and time.

"Single point in space and time" means a single time coordinate and a single position coordinate--the two lightning strikes happen at different position coordinates, in both frames.

At that point, if you take an instantaneous snapshot of the scene when the strikes hit, both passenger and ground observer, in their own respective frames, are located exactly at the same distance from each end of the train, while being perfectly aligned orthogonally. Here, if we could remain at rest

At rest relative to what? Rest and motion have no absolute meaning in relativity, you can only define them relative to particular objects or particular coordinate system. If you are at rest relative to the ground, you are moving relative to the train (and thus moving relative to the train's rest frame), while if you are at rest relative to the train, then you are moving relative to the ground (and thus moving relative to the ground's rest frame).

and let the flashes follow their course, everyone would be happy to see that all happens simultaneously

But the whole point of the thought-experiment is to show that different frames define "simultaneity" differently--if the flashes are simultaneous in the ground frame, they cannot also be simultaneous in the train frame without violating the postulate that every frame should measure the speed of all light rays to be c.

give mirrors to the ground observer and synchronised detector/clocks to the passenger and he will observe that both reflected light flashes stop the clocks simultaneously.

I don't understand, why does the passenger need two clocks? Are they at different positions on the train? And why is he stopping the clocks when he receives the reflected light from the mirrors (and where are the mirrors positioned on the ground?) as opposed to stopping them when he receives the light from the flashes themselves?

Let it roll on and the reflected light waves from both strike impact locations will travel, at c, eventually being perceived simultaneously by the ground observer, who is stationary with respect to the strikes, while the passenger will see a slight difference because, while the light waves travel at c, he is moving away from the back / towards the front strike locations, making it appear that they were not simultaneous when indeed they were.

Why do you say "reflected" light waves? It sounds like you're just talking about the ordinary light waves that proceed directly from the flashes to each observer here, no?

In any case, I think what you're not understanding here is that in relativity there is no objective truth about whether events "were" or "were not" simultaneous, all we can say is whether they happen at the same time coordinate in a given coordinate system. I'll lay out the steps of the argument in order so you can tell me where you disagree with a step:

1. Both strikes happen at the same time coordinate in the ground frame, and both the ground-observer and the train-observer are equidistant from the strikes at the time they occur in the ground frame.

2. In the ground frame, the train-observer is moving towards the position of one strike and away from the position of the other. If we assume the light from each strike heads towards the train-observer at a speed of c in this frame (as is required by the 2nd postulate of relativity), the light from one strike must reach the train-observer before the light from the other.

3. Since all frames must agree about local events, all frames must agree the light from the strikes reaches the train-observer at different times.

4. In the train-observer observer's rest frame, the train-observer is at rest at a fixed position, as are the front and back of the train which are both at an equal distance D from the train-observer.

4. If the time coordinate of the strike at the front is t1 in the train-observer's frame, then assuming the light moves at c in this frame and the distance from the front to the train-observer is D, the light must reach the train-observer at time coordinate t = t1 + D/c in this frame.

5. If the time coordinate of the strike at the back is t2 in the train-observer's frame, then assuming the light moves at c in this frame and the distance from the back to the train-observer is D, the light must reach the train-observer at time coordinate t = t2 + D/c in this frame.

6. If the light from both strikes reaches the train-observer at different times, that must mean t1 + D/c is not equal to t2 + D/c.

7. The only way for them not to be equal is if t1 is not equal to t2. Therefore, the strikes must have happened at different time-coordinates in the train-observer's rest frame.

The ground observer’s predictions are not for the train’s rest frame, they are for the train’s frame as it is in motion.

Again, "in motion" means nothing in itself in relativity. The train is in motion relative to the ground's rest frame, and the ground is in motion relative to the train's rest frame. Also, it's not clear you understand that the phrase "train's rest frame" specifically refers to the inertial coordinate system where the train is at rest, i.e. its position coordinate remains unchanged as the time coordinate varies in this coordinate system.

So, I think that the agreement about preserving local events when viewed from the passenger’s perspective should be that, although they did, she sees that the strike flashes have not reached the stationary ground observer simultaneously, keeping in line with observations made in her own frame.

But in the ground frame, the strikes occurred simultaneously at equal distances from the ground observer, no? Therefore, if we assume the light moves at c in the ground frame, we must predict in the ground frame that the light from each strike reaches the ground observers simultaneously. And this means that the events of both light rays reaching the ground observer happen at the same position and time in the ground frame, so this is a fact about local events coinciding which different frames must agree on. Thus it must also be true in the train frame that the light from each strike reached the ground observer simultaneously.

This will always appear to be true as the passenger is in motion with respect to the reflected light from the ground observer, exactly as it was inside the train with respect to the original strikes.

Still don't understand what you mean by "reflected light"--reflected from where? The thought-experiment as Einstein stated it was only meant to deal with the light traveling directly from the strikes to each observer, not with any light reflected off mirrors.

Maybe it would help to put some numbers on this problem? Suppose that in the ground frame, at t=0 seconds both the ground-observer and the train-observer are right next to each other at position x=0 light-seconds on the x-axis. The strike at the back of the train happens at x=-8, t=0. The strike at the front of the train happens at x=+8, t=0. The train observer is moving in the +x direction at 0.6c, so for example at x=10 seconds in the ground frame he will be at position x=6 light-seconds (in general his position as a function of time will be given by x(t) = 0.6c*t). The light from each strike must move at c = 1 light-second/second in the ground frame, so if the strike at the back happened at x=-8, t=0, that means at t=1 the light from that strike has reached x=-7, at t=2 the light from that strike has reached x=-6, and so forth (in general for this light ray we have x(t) = -8 + 1c*t). And if the strike at the front happened at x=8, t=0 that means at t=1 the light from that strike has reached x=7, at t=2 the light from that strike has reached x=6, etc. (for this light ray we have x(t) = 8 - 1c*t)

Given these numbers, would you agree that at t=8, the light from both strikes will be at position x=0, the same as the position of the ground-observer (who isn't moving in this frame)? And would you agree that at t=5, the train-observer will be at position x=3, and the light from the strike at the front is also at x=3? Finally, would you agree that at t=20, the train-observer will be at position x=12, and the light from the strike at the back is also at x=12?

 

[TcheQ]

Sports Fans -

I am a doctor (MD) so give me a break here.

Blood doesn't move that fast and we cannot prove Fizeau's experiment with it...

If mathematics is your thing, use these calcs:
Train length 10m
Observer at a point equidistant from each point of light (5m)
speed of light=c
train speed =100m/s
how long does it take the light to reach the observer from each point? given velocity=d/t, d=distance, t=time (solve for T)

example solution:
If the train was at rest, t= 5/c for left and right.

 

[stevmg]

I was a math major but branched into statistics and not relativity. I still didn't do brain surgery when I became a doctor so don't worry.

Actually, all this explaining got me back to Einstein Section 9 "Relativity" where he gives his example of the lack of simultaneity in the moving frame (I guess we call that S') while there is simultaneity in S. He doesn't use Lorentz's equations to show it but just gives us a feel for it by describing it. By using the equations that JesseM alluded to:

(x')^2 - (c^2)(t')^2 = x^2 - (c^2)t^2 (I dropped the y, y', z, z' coordinates) one sees the effect of x' on t' for given x and t. Because c is constant there is a non-linear change in x' with a change in t' which makes any attempt at simutaneity impossible. Holding c as constant and having t as a variable that changes, even without knowing what the Lorentz equations are, it would be impossible to alter the x or x' without non-linearly altering the t and this would throw the t off fourse and appear before of after it should by a linear (or Galilean) approach. In other words, by holding c constant, we alter the t. The x is also altered in S' by Lorentz too which makes it even more complicated. The Lorentz equations were derived in a strict mathematical sense from the assumption of constant c. The gamma correction [SQRT(1 - v^2/c^2)] and all the Lorentz equations are obtained by mathematical derivation from that assumption (constant c, all else change may be variable) [Appendix I, Einstein "Relativity"] and is not a "gift from God." Therefore, one cannot separate the explanation of lack of simultaneity because of a constant c from the actual equations (once derived) as they are equivalent in the logical sense (statement A is true if and only if statement B is true.)

 

[JesseM]

Therefore, one cannot separate the explanation of lack of simultaneity because of a constant c from the actual equations (once derived) as they are equivalent in the logical sense (statement A is true if and only if statement B is true.)

I don't think you need both postulates to demonstrate lack of simultaneity, you can demonstrate it with the postulate of constant c without needing to refer to the postulate that the laws of physics are the same in every inertial frame. In any case, just because A logically implies B which logically implies C, that doesn't mean that from a pedagogical point of view it's always good to go through the intermediary of B if there's a way to explain how C follows from A directly. For example, it might be true that we can use the axioms of arithmetic to prove some esoteric theorem from number theory, and then use that theorem to prove that 1+1=2, but if you want to show how 1+1=2 follows from the axioms of arithmetic it'll be a lot less confusing if you choose a more direct route!

 

[stevmg]

True - Einstein does assume that the laws of physics are true in all inertial frames and it is from that that the Lorentz equations are derived (the constancy of c being the central or core theme.) But, as you say, his example of the train and embankment does not require that the laws of physics be the same. Of course, if the laws of physics were different in different inertial frames, I suppose it would be possible for one to offset a non simultaneity by a fluke of a complementary action by a different law of physics in another frame. Of course, Albert did not go into that.

 

[danielatha4]

If for the observer on the platform the time at which the front strike occurs is t=0 then the time the back strike occurs must also be t=0.

According to the Lorentz transformation for time: t’= gamma(t – (vx/c^2))

Then the passenger must have seen the front strike at t’=-gamma(vx/c^2) and the back strike must have occurred at t’=-gamma(vx/c^2)

Where gamma, v, x, and certainly c are not different in either of the time equations to describe the front and back strike.

t-t=0

t’-t’=(-gamma(vx/c^2))-(-gamma(vx/c^2))=0

 

[JesseM]

Where gamma, v, x, and certainly c are not different in either of the time equations to describe the front and back strike.

What do you mean? The front and back strike occurred at different positions in the platform frame, so the two strikes should have different x-coordinates. If you assume the platform observer is at x=0 and the strikes occurred at equal distances from him, then if the front strike occurred at x=X, the back strike must have occurred at x=-X.

 

[danielatha4]

My mistake, so let me make sure I'm going about this the right way.

I'm going to give numbers to all these variables.

t for front strike and back strike = 0
v=.8c
length of train=L=100

X coordinate for front strike = .5L = 50
X coordinate for back strike = -.5L = -50

For front strike:
t' = 0 - ((.8c)(50)/c^2) * gamma
gamma ends up being 5/3
t' = -2.22E-7

Does this mean the front strike for the observer on the train happens before the front strike for the observer on the platform?

For the back strike:
t' = 0 - ((.8c)(-.5L)/c^2) * gamma
where gamma again equals 5/3
t' = 2.22E-7

So the back strike for the train observer happens after the back strike for the platform observer?

And the delay between the front strike and back strike for the observer is 4.44E-7s ?

Did I just correctly confirm why the flashes are not simultaneous in the train frame of reference?

 

[ValenceE]

Hello again dear JesseM,

OK, let’s forget about clocks and mirrors, didn’t really take the time to explain it all and, furthermore, I think I got it wrong in the last part, but on to my main arguments, which are more important, so here are my replies to your comments…

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Why do you say "reflected" light waves? It sounds like you're just talking about the ordinary light waves that proceed directly from the flashes to each observer here, no? yes, I was referring to the flashes as they reflect from the train’s surface, but this is irrelevant as you point out.

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”In any case, I think what you're not understanding here is that in relativity there is no objective truth about whether events "were" or "were not" simultaneous, “

Indeed, here is precisely where I have difficulty. It is only because of SR’s use of the constant speed of light that events as seen from observers in different frames cannot be said to be simultaneous. Every measurement made that involves light (or any EM wave) is based on ‘c’ , so of course if measurements from inertial/rest VS moving frames are compared, the results will differ. However, imo, that does not invalidate simultaneity, it just proves that c is constant.

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all we can say is whether they happen at the same time coordinate in a given coordinate system. I'll lay out the steps of the argument in order so you can tell me where you disagree with a step:

1. Both strikes happen at the same time coordinate in the ground frame, and both the ground-observer and the train-observer are equidistant from the strikes at the time they occur in the ground frame. agreed

2. In the ground frame, the train-observer is moving towards the position of one strike and away from the position of the other. If we assume the light from each strike heads towards the train-observer at a speed of c in this frame (as is required by the 2nd postulate of relativity), the light from one strike must reach the train-observer before the light from the other. agreed

3. Since all frames must agree about local events, all frames must agree the light from the strikes reaches the train-observer at different times. agreed

4. In the train-observer observer's rest frame, the train-observer is at rest at a fixed position, as are the front and back of the train which are both at an equal distance D from the train-observer. agreed

4. If the time coordinate of the strike at the front is t1 in the train-observer's frame, then assuming the light moves at c in this frame and the distance from the front to the train-observer is D, the light must reach the train-observer at time coordinate t = t1 + D/c in this frame. agreed

5. If the time coordinate of the strike at the back is t2 in the train-observer's frame, then assuming the light moves at c in this frame and the distance from the back to the train-observer is D, the light must reach the train-observer at time coordinate t = t2 + D/c in this frame. agreed

6. If the light from both strikes reaches the train-observer at different times, that must mean t1 + D/c is not equal to t2 + D/c. agreed

7. The only way for them not to be equal is if t1 is not equal to t2. agreed
Therefore, the strikes must have happened at different time-coordinates in the train-observer's rest frame. Disagreed, the truth is this is only apparent... Why restrict the conclusion to this one possibility, especially when it’s wrong? Couldn’t the passenger, being aware of SR, knowing that she is in a moving train, wonder if the strikes could have been simultaneous?, t1 and t2 are not equal because the strike impact locations are in motion relative to the train’s rest frame.

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Maybe it would help to put some numbers on this problem? Suppose that in the ground frame, at t=0 seconds both the ground-observer and the train-observer are right next to each other at position x=0 light-seconds on the x-axis. The strike at the back of the train happens at x=-8, t=0. The strike at the front of the train happens at x=+8, t=0The train observer is moving in the +x direction at 0.6c, so for example at x=10 seconds in the ground frame he will be at position x=6 light-seconds (in general his position as a function of time will be given by x(t) = 0.6c*t). The light from each strike must move at c = 1 light-second/second in the ground frame, so if the strike at the back happened at x=-8, t=0, that means at t=1 the light from that strike has reached x=-7, at t=2 the light from that strike has reached x=-6, and so forth (in general for this light ray we have x(t) = -8 + 1c*t). And if the strike at the front happened at x=8, t=0 that means at t=1 the light from that strike has reached x=7, at t=2 the light from that strike has reached x=6, etc. (for this light ray we have x(t) = 8 - 1c*t)

Given these numbers, would you agree that at t=8, the light from both strikes will be at position x=0, the same as the position of the ground-observer (who isn't moving in this frame)? And would you agree that at t=5, the train-observer will be at position x=3, and the light from the strike at the front is also at x=3? Finally, would you agree that at t=20, the train-observer will be at position x=12, and the light from the strike at the back is also at x=12?

Agreed to all of the above, but, for me, in no way does this prove or demonstrate that the flashes didn’t happen simultaneously. All we can be certain about is that the passenger is moving at 0.6c relative to the ground observer and the original strike locations. Also, in this explanation, since both strikes happen at t=0 doesn’t that make them simultaneous for all observers?




Regards,

VE

 

[JesseM]

My mistake, so let me make sure I'm going about this the right way.

I'm going to give numbers to all these variables.

t for front strike and back strike = 0
v=.8c
length of train=L=100

X coordinate for front strike = .5L = 50
X coordinate for back strike = -.5L = -50

For front strike:
t' = 0 - ((.8c)(50)/c^2) * gamma
gamma ends up being 5/3
t' = -2.22E-7

Does this mean the front strike for the observer on the train happens before the front strike for the observer on the platform?

I don't know if it's really physically meaningful to talk about whether an event in one frame happened before or after the same event in a different frame--after all, the placement of the temporal origin is pretty arbitrary, it's just to make the math a little easier that we assume that the origin of one frame lines up with the origin of the other. Normally people just talk about whether one event happened before, after, or simultaneously with another event in a single frame.

For the back strike:
t' = 0 - ((.8c)(-.5L)/c^2) * gamma
where gamma again equals 5/3
t' = 2.22E-7

So the back strike for the train observer happens after the back strike for the platform observer?

And the delay between the front strike and back strike for the observer is 4.44E-7s ?

I'm not sure what units you're using so I can't check the actual numbers, but your equations are correct so that's probably the right answer (and it's definitely correct that the front strike happens before the back strike in the train observer's frame)

Did I just correctly confirm why the flashes are not simultaneous in the train frame of reference?

Yes, this looks like a correct confirmation using the Lorentz transformation.

 

[ValenceE]

(and it's definitely correct that the front strike happens before the back strike in the train observer's frame)


Again, I must disagree... It’s definitely correct that the front strike appears to happen before the back strike in the train observer’s frame.

Both strikes hit at the same time. Relativity is not about time, it’s about motion and the constant speed of light.


VE

 

[TcheQ]

Again, I must disagree... It’s definitely correct that the front strike appears to happen before the back strike in the train observer’s frame.

There are no illusions. It actually happens. If you are still confused, use the example I quoted for stevemg in this thread.

 

[Mentz114]

Agreed to all of the above, but, for me, in no way does this prove or demonstrate that the flashes didn’t happen simultaneously.

You are still talking as if simultaneity is an absolute. In whose frame ?

... since both strikes happen at t=0 doesn’t that make them simultaneous for all observers?

Definately not. The fact that the strikes happen at t=0 only applies to some observers.

You must try to understand that if two events are separated spatially, there is no way to define whether they are 'simultaneous' except by adopting a convention.

If I'm in the middle of the carriage and I send beams to the ends at the same time by my clock, the beams appear to me to reach the ends at the same reading on my clock. This will also be true for anyone on the plane perpendicular to the length of the carriage and passing through my position in the middle.

Other observers will say that the beams did not reach the ends at the same time on their clocks.

Simultaneity is relative.

 

[stevmg]

OMG!

I can't even remember what examples anyone gave me!

The only absolutes are the fact that the flashes occurred and the fact that the train is of a given length. Good old M' (the train observer) sits at the midpoint of the train.

The key elemnt here is that x' and t' (in the moving coordinates) when placed in the Lorentz equation x^2 - ct^2 = (x')^2 - c(t')^2 (again, I droped the y, y' and z, z' coordinates) for given x and given t (the ground reference coordinates) then x' and t' change (parabolic equation) annd for given x' (the train coordinates) the t's are different and not linearly related to the original x and t. Hence half the length forward (a function of x) and half the length back do not add up to the same t's as they would originally. If you use Galilean coordinate with no Lorentz correction, you will get one-to-one equivalence as you do with a straight line. Not so with a parabola.

Now this is right - I know it! Damn it. I've tutored math since I was in high school and college, my own kids, and even my troops when I was active duty and these kids were going to college.

Remember, the laws of physics ARE the same in all reference frames and Lorentz-Einstein states that if this is so, then these equations hold.

 

[JesseM]

In any case, I think what you're not understanding here is that in relativity there is no objective truth about whether events "were" or "were not" simultaneous, “

Indeed, here is precisely where I have difficulty. It is only because of SR’s use of the constant speed of light that events as seen from observers in different frames cannot be said to be simultaneous. Every measurement made that involves light (or any EM wave) is based on ‘c’ , so of course if measurements from inertial/rest VS moving frames are compared, the results will differ. However, imo, that does not invalidate simultaneity, it just proves that c is constant.

But I think you're forgetting the first postulate of relativity, which says that the laws of physics work exactly the same in every inertial frame, so there can be no physical basis for saying one frame's judgments about things like velocity or simultaneously are more correct than any others. Now, as a purely philosophical matter, this doesn't stop you from believing in such a thing as objective truths about simultaneity, in which case there would presumably be only a single "metaphysically preferred" frame in which objectively simultaneous events were actually assigned the same time-coordinate--the problem is, as long as relativity is correct there is absolutely no physical experiment whatsoever that could tell you which frame this is!

Do you at least agree with that much, that unless relativity is correct there can be no experimental method of determining the truth about simultaneity, since different frames differ in their opinions about whether two events happened at the same time-coordinate, and the first postulate says that the laws of physics work exactly the same way in every frame so there can be no experimental basis for picking out one frame as a "preferred" one?

all we can say is whether they happen at the same time coordinate in a given coordinate system. I'll lay out the steps of the argument in order so you can tell me where you disagree with a step:

1. Both strikes happen at the same time coordinate in the ground frame, and both the ground-observer and the train-observer are equidistant from the strikes at the time they occur in the ground frame. agreed

2. In the ground frame, the train-observer is moving towards the position of one strike and away from the position of the other. If we assume the light from each strike heads towards the train-observer at a speed of c in this frame (as is required by the 2nd postulate of relativity), the light from one strike must reach the train-observer before the light from the other. agreed

3. Since all frames must agree about local events, all frames must agree the light from the strikes reaches the train-observer at different times. agreed

4. In the train-observer observer's rest frame, the train-observer is at rest at a fixed position, as are the front and back of the train which are both at an equal distance D from the train-observer. agreed

4. If the time coordinate of the strike at the front is t1 in the train-observer's frame, then assuming the light moves at c in this frame and the distance from the front to the train-observer is D, the light must reach the train-observer at time coordinate t = t1 + D/c in this frame. agreed

5. If the time coordinate of the strike at the back is t2 in the train-observer's frame, then assuming the light moves at c in this frame and the distance from the back to the train-observer is D, the light must reach the train-observer at time coordinate t = t2 + D/c in this frame. agreed

6. If the light from both strikes reaches the train-observer at different times, that must mean t1 + D/c is not equal to t2 + D/c. agreed

7. The only way for them not to be equal is if t1 is not equal to t2. agreed
Therefore, the strikes must have happened at different time-coordinates in the train-observer's rest frame. Disagreed, the truth is this is only apparent...[/color]

In the last comment you seem to be answering a different question than the one I asked. I didn't ask whether the events "really" happened simultaneously in any objective metaphysical sense, I only asked if you agreed the strikes happened at different time coordinates in a particular coordinate system. Since you already agreed that the time-coordinate t1 assigned to one strike is not equal to the time-coordinate t2 assigned to the other, hopefully you would not dispute that they "happened at different time-coordinates in the train-observer's rest frame"?

And if you want to discuss the notion of objective simultaneity, your claim that it's "only apparent" still doesn't make sense, since you have no basis for believing that it's the platform observer's frame, and not the train-observer's frame, whose judgments about simultaneity line up with the metaphysical truth about "real" simultaneity. Isn't it perfectly conceivable that the train-observer's frame is the objectively correct one? Consider this: one could easily design a similar thought-experiment which was almost the same as the one we've been talking about, except instead of assuming from the start that the strikes happened simultaneously according to the platform-observer's time coordinates, we instead assume they happened simultaneously according to the train-observer's time coordinates. Then we could use exactly the same reasoning to show the train-observer would receive light from both strikes simultaneously, but the platform-observer would receive light from the two strikes at different times, indicating that the strikes happened at different times in the platform-observer's coordinate system. Would you still say it is the train-observer's judgment that is only "apparent" in this case, or would you say it's the platform-observer's, or would you concede that even if one judgment is correct and one is apparent, we have no way of knowing which is which short of divine revelation?

Why restrict the conclusion to this one possibility, especially when it’s wrong? Couldn’t the passenger, being aware of SR, knowing that she is in a moving train

Again, in relativity there is no notion that some frames are "moving frames", or that there is one preferred frame that is "really" at rest. Each observer defines themselves to be at rest in their own frame. Since the laws of physics don't distinguish any frame as preferred, then even if there was an objective truth about which objects were at rest and which were in motion (the Newtonian idea of absolute space), there would be absolutely no experimental procedure that could determine which was which.

Both strikes hit at the same time. Relativity is not about time, it’s about motion and the constant speed of light.

Of course relativity is about time! The phrase relativity of simultaneity is a standard one that you will find in any relativity textbook, and Einstein used the phrase himself in chapter 9 of this book. And didn't you already agree that events which happen at the same time-coordinate in one frame happen at different time-coordinates in another? Do you disagree that the first postulate of relativity means that no frame is physically preferred over any other, so there is no basis for judging one frame to be in a state of absolute rest while others are in a state of absolute motion?

 

[stevmg]

I lied -

Post #47 - that's a hyperbola, not a parabola - still the same non-linear relationship.

C'mon - that's analytic geometry and I took that course in September 1959 - January 1960. That's before all of you, I bet, were even born.

 

[yuiop]

The train example discussing non-simultaneity that I'm sure most of you have heard of:



However, wouldn't the passenger see the strikes of lightning at the same time? As she is in an inertial reference frame and is equi-distance from the front and back?

I think the problem with the youtube video mentioned in the OP is that the video segment from 0.27 to 0.29 shows the train as stationary during the time the light spheres expand to reach the trackside observer simultaneously and as a consequence it appears as if the light also reaches the passenger simultaneously. This is unphysical and to someone that is trying to understand the example for the first time, this image is confusing at the subliminal level. I think this video concentrates more on presentation than on the accuracy of the content.

 

[Doc Al]

Yes, the video is misleading and sloppy in that regard.

 

[ValenceE]

Dear JesseM,

6. If the light from both strikes reaches the train-observer at different times, that must mean t1 + D/c is not equal to t2 + D/c. agreed

7. The only way for them not to be equal is if t1 is not equal to t2. agreed
Therefore, the strikes must have happened at different time-coordinates in the train-observer's rest frame. Disagreed, the truth is this is only apparent...
In the last comment you seem to be answering a different question than the one I asked. I didn't ask whether the events "really" happened simultaneously in any objective metaphysical sense, I only asked if you agreed the strikes happened at different time coordinates in a particular coordinate system. Since you already agreed that the time-coordinate t1 assigned to one strike is not equal to the time-coordinate t2 assigned to the other, hopefully you would not dispute that they "happened at different time-coordinates in the train-observer's rest frame"?

Sorry, I mistakenly agreed to the first part of point 7 and forgot to write that “…t1 and t2 are apparently not equal…” in the second part.

It is not T1 and T2 that are not equal, it is the front and back D parameters that are changing. The frame in which the simultaneously emitted light from both strike locations occurred is moving from front to back wrt the train, at the train’s speed, changing both D parameters. This happens while the emitted light is traveling at c, so ; frontD decreases, front flash is seen first, backD increases, back flash is seen later, making the train observer incorrectly conclude that the flashes must not have happened simultaneously.

Do you at least agree with that much, that unless relativity is correct there can be no experimental method of determining the truth about simultaneity, since different frames differ in their opinions about whether two events happened at the same time-coordinate, and the first postulate says that the laws of physics work exactly the same way in every frame so there can be no experimental basis for picking out one frame as a "preferred" one?

Did you mean “…unless relativity is incorrect …” in the above paragraph?
Regardless, I would agree with the rest of it, for now anyway, until I can come up with one...

regards,

VE

 

[TcheQ]

I think the problem with the youtube video mentioned in the OP is that the video segment from 0.27 to 0.29 shows the train as stationary buring the time the expanding light spheres reach the trackside observer simultaneously and as a consequence it appears as if the light also reaches the passenger simultaneously. This is unphysical and to someone that is trying to understand the example for the first time, this image is confusing at the subliminal level. I think this video concentrates more on presentation than on the accuracy of the content.

If the station observer is stationary(:P), and the train travels at a normal speed(100km an hour), how would they draw it?. You can't draw a train moving 10e-6m

 

[JesseM]

Sorry, I mistakenly agreed to the first part of point 7 and forgot to write that “…t1 and t2 are apparently not equal…” in the second part.

It is not T1 and T2 that are not equal, it is the front and back D parameters that are changing.

How can they be changing in the train's rest frame, where by definition both ends of the train, as well as the observer at the center, must have a position coordinate that stays constant over time? Remember, this is just a coordinate system, if you're trying to talk about how distance or time are "really" behaving then you're confusing the issue, D and t1 and t2 refer to nothing more than coordinate labels.

Are you familiar with the basic idea (seen in both Newtonian physics and relativity) that "frames" refer to coordinate systems, and that different frames have different definitions of which objects are at rest relative to that frame and which are moving?

The frame in which the simultaneously emitted light from both strike locations occurred is moving from front to back wrt the train, at the train’s speed, changing both D parameters.

I wasn't talking about the ground frame! Reread question 4 (which you agreed to), D refers specifically to the distance in the train's rest frame:

4. In the train-observer's rest frame, the train-observer is at rest at a fixed position, as are the front and back of the train which are both at an equal distance D from the train-observer.

Do you disagree that in the coordinate system constructed to be the train's rest frame (i.e. one where the coordinate position of any part of the train doesn't vary with coordinate time), the two ends of the train are at a constant and equal distance from the center of the train, and therefore if the lightning strikes locally coincide with the ends of the train, it must be true in this coordinate system (not in any absolute sense) that both strikes happened an equal distance from the center?

Did you mean “…unless relativity is incorrect …” in the above paragraph?

Yes, sorry...I also see I included two separate premise #4's on my list...my proofreading skills obviously leave something to be desired!

 

[yuiop]

If the station observer is stationary(:P), and the train travels at a normal speed(100km an hour), how would they draw it?. You can't draw a train moving 10e-6m

Relativity effects do not become obvious until the relative velocities are a significant fraction of the speed of light. A casual estimate of the speed of the train in the video is about 0.6c or about 180,000 km per hour.

 

[TcheQ]

Relativity effects do not become obvious until the relative velocities are a significant fraction of the speed of light. A casual estimate of the speed of the train in the video is about 0.6c or about 180,000 km per hour.

The observance of disparity in the time-lapse between the signal receival is not a relativity issue.

 

[yuiop]

The observance of disparity in the time-lapse between the signal receival is not a relativity issue.

Good point. Even if length contraction and time dilation did not exist and even if light propogated at (c +/- v) relative to the source and even if there was a medium for light, there would still be a disagreement about the simultaneity of spatially separated events. From this point of view it is surprising that the "relativity of simultaneity" was not predicted long before relativity itself. I guess the only thing that would not allow a disparity would be instantaneous transmission of light. I guess if Gallileo had thought about it really hard when he came up with his ship thought experiment, he could have have come up with the basics of Special Relativity or something very like it.

However, the difference between the speed of light and the speed of typical train (100 km per hour) is so great, that it would be very difficult in practice to detect a time disparity in the arrival times of the signals.

 

[stevmg]

Somebody help me...

If the train is 1/2 "c" (i.e., 150,000 km) long (I know, c is the speed of light and not a distance so that's why I put it in quotes) and the train is moving to the right by the Einsteinian example, and M' is the obsever half way down the train, do the math for me - how long does it take for M' to hit the flash from the forward lightning strike.

Clearly, it would take 0.25 seconds for the light to reach M our stationary ground obserever but do the math and tell me how long will it take the flash to reach our lady observer M' on the moving train? I cannot wrap my brain around the closure speed and the distance. Closure speed is NOT c + 0.5c - impossible, so let's take it from there.

H-E-L-P!

 

[JesseM]

Somebody help me...

If the train is 1/2 "c" (i.e., 150,000 km) long (I know, c is the speed of light and not a distance so that's why I put it in quotes)

Better to say half a light-second long (a light-second being the distance light travels in one second, analogous to a light-year), since 300,000 km is (approximately) the distance light travels in one second.

and the train is moving to the right by the Einsteinian example, and M' is the obsever half way down the train, do the math for me - how long does it take for M' to hit the flash from the forward lightning strike.

Clearly, it would take 0.25 seconds for the light to reach M our stationary ground obserever but do the math and tell me how long will it take the flash to reach our lady observer M' on the moving train? I cannot wrap my brain around the closure speed and the distance. Closure speed is NOT c + 0.5c - impossible, so let's take it from there.

Actually the closing speed here is 1.5c, relativity only says that the velocity of any object or signal taken on its own cannot exceed c in an inertial frame, it doesn't say that the distance between two objects cannot change at a speed greater than c. So since they start out a distance of 0.25 light-seconds apart, with a closing speed of 1.5c the time for them to meet should be 0.25/1.5 = 1/6 of a second in the ground frame.

You could also figure it out without referring to the concept of closing speed at all, just by figuring out the position of both M' and the light signal as a function of time. If M' is at position x=0 at t=0, then since the train is 0.5 light-seconds long and M' is at the middle, the front end of the train should be at x=0.25 light-seconds at t=0, and that's when the lightning strikes. Then the light ray from the strikes is heading in the -x direction at a speed of 1 light-second/second, so its position as a function of time must be x(t) = 0.25 light-seconds - (1 light-seconds/second)*t...without including the units, this can just be written as x(t) = 0.25 - t. You can see that this function tells you that at t=0 the light is at position x=0.25, at t=1 the light has traveled 1 light-second in the -x direction to x=-0.75, and so forth. Meanwhile M' starts at x=0 and heads in the +x direction at 0.5c, so for M' the position as a function of time would be x(t) = (0.5 light-seconds/second)*t, or x(t) = 0.5*t. Then if you want to find the time when both are at the same position, set them equal to each other:

0.25 - t = 0.5*t
Adding t to both sides gives:
0.25 = 1.5*t
So, solving for t gives:
t=0.25/1.5 = 1/6.

 

[stevmg]

JesseM -

You are a bloody genius...

The closing speed is 0.5*c + 1.0*c = 1.5*c. No obect or electromagnetic wave itself can travel greater than c but the concept of greater than c still exists.

Now, say an object is traveling to the right at "just under" c and the light from a flash is traveling to the left at c (because that's how fast light travels in a frame of reference) , then the "closure" speed is virtually 2*c (just a tad under.) Is that right?

Nothing like keeping alive old Newton Galilean coordinates, I guess.

By the way, should we continue this discussion on a new topic somewhere else? I think the train example has been beaten to death but this other aspect of SR can be explored more...

If you start a new topic somewhere else, tell me how to get there because how I got here is still a mystery to me but God is Great and Merciful and got me into this original simultaneity discussion with all of you.

I used to tell my troops that they were to think of me not as their commander but as if I were God and that everything I said came straight from the burning bush. After the usual half hour of laughing we then got down to real business.

Yea - I do believe in God AND evolution - but that's another story...

Steve