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

[JesseM]

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?

Yup, exactly!

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 want to start a new topic about subjects other than this train thought-experiment, there's a "new topic" button at the upper left of the list of thread topics here, so just click that and then you'll see a box where you can write the first post, just like with writing a new post on an existing thread (and there'll also be a line at the top to put in the title of the new thread, that's the only difference).

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.

 

[ValenceE]

Hello all, hello again JesseM,

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:

I know, but I am, as it crucially relates to my point.

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.

Yes, I agree entirely that this is valid and both D’s are equal and remain equal throughout. However, the distance from both front and back flash wave fronts wrt the passenger will vary while both D’s remain equal because;

- Their speed is c and is frame independent.

- Both expand spherically, at c, from the original strike locations.

- Both spherical expansions are part of Earth’s frame, and are moving along from front to back at the train’s speed, when observed from the train-passenger rest frame.

Still calmly sitting at the train’s exact midpoint, the passenger has no alternative other than to see the front flash before the back flash although they did happen simultaneously.

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?

On the contrary, agreed.

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

No problem, you’re not the only one and my understanding/explanation skills definitely can be improved…


regards,

VE

 

[ValenceE]

Hello TcheQ and kev,


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

Reply from kev;

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.

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Interesting read... I'll meditate and ponder a while before replying.


regards,

VE

 

[DrGreg]

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.

Sorry, that doesn't make sense.

Sure, everyone agrees the two signals aren't received simultaneously by the mid-train observer, both under relativistic assumptions and under Galilean assumptions. But the interpretation is very different. Under pre-Einstein Galilean physics, the explanation is that the two flashes occurred simultaneously but the signals traveled at different speeds relative to the observer. Under Einstein's postulates, both signals traveled at the same speed and therefore could not have been emitted simultaneously relative to the observer.

 

[JesseM]

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:

I know, but I am, as it crucially relates to my point.

If you want to talk about some other notion of distance, could you use a different term than "D"? I specifically defined the symbol D to mean the distance between M' and either end of the train in the train frame, so it's confusing if you use it to mean something different (feel free to use some other symbol like d to refer to whatever notion of distance you want to talk about)

Yes, I agree entirely that this is valid and both D’s are equal and remain equal throughout. However, the distance from both front and back flash wave fronts wrt the passenger will vary while both D’s remain equal because;

- Their speed is c and is frame independent.

- Both expand spherically, at c, from the original strike locations.

In each frame, light expands spherically from the position coordinate where a flash happened in that frame. For example, if in the train frame the central observer is at position x=0 on the x-axis of this frame, and the front of the train is at x=D while the back is at x=-D, then the light spheres will remain centered on x=D and x=-D in this frame, i.e. they will remain centered on the front and back of the train. In contrast, in the ground frame the light spheres remain centered on the fixed position on the ground that the strikes happened at, and since the train is moving relative to the ground, the front and back of the train do not remain at the center of the light spheres in this frame.

Do you understand that the two frames disagree about whether the front and back of the train remain at the center of the light spheres caused by the strikes?

Both spherical expansions are part of Earth’s frame, and are moving along from front to back at the train’s speed, when observed from the train-passenger rest frame.

Spherical expansions aren't "part of" one frame or another, different frames view the expanding light sphere differently. If in the train frame the light sphere remained centered on the position on the tracks where the strike happened (i.e. a fixed position in the ground frame), then since that position on the tracks is moving at speed v in the train frame, this would mean that one side of the light sphere was moving at speed c+v and the other was moving at speed c-v, contradicting the second postulate of relativity. The second postulate demands that both sides of the light sphere move at c as measured in that frame, which means the center of the sphere must have a fixed position in that frame.

 

[TcheQ]

Sorry, that doesn't make sense.

Sure, everyone agrees the two signals aren't received simultaneously by the mid-train observer, both under relativistic assumptions and under Galilean assumptions. But the interpretation is very different. Under pre-Einstein Galilean physics, the explanation is that the two flashes occurred simultaneously but the signals traveled at different speeds relative to the observer. Under Einstein's postulates, both signals traveled at the same speed and therefore could not have been emitted simultaneously relative to the observer.

Pre-Lorentz/Einstein musing on simultaneity would have required two assumptions:
1. Light does not travel instantaneously
2. Light travels in a different medium to matter (aether)

The technology wasn't up to speed with the mathematics+science to confirm GR in the early 1900s, let alone more fanciful ideas earlier than this.

I do not doubt at least one scientician thought about this in the 5000 or so years of information-sharing civilization before Einstein. An interesting thought experiment would be to determine when the earliest time it could have been first mused upon, and first tried and tested.

 

[ValenceE]

Hello all, Metz114, JesseM

I have two related questions for you...


Metz114, in post 46, you wrote;

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.

Are you in a carriage that is moving at a constant speed or at rest?

I have no problem with the carriage being at rest, but if the carriage is in motion and the light reaches both ends at the same time, doesn’t that mean that the emitted light has to be co-moving with the carriage?


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JesseM, in post 65, you wrote;

In each frame, light expands spherically from the position coordinate where a flash happened in that frame. For example, if in the train frame the central observer is at position x=0 on the x-axis of this frame, and the front of the train is at x=D while the back is at x=-D, then the light spheres will remain centered on x=D and x=-D in this frame, i.e. they will remain centered on the front and back of the train. In contrast, in the ground frame the light spheres remain centered on the fixed position on the ground that the strikes happened at, and since the train is moving relative to the ground, the front and back of the train do not remain at the center of the light spheres in this frame.

Do you understand that the two frames disagree about whether the front and back of the train remain at the center of the light spheres caused by the strikes?

Yes, I understand your description of it but, in the same manner I asked Metz114;

Are you saying that, in the train’s frame, the strike locations (light sphere centers) and the expanding light spheres are co-moving with the train and train observer?


regards,

VE

 

[JesseM]

Yes, I understand your description of it but, in the same manner I asked Metz114;

Are you saying that, in the train’s frame, the strike locations (light sphere centers) and the expanding light spheres are co-moving with the train and train observer?

What do you mean by "co-moving"? Neither the train nor the train observer are moving in this frame, since this is their rest frame (remember that there is no absolute notion of 'moving' vs. 'rest' in relativity, these terms are always defined relative to one frame or another). The light spheres do remain centered at the (fixed) positions of the two ends of the train in this frame, though.

 

[Mentz114]

Hi ValencE,

I said

"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."

you asked

Are you in a carriage that is moving at a constant speed or at rest?

It makes no difference. Everything I describe takes place in the carriage.

Please note that you must specify a frame in which the carriage may or may not be moving, before you can ask if it is moving.

( Pace Jesse, I think you said the same thing ).

 

[yuiop]

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

I have made an animated gif of the train thought experiment showing both the point of view of the observer on the train and that of the observer on the track in a split screen, that I hope is more accurate than the youtube video mentioned in the OP.

This is the link to the animation: http://i910.photobucket.com/albums/ac304/kev2001_photos/Etrain2e.gif (It might take a minute to load on a slow connection.)

 

[stevmg]

Outstanding!

Now, the key element for dumb me here is this: It was never made clear until I entered this blog that an event that occurs in one time frame occurs in all time frames. JesseM pointed that out. Einstein never made that clear in his explanation in section IX of "Relativity." In the case presented, the event is the reaching of the flash to the observer on the train, not the ground. So, in the inertial time frame, the flashes meet at different times (B-flash before A-flash) so in all time frames, the same must be true. The ground observer is irrelevant in this case, just a point of reference but he has no play in the relativity of simultaneity that Einstein wants to present. Your animation is quite "spot on" I wish that this had made clearer to me initially but that's what these blogs are for.

Steve G
The light reaching

 

[Doc Al]

I have made an animated gif of the train thought experiment showing both the point of view of the observer on the train and that of the observer on the track in a split screen, that I hope is more accurate than the youtube video mentioned in the OP.

Nice job, kev!

 

[DrGreg]

Kev's great animation presents a good opportunity to learn about space-time diagrams. I've attached some to this post, depicting just what kev's animation shows and with the same colour scheme. You might like to compare the two side-by-side to see how space-time diagrams work. Time goes vertically up and distance goes horizontally.

Note my diagrams are not to scale and doesn't show exactly the same velocities as kev, so don't get out a ruler to measure distances from them!

 

[stevmg]

To JesseM or anyone who can help... Using Einstein's Train Example

If the train is 1-light-sec long, and is moving at 0.5c to the right, and the observer on the train is clearly at 0.5lightsec at the midpoint

1) What is the time difference between observation of the flash from B and A using the Earth as the reference frame (clearly B is seen before the flash from A)

2) What is the time difference between observation of the flash from B and A using the train as the reference frame.

I'd like to see how the calculations are carried out.

 

[Saw]

In case it helps, I made the attached ST diagram for a case that considers v=0.5c, as you asked, although the train is 2 l-s long (it's each half what is 1 l-s long). It depicts a thought experiment proposed by Brian Greene where a referee on the train fires light signals to duellers on the back and on the front of the train and, when warned by their respective signals, the duellers fire lasers to their opponents.

In the train frame, Back and Front receive their signals simultaneously (at t=1 s) and are fired by the other's laser also simultaneously (at t=3s). The difference is 2 s.

In the ground frame, Back receives his signal earlier (at t=0.577 s) and Front later (at t = 1.732 s). Back is fired at t=2.886 s and Front at t = 4.041 s. The difference is 2.309 s.

There are many ways to get the numbers. One is calculating first in the rest frame of the train, where obviously light takes 1 s to traverse each 1 l-s distance. Then you apply the Lorentz Transformations to get the coordinates in the ground frame.

Edit: it appears that the system does not attach the drawing because it's already in https://www.physicsforums.com/showpost.php?p=2123480&postcount=22"

 

[Doc Al]

To JesseM or anyone who can help... Using Einstein's Train Example

If the train is 1-light-sec long, and is moving at 0.5c to the right, and the observer on the train is clearly at 0.5lightsec at the midpoint

1) What is the time difference between observation of the flash from B and A using the Earth as the reference frame (clearly B is seen before the flash from A)

2) What is the time difference between observation of the flash from B and A using the train as the reference frame.

I'd like to see how the calculations are carried out.

You asked a related question https://www.physicsforums.com/showpost.php?p=2599684&postcount=33", to which I responded. But you never followed up.

 

[TcheQ]

Question 1) relies on question as 2). As other's have stated, it a simple coordinate transfer once 2) is derived. Also you didn't state how long the train is when at rest, but I'm going to assume it's 1ls long in it's own frame (it really doesn't matter since I'm going to use train length =L, however it does matter for question 1), which i am not going to answer)

If the train is 1-light-sec long, and is moving at 0.5c to the right, and the observer on the train is clearly at 0.5lightsec at the midpoint

2) What is the time difference between observation of the flash from B and A using the train as the reference frame.

I'd like to see how the calculations are carried out.

train length L
train velocity v
unknown time t1, t2
I am going to place the observer at 0,
the back of the train at -L/2
the front of the train L/2
observer O travels at v
light travels at c to the right, -c to the left
light1 from the back of the train reaching the observer -L/2 +ct1=vt1
light2 from front of the train reaching the observer L/2-ct2=vt2
solve for t1, t2

-L/2=vt1-ct1, L/2-ct2=vt2
-L/2=t1(v-c), L/2=t2(v+c)
t1=L/2*1/(c-v), t2=L/2*1/(c+v)
subsitute your values
L=1ls, v=.5c, c=c
t1=ls/c, t2=ls/3c
t2-t1=2/3*ls/c=.667s