# Einstein's Train

stevmg
This subject has had other threads but I think specific answers to the questions that I will pose will clear up many points of confusion.

To begin with, Einstein's train in section IX. of "Relativity" goes like this:

Given a train of length AB proceeding from left to right at velocity v. There is an observer in the train at midpoint M'. On the ground (Einstein calls this an "embankment") is an observer at M. Just as M' passes M, lightning hits the front and back of the train simultaneously in the (inertial) frame of reference of the ground. He then states that the observer at M will see the two flashes, from B and A respectively at the same time because it takes light, traveling at c the same time time to traverse the equal distances from B to M and A to M. He also states that the observer on the train at M' will see the B-flash before the A-flash because the train is moving towards to the light source at B and thus meets up with the light earlier than the light from source A catches up with him.

train A___________________M'___________________B ---> v

M (assume M is right opposite M' at the time of the lightning strikes

......................A___________________M'_____________________B

.........................................M (notice how M' has moved to the right after the lightning stirkes)
I had to put in the "......" to move the line to the right and ignore the blank __" "__ as it should be filled in

For the sake of illustration, let us say the train is 2 lt-sec long, thus AM' and M'B are both 1 lt-sec. Assume the velocity of the train is 0.5c to the right, the flashes from B and A would both meet M (the non-moving observer) at 1.0 lt-sec/(1.0*c) = 1 second after the flashes actually occurred)

M' (the observer on the train moving to the right at 0.5c) would meet the flash from B by this calculation:

0.5c*t1 + 1.0c*t1 = 1 lt-sec. Thus 1.5*t = 1.0 lt sec.
t1 = 1lt-sec/1.5c = 0.6666... sec (2/3 of a second after the flash from A)

The flash from A, or the back of the train, would catch up with M' by this calculation:

t2*c = 1 lt-sec + 0.5c*t2. Thus 0.5*t2 = 1 lt-sec
t2= 2 sec (after the flash from B)

So, instead of observer M' "seeing" the flashes simultaneously, there would be a 1 and 1/3 or 1.3333... second time difference between "seeing" the flash from B followed by the flash from A. [We got that by subtracting 1.6666.... from 2.0. How did we get that? From the flashes- which were simultaneous in the ground frame of reference - it took 2 seconds for the second event (seeing flash A) to happen and 1.6666 sec for the first event (seeing flash B) which leaves a time difference of 1.3333... seconds]

We call the frame of reference (FOR) of the ground or "embankment" as an "inertial frame" because there is no acceleration/deceleration or curvilinear motion in this frame.

Now, switching to the train as an inertial frame of reference (again, constant velocity of the frame with reference to the ground FOR - 0.5c, no acceleration/deceleration, no curvilinear motion) we know that the observer M', although not moving in the train frame will still "see" the flash from B ahead of the flash from A, by the law of lack of simultaneity under relativity for different frames looking at the same local event.

However, what would be the time differences in the train FOR?

A) In the train FOR what would be the time of origin of the flash from B prior to M' "seeing it?" (negative numbers ok) Would it be -1 sec because he (M') is not moving in his frame and it would take 1 second for the light to reach M' (1 lt-sec/c)?

B) Likewise, what would be the time of origin (negative numbers ok) of the flash from A prior to M' "seeing" it. Again, I would assume it would be 1 second prior.

C) What would be the time interval as far as M' is concerned between "seeing" the flashes of B then A.

Would it be the time dilation formula:
1.3333..../SQRT(1 - (0.5)2) = 1.3333..../SQRT(0.75) = 1.1547... seconds?

Now, I know this has been discussed before in a lot of threads and a lot of references to a lot of posts and a lot of "you do it" admonitions have been stated to the inquirers, but, if someone, (preferably a science advisor) could go through these calculation enumerated above as is and either confirm them or correct them, this would immeasurably contribute to the understanding of Lorentz's contribution to relativity and how to actually use it!

H-E-L-P!

Last edited:

This subject has had other threads but I think specific answers to the questions that I will pose will clear up many points of confusion.

To begin with, Einstein's train in section IX. of "Relativity" goes like this:

Given a train of length AB proceeding from left to right at velocity v. There is an observer in the train at midpoint M'. On the ground (Einstein calls this an "embankment") is an observer at M. Just as M' passes M, lightning hits the front and back of the train simultaneously in the (inertial) frame of reference of the ground. He then states that the observer at M will see the two flashes, from B and A respectively at the same time because it takes light, traveling at c the same time time to traverse the equal distances from B to M and A to M. He also states that the observer on the train at M' will see the B-flash before the A-flash because the train is moving towards to the light source at B and thus meets up with the light earlier than the light from source A catches up with him.

train A___________________M'___________________B ---> v

M (assume M is right opposite M' at the time of the lightning strikes

......................A___________________M'_____________________B

.........................................M (notice how M' has moved to the right after the lightning stirkes)
I had to put in the "......" to move the line to the right and ignore the blank __" "__ as it should be filled in

For the sake of illustration, let us say the train is 2 lt-sec long, thus AM' and M'B are both 1 lt-sec. Assume the velocity of the train is 0.5c to the right, the flashes from B and A would both meet M (the non-moving observer) at 1.0 lt-sec/(1.0*c) = 1 second after the flashes actually occurred)
You're forgetting about length contraction here. If the train is 2 lt-seconds long in its own rest frame, then since it's moving at 0.5c in the rest frame of M, in this frame it is shrunk by a factor of $$\sqrt{1 - 0.5^2}$$ = 0.86603. So, in the rest frame of M, M will receive the light from the flashes only 0.86603 seconds after the flashes actually occurred.

Alternatively, you might want to say the train is 2 lt-seconds long in the frame of M, in which case the train will be longer in its own frame (in the rest frame of M'), 2/0.86603 = 2.3094 light-seconds long, or 1/0.86603 = 1.1547 light-seconds from M' at the middle to either end of the train.
stevmg said:
M' (the observer on the train moving to the right at 0.5c) would meet the flash from B by this calculation:

0.5c*t1 + 1.0c*t1 = 1 lt-sec. Thus 1.5*t = 1.0 lt sec.
t1 = 1lt-sec/1.5c = 0.6666... sec (2/3 of a second after the flash from A)
OK, let's assume the train is 2 light-seconds long in the frame of the ground observer M. In this case the position of M' as a function of time is x(t) = 0.5c*t, and the position of the light as a function of time is x(t) = 1 light-second - 1c*t, so we set them equal and get the equation above...looks right to me.
stevmg said:
The flash from A, or the back of the train, would catch up with M' by this calculation:

t2*c = 1 lt-sec + 0.5c*t2. Thus 0.5*t2 = 1 lt-sec
t2= 2 sec (after the flash from B)
For M' we again have x(t) = 0.5c*t, for the other light beam we have x(t) = -1 light-second + 1c*t, so yeah, setting them equal will give the above.
stevmg said:
So, instead of observer M' "seeing" the flashes simultaneously, there would be a 1 and 1/3 or 1.3333... second time difference between "seeing" the flash from B followed by the flash from A. [We got that by subtracting 1.6666.... from 2.0. How did we get that? From the flashes- which were simultaneous in the ground frame of reference - it took 2 seconds for the second event (seeing flash A) to happen and 1.6666 sec for the first event (seeing flash B) which leaves a time difference of 1.3333... seconds]

We call the frame of reference (FOR) of the ground or "embankment" as an "inertial frame" because there is no acceleration/deceleration or curvilinear motion in this frame.

Now, switching to the train as an inertial frame of reference (again, constant velocity of the frame with reference to the ground FOR - 0.5c, no acceleration/deceleration, no curvilinear motion) we know that the observer M', although not moving in the train frame will still "see" the flash from B ahead of the flash from A, by the law of lack of simultaneity under relativity for different frames looking at the same local event.

However, what would be the time differences in the train FOR?

A) In the train FOR what would be the time of origin of the flash from B prior to M' "seeing it?" (negative numbers ok) Would it be -1 sec because he (M') is not moving in his frame and it would take 1 second for the light to reach M' (1 lt-sec/c)?
Like I said above, if the distance from M' to the end of the train is 1 light-second in the rest frame of the embankment observer M, then the distance from M' to the end of the train will actually be 1.1547 light-seconds in the rest frame of M', since the length of the train is shrunk in the embankment frame due to length contraction. But given this correction, it is true that in the rest frame of M', the time of the flash itself must have happened 1.1547 second before the time that M' received the light from the flash, since M' was not moving in this frame. However, that doesn't mean the time-coordinate of the flash was t'=-1.1547 seconds, because the time that M' received the light from the flash was not t'=0 seconds (at least not if you use the usual convention that M and M' both had their clocks set to 0 at the moment they were next to one another). Since M' received light from the flash at B at a time of 0.666... seconds in the frame of M, and since the clock of M' was slowed down by a factor of 0.86603 in the frame of M, that means the clock of M' must have only read 0.666... * 0.86603 = 0.57735 when the light from flash B reached him. Thus in the rest frame of M', the time of the flash itself must have been:
t' = 0.57735 - 1.1547 = -0.57735.

You can also get this answer using the Lorentz transformation. If you know that in the frame of M, the flash happened at x=1 light-second, t=0 seconds, then we can transform into the frame of M':

x' = gamma*(x - vt) = (1 - 0.5*0)/0.86603 = 1.1547
t' = gamma*(t - vx/c^2) = (0 - 0.5*1)/0.86603 = -0.57735
stevmg said:
B) Likewise, what would be the time of origin (negative numbers ok) of the flash from A prior to M' "seeing" it. Again, I would assume it would be 1 second prior.
It would 1.1547 seconds prior to the time that M' saw it. In this case, M' saw it at t' = 2 * 0.866025 = 1.73205 seconds, so it happened at t' = 1.73205 - 1.1547 = 0.57735 seconds in the M' frame. Again you could get the same answer by plugging x=-1, t=0 into the Lorentz transformation.
stevmg said:
C) What would be the time interval as far as M' is concerned between "seeing" the flashes of B then A.
As noted above, M' receives the light from flash B at 0.57735 seconds and the light from flash A at 1.73205 seconds, so the interval is 1.73205 - 0.57735 = 1.1547 seconds.
stevmg said:
Would it be the time dilation formula:
1.3333..../SQRT(1 - (0.5)2) = 1.3333..../SQRT(0.75) = 1.1547... seconds?
Yes, that's another valid way of calculating it.

stevmg
There!

I had to see the actual full-fledged calculation to understand it all. I will now go over it all slowly to completely grasp the concept in a "working" manner. That is why I asked for a full confirmation (hence, your corrections) and calculations to really understand what was what.

Both DaleSpam and you have been most helpful in having me understand all this - bit by bit. Not that I am ever going to need it in the "real" world, but this is a curiousity I have had since it was touched upon in college.

I also feel that the discipline of learning something new does help one in the "real" world even if the specific subject matter addressed is never actually encountered.

Again, thank you so much.

stevmg
P.S. -

Caught another error in what I said:

"So, instead of observer M' "seeing" the flashes simultaneously, there would be a 1 and 1/3 or 1.3333... second time difference between "seeing" the flash from B followed by the flash from A. [We got that by subtracting 1.6666.... from 2.0. How did we get that? From the flashes- which were simultaneous in the ground frame of reference - it took 2 seconds for the second event (seeing flash A) to happen and 1.6666 sec for the first event (seeing flash B) which leaves a time difference of 1.3333... seconds]"

Ditz-brain here (me) can't subtract 1.6666... from 2.0... I stated that would equal 1.333... when that subtraction would equal 0.333... Sorry for that stupid error but, of course, it did not change that my application of the time motion and relativity equations had to be corrected. Again, I will review your corrections slowly and on paper where I can think better.

Steve G

stevmg
P.P.S. -

I really need help. Again, below is the quote that I used which was incorrect.

"So, instead of observer M' "seeing" the flashes simultaneously, there would be a 1 and 1/3 or 1.3333... second time difference between "seeing" the flash from B followed by the flash from A. [We got that by subtracting 1.6666.... from 2.0. How did we get that? From the flashes- which were simultaneous in the ground frame of reference - it took 2 seconds for the second event (seeing flash A) to happen and 1.6666 sec for the first event (seeing flash B) which leaves a time difference of 1.3333... seconds]"

I stated we got that by "subtracting 1.666... from 2.0" when I should have stated 0.666... from 2.0 = 1.333.... The 1.333... was correct after all, hence it works out with the time dilation formula as you pointed out - assuming that the original length of the train of 2.3094 lt-sec in the M' FOR or 2.0 lt-sec in the M FOR

If you ever see me in the street I give you full permisssion to just shoot me.

Last edited:
P.P.S. -

I really need help. Again, below is the quote that I used which was incorrect.

"So, instead of observer M' "seeing" the flashes simultaneously, there would be a 1 and 1/3 or 1.3333... second time difference between "seeing" the flash from B followed by the flash from A. [We got that by subtracting 1.6666.... from 2.0. How did we get that? From the flashes- which were simultaneous in the ground frame of reference - it took 2 seconds for the second event (seeing flash A) to happen and 1.6666 sec for the first event (seeing flash B) which leaves a time difference of 1.3333... seconds]"

I stated we got that by "subtracting 1.666... from 2.0" when I should have stated 0.666... from 2.0 = 1.333.... The 1.333... was correct after all, hence it works out with the time dilation formula as you pointed out - assuming that the original length of the train of 2.3094 lt-sec in the M' FOR or 2.0 lt-sec in the M FOR.
What do you need help with here, just confirmation? It looks like you've got it...the time between the light rays reaching M' as viewed in the M frame is indeed 1.333... seconds under the assumption the train is 2 light-seconds long in the M frame, so to find the time between the rays reaching M' as measured by his own clock, just multiply 1.333... by sqrt(1 - 0.5^2) (not divide it as you wrote in the original post, I didn't catch that before) giving 1.1547 seconds.

stevmg
JesseM -

Thanks for the re-confirmation of my calculations (including the adjusted length of the train in the M' FOR)

The help I was referring to in the last P.P.S. was help in getting me a new brain as the dumb mistypings and subtraction errors were idiotic.

I DID give you permission for the future to just shoot me.

stevmg
Summarized:

Train (AB) 2.3094 lt-sec long traveling to the right at 0.5 c. Observer M' is at the midpoint of the train 1.1547 lt-sec from the front and back of the train. M is a ground observer next to the tracks. The clocks on the ground and train start ticking just as M' goes by M. Also at that moment lightning strikes the front (B) and back (A) simultaneously in the M frame of reference FOR. (Look at the two down "arrows $$\downarrow$$")

train $$\downarrow$$A___________________M'___________________B$$\downarrow$$ ---> 0.5c

M (assume M is right opposite M' at the time of the lightning strikes) and the train moves onto the right while light emanates from where A and B were at the beginning and travels towards the center where M is.

......................A___________________M'______ _______________B

.........................................M
I had to put in the "......" to move the line to the right and ignore the blank __" "__ as it should be filled in

Set t0 = 0.0 sec (M or ground FOR) at the instant of M' crossing M and the lightning strikes
Set t'0 = 0.0 sec (M' or train FOR) at the instant of M' crossing M and the lightning strikes

t1 = time on ground clock that the lightning from B passes M' as both approach each other.

By the length contraction principle, L [distance in ground FOR] = (1/$$\gamma$$)*L' [distance in train FOR]. Thus:

L = SQRT(1 - (0.5c)^2[/]/c^2) * 2.3094 = SQRT(0.75) * 2.3094 = 2.0 lt-sec

Thus M' is 2.3094/2 = 1.1547 lt-sec from A' and B' in the train FOR
M is 2.0/2 = 1.0 lt-sec from A and B in the ground FOR

t1 = time on ground FOR when lightning flash from B travels backwards towards A and hits the observer M' in the middle of the moving train.
t1*c + t1*0.5c = (2 lt-sec - 1 lt-sec) [front of train where B was originally at time of lightning strike - midddle of train where M' was originally)

t1*(1.0 + 0.5)c = 1 lt-sec
t1 = 2/3 sec = 0.666... sec

and...

t2 = time on ground FOR when lightning flash from A catches up with and hits the observer M' in the middle of the moving train.

1lt-sec + t2*0.5c = t2*c
1lt-sec = 0.5c*t2
t2 = 2.0 sec

This is all in the ground FOR

The $$\Delta$$t = 2.0 sec - 0.333... sec = 1.666... sec which is the difference in time that the observer M' sees the lightning strikes between the "front" (B) and the "back" (A) IN THE GROUND FOR.

To convert these to the train FOR we use the time dilation formula
$$\Delta$$t' = (1/$$\gamma$$)*$$\Delta$$t:

t1' = SQRT(0.75)*0.666... = 0.866...*0.666 = 0.57735
t2' = SQRT(0.75)*2.0... = 0.866...*2.0 = 1.7321

The $$\Delta$$t = 1.7321 - 0,5678735 = 1.1547... sec which is the difference in time that the observer M' sees the lightning strikes between the "front" (B) and the "back" (A) IN THE TRAIN FOR.

This is all in the train FOR

Remember that the observer M on the ground sees the lightning flashes simultaneously as the light from B and A has to travel 1.0 lt-sec for both and each flash would take exactly 1.0 sec to reach M from B and A respectively - i,e, - there is no time difference between the B flash and A flash meeting M.

*********************************************************************

I think this is an excellent teaching example of simple relativity covering time dilation, length contraction and how to work with these concepts to make them come out. It also specifically covers the demonstration of Einstein's Section IX in "Relativity" where he discusses the lack of maintenance of simultaneity in different frames of reference. Of course, in "Relativity," the Lorentz equations which were used in this example were discussed after the section on simultaneity but it does demonstrate the situation very well when looking back.

Again, thank you JesseM