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.(adsbygoogle = window.adsbygoogle || []).push({});

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 velocityv. 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*t_{1}+ 1.0c*t_{1}= 1 lt-sec. Thus 1.5*t = 1.0 lt sec.

t_{1}= 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:

t_{2}*c = 1 lt-sec + 0.5c*t_{2}. Thus 0.5*t_{2}= 1 lt-sec

t_{2}= 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- whichsimultaneous 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]were

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 thetrainas aninertialframe 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"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.still

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') isnot moving in his frameand 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!

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# Einstein's Train

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