# Modern Relativity question

use Special Relativity Just help me confirm some of these answers. I have doubts on numers 1,

Train goes (3/5)c relative to tracks. When the back end is opposite point A on the tracks, and the front end if opposite B then lightning strikes the two ends simulatneously according to a passenger on the train. According to passengers the train is 60m long

1) Which flash occured first from the viewpoint of an obserer standing by the tracks?

WEll this is the opposite of simultaneity. In order for someone to see both flashes on the train simulateneously, then the flash from point B is the one which occured first from the stationary observers point of view.

2) How long is the movinbg train from the obserers point of view? Since 'moving objects appear smaller', then observed length L = 60 (root (1-9/25)) which is 48.

3) How far apart are points A and B on the tracks form the viewpoint of an obserer on the track?
err since the two flashes did occur simultaneously as per the length of the train which he observed then the distance would haev been 48m

4) How far apart are A and B from the viewpoint of observer on train?
Since it did occur corresponding to the train's length it would be 60m?

5) According to the observers on the train how much time will tick off one of the tack clocks during the time 20 secs elapse on the train clocks?
Should be longer than 20 secs. So it would be 20 times gamma

6) A window on the side of the train has an area of 10 sq metres according to a seated passenger on the train. What is the area of the moving window from the observer on the track's perspective?
This is using 2 dimension analysis. Wouldi assume that the window is square or rectangular? But only the lenght would contract and not te height. If L = 2 and H = 5 then L would reduce to 1.6m and thus areai s 8 sq m and if it was vice versa then it would be the same, so it doesnt make a diff, does it?

7) (LAST ONE!) Suppose the train came to rest, then waht would the observer on the tracks say the lenght of the train was?
waht it really is, 60!

your input into some of the wrong ones such as 1 (i think), 3 and 4 would be greatly appreciated!

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nrqed
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stunner5000pt said:
use Special Relativity Just help me confirm some of these answers. I have doubts on numers 1,

Train goes (3/5)c relative to tracks. When the back end is opposite point A on the tracks, and the front end if opposite B then lightning strikes the two ends simulatneously according to a passenger on the train. According to passengers the train is 60m long

1) Which flash occured first from the viewpoint of an obserer standing by the tracks?

WEll this is the opposite of simultaneity. In order for someone to see both flashes on the train simulateneously, then the flash from point B is the one which occured first from the stationary observers point of view.
It's the lightning at the *back end* which occured first (so at A). One way to see this is to set up the equations (define your two events, etc etc). That's the *safe* way to do it. But if you are used to visualize SR problems, it can be seen directly. If the lightning occured at the same time in the frame of the train, the light emitted from the two lightning strikes will reach the middle of the train at the same time. Now, this particular event (two light beams reaching the middle of the train at the same time) is something that everybody (in all frames) will agree on since it occurs at a single spacetime point.

Now, consider this from the point of view of the tracks. A lignthning strikes the front end and a light beam travels toward the center of the train. Same thing for the lightning strike at the back end of the train. Which one occured first? Well the light emitted from the strike at the back end has to do some catching up with the train since this one is moving away from the event. In order for this beam to reach the center of the train at the same instant as the light coming from the front strike, it must have been emitted *earlier* than the light coming from the lightning strike at the front.

2) How long is the movinbg train from the obserers point of view? Since 'moving objects appear smaller', then observed length L = 60 (root (1-9/25)) which is 48.
Right.

3) How far apart are points A and B on the tracks form the viewpoint of an obserer on the track?
err since the two flashes did occur simultaneously as per the length of the train which he observed then the distance would haev been 48m
No. The two flashes did not occur simultaneously from the point of view of someone on the tracks! That's what question 1 was all about.
To answer this, you must setup the events and use the Lorentz transfos. It's straightforward. Let me know if you have difficulties with this.

4) How far apart are A and B from the viewpoint of observer on train?
Since it did occur corresponding to the train's length it would be 60m?
yes
5) According to the observers on the train how much time will tick off one of the tack clocks during the time 20 secs elapse on the train clocks?
Should be longer than 20 secs. So it would be 20 times gamma
Right.
6) A window on the side of the train has an area of 10 sq metres according to a seated passenger on the train. What is the area of the moving window from the observer on the track's perspective?
This is using 2 dimension analysis. Wouldi assume that the window is square or rectangular? But only the lenght would contract and not te height. If L = 2 and H = 5 then L would reduce to 1.6m and thus areai s 8 sq m and if it was vice versa then it would be the same, so it doesnt make a diff, does it?
Correct answer. One size is reduced by a factor of 0.8 so the area is reduced by the same factor.
7) (LAST ONE!) Suppose the train came to rest, then waht would the observer on the tracks say the lenght of the train was?
waht it really is, 60!
yes, but you should not say "what it REALLY is". The length of the train *REALLY* is 48 meters as measured from the tracks when it is moving at 0.6c!

your input into some of the wrong ones such as 1 (i think), 3 and 4 would be greatly appreciated!

Hope this helps. Let me know if you need help with using teh Lorentz transfos.

Pat

How far apart are points A and B on the tracks form the viewpoint of an observer on the track

I would take the Lorentz TE' into consideration, but i'm a little confused on where to start with them. Ok let the moving tain be the S frame and the stationary observer be the S' frame.

Then from the S' t' = 1.25 (t - 3x/5c)
x' = 1.25 (x - 3t/5c)

but i'm confused as to how to interpret my data...

Doc Al
Mentor
Allow me to rewrite the Lorentz transformation that you need like so:
$$\Delta x' = \gamma (\Delta x - v \Delta t)$$
or, the inverse transform,
$$\Delta x = \gamma (\Delta x' + v \Delta t')$$
where I use primes to represent the train frame.

To use it, think this way. We are given measurements made in the train frame: $\Delta x' = L_0$ while $\Delta t' = 0$. This means that the train frame measured the distance between points A and B to be equal to the length of the train (which I call $L_0$). Of course the position measurements were made at the same time ($\Delta t' = 0$) according to the train.

So: $$\Delta x = \gamma L_0$$

(This answer should be "obvious" since the length of A-B must be just right so that its contracted length---according to the train frame---equals $L_0$.)

Doc Al said:
Allow me to rewrite the Lorentz transformation that you need like so:
$$\Delta x' = \gamma (\Delta x - v \Delta t)$$
or, the inverse transform,
$$\Delta x = \gamma (\Delta x' + v \Delta t')$$
where I use primes to represent the train frame.

To use it, think this way. We are given measurements made in the train frame: $\Delta x' = L_0$ while $\Delta t' = 0$. This means that the train frame measured the distance between points A and B to be equal to the length of the train (which I call $L_0$). Of course the position measurements were made at the same time ($\Delta t' = 0$) according to the train.

So: $$\Delta x = \gamma L_0$$

(This answer should be "obvious" since the length of A-B must be just right so that its contracted length---according to the train frame---equals $L_0$.)
so what you're saying is that despite the observer seeing the train move past with a length of 48 m apparently, the distance between the ppoints A and B would not appear to be any different from the real length of the train. So that means the points A and B are 60 m apart??

Doc Al
Mentor
stunner5000pt said:
so what you're saying is that despite the observer seeing the train move past with a length of 48 m apparently, the distance between the ppoints A and B would not appear to be any different from the real length of the train. So that means the points A and B are 60 m apart??
No.

The observer on the ground will measure the length of the train to be 48m. This measurement is not the length of the track between A and B, since the ends of the train do not match up with A and B at the same time according to the ground observers.

Observers on the train will measure the length of the track between A and B to be equal to 60m, the length of the train, since according to them the ends of the train do match up with A and B at the same time. But this is the measurement of A-B according to the train. The distance between A and B according to the ground observers is $\Delta x = \gamma L_0$ = 1.25*60 = 75m.

Doc Al said:
No.

The observer on the ground will measure the length of the train to be 48m. This measurement is not the length of the track between A and B, since the ends of the train do not match up with A and B at the same time according to the ground observers.

Observers on the train will measure the length of the track between A and B to be equal to 60m, the length of the train, since according to them the ends of the train do match up with A and B at the same time. But this is the measurement of A-B according to the train. The distance between A and B according to the ground observers is $\Delta x = \gamma L_0$ = 1.25*60 = 75m.
That first statement of yours clinched the idea for me. Thank you very very much!
true the groun observer will see the points more apart than the real lengh of the train.

Thank you so much !