Calculating Time Difference and Length Contraction in Train Overtaking Scenario

[Amrator]

Homework Statement

Train A has proper length $L$. Train B moves past A (on a parallel track, facing the same direction) with relative speed $\frac{4c}{5}$ (as measured by either train; so each one sees the other move at $\frac{4c}{5}$). The length of B is such that A says that the fronts of the trains coincide with at exactly the same time as backs coincide. What is the time difference between the fronts coinciding and the backs coinciding, as measured by B? Solve this in two ways: (a) by using length contraction, and (b) by using the rear-clock-ahead effect (among other things).

Relevant Equations

$$t_{observed} = \gamma t_{proper}$$
$$L_{observed} = \frac{L_{proper}} {\gamma}$$
$$t_R - t_L = \frac{Lv}{c^2}$$
$$V = \frac{u+v}{1+\frac{uv}{c^2}}$$

a) So I think I solved a. "The length of B is such that A says that the fronts of the trains coincide with at exactly the same time as backs coincide." This tells us that in A's frame, B has the same length as A. So $L_{observed}$ in A's frame is simply $L$. To obtain the proper length of B, you would then simply multiply $L$ by $\gamma$. In B's frame, the length of A is contracted by $\frac{3}{5}$. Now, to obtain the time difference, we subtract the length contraction of A from the proper length of B, and we divide that result by the speed of A. Doing so gives us $\frac{4L}{3c}$.

b) We have to use the rear-clock-ahead effect which is $t_R - t_L = \frac{Lv}{c^2}$. The rear-clock-ahead effect basically states that if a train with length $L$ moves with speed$v$ relative to you, then you observe the rear clock reading $\frac{Lv}{c^2}$ more than the front clock, at any given instant. So we have to add it to some other time to obtain the time difference. The issue I don't know how to get that time. At the initial time, the front clock will read $-\frac{Lv}{c^2}$.

 

[PeroK]

Homework Statement:: Train A has proper length $L$. Train B moves past A (on a parallel track, facing the same direction) with relative speed $\frac{4c}{5}$ (as measured by either train; so each one sees the other move at $\frac{4c}{5}$). The length of B is such that A says that the fronts of the trains coincide with at exactly the same time as backs coincide. What is the time difference between the fronts coinciding and the backs coinciding, as measured by B? Solve this in two ways: (a) by using length contraction, and (b) by using the rear-clock-ahead effect (among other things).
Relevant Equations:: $$t_{observed} = \gamma t_{proper}$$
$$L_{observed} = \frac{L_{proper}} {\gamma}$$
$$t_R - t_L = \frac{Lv}{c^2}$$
$$V = \frac{u+v}{1+\frac{uv}{c^2}}$$

a) So I think I solved a. "The length of B is such that A says that the fronts of the trains coincide with at exactly the same time as backs coincide." This tells us that in A's frame, B has the same length as A. So $L_{observed}$ in A's frame is simply $L$. To obtain the proper length of B, you would then simply multiply $L$ by $\gamma$. In B's frame, the length of A is contracted by $\frac{3}{5}$. Now, to obtain the time difference, we subtract the length contraction of A from the proper length of B, and we divide that result by the speed of A. Doing so gives us $\frac{4L}{3c}$.

`

I think you have calculated the wrong thing here. I think the question asks how long it takes the trains to completely pass each other: from when their fronts coincide until their rears coincide.

To do this using method b) you need to use coordinates and generally be more formulaic in your calculations.

 

[Amrator]

`

I think you have calculated the wrong thing here. I think the question asks how long it takes the trains to completely pass each other: from when their fronts coincide until their rears coincide.

To do this using method b) you need to use coordinates and generally be more formulaic in your calculations.

But they're facing the same direction.

 

[PeroK]

But they're facing the same direction.

Ah, yes. Sorry.

 

[PeroK]

b) We have to use the rear-clock-ahead effect which is $t_R - t_L = \frac{Lv}{c^2}$. The rear-clock-ahead effect basically states that if a train with length $L$ moves with speed$v$ relative to you, then you observe the rear clock reading $\frac{Lv}{c^2}$ more than the front clock, at any given instant. So we have to add it to some other time to obtain the time difference. The issue I don't know how to get that time. At the initial time, the front clock will read $-\frac{Lv}{c^2}$.

The idea is this. The ends being adjacent are simulatenous events in frame A. If there are two synchronised (in frame B) clocks at either end of train B, then these are not synchronised in frame $A$. But, they do show the time in frame B of the two events (fronts adjacent and rears adjacent).

The difference in frame B is, therefore, equal to the lack of synchronisation as measured in frame A.

You need to check the particular formula for the rear-clock-ahead. And be careful about what length to use.

It's also worth using the Lorentz Transformation to solve this problem.

 

[Amrator]

`

I think you have calculated the wrong thing here. I think the question asks how long it takes the trains to completely pass each other: from when their fronts coincide until their rears coincide.

To do this using method b) you need to use coordinates and generally be more formulaic in your calculations.

I apologize @PeroK, but could you elaborate for part b?

 

[PeroK]

I apologize @PeroK, but could you elaborate for part b?

I just beat you to it.

 

[Amrator]

The idea is this. The ends being adjacent are simulatenous events in frame A. If there are two synchronised (in frame B) clocks at either end of train B, then these are not synchronised in frame $A$. But, they do show the time in frame B of the two events (fronts adjacent and rears adjacent).

The difference in frame B is, therefore, equal to the lack of synchronisation as measured in frame A.

You need to check the particular formula for the rear-clock-ahead. And be careful about what length to use.

It's also worth using the Lorentz Transformation to solve this problem.

Why is that?

 

[PeroK]

Why is that?

That's the relativity of simultaneity. If I were teaching SR I would teach that first so the students would never forget it!

 

[Amrator]

The idea is this. The ends being adjacent are simulatenous events in frame A. If there are two synchronised (in frame B) clocks at either end of train B, then these are not synchronised in frame $A$. But, they do show the time in frame B of the two events (fronts adjacent and rears adjacent).

The difference in frame B is, therefore, equal to the lack of synchronisation as measured in frame A.

You need to check the particular formula for the rear-clock-ahead. And be careful about what length to use.

It's also worth using the Lorentz Transformation to solve this problem.

I don't know. I'm lost. I tried doing $\frac{Lv}{c^2} + \frac{L}{\gamma^{2}v}$ But that didn't give me the same answer as a...

 

[PeroK]

I don't know. I'm lost. I tried doing $\frac{Lv}{c^2} + \frac{L}{\gamma^{2}v}$ But that didn't give me the same answer as a...

Try this:

http://www.people.fas.harvard.edu/~djmorin/Relativity Chap 1.pdf

Section 1.3. The first thing to learn about SR!

 

[TSny]

Your answer to (a) looks good.

b) We have to use the rear-clock-ahead effect which is $t_R - t_L = \frac{Lv}{c^2}$. The rear-clock-ahead effect basically states that if a train with length $L$ moves with speed$v$ relative to you, then you observe the rear clock reading $\frac{Lv}{c^2}$ more than the front clock, at any given instant.

Yes

So we have to add it to some other time to obtain the time difference.

Why do you say this? Doesn't $\frac{Lv}{c^2}$ (with the appropriate L) already give you the answer?