# General Relativity Problem (frames of reference)

1. Sep 8, 2016

### RJLiberator

1. The problem statement, all variables and given/known data
A clock moving at v = (3/5)c reads 12:00 as it passes us in our frame of reference, how far away will it be (in light hours) when it reads 1:00.

2. Relevant equations
I denote a prime to mean the reference frame of the clock at rest. I use regular lettering to denote 'our' frame of reference.

c = speed of light

t' = t/sqrt(1-v^2/c^2)
L' = L*sqrt(1-v^2/c^2)

L = v*t
3. The attempt at a solution

Alright, I use Lorentz contraction to solve this problem.

First I find the length L using v and t = 1 hr. L = (3/5)c*1 hr = (3/5)c light hours.
Now, I use Lorentz contraction to find L'
L' = (3/5)c*sqrt(1-9/25)
L' = 12c/25

2. Sep 8, 2016

### SammyS

Staff Emeritus
First of all, 12c/25 has units of speed, not distance.

When one hour elapses in "our" frame of reference, the clock has traveled 3/5 light⋅hour (in our frame).

From "our" point of view the moving clock runs slower than normal, so when the clock reads 1:00, it will have traveled farther than 3/5light⋅hours.

3. Sep 10, 2016

### RJLiberator

Thank you for your assistance SammyS.

I see the difference now in light hours. When saying 12c/25, that is incorrect as it is a speed. I should have said 12/25 light hours. Noted.

In our FOR it is clear that the clock traveled a distance of 3/5 light hours.
So I can use time dilation here to see:

$$t' = \frac{1hr} {\sqrt{1-\frac{\frac{9c^2} {25}} {c^2}}}$$
$$t' = 1.25 hr$$

Therefore, we say the distance will be 1.25 hr * 3c/5 = 3/4 light hours.

3/4 light hours > 3/5 light hours

Is that now a correct line of reasoning?

4. Sep 10, 2016

### SammyS

Staff Emeritus
Yes. That's what I get.