# Basic Special Relativity

1. May 16, 2017

### ChrisJ

Its been a few years since I have done any special relativity so I am a bit rusty, need help with either my working/understanding or if correct, making sense of my answer. This is not CW, just a question from a past exam paper that I am using as preparation.

1. The problem statement, all variables and given/known data

Street light A and B are situation 4km apart, and according to clocks on the ground were switched on at exactly the same time. According to an observer on a "train" moving from A to B at 0.6c, which light switched on first? How much later, in seconds, did the second light go on?

2. Relevant equations
$\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$
$x' = \gamma \left( x-vt \right)$
$t' = \gamma \left( t - \frac{vx}{c^2} \right)$

3. The attempt at a solution

In this course we have used the matrix form of the Lorentz Transform as standard but its easier/quicker to code the TeX without it as I am not used to matrices in TeX.

But that method made me associated the coordinates [ in (t,x) ] for A as (0,0) and for B as (4km,0) . Using these two coordinates I then transformed them into the prime (observers) frame.

But first to make things quicker I just found $\gamma = \frac{1}{\sqrt{1-v^2/c^2}} = \frac{1}{ \sqrt{1-0.6^2} } = 1.25$

For A:
$x' = \gamma \left( x-vt \right) = \gamma \left(0-0.6c(0) \right) = 0$
$t' = \gamma \left( t - \frac{vx}{c^2} \right) = \gamma \left( 0 - \frac{(0.6c)(0)}{c^2} \right) = 0$

For B:
$x' = \gamma \left( x-vt \right) = 1.25 \left(4000-0.6c(0) \right) = 5000$m
$t' = \gamma \left( t - \frac{vx}{c^2} \right) = 1.25 \left( 0 - \frac{(0.6)(4000)}{c} \right) = -1 \times 10^{-5} \textrm{s} = -10 \mu s$

So is it that for the observer light B turned on first and turned on $10 \mu s$ before A did?

Any help appreciated :)

2. May 16, 2017

### TSny

Looks good to me.