Basic Special Relativity

  • Thread starter ChrisJ
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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. Homework Statement

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?

Homework 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) ##

The Attempt at a Solution


[/B]
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 :)
 

Answers and Replies

  • #2
TSny
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Looks good to me.
 

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