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Introductory Physics Homework Help
Which light turned on first and how much later did the second one turn on?
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[QUOTE="ChrisJ, post: 5763740, member: 600992"] 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. [B] 1. Homework Statement [/B] 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? [h2]Homework Equations[/h2] ##\gamma = \frac{1}{\sqrt{1-v^2/c^2}}## ##x' = \gamma \left( x-vt \right) ## ## t' = \gamma \left( t - \frac{vx}{c^2} \right) ## [h2]The Attempt at a Solution[/h2] [/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[B]. [/B] 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 :) [/QUOTE]
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Which light turned on first and how much later did the second one turn on?
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