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Special Relativity: Bullets on a train
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[QUOTE="six7th, post: 4522803, member: 351050"] Ok I see that this is true if the two events are one shot being fired and another being fired. So I now have: Δt = γΔt' = γ[itex]\frac{l'}{u'}[/itex] The distance traveled by a bullet in the ground frame is therefore: uΔt = uγ[itex]\frac{l'}{u'}[/itex] Using the velocity addition formula this gives uΔt = [itex]\frac{v+u'}{1+\frac{uv}{c^2}}\frac{\gamma l'}{u'}[/itex] Now in the ground frame the train also travels vΔt, so subtracting this from uΔt will give us the actual distance of the bullet traveled in the ground frame. This gives: [itex]\gamma^2 l\frac{v+u'}{u'+\frac{vu'^2}{c^2}} - \gamma \frac{vl'}{u'}[/itex] [itex]\gamma^2 l(\frac{v+u'}{u'+\frac{vu'^2}{c^2}} - \frac{v}{u'})[/itex] Which is much larger than l when v tends to c, which cannot be correct so I must be missing some other step. Edit: Just realized I actually needed to simplify some more. And the [itex]\gamma^2[/itex] cancels out to give: [itex] distance = \frac{u'^2}{u'^2+\frac{vu'^3}{c^2}}l[/itex] Which is always less than l, for all c and u. [/QUOTE]
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Special Relativity: Bullets on a train
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