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# This basically follows from problem 3 above note

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Unformatted text preview: e the speed of light is 1 in all reference frames. We now solve for ∆t and conclude that we have ∆t = √ 2d 1 − v2 X2 = 2d √ 1 , 0, 0, −v 1 − v2 (22) It is now clear that we have ∆t2 = ∆t 2 − ∆z 2 for E 1 and E 2, which is what we set out to prove. (Note that ∆x, y, z, x , y = 0 for E 1 and E 2 anyway.) Ok, now we want to argue that this statement actually holds for all timelike separated events. This basically follows from problem 3 above; note however that this feels vaguely like cheating, as if we were so conﬁdent of the form of the Lorentz transformation this entire problem is trivial anyway and we don’t need to muck about with anachronisms like light clocks. So to make life fun let us instead imagine we do not know the Lorentz boosts and are trying to prove this using only thought experiments. This is still possible; consider a new frame F with a diﬀerent boost velocity u. In this frame the event E 2 has coordinates 1 X2 = 2d √ , 0, 0, −u (23) 1 − u2 Now let’s ask; by choice of d and u can we put this event anywhere in the (∆t ,...
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