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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
X2 = 2d √
, 0, 0, −u
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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