Posted by **kim** on Monday, December 7, 2009 at 11:10pm.

Two events in S are separated by a distance D=x2 - x1 and a time T = t2 - t1 . a) Use the Lorenrz transformation to show that in frame S' , which is moving with speed v relative to S, the time separation is t2 - t1 = ã(T - vD/c2 ). b) Show that the events can be simultaneous in frame S' only if D is greater than cT. c) If one of the events is the cause of the other, the separation D must be less than cT since D/c is the smallest time that a signal can take to travel from x1 to x2 in frame S. Show tht if D is less than cT, t2' is greater than t1' in all refence frames. d) Suppose that a signal could be sent with speed c'>c so that in frame S the cause precedes the effect by the time T=D/c'. Show that there is then a reference frame moving with speed v less than c in which the effect precedes the cause.

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