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.

## Answer This Question

## Related Questions

- physics - Two events in S are separated by a distance D=x2 - x1 and a time T = ...
- College Physics - Use the relativistic coordinate transformation (x, y, z, t) &#...
- PHY 2054 - Use the relativistic coordinate transformation (x, y, z, t) − (...
- physics - Two ﬁrecrackers explode at the same place in a rest frame with ...
- Physics - Two events that are simultaneous in one frame of reference will be: A...
- Physics - - How do you calculate proper time (Tau) between two events with ...
- physics - Can someone please explain to me why or how this statement is true? I ...
- Physics - How would you change the following Galilean transformation equations ...
- Alg 2 - a rectangular picture frame measures 8 inches by 4 inches. You want to ...
- physics - as Sue runs, her knee joint passes through the following vertical ...

More Related Questions