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Then U x¢ = 0, U y¢ = c, U z¢ = 0 = Now Ux = Uy = Uz = \ U= U x¢ + v 1+ v U x¢ c2 U y¢ G çæ1 + v U x¢ ÷ö è ø c2 U z¢ G çæ1 + v U x¢ ÷ö è ø c2 U x2 + U y2 é = êv2 + c2 ë Q U x¢ = 0 c G Q U y¢ = c, U x¢ = 0 =0 Q U z¢ = 0 = v = + U z2 = U x2 + U y2 æ c2 ö = ç v2 + ÷ è G2 ø 1/2 æ v2 ö ù çè1 - 2 ÷ø ú c û = [v2 + c2 – v2]1/2 = c 1/2 24 A Primer of Special Relativity (C) ? plus ? = ?! If we substitute U x¢ = c and v = c in Eqn. (1a), we get Ux = U x¢ + v 1+ v U x¢ c2 = c+c 1+ c c c2 = c. The student should convince himself that relative velocity of two objects or two frames or an object and a frame cannot exceed c.

1 S¢ moves relative to S with speed v along the positive X direction. We define an event as an occurrence which takes place at some instant t at a point (x, y, z), for example arrival of a particle at time t at a point (x, y, z) or an electric bulb at (x, y, z) flashing at time t etc. Suppose a clock at rest at (x¢, o, o) in frame S¢ sends one light flash at time t1¢ and the next one at a later time t2¢ . In S¢, the time between the two events (consecutive flashes) is T0 = t2¢ - t1¢. T. è c2 ø Hence the time interval between the two events (two consecutive flashes) as measured in S is T = t2 – t1 = G t2¢ - t1¢ = GT0 That is T= T0 æ v2 ö çè1 - 2 ø÷ c ...

Show that the baby was born in the hospital. 29. A stationery missile explodes and breaks into equal pieces. The two pieces move with velocities c/2, one to the right and the other to the left. The piece moving to the right again explodes into two pieces such that with respect to its own rest frame, the resulting two pieces take off with velocities c/2 one to the right and the other to the left. Find the velocities of these last two pieces with respect to earth. (Ans. 4/5 c and zero) 30. If S¢ has velocity c relative to S, show that all bodies moving relative to S with speeds less than c have a speed c as observed from frame S¢.

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Black holes, theory and observation by Hehl


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