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The rod AB oriented parallel to the x^' axis of the reference frame K^' moves in this frame with a velocity v^' along its y^' axis. In its turn, the frame K^' moves with a velocity V relative to the frame K as shown in figure. Find the angle theta between the rod and the x axis in the frame K. |
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Answer» Solution :In `K^'` the coordinates of A and B are `A:(t^', 0, -v^'t^', 0), B: (t^', L, -v^'t^', 0)` After performing Lorentz transformation to the frame K we get `A: tgammat^' B:t =gamma(t^'+(Vl)/(c^2))` `x=gammaVt^'` `x=gamma(l+Vt^')` `y=v^'t^'` `y=-v^'t^'` `z=0` `z=0` By TRANSLATING `t^'rarrt^'-(Vl)/(c^2)`, we can WRITE the coordinates of B as `B: t=gammat^'` `x=gammal(1-V^2/c^2)+Vt^'gamma=lsqrt(1-v^2/c^2)+Vt^'gamma` `y=-v^'(t^'-(Vl)/(c^2))`, `z=0` Thus `Deltax=lsqrt(1-(V^2/c^2))`, `Deltay=(v^'Vl)/(c^2)` HENCE `TANTHETA^'-(v^'V)/(c^2sqrt(1-(v^'V)/(c^2))` 
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