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The rod A^'B^' moves with a constant velocity v relative to the rod AB (figure). Both rods have the same proper length l_0 and at the ends of each of them clocks are mounted, which are synchronized pariwise: A with B and A^' with B^'. Suppose the moment when the clock B^' gets opposite the clock A is taken for the beginning of the time count in the reference frames fixed to each of the rods. Determine: (a) the readings of the clocks B and B^' at the moment when they are opposite each other, (b) the same for the clocks A and A^'. |
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Answer» Solution :At the instant the picture is taken the coordinates of `A,B,A^',B^'` in the rest frame of AB are `A:(0,0,0,0)` `B:(0,l_0,0,0)` `B^':(0,0,0,0)` `A^':(0_1-l_0sqrt(1-v^2//c^2),0,0)` In this frame the coordinates of `B^'` at other TIMES are `B^': (t, vt, 0, 0)`. So `B^'` is opposite to B at time `t(B)=l_0/v`. In the frame in which `B^'`, `A^'` is at rest the time corresponding this is by Lorentz transformation. `t^0(B^')=(1)/(SQRT(1-v^2/c^2))(l_0/v-(vl_0)/(c^2))=l_0/vsqrt(1-v^2//c^2)` Similarly in the rest frame of A, B, te coordinates of A at other times are `A^':(t, -l_0sqrt(1-v^2/c^2)+vt, 0,0)` `A^'` is opposite to A at time `t(A)=l_0/vsqrt(1-v^2/c^2)` The corresponding time in the frame in which `A^'`, `B^'` are at rest is `t(A^')=gammat(A)=l_0/v` 
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