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A rod AB oriented along the x axis of the reference frame K moves in the positive direction of the x axis with a constant velocity v. The point A is the forward end of the rod, and the point B its near end. Find: (a) the proper length of the rod, if at the moment t_A the coorniate of the point A is equal to x_A, and at the moment t_B the coordinate of the point B is equal to x_B, (b) what time interval should separate the markings of coordinates of the rod's ends in the frame K for the difference of coordinates to become equal to the proper length of the rod. |
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Answer» SOLUTION :(a) In the REFERENCE FRAME K with respect to which the rod is moving with VELOCITY v, the COORDINATES of A and B are `A: t, x_A+v(t-t_A),0,0` `B: t, x_B+v(t-t_B), 0,0` Thus `l=x_A-x_B-v(t_A-t_B)=l_0sqrt(1-beta^2)` So `l_0=(x_A-x_B-v(t_A-t_B))/(sqrt(1-v^2//c^2))` (b) `+-l_0-v(t_A-t_B)=l=l_0sqrt(1-v^2//c^2)` (since `x_A-x_B` can be either `+l_0` or `-l_0`.) Thus `v(t_A-t_B)=(+-1-sqrt(1-v^2//c_2))l_o` i.e. `t_A-t_B=l_0/v(1-sqrt(1-v^2/c^2))` or `t_B-t_A=l_0/v(1+sqrt(1-v^2//c^2))` 
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