Angle of a rod (relativity)

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I don't know how to edit the title but the problem is solved.



One needs two simultaneous events in frame S not in S'.
(Just in case someone tried to figure it out)


1. The problem statement, all variables and given/known data

Hi,

this is not a homework, but an exercise of the book Introduction to Special Relativity by Wolfgang Rindler
(Oxford Science Publications):

In S' a straight rod parallel to the x'-axis moves in the y'-direction with constant velocity u.
Show that in S the rod is inclined to the x-axis at an angle

correction:

[tex]\theta=-arctan( \frac{\gamma u v }{c^2})[/tex]



2. Relevant equations

LT:

[tex] (1)~ x=\gamma (x' + v t')[/tex]
[tex] (2)~ t = \gamma(t' + \frac{v x'}{c^2})[/tex]


3. The attempt at a solution

The x and x' axis of S and S' are parallel. The beginning point of the rod may be placed
in x'=y'=0 at t'=0.
For t'=0 I get two events in S for the beginning and end point of the rod, which are [with (1) and (2)]

[tex]P_1 : x=0, t=0[/tex]

[tex]P_2 : x=\gamma L, t=\gamma \frac{v L}{c^2}[/tex]


And u_y' = u transforms like

[tex]\frac{u}{\gamma (1 + \frac{v u_x'}{c^2})}[/tex]

with u_x' = 0 this is

[tex]u_y = \frac{u}{\gamma} [/tex]


so with

[tex]\Delta y = u_y t = \frac{u v L}{c^2}[/tex]

[tex]\tan(\theta) = \frac{\Delta y}{\Delta x} = \frac{u v}{\gamma c^2} [/tex]

which is not the same as the solution given above.

Thanks.
 
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