Angle of a rod (relativity)

[kron]

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)

Homework Statement

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:

$$\theta=-arctan( \frac{\gamma u v }{c^2})$$

Homework Equations

LT:

$$(1)~ x=\gamma (x' + v t')$$
$$(2)~ t = \gamma(t' + \frac{v x'}{c^2})$$

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)]

$$P_1 : x=0, t=0$$

$$P_2 : x=\gamma L, t=\gamma \frac{v L}{c^2}$$And u_y' = u transforms like

$$\frac{u}{\gamma (1 + \frac{v u_x'}{c^2})}$$

with u_x' = 0 this is

$$u_y = \frac{u}{\gamma}$$so with

$$\Delta y = u_y t = \frac{u v L}{c^2}$$

$$\tan(\theta) = \frac{\Delta y}{\Delta x} = \frac{u v}{\gamma c^2}$$

which is not the same as the solution given above.

Thanks.

 

[BigJDubs]

SolutionWe can use the Lorentz transformation to calculate the coordinates of the rod in S frame. Let's consider P1 and P2 as the beginning and end points of the rod in S' frame.P1 : x'=0, y'=0, t'=0P2 : x'=L, y'=u*t', t'=tUsing the Lorentz Transformations, we can calculate the coordinates of P1 and P2 in S frame.P1 : x=0, y=0, t=0P2 : x=\gamma L, y=\gamma u t - \frac{\gamma u v L}{c^2}, t=\gamma t + \frac{v L}{c^2}Now we can calculate the angle of the rod in S frame.\tan(\theta) = \frac{\Delta y}{\Delta x} = \frac{\gamma u t - \frac{\gamma u v L}{c^2}}{\gamma L} \tan(\theta) = \frac{u t - \frac{u v L}{c^2}}{L} \tan(\theta) = \frac{u v}{\gamma c^2} So \theta=-arctan( \frac{\gamma u v }{c^2})