Special Relativity - Angle Transformations

[Rubber Ducky]

Homework Statement

A rod of length $L_0$ moves with a speed $v$ along the horizontal direction. The rod makes an angle of $θ_0$ with respect to the x'-axis.

(a) Show that the length of the rod as measured by a stationary observer is given by

$$L=L_0\sqrt{1-\frac{v^2}{c^2}cos^2θ_0}$$


(b) Show that the angle that the rod makes with the x-axis is given by the expression
$$tanθ=γtanθ_0$$

(Take the lower end of the rod to be at the origin of the primed coordinate system.)

Homework Equations

$$γ=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

$$L=\frac{L_0}{γ}$$

${L_0}^2=(x')^2+(y')^2$ and $L^2=x^2+y^2$

The Attempt at a Solution

Let x and y be the rod's length and height (picture the rod forming the hypotenuse of a right triangle):

$x'=L_0cosθ_0$

There is no movement in the y (or y') direction, so $y'=y=L_0sinθ_0$

Meanwhile, the x component will contract in the non-prime reference frame, so $x=\frac{x'}{γ}=\frac{L_0cosθ_0}{γ}$

Thus $L^2=x^2+y^2=\frac{L_0^2cos^2θ_0}{γ^2}+L_0^2sin^2θ_0$

The algebra gets messy at this point, and I'm not sure what methods I should be using to yield the required form. I looked at my trig identities but none really seemed to fit the situation. And hopefully I haven't made a silly error in the physics side of things!

 

[TSny]

Thus $L^2=x^2+y^2=\frac{L_0^2cos^2θ_0}{γ^2}+L_0^2sin^2θ_0$

The algebra gets messy at this point, and I'm not sure what methods I should be using to yield the required form.

Write out the γ² factor in terms of v/c and simplify. It's not too bad.