# Angle transformation under Lorentz transformation

The angle transformation under Lorentz transformation:

$$\cos\theta = \frac{\cos\theta' + \beta}{1 + \beta\cos\theta},\quad \sin\theta = \frac{\sin\theta'}{\gamma^2(1 + \beta\cos\theta')}$$

is normally derived from the velocity transformation:

$$u_\parallel = \frac{u_\parallel' + v}{1+(v/c^2)u_\parallel'},\quad u_\perp = \frac{u_\perp/\gamma}{1+(v/c^2)u_\parallel'}$$

and then

$$\tan\theta = \frac{u_\perp}{u_\parallel} = \frac{u_\perp'/\gamma}{u_\parallel' + v} = \frac{u'\sin\theta'}{\gamma(u'\cos\theta + v)}$$

letting $$u' = c$$ and get:

$$\tan\theta = \frac{\sin\theta'}{\gamma(\cos\theta'+\beta)}$$

$$x = \gamma(x'+\beta ct'),\quad y = y'$$

and:

$$\tan\theta = \frac{y}{x} = \frac{y'}{\gamma(x'+\beta ct')}$$

which leads to an annoying $$ct'$$ here

Can someone tell me why we must use velocity transformation and not length transformation to get the angle transformation?

• Why not start with "time transformation"? Velocity is not any kind of "associated quantity" with length. Oct 29, 2023 at 11:05
• time transformation is associated with length Oct 30, 2023 at 5:47

I think, since you are only considering 2D (as oposed to 3D), i.e. a component of the velocity that is perpendicular to the Lorentz transformation and another that is parallel with it, you are allowed to say that $$x'=r'\cos\theta'\\y'=r'\sin\theta'$$ for some distance $$r'$$. The latter distance, then, is going to be the distance that satisfies $$ct'=r'$$, and therefore you will have the Eq. in your original post $$\tan\theta=\frac{\sin\theta'}{\gamma(\cos\theta'+\beta)}$$