# How angles transform under Lorentz transforms - a thought experiment

The speed of light, not the velocity of light, is Lorentz-invariant.

Wikipedia discusses the relativistic aberration of light.

You're right that the (spatial) angle of the light beam in $$S$$ is not $$θ'$$. But it isn't $$θ$$ either.

If you look at constant-coordinate-time slices of $$S$$, the pipe is moving to the right (while pointing diagonally), while the light is moving diagonally. If the light moved diagonally at the same angle as the pipe is pointing, the pipe would "leave it behind" and it would hit the wall. The angle of the light's motion has to be closer to the horizontal for it to avoid hitting the pipe. If you work out the necessary angle (keeping in mind that the speed must be $$c$$), you'll get the relativistic aberration formula.

When dealing with the direction light travels, it's best to use the wave vector $${\bf k}$$, and it's 4-vector form as the 4-gradient of the Lorentz scalar phase:

$$k^{\mu} \equiv \partial^{\mu}\phi = (\frac 1 c\frac{\partial\phi}{\partial t}, {\bf \nabla}\phi)$$

For a plane wave:

$$A(x,t) = Ae^{i\phi(x, t)}=Ae^{ik^{\mu}x_{\mu}}=Ae^{i({\bf k \cdot x} -\omega t)}$$

is:

$$k^{\mu} =(\omega/c, {\bf k}) = (\omega/c, k\sin{\theta},0,k\cos{\theta}) = k(1, \sin{\theta}, 0, \cos{\theta})$$

Here $$\theta$$ is the angle in unprimed frame.

A boost along $$-z$$ at $$v$$ is a straightforward Lorentz transformation:

$$\omega' = \gamma(\omega+\frac v c k\cos{\theta})$$ $$k_x' = k_x = k\sin{\theta}$$ $$k_y'=0$$ $$k_z'= k_z + \frac v {c^2}\omega = k\gamma(\cos{\theta}+\beta)$$

The primed angle is then:

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

which is the standard result for stellar aberrations. (Note: you also get the relativistic Doppler shift from $$\omega'/\omega$$).