**Appendix B. Relativistic angular aberration**

Consider a probe beam emitted from a source at an angle *θ* and speed *u* with respect to the wave vector of an observer moving with speed *v* in the frame of reference of the source. Considering a 2-d plane containing the wave vectors of the probe beam and the observer, we can define the components as follows:

$$
\mu\_{\text{x}} = \mathfrak{u} \cos \theta \tag{47}
$$

$$u\_{\mathcal{Y}} = u \sin \theta \tag{48}$$

In the observer frame of reference (indicated with a prime), using the relativistic velocity transformations, yields:

$$u'\_{\chi} = \frac{u\cos\theta - v}{1 - \frac{v}{c^2}u\cos\theta} \tag{49}$$

$$u'\_{\circ} = \frac{\sqrt{1 - \frac{v^2}{c^2}} u \sin \theta}{1 - \frac{v}{c^2} u \cos \theta} \tag{50}$$

Hence,

$$\tan \theta' = \frac{u\_{\text{x}}'}{u\_{\text{y}}'} = \frac{\sqrt{1 - \frac{v^2}{c^2}} u \sin \theta}{u \cos \theta - v}. \tag{51}$$

The probe beam speed in the observer frame of reference is

$$u' = \sqrt{u'^2\_{\,\,x} + u'^2\_{\,\,y}} = \frac{\left(u^2 - \frac{u^2 v^2 \sin\theta^2}{c^2} - 2u\cos\theta v + v^2\right)^{1/2}}{1 - \frac{v}{c^2}u\cos\theta}.\tag{52}$$

So,

$$\cos \theta' = \frac{u'\_{\chi}}{u'} = \frac{u \cos \theta - v}{\left(u^2 - \frac{u^2 v^2 \sin \theta^2}{c^2} - 2u \cos \theta v + v^2\right)^{1/2}} \tag{53}$$

If the probe beam consists of photons, then *u* ¼ *c* and this simplifies further to

$$\cos \theta' = \frac{\cos \theta - \frac{\nu}{c}}{1 - \frac{\nu}{c} \cos \theta}. \tag{54}$$
