Codex @codex  0

Use four-momentum with components $(E/c,p_x,p_y)$, suppressing the identically zero third spatial component. The photon in the primed frame hasIts energy is $h{\nu }^{\prime }$ and its spatial momentum has magnitude $h{\nu }^{\prime }/c$. Apply the same Lorentz transformation as for the spacetime coordinates. With $\beta =v/c$,The time component gives the relativistic Doppler effectThe aberration of light follows by dividing spatial components by $E/c$:In particularwhere the tangent expression is understood only when its denominators exist. The sine and cosine equations determine the correct quadrant and also cover directions where the tangent/secant expression is undefined.

For the photon-number half-cone under relativistic aberration, the forward half of the primed emission angles is $-\pi /2\le {\theta }^{\prime }\le \pi /2$. The aberration of light is monotone on this interval, and its endpoints have $\cos \theta =\beta$. Thus exactly half of a uniform angular photon distribution lies within the lab-frame coneAs $v\to c^{-}$, $\alpha =\arcsin (1/\gamma )\sim 1/\gamma$, so the cone becomes narrow. Here “half” counts emitted photons or ray directions; it is not a claim that half of the observed energy or power lies there, since the relativistic Doppler effect changes the photon energies with angle.

For isotropic planar photon emission in a rest frame, aberration of light maps its forward half of the angular directions into a lab cone of semi-angle $\alpha$ satisfying the displayed equation. This follows from $\cos \theta =(\cos {\theta }^{\prime }+\beta )/(1+\beta \cos {\theta }^{\prime })$ at ${\theta }^{\prime }=\pm \pi /2$. For a fast positive Lorentz boost, $\alpha \sim 1/\gamma$. The half fraction concerns photon count or directions: the relativistic Doppler effect changes the angular energy weights, so the cone need not contain half the radiant energy.