= Solution
Choose the photon direction in the $x'$–$y'$ plane. Its <four-momentum> in the stellar rest frame is
$$
p'^\mu=\frac{h\nu'}c(1,\cos\theta',\sin\theta',0).
$$
Let $\beta=v/c$, $\gamma=(1-\beta^2)^{-1/2}$. The <Lorentz transformation> to $S$ gives
$$
p^0=\gamma(p'^0+\beta p'^x),\quad
p^x=\gamma(p'^x+\beta p'^0),\quad p^y=p'^y,\quad p^z=p'^z.
$$
Using $p^0=h\nu/c$ and $p^x=(h\nu/c)\cos\theta$ yields the <relativistic Doppler effect> and <relativistic aberration>:
$$
\boxed{\nu=\gamma\nu'(1+\beta\cos\theta'),\qquad
\cos\theta=\frac{\cos\theta'+\beta}{1+\beta\cos\theta'}.}
$$
For the oppositely directed transverse photons, their rest-frame spatial momenta cancel and their total energy is $E=2h\nu'$. Boosting this combined four-vector gives laboratory energy $\gamma E$ and momentum
$$
\boxed{\mathbf p=\frac{\gamma Ev}{c^2}\,\widehat{\mathbf x}
=\frac{Ev}{c^2\sqrt{1-v^2/c^2}}\,\widehat{\mathbf x}.}
$$
The star loses rest mass $E/c^2$. At unchanged speed its momentum therefore decreases by $\gamma(E/c^2)v$, exactly the photon momentum. Its laboratory energy loss is likewise $\gamma E$. There is no rest-frame recoil from the symmetric emission, and constant speed is consistent with <four-momentum conservation> because the stellar mass decreases.
Back to article page
