Why is radiation under Poynting-Robertson drag anisotropic? According to the Wikipedia article on Poynting-Robertson drag, the reason solar radiation slows an orbiting object is because the re-radiation of photons by the object is anisotropic in the frame of the star, as given in this picture:

Why would that radiation be anisotropic, though? Wouldn't the electrons of the object radiate light more or less randomly, especially if the object was not very reflective or smooth?
 A: They do radiate light randomly in all directions -- in the object's own reference frame. But not from the Sun's reference frame. The effect in general is called "relativistic beaming."
Here's the clearest derivation that I know. Take the Pauli matrices $\sigma_i$ and adjoin the identity matrix to them as $\sigma_0 = I$. Now take a four-vector $v^\mu$, and form the 2x2 Hermitian matrix $V = \sigma_\mu v^\mu.$ You can prove that in fact $\det V = v_\mu v^\mu$ in the $(+\,-\,-\;-)$ metric, with $\operatorname{Tr}V = 2 v^0.$ Lorentz transforms therefore preserve determinants; we can classify them as the "special" Lorentz transforms $V \mapsto L V L^\dagger$ with $\det L = \pm 1,$ where the $-$ doesn't matter and can be safely ignored, and some other operations which can't be done this way, like the 4-reversal $V\mapsto -V$ or the parity transform $V \mapsto (\det V) ~ V^{-1},$ which reverses the sign of the spatial indices but not the time index.
That won't be completely clear after just reading one paragraph, but actually start to write out the expression for $V$, $$V = \begin{bmatrix}v^0 + v^3 & v^1 -i v^2\\v^1 + i v^2 & v^0 - v^3\end{bmatrix}$$ and derive for example that the Lorentz transform of a boost with rapidity $w = \tanh^{-1}\beta$ in the $+z$-direction is $$L = \begin{bmatrix}e^{w/2}&0\\0&e^{-w/2}
\end{bmatrix},$$
while the unitary matrices provide rotations etc., and you will get it soon enough.
Suppose we look at null vectors: the condition $\det V = 0$ forces $V$ to be a projection, in other words, $$V = \begin{bmatrix}\alpha \\ \beta \end{bmatrix}\begin{bmatrix}\alpha^* & \beta^* \end{bmatrix} = \begin{bmatrix}\alpha^* ~\alpha & \alpha ~\beta^* \\ \alpha^*~ \beta & \beta ~\beta^*\end{bmatrix}$$In particular, for null future-pointing vectors $r^\mu = (r, \vec r)$ we can interpret $(v^1 + i v^2) / (v^0 - v^3)= \alpha^* \beta / (\beta^* \beta) = (\alpha/\beta)^*$ as the stereographic projection of $\vec r$ onto $\mathbb C.$ So we're not off in la-la land; this is a real way to talk about where the light is going. In particular, the complex point $\alpha/\beta = 0$ is the projection of the point $(x, y, z) = (0, 0, -r)$ while the projection of the point $(0, 0, +r)$ is complex-$\infty$.
Now we just wrote our $\det V = 0$ matrix $V$ in the form $v ~ v^\dagger$ where $v = [\alpha;\;\beta];$ this object is a so-called spinor which represents the null vector. The action of the boost $V \mapsto L V L^\dagger$ is the same as the effect $v \mapsto L v$. In turn this quantity $\gamma = \alpha / \beta$ that tells us where the light ray is directed (in stereographic coordinates) gets mapped by the Lorentz matrix $\begin{bmatrix} a&b\\c&d\end{bmatrix}$ to the new value $\gamma' = (a~\alpha + b~\beta)/(c~\alpha + d~\beta).$
For the above Lorentz boost matrix the result you get is $\gamma' = e^w ~ \gamma.$ 
So, imagine a uniform spherical density of outgoing null vectors from a stationary sphere. Imagine these as dots on the complex plane. We boost into a frame where that sphere is going towards $+z$ with speed $c \tanh w$. The dots all move from some $\gamma$ to some new $e^w~\gamma$, being "scaled" outwards in magnitude. This amounts to them moving homogeneously away from $z = -r$ and towards $z = +r$, which is in the direction of the motion.
That is relativistic beaming. Light which appears locally isotropic is "beamed" into the same direction as the particle is moving via the Lorentz boost.
