A particle moving in one direction enduring radiation perpendicular to its direction of motion will feel a force in the opposite direction to its motion. This force is the Poynting Robertson effect, and it is a relativistic effect. It can be easily understood if one thinks of the problem in the particle's reference frame. In that frame the photons travel at a different angle. Unfortunately, the interpretation is not that simple in the lab frame. There is a common misconception that the braking is due to an asymmetric emission by the particle. However, as we show below, the phenomenon can be explained without emission at al.

Rest Frame Edit

Suppose a particle is moving in the $ \hat{x} $ direction with a velocity $ v \ll c $. Suppose further that it is in a radiation field with photons moving in the $ \hat{y} $ direction. In the particle's rest frame the 4 - momentum (actually 3 momentum because we only consider 2 spatial dimensions)

$ \left(\begin{matrix} \epsilon' \\ {p'}_x \\ {p'}_y \end{matrix} \right ) = \left( \begin{matrix} \gamma & \gamma \beta & 0 \\ \gamma \beta & \gamma & 0 \\ 0 & 0 & 1 \end{matrix} \right ) \left(\begin{matrix} \epsilon \\ 0 \\ \epsilon \end{matrix} \right ) = \epsilon \left(\begin{matrix} \gamma \\ \gamma \beta \\ 1 \end{matrix} \right ) $

The number of incident photons within a time interval is $ n c \Delta t R^2 $, where $ n $ is their density, $ c $ is the speed of light, $ \Delta t $ is the time interval and $ R $ is the radius of the particle. All in all, the change in the x component of the momentum is

$ \Delta {p'_x} = - \frac{\epsilon}{c} \beta n c \Delta R^2 = - \frac{v}{c^2} f = -\left( \frac{R}{r} \right)^2 \frac{L}{c^2} \sqrt{\frac{G M}{r}} \Delta t $

where $ f $ is the energy flux of the photons. The last transition assumes the particle circles a star at a distance $ R $. The mass of the star is $ M $ and its luminosity is $ L $. The force is therefore given by

$ F_{PR} = \frac{\Delta {p'}_x}{\Delta t} \approx -\left( \frac{R}{r} \right)^2 \frac{L}{c^2} \sqrt{\frac{G M}{r}} $

Lab Frame Edit

Again, we have a particle moving in the x direction, and radiation moving in the y direction. When the particle absorbs a photon its mass increases from $ m $ to $ m + \epsilon/c^2 $. Since the momentum in the x direction does not change, the velocity decreases to $ \frac{m v}{m+\epsilon/c^2} \approx v \left( 1 - \frac{\epsilon}{m c^2}\right) $. Hence the absorption of every photon decreases the velocity of the particle. Multiplying by the photon flux and the cross section of the particle leads to the same expression as in the previous section. This derivation assumes that the emission from the particle is symmetric, so it does not accelerate the particle, contrary to a common misconception.