What is the radiation field of a black body with temperature $T$ translating at speed $v$?

Question

Suppose I have a spherical black body at temperature $T$ with radius $R$. When stationary in a vacuum at zero Kelvin, it produces an incoherent radiation field according to Planck's law $$ B_\nu = \frac{2 h\nu^3}{c^2} \frac{1}{e^{-h\nu/kT}-1} $$ producing a spectral flux given by $$ F_\nu = \frac{2 h\nu^3}{c^2} \frac{2\pi}{e^{-h\nu/kT}-1} \frac{R^2}{r^2}. $$ The factor of $2\pi$ is the number of Steradians each element of the sphere's surface radiates into, and the final factor in the product is the ratio of the area of the sphere to the area of the imaginary sphere the radiation has been spread out on.

So far, so good. This is basic physics.

Now, if we change to a reference frame moving with velocity $-\vec{v}$, it sees the sphere moving at $\vec{v}$. What radiation pattern is seen in that moving frame, and why doesn't the radiation produce a net force on the sphere?

Having not bothered with the derivation, yet, I would expect the Dopper shift to definitely be relevant. Since the radiation is incoherent, though, does that make it so the headlight effect doesn't kick in?

The frame where the sphere is stationary certainly doesn't see it accelerating, so it cannot be accelerating in the moving frame, either. Both the headlight and Doppler effect would, naively, cause an imbalance of forces that would tend to decelerate the sphere. Since that is, obviously, not the case, what factor counter-balances the Doppler (and possible headlight) effect?

Answer 1

The radiation does produce a net force on the sphere. The sphere's velocity is constant, but its mass is decreasing, so its momentum is decreasing, and $F=dp/dt$.

The Doppler and headlight effects are both present and give the light emitted in the forward direction a larger (three-)momentum than the light emitted in the reverse direction. The reaction force on the sphere is in the backward direction and decreases its momentum without affecting its velocity.

Answer 2

Longitudinal radiation is Doppler shifted with

$$\nu' {\displaystyle ={\sqrt {\frac {1-\frac{v}{c} }{1+\frac{v}{c} }}}\,\nu}$$

If that relation is put into the first relation $$B_\nu = \frac{2 h\nu^3}{c^2} \frac{1}{e^{-h\nu/kT}-1}$$

it becomes $$B_\nu' = \frac{2 h \left(\frac {1-\frac{v}{c} }{1+\frac{v}{c} }\right)^\frac{3}{2} \nu^3}{c^2} \frac{1}{e^{-h \sqrt {\frac {1-\frac{v}{c} }{1+\frac{v}{c} }} \nu/kT}-1}$$

A series expansion with higher terms skipped might be more intelligible.

Integrate this function over the surrounding sphere to find the net radiation pressure. Treat the velocity as vectorial and only include the result from a radiation direction which is parallel to the velocity. The other radiation component perpendicular to the velocity should be calculated from the radiation transformed with transverse doppler shift.

Answer 3

A better way is to Doppler shift the wavelengths before deriving Planck's law.