Apparently the CMB is slowing down all objects moving relatively to it. Is the same true for thermal radiation?

1/ Let's have two object with a non-zero temperature that first move towards each other, pass by and then continue moving apart.

2/ Both objects emit thermal radiation. When objects move towards each other they receive the thermal radiation from the other object as blue shifted. When they move apart they receive the thermal radiation from the other object as red shifted.

3/ The thermal radiation will exert "pressure" on the other object. The blue photons will have more momentum than the red photons, thus after the objects pass by they will end up having lower relative speed.

4/ Additionally, if our moving objects are not just particles but bit bigger - let's say marbles - then the blue photons will heat up one side of the object more than the red photons. The hotter side will then emit more photons against the direction of movement, further slowing down the marble. (Radiation pressure by emission)


5/ Let's say we have just one object randomly bouncing in a perfectly elastic box. If we keep the box at a stable temperature, will the above described effect eventually stop the object in the box?

6/ Lat's say we have lots of moving objects - even infinitely many in infinitely large space, like the universe. Will this thermal radiation pressure over time slow down the relative movement of all these objects?

  • $\begingroup$ Some of the questions do not require that radiation be thermal (it is a matter of momentum transfer), while others suggest that the objects are not really in thermal equilibrium (one side is hotter), that is radiation is "thermal" but not really black body radiation (see here). Finally, the frequency shift is not really important at non-relativistic speeds. $\endgroup$
    – Roger V.
    Commented Feb 9, 2022 at 11:12
  • $\begingroup$ @RogerVadim sorry, I do not really understand. $\endgroup$ Commented Mar 18, 2022 at 14:10

1 Answer 1


Let's model the system as a flat object with two surfaces. The flux incident upon a surface due to blackbody radiation, will be $\sigma T^4$, per unit area, where $T$ is the apparent temperature of the radiation.

If you move through the blackbody radiation field with a speed $\beta=v/c$ and Lorentz factor $\gamma$ then the apparent temperature in the forward direction is $T \gamma (1+\beta)$ and the temperature in the opposite direction is $\gamma (1 - \beta)$.

Thus, if you have a flat square with surface vector coincident with its velocity vector, then the net force per unit area exerted by the blackbody radiation will be $$ f = \frac{\sigma}{c}T^4 \gamma^4 \left( (1+\beta)^4 - (1 - \beta)^4 \right)\, , $$ where $T$ is the blackbody radiation temperature in its own reference frame and $v$ is the speed with respect to that reference frame. A low speed approximation to this would be $$ f \simeq \frac{8\sigma\beta}{c}T^4 $$ and would act to slow the object down and is linear in terms of velocity.

If the object is of mass $m$ and surface area $A$, then $$\frac{dv}{dt} = -\frac{8A\sigma\beta}{mc}T^4\ . $$ Thus the velocity decays exponentially with a timescale of $\tau=mc^2/8A\sigma T^4$.

If the body heats up and emits radiation itself then this could be modified. If the temperature on both sides is uniform, there is no effect. If the temperature on each side matches the apparent temperature of the blackbody radiation, then it will be like the surfaces act as a mirror and it will double the momentum change imparted by the radiation and halve the velocity decay timescale to $\tau/2$

If the object bounces around in a box though that raises the interesting possibility of two regimes. If the timescale between bounces is much longer than the timescale on which the object can change its temperature then I think the above analysis applies and the speed decay timescale is $\tau/2$. However, if the bounce timescale is shorter than the thermal timescale of the object then I think it will just end up being at the average temperature $T$ and the timescale for the speed decay is $\tau$.


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