1/ Object moving relative to the CMB frame of reference will see the CMB blue shifted where it is heading and red shifted where it came from.


2/ The blue photons ahead should have more momentum than the red photons behind, thus projecting a force opposite to the direction of the movement.


3/ Also, the blue photons ahead should have more energy than the red photons behind, thus heating the front part of the object more than the back part. Unless the heat is perfectly dissipated within the object, which should be against the laws of thermodynamics, the forward part will radiate more thermal energy, thus projecting a force opposite to the direction of the movement.


6/ How big is this effect on a cosmological scale? Let's say how much is Earth's rotation slowed down by CMB?

EDIT: I moved points 4 and 5 to a separate question.


2 Answers 2


Yes, CMB photons produce drag on anything moving relative to the CMB. However, the amount is extremely small, far below anything we can measure.

As a rough estimate, the temperature of the CMB is $T \approx 3 \ \text{K}$, so in natural units ($\hbar = c = 1$), the rate of CMB photon momentum flow of CMB photons per unit area $$\frac{dp}{dA \, dt} \sim T^4.$$ For an object moving with relative speed $v$, the fraction of this momentum captured is of order $v/c$, where $v/c \sim 10^{-4}$ for the Earth's orbit. This corresponds to a drag force on the Earth of $$F \sim \frac{v}{c} \frac{dp}{dt} \sim \frac{v}{c} \, T^4 R_E^2 \sim 10^{-3} \ \mathrm{N}$$ which is microscopic, and negligible even if you sum it over the entire history of the Earth.

In particular, the CMB drag force is completely overwhelmed by the drag from solar radiation, which is about $10^8$ times more intense than the CMB radiation. (This drag from solar radiation is important for small particles of dust, which leads to the Poynting-Robertson effect.)

  • $\begingroup$ What's $R_{E}$? $\endgroup$ Commented Feb 8, 2022 at 0:40
  • 4
    $\begingroup$ While the drag is negligible for macroscopic objects, sufficiently energetic cosmic rays are cooled by pion photoproduction when protons scatter from blueshifted CMB photons. $\endgroup$
    – rob
    Commented Feb 8, 2022 at 0:55
  • $\begingroup$ @MaximalIdeal The radius of the Earth. $\endgroup$
    – knzhou
    Commented Feb 8, 2022 at 2:36
  • $\begingroup$ Could it be more significant at the scale of galaxies? $\endgroup$
    – Barmar
    Commented Feb 8, 2022 at 16:20
  • 1
    $\begingroup$ @Barmar I don't think so... even the tenuous interstellar medium contains much much more momentum/energy than the CMB. $\endgroup$
    – knzhou
    Commented Feb 8, 2022 at 18:10

The flux incident upon a surface, due to blackbody radiation, will be $\sigma T^4$, per unit area, where $T$ is the apparent temperture of that 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 CMB will be $$ f = \frac{\sigma}{c}T^4 \gamma^4 \left( (1+\beta)^4 - (1 - \beta)^4 \right)\, , $$ where $T$ is the CMB 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 $$ f \simeq \frac{8\sigma\beta}{c}T^4 \ \ \ \ {\rm or}\ \ \ \ \frac{8 \sigma}{c}T^3\, \Delta T\ ,$$ where $\Delta T$ would be the semi-amplitude of the apparent temperature variation of the CMB in the moving frame of reference.

Earth travels at about 370 km/s with respect to the CMB (it varies with time of year), resulting in $\Delta T \sim 0.0034$ K. If we just model the Earth as "flat" with an area of $\pi R_e^2$, then the net force exerted by the CMB is only 0.01 N.

  • $\begingroup$ +1 Great answer. Just wanted to tack on a calculation for OP's inquiry on Earth's rotation: treating Earth as a rotating disc, one can compute the drag torque by integrating $$\tau=\biggr|\int \vec r \times \vec f dA \biggr|=\frac{4\pi^2\sigma R_E^4}{(1\text{ day})\cdot c^2}T^4 \approx 40\text{ N}\cdot\text{m}.$$ This corresponds to a rotational deceleration of about $$5\cdot 10^{-37}\frac{\text{rad}}{\text{s}^2}\approx 2\cdot 10^{-25}\frac{\text{rotations}/\text{day}}{\text{year}}$$ So it would take ~$10^{24}$ years for this drag to slow the Earth's rotation by $20 \%$, absent other effects. $\endgroup$
    – jawheele
    Commented Feb 9, 2022 at 22:34

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