# Is the fact that we're moving with a certain speed with respect to the CMB special-relativity consistent?

As a sidenote to an exercise about the aberration of CMB at the dipole level, which scope was to find the peculiar velocity we have with respect to the cosmic background (assuming the doppler effect derived from it is the only source of dipole aberration), my professor left the following question:

Is the fact that we're moving with a certain speed with respect to the CMB special-relativity consistent?

My calculations were as follows:

Apparently, the difference in the temperature of the CMB due to the dipole is $$\Delta T_{l=1}=3.372\cdot 10^{-3}K$$, and choosing our axis so that the spherical harmonic with $$m=0$$ is oriented with the dipole, it's easy to see that, at the dipole order, temperature will be written: $$T(\theta)=T_0(1+\frac{v}{c}cos\theta)$$ So I just took $$\frac{\Delta T_{l=1}}{T} = \frac{v}{c} cos\theta$$, and this leads to $$v\approx371\frac{km}{s}$$.

I don't see how this such small speed could not be consistent with special relativity, but I'm suspicious that answering "Yes it is consistent" would be too easy to be true... Is my reasoning correct? Thanks!

## 1 Answer

The actual speed with which we are moving relative to the CMB is unimportant; we could be moving at $$0.99c$$ and special relativity would still apply, since that theory only requires a velocity that is less than $$c$$. It's like the Newtonian formula for calculating kinetic energy - it works regardless of whether your speed is $$371 km/s$$ (an extremely fast speed by terrestrial standards) or $$0.001 m/s$$.

Your professor is probably asking something else entirely: special relativity implies all reference frames (in the absence of gravity) are equal, yet we are clearly moving with respect to the CMB. In other words, can we conclude that the frame in which the CMB is stationary is special? If we can, does that violate special relativity?

The answer to whether the CMB rest frame violates special relativity is obviously "no" (or we'd have rejected special relativity!), but the explanation is quite deep, and drives at the heart of what exactly is meant by all reference frames are equal. You might want to have a think about it before looking at this question.