# Finding the CMB temperature anisotropies contributed at different orders due to a uniform boost

I've just started reading about the physics and features of Cosmic Microwave Background. I came across this pdg review titled Cosmic Microwave Background by D. Scott and G. F. Smoot says that

At every point in the sky, one observes a blackbody spectrum, with temperature $\color{blue}{T(\theta)}$.

Certainly, the satellites do not point "thermometers" at various points in the sky. However, I have a crude idea (may be wrong) about how they do it. Presumably, the detectors (in the satellites those map the temperature of the CMB sky) basically measure the flux of blackbody photons of various frequencies at a given direction $\hat{\textbf{n}}$ or equivalently, $(\theta,\phi)$. Then, from the intensity versus wavelength plot, read off the temperature $T$ of the CMB at each point.

The temperature $T(\theta,\phi)$ at any point on the CMB sky should be a function of that coordinate of the sky $(\theta,\phi)$. This is why one expands the temperature in terms of spherial harmonics as: $$T(\theta,\phi)=\sum\limits_{l,m}a_{lm}Y_{lm}(\theta,\phi).\tag{1}$$

However, in the reference mentioned above claims that the temperature becomes a function of $\theta$ only. In particular, it gives the following expression $$T(\theta)=T_0(1-\beta^2)^{1/2}(1-\beta\cos\theta)^{-1}\tag{2}$$ where $\theta$ is probably the angle between any photon momentum and the direction of uniform boost. Where did the azimuthal angle $\phi$ disappear?

Edit in response to @probably_someone's answer Formula (2), if I understood it correctly, is a special case of (1) when the anisotropy in temperature of CMB is solely due to a uniform motion of the observer w.r.t the CMB.

Does it mean that I can use equation (1), and set $\phi=0$ (without any loss of generality) to understand the effect of anisotropy caused by a uniform boost and read off the contributions at different orders in $l$? Is this the way to reconcile (1) and (2)?

• I am voting to close this question as it is lacking basic material that's necessary to formulate what is being asked. The question needs to stand on its own, and not rely on external resources to make sense. – Emilio Pisanty Jun 29 '18 at 18:26
• @EmilioPisanty You've already lifted the close vote. – SRS Jul 1 '18 at 16:36

## 2 Answers

When the authors use $T(\theta)$, they are specifically referring to the case of the temperature observed by an observer moving at a constant velocity. This situation has axial symmetry, which implies that only the angle relative to the direction of motion $\theta$ will significantly matter.*

*In reality, the situation is not exactly axially symmetric, because the CMB has fluctuations on the order of $10^{-5}$, but the effect of the Lorentz boost will usually far outstrip these effects.

• That would mean that $\theta$ not the polar angle on the sky but the angle between any CMB photon and the observer. Is that right? @probably_someone – SRS Jun 29 '18 at 15:47
• @SRS $\theta$ is the polar angle on the sky, where the pole is along the direction of motion. – probably_someone Jun 29 '18 at 15:50
• @SRS Must be the motion of the satellite determining the axis, see page 4 here web.vu.lt/ff/t.gajdosik/files/2013/12/ve_gr7.pdf "7.General relativity CMB – anna v Jun 30 '18 at 13:25

The temperature of CMB is measured by its frequency distribution (as you described in the question). Let's assume the motion of the detector relative to CMB is uniform at least during the measurement at each particular angle.

The measured frequencies are affected by the relativistic Doppler effect at the detector velocity. The Doppler shift consists of the part in the direction of motion and the transverse part (perpendicular to the direction of motion). For symmetry reasons, the Doppler shift does not depend on the azimuthal angle.

Thus for every velocity $v$ and angle $\theta$, we must apply the Doppler shift formula (2) from the "Motion in an arbitrary direction" section of: https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

Note that this formula matches exactly the formula (2) in your question, provided the peak frequency is proportional to the temperature according to the Wien's displacement law: https://en.wikipedia.org/wiki/Wien%27s_displacement_law

Finally, the change of the detector velocity between two measurements can be easily obtained knowing the orbit parameters of the satellite. Provided the extreme uniformity of the CMB (with the dipole removed), I would expect that such adjustments were made (although I don't know for sure). In other words, $\beta=\dfrac{v}{c}$ is different for every $\theta$ (which also in turn must be adjusted for the angle of the satellite trajectory at the time of the measurement).

Obviously the relativistic Doppler shift anisotropy does not represent a single spherical harmonic. Therefore it is productive to remove this anisotropy first by using the formula (2) before analyzing the intrinsic CMB anizotropies in terms of spherical harmonics using the formula (1).

• Minor comment to the post (v1): In the future please link to desktop Wikipedia pages rather than mobile Wikipedia pages, i.e. without the "dot emm". – Qmechanic Jul 3 '18 at 8:01