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In the book 'An introduction to science of cosmology' by Raine and Thomas, they have two equations $$ \frac{\Delta T}{T} = \frac{v}{c}\cos\theta'\qquad (4.14) $$ $$ T = (3.372±0.014)×10^{-3}\cos\theta' K \qquad (4.15) $$ From these two relations the velocity of solar system relative to CMB has been written as $371\pm 1$ km/s in the direction $(l = 264.14\pm 0.3^{\circ},\ b = 48.26\pm 0.3^{\circ})$. How do we arrive at this velocity and direction.

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  • $\begingroup$ CMB dipole anisotropy... so what's the question? $\endgroup$ – CuriousOne Sep 9 '15 at 18:05
  • $\begingroup$ You have to numerically fit the predicted anisotropy to the measured data. That may sound easy but I am fairly sure that it's not. The CMB data set has a number of non-trivial systematic error sources that need to be controlled carefully. You will have to read the full data analysis papers to understand how the collaboration did it and it's usually a good idea to contact the team members responsible for such an analysis to discuss the details if you want to do your own analysis or use the result for non-trivial research of your own. $\endgroup$ – CuriousOne Sep 9 '15 at 18:24
  • $\begingroup$ @CuriousOne I think measuring just the dipole is probably much easier - it is very "loud" compared to the rest of the signal, and is the first thing that needs to be removed to do any other analysis. Probably to get very accurate removal requires a lot of care, but just getting a velocity to within 1km/s wouldn't be too bad, I think. $\endgroup$ – Kyle Oman Sep 9 '15 at 18:49
  • $\begingroup$ @KyleOman: Could be. 1km/s out of 370 sounds like a pretty good precision to me... but maybe the signal is not sensitive to the galactic background. My view on these things is a bit "clouded" since I got to hear an earful from a postdoc once who spent what little career she had in the field on cataloguing galactic gas clouds which needed to be excluded from the CMB analysis... there was very little scientific honor to be had from that and it was still absolutely vital to have as complete a set to remove systematic errors from the analysis. This may be less crucial for the dipole. $\endgroup$ – CuriousOne Sep 9 '15 at 23:07
  • $\begingroup$ @CuriousOne Yeah the signal looks something like *****DIPOLE*****, quadrupole, octupole, hexadecapole, ... You can see the raw CMB map here. You can kind of see a bit of the galactic foreground along the equator, but overwhelmingly you see the DIPOLE ;) $\endgroup$ – Kyle Oman Sep 10 '15 at 3:29
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If we are moving relative to the CMB then CMB photons will be doppler shifted (and the amount depends on the direction relative to the CMB frame). This induces a dipole in the temperature fluctuation map

$$T(\theta) = T_0(1+v/c\cos\theta)$$

to first order in $v$, which we can measure. The direction and velocity can for example be found by fitting a dipole to CMB temperature maps produced by e.g. WMAP and Planck. For more details on how it's acctually done in practice you see for example the Planck article on the Doppler boosing of the CMB.

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First, you make a measurement of the CMB temperature map. With the average temperature subtracted off, this looks like:

enter image description here

The main feature to see here is a clear dipole, the signature of the doppler shift caused by the motion of the detector relative to the CMB rest frame. There's also a bit of fuzz visible along the equator from galactic foreground sources. The actual cosmologically interesting CMB signal is buried under the overwhelming dipole and foregrounds.

To get a measurement of the velocity, one simply fits a dipole to the map (with the monopole, or mean temperature, already subtracted off), which looks like your equation 4.14. Your 4.15 looks a bit fishy to me out of context, because if $\theta'$ ranges from $0$ to $2\pi$, you'll get negative values for temperature. I don't think it's actually needed, in any case. Once the dipole is fit you'll know which direction the motion is in (I'm guessing $\theta'$ is defined as an angle offset from some reference angle), and its magnitude ($v/c$). From there it's just a matter of converting to galactocentric coordinates $(l,b)$.

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