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I've read many posts, books, articles and so on about the CMB power spectrum (this page for example http://background.uchicago.edu/~whu/intermediate/map5.html) to try to understand it, but I seem to still have very basic questions about it that I can't answer. enter image description here

I read everywhere that the CMB power spectrum is the expantion the CMB temperature anisotropy map in terms of spherical harmonics. So my first question is, why are we expanding it in terms of spherical harmonics ? I know what it means mathematically (I have read a lot from this page https://en.wikipedia.org/wiki/Spherical_harmonics and this page https://www.roe.ac.uk/ifa/postgrad/pedagogy/2006_tojeiro.pdf), but I don't understand the meaning of doing that, why we do that and what does it bring.

Then, I've read from the above site the following:

For example l=10 corresponds to roughly 10 degrees on the sky, l=100 corresponds to roughly >1 degree on the sky."

1 degree starting from where and in which direction? And I know the formula relating l and angles but again, I don't understand the meaning of it. How can a l be an angle ? And how can I see this angle on the sky (or can I see this angle ? If not what is this 1 degree angle telling me ?)?

I struggle a lot trying to understand that. Thank you for your help.

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    $\begingroup$ Are you familiar with Fourier Transforms? It's the same really, one for periodic stuff on a sphere, the other for periodic stuff in time $\endgroup$
    – rfl
    Jan 19, 2023 at 16:52
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    $\begingroup$ I'm a bit familiar but basic bachelor of physics stuff. What could I read in order to try to understand this ? $\endgroup$
    – Apinorr
    Jan 19, 2023 at 16:54
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    $\begingroup$ Somewhere I saw an animation from the deconvolution of a simple map of Earth into spherical harmonics. If you can find that it'll clear this up immediately... can't find it real quick right now though $\endgroup$
    – rfl
    Jan 19, 2023 at 17:38
  • $\begingroup$ How can a l be an angle ? $\ell$ isn’t an angle; it’s an integer. $\endgroup$
    – Ghoster
    Jan 19, 2023 at 18:16
  • $\begingroup$ @Ghoster Note however that $360⁰/(\ell+1)$ is angle-ish. The monopole $\ell=0$ covers the entire sky; the dipole $\ell=1$ has two features separated by a half-turn; the quadruple $\ell=2$ has features at pole-equator-pole; and so on. $\endgroup$
    – rob
    Jan 19, 2023 at 21:45

1 Answer 1

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Ultimately the goal of CMB observations is to understand that statistical distribution of temperature fluctuations on the sky.

There are quite a few things to unpack there.

First, the average temperature of the microwave background radiation is about $2.73\ {\rm K}$. This has been known since the early 90s with COBE.

The fluctuations in temperature are deviations from this average value. Typically one looks at a normalized fluctuation, defined like so \begin{equation} \delta(\hat{n}) = \frac{T(\hat{n}) - \bar{T}}{\bar{T}} \end{equation} where $T(\theta, \phi)$ is the observed temperature in the direction of the normal vector $\hat{n}$, $\bar{T}=2.73\ {\rm K}$ is the average temperature of the CMB, and $\delta$ is the normalized temperature fluctuation.

Now, we want to understand $\delta$. There are a few things we can say without a lot of work. First, the average value of $\delta$ across the sky is zero. This simply follows from how we defined it. Second, the average value of $|\delta|$ -- ie, of the magnitude of delta -- is about $10^{-5}$. This means that in a typical direction, $T(\hat{n})$ will only start deviating from $\bar{T}$ after about 5 decimal places. That is wild, and in fact the agreement is better than the number of significant digits I gave you for $\bar{T}$ above.

To talk in more detail, we need to understand that $\delta$ is a random variable. When we look in a given direction on the sky, theory does not predict exactly what value of $\delta$ we will see in that direction. Instead, theory predicts a probability distribution from which $\delta(\hat{n})$ in our Universe is drawn.

In fact, the probability distribution for $\delta(\hat{n})$ is -- as far as anyone knows -- a multivariate Gaussian distribution. In other words, the pdf $p(\delta(\hat{n}))$ has the form \begin{equation} p(\delta(\hat{n})) \propto \exp\left(- \int d^2 \hat{n} \int d^2 \hat{n}' \left[ \delta(\hat{n}) \times \Sigma(\hat{n}, \hat{n}')^{-1} \times \delta(\hat{n}') \right] \right) \end{equation} where the "matrix" $\Sigma(\hat{n}, \hat{n}')$ is the covariance between directions $\hat{n}$ and $\hat{n}'$.

There is a lot going on in this expression. What it captures is that we expect the temperature fluctuation in some directions to be correlated with the temperature fluctuation in other directions. The pattern of these correlations is captured in the covariance matrix. The goal of CMB analysis is therefore to use data to estimate the covariance matrix. Furthermore, theory allows us to compute the covariance matrix given an initial distribution of density perturbations (which is an output of inflation theory), and a model of how those density perturbations evolve through the history of the Universe (which is quite complicated but well-understood and based on established physics).

Now, the covariance matrix as I've written it is a very complicated object, depending in an arbitrary way on two sky directions $\hat{n}$ and $\hat{n}'$. In fact, it turns out we can make some simplifications.

First, because of there is no preferred direction on the sky, it turns out that the covariance matrix is a function only of the relative sky position $\hat{n}-\hat{n}'$, rather than of the two sky positions independently $$ \Sigma(\hat{n}, \hat{n}') = \tilde{C}(\theta) $$ and $\tilde{C}(\theta)$ is the correlation function, and \hat{n}\cdot \hat{n}' = \cos(\theta) (ie, $\theta$ is the angle between $\hat{n}$ and $\hat{n}'$). (I've put a twiddle on it because my normalization is probably different than what is used in the literature).

The correlation function answers the following question: suppose we are given two points on the sky are separated by an angle $\theta$. If we measure the temperature fluctuation at point $1$, what do we expect the temperature fluctuation at point $2$ (separated from $1$ by an angle $\theta$) to be? If the correlation function at $1$ is large, then we expect that the value of the temperature fluctuation at $2$ will be strongly correlated with the value of the temperature fluctuation separated by a direction $1$. If the correlation function is small, we expect these fluctuations to be uncorrelated.

The second simplification we can make is that we can actually choose a basis so the covariance matrix so it is completely diagonal. This is the secret sauce behind using the spherical harmonic decomposition.

The reason for this second simplification comes from theory. It comes from our believe that the Universe is statistically isotropic. This means that we don't believe there are special directions. (You can, and people have, relaxed this assumption and check if the CMB does have special directions; to date no one has found compelling evidence that the Universe has a special direction at a fundamental level, although there are some anomalies in the data that people debate the significance of).

This second simplification means that we can trade the correlation function $C(\theta)$ for a simpler power spectrum $\tilde{C}_\ell$. It would be too much work to explain exactly how this works in detail, but I will at least try to show the basic idea. By first defining the spherical harmonic transform of the temperature fluctuations \begin{equation} \delta_{\ell, m} = \int d^2 \hat{n} Y_{\ell m}(\hat{n}) \delta(\hat{n}) \end{equation} We can express the probability distribution for the $\delta_{\ell m}$ directly in terms of quantities $\tilde{C}_{\ell}$ \begin{equation} p(\delta_{\ell, m}) \propto \exp\left(- \sum_{\ell, m} \left[ \delta_{\ell, m} \times \frac{1}{\tilde{C}_{\ell}} \times \delta_{\ell, m} \right] \right) \end{equation} and again I've included a fudge-factor "twiddle" superscript in my expression $\tilde{C}_\ell$ to emphasize that there is probably a (possibly $\ell$-dependent) normalization difference between my expression and what is actually used in the literature.

This distribution is much simpler than the one we originally wrote for $\delta(\hat{n})$. It actually doesn't depend on $m$; each $m$ mode for a given $\ell$ is an independent measurement of $C_{\ell}$. Furthermore, the power spectrum is a function of only a single integer $\ell$, rather than our original completely general expression which depended on four real numbers.

The fact that $C_{\ell}$ only depends on one $\ell$ reflects the fact that there is no covariance between different values of $\ell$. We can measure the variance in each $\ell$ independently. Effectively, we have found a basis in which the covariance function is diagonal. This allows us to directly measure the eigenvalues of the covariance function. Our goal from the very beginning was to find a way to measure the properties of the probability distribution governing $\delta(\hat{n})$, which boiled down to measuring the covariance function. We have now found out that we can learn all the information contained in the covariance function by simply measuring the $C_\ell$ for each value of $\ell$.

The final point is that there is an approximate (not exact) correspondence between $\ell$ and the angular separation. If $C_\ell$ is large for some $\ell$, it means that the covariance function is large when two points are separated by an angle approximately equal to $\pi/\ell$. So, when people say there is an excess of power at $\ell=100$, corresponding to about 1 degree, they mean that if you pick a point on the sky and find that it has a large value for $\delta$, then you are very likely to find other points separated by 1 degree which also have a large value for $\delta$. Generally, the value of $\delta$ at points separated by $1$ degree will tend to be strongly correlated.

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    $\begingroup$ First of all, thank you very much for the complete and detailed answer. I need to diggest everything a bit but I thing it make much more sense now. I'm wondering how I could read more about these kinds of explanation. Do people know that from reading the sources or how do they know that ? Because it's not that easy. $\endgroup$
    – Apinorr
    Jan 20, 2023 at 12:46
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    $\begingroup$ @Apinorr It takes a long time to build intuition in complex subjects like this. I think the best way to learn is to have connections with people. If you know a professor who works in this field, see if you can do research or an independent study with them. Try to attend any seminars or colloquia relevant to cosmology. Form a study group with other students interested in this topic. It's hard to learn on your own. But, there are also a lot of good books and reviews out there. Find one that matches your style. I liked Physical Foundations of Cosmology by Mukhanov, but it's quite mathematical. $\endgroup$
    – Andrew
    Jan 20, 2023 at 14:58
  • $\begingroup$ Okay I understand better now, I'll do that, thank you very much for your help. $\endgroup$
    – Apinorr
    Jan 23, 2023 at 9:58

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