13
$\begingroup$

I was wondering, if anyone could guide me through the different peaks of this spectrum (See below)? I've been reading and reading numerous pages about this, but I can't seem to get my head around this.

I know that the high peak tells us the curvature of the Universe, but I'm not sure why and how, and where it comes from. My somewhat conclusion to where it comes from is, that it comes from "sound waves" from the early Universe, before photons decoupled. So basically, due to fluctuations in density (From even earlier Universe), these more dense areas contract and raises the temperature of the photons in the same area. This heats up the area, giving the photons more energy, and from that more radiation pressure. This low-to-high-pressure region makes some kind of sound wave (If I'm not mistaken), which, somehow, translates into the high peak - but again, I'm not quite sure why and how.

The other peaks, I'm really just confused about, and I'm really not sure what they tell me, and, physically, where they come from.

I know it might be a long answer, but I had to try, since I haven't been able to figure it out myself.

Thanks in advance. CMB Power Spectrum

$\endgroup$

1 Answer 1

4
$\begingroup$

As a first step, it would be good to review what this multipole moment $l$ means. The CMB is of course coming at us from a spherical shell around us. We therefore see a spherical projection, and we would like to quantify this much like we decompose a plane wave in trigonometric functions.

To do this on a sphere, we use spherical harmonics instead. These functions are the eigenfunctions of the angular part of the Laplace equation, and you might know them from quantum mechanics in which we use them to solve the three dimensional Schrodinger equation for e.g. the H-atom. They satisfy, for instance, the orthonormality requirement and can be used much like trigonometric functions in real space.

$l$ is a lable for the spherical harmonics, and can be compared to the wave number in plane waves: it is a measure (inversely proportional to) the size of perturbations. Like the wavenumber is inversely proportional to wavelength.

You are quite right that sound waves are involved: Just like sound travelling as density waves in air here on Earth, the photon-baryon fluid (which were coupled before decoupling) are influenced by such waves. Some of these with a particular wavelength (which, remember, we will see as an angular size on the sky) may be caught at a extremum, such that we see a lot of structure at these particular scales.

Whether a scale will be caught at a maximum or not, depends on its oscillation time until recombination. When atoms recombine, photons will not collide frequently anymore and these plane waves cease to exist.

Now, to finally answer your question: how does the spatial curvature of the universe fit into this mix? The first peak corresponds to the largest scale that has been able to reach a maximum at the time of recombinations. This is related to the horizon size at the time: scales that have not entered the horizon have not started oscillating yet (because that is the meaning of the cosmic horizon: scales larger than this cannot 'communicate'.

Now, since we know fairly precisely how long after the big bang decoupling occured and so on, we can calculate fairly precisely how large the horizon should be at this time and this should correspond to a very well-defined angular scale in the sky, if the universe was Euclidian (spatially flat). However, if the space is in fact negatively (positively) curved, the projected size of the first peak (the horizon size at decoupling) would be smaller (bigger) than expected: a direct indication and measure of curvature effect!

You can compare this to a magnifying glass: if you observe a bug to be x-times as big as its (known) size, you can calculate the magnifying power of the glass -- in this analogy, the magnifying power is comparable to the spatial curvature of our universe.

Hope this helps!

$\endgroup$
4
  • $\begingroup$ Your answer is nice but about confused with a few statements. You said, "it is a measure (inversely proportional to) the size of perturbations" Which quantity do you refer to when you say "the size of perturbations"? Do you mean the absolute square of $a_{lm}$ in $\Delta T= \sum a_{lm}Y_{lm}$?If so, why do you say that $l$ is inversely proportional to the size of perturbations? Or do you mean something else by the size of perturbations? @user1991 $\endgroup$
    – SRS
    Commented Oct 10, 2018 at 7:49
  • 1
    $\begingroup$ (Long time ago!) What I meant is that $l$ is inversely proportional to the angular size (marginalized over $m$) $\endgroup$
    – user1991
    Commented Oct 10, 2018 at 7:54
  • $\begingroup$ Okay. Thanks. "a particular wavelength (...) may be caught at a extremum" I also did not understand this part. What do you mean by a particular wavelength being caught at a maximum? $\endgroup$
    – SRS
    Commented Oct 10, 2018 at 7:55
  • $\begingroup$ The idea there is that before recombination, primordial oscillations on all scales are rarefying and expanding, rarefying and expanding - as waves do. At recombination, these waves freeze out. At some scales, the wave will have just reached its maximum amplitude amplitude at that moment, resulting at a lot of visible structure at that scale. At other scales, the oscillations goes through the equilibrium position, resulting in no particular structure at that scale. That's what I meant with 'caught at an extremum'. (is that a little clearer?) $\endgroup$
    – user1991
    Commented Oct 10, 2018 at 8:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.