# Baryonic acoustic oscillations: Why are there standing waves in the CMB?

On page three of the following http://www.quantumfieldtheory.info/CMB.pdf, Klauber talks about the formation of standing waves from acoustic vibrations in the early universe.

He claims that they form in a region equal to the length that sound waves can travel in the given time.

I do not understand why there ought to be standing waves here. Surely that would only happen if the sound waves reflect back off of some boundary such that some forwards-traveling waves interfere with backwards-going waves.

Maybe this will help (too long for a comment) :

Gravity tries to compress the fluid in potential wells.

Photon pressure resists compression resulting in acoustic oscillations

System is equivalent to a mass on a spring falling under gravity

This video with water waves from sound may give intuition : The plasma waves in the early universe come from the compression/decompression in an effective gravitational well . The argument about the available space to the acoustic wave is equivalent to why in the water one observes these wavelengths and not all: the ones out of phase disappear from destructive interference.

• Hey thanks for this. What you've sent has thrown up a lot of follow-up questions. Firstly, in the link that hsnee sent (Martin White's introduction to BAOs), it would appear that gravity was not particularly significant until the time of recoupling. However, in your link, it is the major cause of oscillations. Secondly: Surely the standing waves in the water occur due to the sides of the box, whereas the universe has no sides. So how do the standing waves occur. – Kris Mar 23 '16 at 13:53
• @Kris well, inflatons are all about gravity, gravity is very strong because of the small size of the universe. for b) it is the gravitational well that makes the effective sides. Think of an isolated ball of water in space. It will be able to oscillate while just held by the gravitational well of its mass. – anna v Mar 23 '16 at 14:37
• @annav - This explanation is not satisfactory. A ball of water in space can only exist in a pressurized environment (it would quickly evaporate in a vacuum). The surface tension in this analogy acts as the boundary of a box, allowing for standing waves. I would like to see the OP's question answered as I have the same one: how can a standing wave be created without a boundary? – user32023 Oct 16 '16 at 16:23
• @DonaldRoyAirey Take the shrodinger equation with a potential well. The atom is in vacuum. Standing waves are called "orbitals" en.wikipedia.org/wiki/Schr%C3%B6dinger_equation . The early universe is accepted as quantum mechanical , – anna v Oct 16 '16 at 16:31

In the early universe, the Universe was a hot plasma of photons, protons and electrons (ionised Hydrogen). Now we perturb the universe, the perturbation travels as a sound wave, at the speed of sound in the medium. Now there are two possibilities here

1. The perturbation would cause gravitational collapse to start. This is not very sensible when we consider that the Universe is still a hot plasma. So:
2. The perturbation creates a standing wave, because the texture of the hot plasma requires it to stay uniform in all places other than the perturbation (which is a small excess) -- this is the boundary condition.

The pressure drives the hot plasma to travel at (nearly) the speed of light. When the Universe cools down (the Universe is around 100,000 years old now), the photons and baryons decouple, the photons continue to travel at the speed of light, while the slow down, forming a peak (excess of baryons) at 100 $Mpc/h$

For more information (and plots) I would recommend Martin White's introduction to BAOs here

• I don't understand how a standing wave can travel. Also, how did it become a standing wave to begin with? – Kris Mar 23 '16 at 0:13
• Sorry, I wasn't cautious enough to choose the right terms when I said travels as a standing wave, I've edited this. As for how it becomes a standing wave, it's down to the fact that all the other regions in the universe outside the perturbation are required to stay uniform, rather than react to the perturbation, which acts as nodes on the sides of the wave, causing it to become a standing wave. – hsnee Mar 23 '16 at 18:02
• Thanks, I've been thinking about BAOs on-and-off for the past few days and have attempted to answer my own question below. Please give it a read and tell me if I've misunderstood. – Kris Mar 23 '16 at 18:45

This is my attempt at an answer to my own question, based upon the helpful responses I have been given so far. I am hoping it may be critiqued until it is correct if it is not already.

The initial condition (set by inflation) is one in which the universe is inhomogeneously filled with baryons, photons and dark matter. The interactions of photons and baryons causes a pressure in this mixture for these two components. The dark matter feels no pressure.

Gravitational forces cause collapse of the baryon-photon gas into over-dense regions. When the density of these regions becomes large, the pressure has increased enough to expel the material out from the region again. This process repeats itself until the time of recombination, where the baryon-photon interactions stop, the pressure decreases and there is only gravitational collapse.

Before recombination, the universe is filled with over-dense regions at all scales surrounded by complicated oscillations of the type described above. For any one of these regions, at recombination the maximum distance a perturbation of the density can have traveled from the source is the distance a sound wave can travel by the time recombination occurs. We can call this L.

In general, a complicated three-dimensional density wave surrounds this region. Simplifying to one dimension, there is a complicated wave with furthest extent just L. A Fourier series may be used to describe this wave (series rather than integral due to the finite extent). Each harmonic of the Fourier series is then a standing wave. The squared coefficients of which are plotted in the CMB power spectrum.