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I'm trying to calculate the sound horizon from the start of time to decoupling. To do this I need to know the speed of sound and how it changes as the universe grows. The speed of sound in a fluid is:$$c_s^2=\frac{\partial p}{\partial \rho}$$Where $c_s^2$ is the speed of sound, $p$ is the pressure, $\rho$ is the density. I think I have a handle on calculating the density:$$\rho(z)=\Omega_b h^2 \rho_{crit} (z+1)^3 \space g\space m^{-3}$$But I have no idea how you calculate the pressure. I'm assuming that the pressure was mainly photonic up to the time of decoupling but I'm having trouble finding reference material.

Also, intuitively I would think that the pressure would fall in exact proportion to density since they're both related to the change in volume. So is it enough just to find the pressure at the time of decoupling and divide by the density and use it as a constant?

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Before recombination baryons and photons are highly coupled and act like a single fluid, so the density is $\rho = \rho_{\rm bar} + \rho_\gamma$, with the usual scaling $\rho_{\rm bar} \sim (1 + z)^{3}$ and $\rho_{\gamma} \sim (1 + z)^{4}$. However, the pressure is the same for both components $P = P_{\rm bar} = P_{\gamma}$, this will lead to

$$ c_s^2 = \frac{c^2}{3}\left[\frac{3}{4}\frac{\rho_{\rm bar}}{\rho_\gamma} + 1\right]^{-1} \tag{1} $$

To derive this expression remember that in an uniform field $P_{\gamma} = \rho_\gamma c^2 / 3$ so that the adiabatic speed of sound is

$$ c_s^2 = \left( \frac{\partial P}{\partial \rho}\right)_S = \frac{c^2}{3} \left.\frac{\partial \rho_\gamma}{\partial(\rho_\gamma + \rho_{\rm bar})}\right|_S = \frac{c^2}{3}\left[1 + \left(\frac{\partial \rho_{\rm bar}}{\partial \rho_\gamma}\right)_S \right]^{-1}\tag{2} $$

We need to calculate $(\partial \rho_{\rm bar}/\partial \rho_\gamma)_S$. And to do that we can use the fact that $\rho_{\rm bar}\sim a^{-3}$ and $\rho_{\gamma}\sim a^{-4}$ so

$$ \frac{\partial \rho_\gamma}{\rho_\gamma} = -4 \frac{\partial a}{a} ~~~\mbox{and}~~~\frac{\partial \rho_{\rm bar}}{\rho_{\rm bar}} = -3 \frac{\partial a}{a}~~~\Rightarrow~~~ \frac{\partial \rho_{\rm bar}}{\partial \rho_\gamma} = \frac{3}{4}\frac{\rho_{\rm bar}}{\rho_\gamma} \tag{3} $$

If you replace (3) in (2) you will get (1)

caverac
  • 6,744
  • Why is the pressure the same? –  Dec 13 '18 at 14:31
  • @BenCrowell They are coupled through Compton scattering, pressure is the same – caverac Dec 13 '18 at 14:34
  • $\rho_{\gamma}$ is quoted as roughly $410\space photons\space c^{-3}$. However, $\rho_b$ is in units of $g\space cm^{-3}$ or some equivalent of mass per cubic space, so how do I divide mass density by photon density (since photons have no mass)? –  Dec 13 '18 at 14:35
  • @DonaldAirey You can still calculate its energy density even if they don't have mass, the number you are looking for is $\rho_\gamma = 7.8\times 10^{-34} (1 + z)^4{\rm g~cm}^{-3} $ – caverac Dec 13 '18 at 15:15
  • wow, that was fast ! I saw the fix in real time ! :-) – Cham Dec 13 '18 at 15:39
  • Can you show how the pressure drops out of the equation? That is, does your formula above derive from the fluid formula in my original post? –  Dec 13 '18 at 16:06
  • @DonaldAirey Sure, I updated the answer – caverac Dec 13 '18 at 16:35
  • @caverac - Excellent. Now, for the last question. Since the effects of volume are included in your formula, can the speed of sound be considered constant for all intents and purposes (obviously no-one can compute the speed of sound during the plank or inflation epochs)? –  Dec 13 '18 at 16:42
  • @DonaldAirey No, you can't. $\rho_\gamma = \rho_{\gamma, 0}(1 + z)^{-4}$ and $\rho_{\rm bar} = \rho_{{\rm bar}, 0}(1 + z)^{-3}$, that means $c_s$ will change with time! – caverac Dec 13 '18 at 16:44
  • If $\Omega_b h^2 = 0.022$ (from Plank), then $\Omega_b = 0.022 / (67.3/100)^2 = 0.049$. Then present day baryon density is $\Omega_b \rho_{crit} = 4.13\times 10^{-31}\space g\space cm^{-3}$. So would you agree that $\rho_b(z) = 4.13\times 10^{-31}(1+z)^3 \space g\space cm^{-3}$? –  Dec 14 '18 at 14:09
  • @DonaldAirey That's about right – caverac Dec 14 '18 at 15:33
  • @caverac - Is there any chance you have a reference for your formula $\rho_\gamma = 7.8\times 10^{-34}(1+z)^4\space g\space cm^{-3}$? –  Dec 15 '18 at 20:43
  • @DonaldAirey You can look at it here https://physics.stackexchange.com/questions/94181/where-is-radiation-density-in-the-planck-2013-results – caverac Dec 15 '18 at 20:49
  • @caverac - Hey, I'm double checking these numbers here because I can't get them to agree with a statement on another post that declares the photon density was roughly the same as the baryon density at recombination. Looking at the article you referenced, you used the total radiation density $\rho_{\gamma} + \rho_{vu}$ in the formula for sound velocity. It looks like the velocity of sound is related to the photon density, not the radiation density. Did you mean to pull in the value $\rho_{\gamma} = 4.64511\times 10^{-31}\space g\space cm^{-3}$? –  Jan 11 '19 at 19:01