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I'm reading through Modern Cosmology by Dodelson and Schmidt 2nd edition on my own, and at the start of Section 4.2 the book says that we can compute the baryon/photon ratio at the time of big bang nucleosynthesis and get $$ \eta_{\text{b}} \equiv \frac{n_{\text{b}}}{n_{\gamma}} = 6.0\times 10^{-10} \left(\frac{\Omega_{\text{b}}h^{2}}{0.022}\right). $$ I'm trying to figure out how to get to this result.

Attempt

Note that all calculations are done in units having $\hbar = c = k_{B} = 1$.

  • From (4.5) in the book, $n_{\gamma} = 2T^{3}/\pi^{2}$.
  • I took the baryon number density to be $n_{\text{b}} \approx \rho_{\text{b}} / (\text{mass of proton})$.
  • From (2.72) in the book, $\rho_{\text{b}} = \Omega_{\text{b}}\rho_{\text{cr}}a^{-3(1+w)}$ where I took $w=0$ for baryons.
  • The critical density is $$ \rho_{\text{cr}} = \frac{3H_{0}^{2}}{8\pi G} = \frac{3h^{2}}{8\pi G}(100\text{ km s$^{-1}$ Mpc$^{-1}$})^{2} = \frac{3h^{2}}{8\pi G}\frac{1}{(0.98\times 10^{10} \text{years})^{2}}. $$
  • Lastly, I used the fact that $a\propto 1/T$ to get $a = T_{0}/T$ where $T_{0} = 2.73\text{ K}$ today.

Putting all this together, we get \begin{align*} \frac{n_{\text{b}}}{n_{\gamma}} &= \frac{\rho_{\text{b}}/m_{\text{p}}}{2T^{3}/\pi^{2}} = \frac{\pi^{2}\rho_{\text{b}}}{2T^{3}m_{\text{p}}} = \frac{\pi^{2}\Omega_{\text{b}}\rho_{\text{cr}}a^{-3}}{2T^{3}m_{\text{p}}} = \frac{\pi^{2}\Omega_{\text{b}}\rho_{\text{cr}}}{2T_{0}^{3}m_{\text{p}}} \\[1.2ex] &= \frac{\pi^{2}\Omega_{\text{b}}}{2T_{0}^{3}m_{\text{p}}}\frac{3h^{2}}{8\pi G}\frac{1}{(0.98\times 10^{10} \text{years})^{2}} \\[1.2ex] &= \Omega_{\text{b}}h^{2} \,\cdot\, \frac{\pi^{2}}{2T_{0}^{3}m_{\text{p}}}\frac{3}{8\pi G}\frac{1}{(0.98\times 10^{10} \text{years})^{2}}. \end{align*} Now plugging everything in, using WolframAlpha, and multiplying by whatever factors of $\hbar$, $c$, and $k_{B}$ are needed to get rid of the units, I get a factor of $\approx 3.26\times 10^{-8}$.

The factor should be $6.0\times 10^{-10}/0.022 \approx 2.7\times 10^{-8}$. This in the same order of magnitude, but it is clearly off. What did I do wrong? What is the right way to do the calculation?

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1 Answer 1

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Ahh, I see what I did wrong. At the first bullet point, I invoked (4.5), which is an approximation. It involves an approximation of the distribution by a Boltzmann distribution for a classical dilute gas. This is not what we need for photons.

What we need instead is the Bose-Einstein distribution. In that case, we have \begin{align*} n_{\gamma} &= 2\int \frac{d^{3}p}{(2\pi)^{3}} \frac{1}{e^{p/T} - 1}. \end{align*} This is an integral over 3D momentum space. Switch to spherical coordinates whose volume element is $d^{3}p = p^{2}\sin\theta \, dp\,d\theta\, d\phi = p^{2}dp\, d^{2}\Omega$ to get \begin{align*} &= 2\int \frac{p^{2} dp\, d^{2}\Omega}{(2\pi)^{3}} \frac{1}{e^{p/T} - 1} \\[1.2em] &= 2\int d^{2}\Omega \int_{0}^{\infty} \frac{p^{2} dp}{(2\pi)^{3}} \frac{1}{e^{p/T} - 1} \\[1.2em] &= 2\cdot 4\pi \int_{0}^{\infty} \frac{p^{2} dp}{(2\pi)^{3}} \frac{1}{e^{p/T} - 1} \\[1.2em] &= \frac{8\pi}{(2\pi)^{3}} \int_{0}^{\infty} dp\, \frac{p^{2}}{e^{p/T} - 1} \\[1.2em] &= \frac{8\pi}{(2\pi)^{3}} T^{3} \int_{0}^{\infty} dx\, \frac{x^{2}}{e^{x} - 1} \\[1.2em] &= \frac{T^{3}}{\pi^{2}} \cdot 2\zeta(3) \\[1.2em] &= \frac{2T^{3}}{\pi^{2}}\cdot \zeta(3). \end{align*}

We see that the new $n_{\gamma}$ has an extra factor of $\zeta(3)$ in it. In my calculation this extra factor of $\zeta(3)$ goes into the denominator. When I do this, WolframAlpha gives me $$ 2.712\times 10^{-8} $$ as expected.

I'm not confident in my work (I am self-studying and I have no guidance), so if anyone (who knows their stuff better than me) is reading over this, please tell me whether or not my work here is solid. Thank you.

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