In Daniel Baumann's cosmology book Eq. (3.10) and cosmology lecture notes Eq. (3.2.18) he states that the pressure in the early universe can be defined as $$P=\frac g{(2\pi)^3}\int d^3 p\ f(p)\times\color{red}{\frac{p^2}{3E}}\tag{3.2.18}$$ Where
- $g$ is the internal degrees of freedom
- $f(\mathbf{x,p},t)$ is the phase space distribution function which reduces to $f(p)$ due to assuming homogeneity and isotropy and suppressing the time dependence.
- $E(p) = \sqrt{p^2+m^2}$
The red term is the one I'm having trouble understanding. In the lecture notes he follows this equation with a box and figure (reproduced here) outlining where this term comes from. His argument goes like this:
- Suppose we are in a box volume $V=L^3$ and consider a small area element of size $dA$ of this box with unit normal vector $\hat{\mathbf n}$
- Any particle with velocity $|v|$ that hits this $dA$ in time interval $[t,t+dt]$ must have been in a spherical shell with radius $R=|\mathbf{v}|t$ and thickness $|\mathbf{v}|dt$
- A volume element of this spherical shell $dV$ spanned by a solid angle $d\Omega^2$ can be written as $$dV = {R^2 d\Omega^2}\times |\mathbf{v}|dt$$ which roughly can be understood as $dV=$Area on surface of shell spanend by $d\Omega^2$ $\times$ thickness of the shell
- Multiplying $dV$ by the phase space density $g / (2\pi)^3\times f(E(|\mathbf{v}|))$ gives us $dN$, the number of particles in a small element of the spherical shell with energy $E(|\mathbf{v}|)$ per unit volume of momentum space. $$dN = \frac{g}{(2\pi)^3}\times f(E)\times R^2|\mathbf{v}|dt\ d\Omega^2\tag{3.2.19}$$
- (This is the part that confuses me so I'll quote exactly what he says here): Not all particles in $dV$ reach the target, only those with velocities directed to the area element. Taking into account the isotropy of the velocity distribution, we find that the total number of particles striking the area element $dA \hat{\mathbf{n}}$ with velocity $\mathbf{v}=|\mathbf{v}|\hat{\mathbf{v}}$ is $$dN_A = {\color{red}{\frac{|\hat{\mathbf{v}}\cdot \hat{\mathbf{n}}|dA}{4\pi R^2}}}\times dN = \frac{g}{(2\pi)^3}f(E)\times \frac{|\mathbf{v}\cdot\hat{\mathbf{n}}|}{4\pi}dA\ dt\ d\Omega\tag{3.2.20}$$
- From here we recall that pressure can be thought of as change of momentum per unit area per unit time. If a particle bounces elastically off the wall then it transfers $2|\mathbf{p}\cdot \hat{\mathbf{n}}|$ momentum to the wall. So the contribution to the pressure $dP$ by some particles $dN_A$ is $$\begin{align} dP(|\mathbf{v}|) &= \int \frac{2|\mathbf{p}\cdot\hat{\mathbf{n}}|}{dA\times dt}dN_A\\ &=\frac{g}{(2\pi)^3}f(E)\times \frac{1}{2\pi}\int |\mathbf{p}||\mathbf{v}| |\hat{\mathbf{v}}\cdot \hat{\mathbf{n}}|^2d\Omega^2\\ (\textrm{using }|\mathbf{v}|=|\mathbf{p}|/E)&=\frac{g}{(2\pi)^3}f(E)\times \frac{p^2}{2\pi E}\int_0^{\color{red}\pi}\int_0^\pi \cos^2\theta \times \sin\theta\ d\theta\ d\phi\\ &= \boxed{\frac{g}{(2\pi)^3}f(E)\times \frac{p^2}{3E}} \end{align}$$ where the red integration limit comes from integrating over the part of the hemisphere of the spherical shell inside the box
Step 5 is the part I'm having trouble understanding and specifically the red term in Eq. (3.2.20). $${\color{red} {|\hat{\mathbf{v}}\cdot \hat{\mathbf{n}}|\times \frac{dA}{4\pi R^2}}}$$ The way I'm trying to understand this term is like this. Imagine you are a particle in the spherical shell of radius $R$ (e.g. the shaded region in the figure). After some time $t$ I could be anywhere on a sphere of radius $R$ centered on my current location. The $dA / 4 \pi R^2$ factor is the probability that the direction I randomly decide to go in will land me inside the area element $dA$. But from this context I don't understand: why there is an $|\hat{\mathbf{v}}\cdot \hat{\mathbf{n}}|$ factor flying around?
