# Boltzmann Equation-where'd the 3 come from?

Boltzmann Equation--where'd the 3 come from?

Greetings, I'm a little bit confused about the derivation for the Boltzmann equation for a particle in thermal equilibrium in the Friedman-Robertson-Walker metric. I've been following the exposition in Kolb and Turner, The Early Universe p. 116. I reproduce all the relevant results here.

In particular:
We are given (K&T, eq 5.5) that for a phase space distribution $$f$$ the form of the Liouville operator in the FRW model is given by:

$$\hat{\mathbf{L}}[f(E,t)] = E\frac{\partial f}{\partial t}-\frac{\dot{R}}{R}|\mathbf{p}|^2\frac{\partial f}{\partial E}$$

Further, the number density $$n$$ is given by an integral over momenta (K&T eq. 5.6):

$$n(t) = \frac{g}{(2\pi)^3}\int d^3p f(E,t)$$

where $$g$$ is the number of internal degrees of freedom.

The Boltzmann equation, $$\hat{\mathbf{L}}[f]= \mathbf{C}[f]$$, can then be written out by plugging in the above equation for the Liouville operator on the left hand side.

We can then divide by $$E$$, multiply by $$\frac{g}{(2\pi)^3}$$, and perform a momentum space integral to express the Boltzmann equation in terms of $$n$$.

Kolb and Turner write the result as:

$$\frac{dn}{dt} + 3\frac{\dot{R}}{R}n = \frac{g}{(2\pi)^3}\int\textbf{C}[f]\frac{d^3p}{E}$$

I'm confused by the factor of 3 in the seccond term and am not sure how this is resolved. I'm also not sure how to treat the energy in the momentum integral--I assume that since $$E^2=\mathbf{p}^2+m^2$$, one can rewrite the momentum integral in spherical coordinates where the function $$f$$ is a function of the radial coordinate alone. I assume some integration by parts is necessary, but this still does not account for the factor of 3.

Any help would be appreciated,

Best,
Flip

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Garth
Gold Member
fliptomato said:
I assume some integration by parts is necessary, but this still does not account for the factor of 3.
I haven't read Kolb and Turner's book, but I think you'll find the factor 3 and the rest of that term comes from:
$$\frac{1}{R^3}\frac{d}{dt}(R^3)n$$

Garth

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A bit late for the one who asked, but perhaps it helps those who have the same problem

$$E\frac{\partial f}{\partial t}-\frac{\dot{R}}{R}|\mathbf{p}|^2\frac{\partial f}{\partial E} = \hat{\mathbf{C}}[f(E,t)]$$
divide by E
$$\frac{\partial f}{\partial t}-\frac{\dot{R}}{R}|\mathbf{p}|^2\frac{\partial f}{E\partial E} = \hat{\mathbf{C}}[f(E,t)]\frac{1}{E}$$
integrate
$$\frac{g}{(2\pi)^3}\int d^3p\frac{\partial f}{\partial t}-\frac{\dot{R}}{R}|\mathbf{p}|^2\frac{\partial f}{E\partial E} = \frac{g}{(2\pi)^3}\int\hat{\mathbf{C}}[f(E,t)]\frac{d^3p}{E}$$
$$n(t)-\frac{\dot{R}}{R}\frac{g}{(2\pi)^3}\int d^3p|\mathbf{p}|^2\frac{\partial f}{E\partial E} = \frac{g}{(2\pi)^3}\int\hat{\mathbf{C}}[f(E,t)]\frac{d^3p}{E}$$
use
$$E\partial E=p\partial p$$
$$n(t)-\frac{\dot{R}}{R}\frac{g}{(2\pi)^3}\int 4\pi dp|\mathbf{p}|^2|\mathbf{p}|^2\frac{\partial f}{p\partial p} = \frac{g}{(2\pi)^3}\int\hat{\mathbf{C}}[f(E,t)]\frac{d^3p}{E}$$
$$n(t)-\frac{\dot{R}}{R}\frac{g}{(2\pi)^3}\int 4\pi dpp^3\frac{\partial f}{\partial p} = \frac{g}{(2\pi)^3}\int\hat{\mathbf{C}}[f(E,t)]\frac{d^3p}{E}$$
integration by part:
$$n(t)+\frac{\dot{R}}{R}\frac{g}{(2\pi)^3}\int 4\pi dp3p^2f= \frac{g}{(2\pi)^3}\int\hat{\mathbf{C}}[f(E,t)]\frac{d^3p}{E}$$
$$n(t)+3\frac{\dot{R}}{R}\frac{g}{(2\pi)^3}\int 4\pi dpp^2f= \frac{g}{(2\pi)^3}\int\hat{\mathbf{C}}[f(E,t)]\frac{d^3p}{E}$$
$$n(t)+3\frac{\dot{R}}{R}n(t)= \frac{g}{(2\pi)^3}\int\hat{\mathbf{C}}[f(E,t)]\frac{d^3p}{E}$$

Best,
stanix