In the homogeneous and isotropic FRW Universe, the collisionless Boltzmann equation is given by $$ E\frac{\partial f}{ \partial t}-\frac{\dot{a}}{a}|\textbf{p}|^2\frac{\partial f}{\partial E}=0\tag{1}$$ where the phase space distribution $f$ is a function of the particle's energy $E$ and time $t$. See page 116 of The Early Universe by E. Kolb and M. Turner. The number density of the particle species at any time is given by $$n(t)=\frac{g}{(2\pi)^3}~\int d^3\textbf{p} f(E,t).\tag{2}$$ In Kolb and Turner's book, it is mentioned that by using (2) and doing an integration by parts, Eq.(1) can be reduced to $$\frac{dn}{dt}+3\frac{\dot{a}}{a}n=0.\tag{3}$$

How do we derive Eq.(3)?

My attempt

Taking derivatives of (2) w.r.t $t$ and $E$, we find, $$\frac{dn}{dt}=\frac{g}{(2\pi)^3}\int d^3\textbf{p}~\frac{\partial f}{\partial t},\tag{4}$$ $$\frac{dn}{dE}=\frac{g}{(2\pi)^3}\int d^3\textbf{p}~\frac{\partial f}{\partial E}\tag{5}.$$ Eq.(4) trivially gives the first term of Eq.(3) upon integration by $d\Pi=\frac{g}{(2\pi)^3}\frac{d^3\textbf{p}}{2E}$. But the second term becomes $$-\frac{\dot{a}}{a}\int \frac{|\textbf{p}|^2}{2E}\frac{\partial f}{\partial E}\frac{g~d^3\textbf{p}}{(2\pi)^3}.$$

Any help?

  • $\begingroup$ What's the dependence of $E$ on $\mathbf p$? $\endgroup$
    – John Donne
    Nov 11, 2018 at 22:06
  • $\begingroup$ @JohnDonne It's the relativistic dispersion relation $E=\sqrt{\textbf{p}^2+m^2}$. $\endgroup$
    – SRS
    Nov 11, 2018 at 22:17

1 Answer 1


Rewrite $(1)$ as $$\frac{\partial f}{ \partial t}-\frac{\dot{a}}{a}\frac{|\textbf{p}|^2}{E}\frac{\partial f}{\partial E}=0\tag{1}$$ Now integrate with respect to $d^3\textbf{p}$: $$\frac{dn}{dt}-\frac{\dot{a}}{a}\frac{g}{(2\pi)^3} \int d^3\textbf{p}\frac{|\textbf{p}|^2}{E}\frac{\partial f}{\partial E}=0$$ We must express the second integral in terms of $n$. The integrand is rotationally symmetric so $$\int d^3\textbf{p}\frac{|\textbf{p}|^2}{E}\frac{\partial f}{\partial E}=4\pi\int d\mathrm{p} \frac{p^4}{E}\frac{\partial f}{\partial E}=4\pi\int d\mathrm{p}\, p^3\frac{\partial f}{\partial p}\\ $$ where in the last equality we used the chain rule with the dispersion relation $E = \sqrt{p^2+m^2}$. Now integrating by parts: $$4\pi\int d\mathrm{p}\, p^3\frac{\partial f}{\partial p} = -3 \times 4\pi\int d\mathrm{p}\, p^2 f = -3 \int d^3\textbf{p} f = -3n \frac{(2\pi)^3}{g}$$ Putting this back in you get $(2)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.