I'm reading this text for Boltzmann's equation derivation, which works with a 6N dimensional system, being N the number of particles of the system (each particle has a 3D position vector r and a 3D momentum vector p). At Eq. 2.10 (page 5 of pdf https://www.damtp.cam.ac.uk/user/tong/kintheory/two.pdf) we have the solution of the previous integral, but I don't see this step. Let's take only the first term, so we come from:

$$-N\int\prod_{i=2}^Nd^3r_id^3p_i\sum^N_{j=1}\frac{\vec{p}_j}{m}\cdot\frac{\partial f}{\partial \vec{r}_j}$$

Here f depends on the 6N dimensions. The text says that all the summs except for j=1 are zero, because if you integrate by parts the derivatives which depend on r are with respect p and vice-versa, but I cannot see it. For example, let's take j=2:

$$-N\int\prod_{i=2}^Nd^3r_id^3p_i\frac{\vec{p}_2}{m}\cdot\frac{\partial f}{\partial \vec{r}_2}=-N\int\prod_{i=3}^Nd^3r_id^3p_i\int d^3p_2\int d^3r_2\frac{\vec{p}_2}{m}\cdot\frac{\partial f}{\partial \vec{r}_2}$$

If we do by parts the last integrand we have:

$$\int d^3r_2\frac{\vec{p}_2}{m}\cdot\frac{\partial f}{\partial \vec{r}_2}=\frac{\vec{p}_2}{m}\cdot f-\int d^3r_2\frac{1}{m}\frac{d^3\vec{p}_2}{dr^3}\cdot f$$

The second term is zero. However, why the first term becomes zero?

  • 2
    $\begingroup$ Limits of integration? $\endgroup$ Mar 25, 2021 at 5:39

1 Answer 1


At the end I got an answer. f is a probability distribution and to be normalizable it must be 0 at infinity. Hence that term is also zero.


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