I'm studying kinetic theory by KineticTheory MichaelmasTerm2012 [1].

Let me bring some context in here.

The author have introduce the Lioivelle equation for the distribution $f(q_1,...,q_N,p_1,...,p_N,t)$ of N particle statistical system out of thermodynamic equilibrium: $$\frac{\partial f}{\partial t}=\{H,f\}$$

Then we learn the way of obtaining one-particle-distribution: $$f_1(q,p,t)=N\int\prod^{N}_{n=2}dq_ndp_nf(q,..., q_N,p,...,p_N,t)$$

After that we proceed to derive the equation governing one-particle-distribution by substituting the above integral in the Lioivelle equation:

$$\frac{\partial f_1}{\partial t}=\int\prod^{N}_{n=2}dq_ndp_n\{H,f\}=\int\prod^{N}_{n=2}dq_ndp_n\sum^N_k\left[\frac{\partial f}{\partial p_k}\frac{\partial H}{\partial q_k}-\frac{\partial f}{\partial q_k}\frac{\partial H}{\partial p_k}\right]$$

We should take into consideration hamiltonian of N-particle mechanical system action under force field $\bar{F}=\nabla V$ interacting with each other $$H=\frac{1}{2m}\sum_n(p^2_n) +\sum_n V(q_n)+\sum_{n<k}U(\lVert q_n-q_K\rVert) $$.

With this the equation becomes $$\frac{\partial f_1}{\partial t}=\int\prod^{N}_{n=2}dq_ndp_n\sum^N_k\left[\frac{\partial f}{\partial p_k}\left(\frac{\partial V(q_k)}{\partial q_k} + \sum_{i<j}\frac{\partial U(\lVert q_i - q_j \rVert)}{\partial q_k}\right) - \frac{\partial f}{\partial q_k}\frac{p_k}{m}\right]$$

And I have hard times understanding the following step: the author claims to do the integration by parts on each $dq_ndp_n$ pair which vanishes the outter sum (remains the only term we cannot integrate upon) and reduces the equation to the following: $$\frac{\partial f_1}{\partial t}=\int\prod^{N}_{n=2}dq_ndp_n\left[\frac{\partial f}{\partial p_1}\left(\frac{\partial V(q_1)}{\partial q_1} + \sum_{k=2}^N\frac{\partial U(\lVert q_1 - q_j \rVert)}{\partial q_1}\right) - \frac{\partial f}{\partial q_1}\frac{p_1}{m}\right]$$ I don't get the technique the author used in integrating by parts, I haven't succeed in straightforward calculations, so I definitely somethin missing.


1 Answer 1


Let's just do $N=2$ as an example that will illustrate the general idea.

Let's write $$ \frac{\partial f_1}{\partial t} = \int dq_2 dp_2 \left[ \frac{\partial f}{\partial p_1} F_1(q_1) + \frac{\partial f}{\partial p_2} F_2(q_2) + \frac{\partial f}{\partial q_1}G_1(p_1) + \frac{\partial f}{\partial q_2}G_2(p_2) \right] $$ where $$ F_k(q_k) \equiv \frac{\partial V}{\partial q_k} + \sum_{i < j} \frac{\partial U(\|q_i - q_j\|)}{\partial q_k} $$ and $$ G_k(p_k) \equiv -\frac{p_k}{m} $$ Now, let's consider the two terms we can integrate by parts. First, we have the term with the $p_2$ derivative: $$ \frac{\partial f_1}{\partial t} \supset \int dq_2 dp_2 \frac{\partial f}{\partial p_2} F_2(q_2) = {\rm B.T.} - \int dq_2 dp_2\ f \frac{\partial F_2(q_2)}{\partial p_2} $$ where $\supset$ means "the left hand side can be written as a sum of terms including the one on the right hand side", and "B.T." means "boundary term".

Now, $$ \frac{\partial F_2(q_2)}{\partial p_2} = 0 $$ since $p_2$ and $q_2$ are independent coordinates on phase space, therefore the second term above vanishes. Meanwhile, the boundary term vanishes because in order to be normalizable, the distribution $f$ must die off at asymptotically at large values of $p_2, q_2$. As a result, the entire right hand side of the $\supset$ symbol above is simply zero.

By a similar argument, the term involving $\frac{\partial f}{\partial q_2} G_2(p_2)$ vanishes.

Therefore, we are left with the expression you ended up with: $$ \frac{\partial f_1}{\partial t} = \int dq_2 dp_2 \left[\frac{\partial f}{\partial p_1} F_1(q_1) + \frac{\partial f}{\partial q_1}G_1(p_1)\right] $$

This argument can be trivially generalized to arbitrary $N$.

  • $\begingroup$ Yeah, that's it! The clue I was looking for: the distribution should vanish at boundary in order to be normalaizable. I had similar thought, but I must have finded external proof in order to be sure $\endgroup$ Commented Oct 5, 2023 at 22:46

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