Let $x_i$ is a position vector (for simplicity in 1D) of an $i$-th particle. $V(x_i,x_j)=\phi(|x_i - x_j|)$ is some function that depdends only on the distance between the two particles. I would like to show the following:

$$ \frac{\partial}{\partial x_k} \sum_{1\leq i < j \leq N} V(x_i,x_j)=\sum_{j\neq i} \frac{\partial V(x_i,x_j)}{\partial x_i} $$

when I try that and explicitly write out the sums I only get to: $$ \frac{\partial}{\partial x_k} \sum_{1\leq i < j \leq N} V(x_i,x_j)=\sum_{j=2}^N\sum_{i=1}^{j-1} \left( \delta_{ik} \frac{\partial V(x_i,x_j)}{\partial x_k} + \delta_{jk} \frac{\partial V(x_i,x_j) }{\partial x_k}\right) $$

And i am stuck with this. I don't see how to turn these sums in the the single one. Specifically there is the issue of the first sum starting from 2 and the other is related to the second term which, after performing the derivative, will carry opposite sign to the first one.

I think, i am doing something very dumb but i just cannot see what.

Note: this is problem arises when one tries to derive Boltzman kinetic equation using the BBGKY hierarchy, $V$ is the potential energy of particle-particle interaction and the above needs to be evaluated when one wants to get momentum from the Hamiltonian.

To be more explicit, starting from the Liouville's equation:

$$ \frac{\partial f_N }{\partial t} + \sum_{i=1}^N \left(\frac{\partial f_N}{\partial x_i}\frac{\partial H_N}{\partial p_i} - \frac{\partial f_N}{\partial p_i}\frac{\partial H_N}{\partial x_i}\right) = 0 $$

with $H_N=\sum_{i=1}^N \frac{p_i^2}{2m} + \Phi(x_1,x_2,\dots,x_N,t)$, $\Phi(x_1,\ldots,x_N,t)=\sum_{1\leq i < j \leq N} V(x_i,x_j)$ and $f_N(x_1,p_1,\ldots,x_N,p_N)$ is the probability distribution function over phase space.

Reduced distribution function is introduced as: $$ f_s(x_1,p_1,\ldots,x_s,p_s) = A_s\int f_N dx_{s+1}dp_{s+1}\cdots dx_Ndp_N $$ where the integral is over the volume of phase space.

And it is to be shown under the assumptions that $f_N$ vanishes on the border of phase space and that $f_N$ is a symmetric function of the coordinates of particles and $V$ is the volume of the configuration space, that with this the Liouville equation for the reduced distribution function can become:

$$ \frac{\partial f_s}{\partial t} + \sum_{i=1}^N\frac{p_i}{m}\frac{\partial f_s}{\partial x_i} - \sum_{i=1}^s\sum_{j=1,j\neq i}^s\frac{\partial V(x_i,x_j)}{\partial x_i}\frac{\partial f_s}{\partial p_i}-\frac{N-s}{V}A_s\sum_{i=1}^s\int \frac{\partial V(x_i,x_{s+1})}{\partial x_i} \frac{f_{s+1}}{\partial p_i} dx_{s+1}dp_{s+1}=0 $$

The first two terms follow immediatelly after integrating the Liouville equation for $f_N$. The first term follows immeditally, the second term is the consequence of $\sum_{i=1}^N \int\left(\frac{\partial f_N}{\partial x_i}\frac{\partial H_N}{\partial p_i}\right)dx_{s+1}dp_{s+1}\cdots dx_Ndp_N$ which is written as $\sum_{i=1}^s+\sum_{i=s+1}^N$, using perpartes on the second one results in zero using the assumption that $f_N$ vanishes on the border, the $\sum_{i=1}^s$ sum then immediatelly gives the second term in the eq. for $f_s$.

Splitting the sum:

$\sum_{i=1}^N \int \left(\frac{\partial f_N}{\partial p_i}\frac{\partial H_N}{\partial x_i}\right)dx_{s+1}dp_{s+1}\cdots dx_Ndp_N$

in the same manner, the $\sum_{i=s+1}^N$ one should be again zero as above and one is left only with the term:

$\sum_{i=1}^s\sum_{1\leq j < k \leq N} \int \left(\frac{\partial f_N}{\partial p_i}\frac{\partial V(x_j,x_k)}{\partial x_i}\right)dx_{s+1}dp_{s+1}\cdots dx_Ndp_N$

where already $\frac{\partial H_N}{\partial x_i}=\sum_{1\leq j < k \leq N}\frac{\partial V(x_j,x_k)}{\partial x_i}$ is used. This can be written explicitly as:

$$ \sum_{i=1}^s\sum_{k=2}^N\sum_{j=1}^{k-1} \int \left(\frac{\partial f_N}{\partial p_i}\frac{\partial V(x_j,x_k)}{\partial x_i}\right)dx_{s+1}dp_{s+1}\cdots dx_Ndp_N=\\ \sum_{i=1}^s\sum_{k=2}^s\sum_{j=1}^{k-1} \int \left(\frac{\partial f_N}{\partial p_i}\frac{\partial V(x_j,x_k)}{\partial x_i}\right)dx_{s+1}dp_{s+1}\cdots dx_Ndp_N + \\ \sum_{i=1}^s\sum_{k=s+1}^N\sum_{j=1}^{k-1} \int \left(\frac{\partial f_N}{\partial p_i}\frac{\partial V(x_j,x_k)}{\partial x_i}\right)dx_{s+1}dp_{s+1}\cdots dx_Ndp_N = \\ \sum_{i=1}^s\sum_{k=2}^s\sum_{j=1}^{k-1} \frac{\partial V(x_j,x_k)}{\partial x_i}\frac{\partial}{\partial p_i}\int f_N dx_{s+1}dp_{s+1}\cdots dx_Ndp_N + \\ \sum_{i=1}^s\sum_{k=s+1}^N\sum_{j=1}^{k-1} \int \left(\frac{\partial f_N}{\partial p_i}\frac{\partial V(x_j,x_k)}{\partial x_i}\right)dx_{s+1}dp_{s+1}\cdots dx_Ndp_N =\\ \sum_{i=1}^s\sum_{k=2}^s\sum_{j=1}^{k-1} \frac{\partial V(x_j,x_k)}{\partial x_i}\frac{\partial f_s}{\partial p_i} + \sum_{i=1}^s\sum_{k=s+1}^N\sum_{j=1}^{k-1} \int \left(\frac{\partial f_N}{\partial p_i}\frac{\partial V(x_j,x_k)}{\partial x_i}\right)dx_{s+1}dp_{s+1}\cdots dx_Ndp_N $$

which then leads to the last two terms in the equation for $f_s$ but from here on I cannot get anywhere with either term.

In every literature, I have looked the transition from $f_N$ to $f_s$ is skipped as if absolutely obvious.


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