# derivative of a sum - derivation of Boltzman equation

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.