# The work function and mutual forces between particles

#### dRic2

Gold Member
Problem Statement
Let us assume that the work function of an assembly of free particles, subject to mutual forces between these particles, depends only on the relative coordinates
$$\xi_{ik} = x_i - x_k$$
$$\eta_{ik} = y_i - y_k$$
$$\zeta_{ik} = z_i - z_k$$
of any given particle $P_i$ and $P_k$:
$$U = U(\xi_{ik}, \eta_{ik}, \zeta_{ik})$$
Let the coordinate $x_i$ be varied by $\delta x_i$, thus obtaining the x-component of the force acting at $P_i$. Show that the quantity:
$$X_{ik} = \frac {\partial U}{\partial \xi_{ik}}$$
can be interpreted as the x-component of the force on $P_i$ due to $P_k$.
Relevant Equations
.
I tried to apply the chain rule

$$X_{ik} = \frac {\partial U}{\partial \xi_{ik}} = \frac {\partial U}{\partial x_{i}} \frac {\partial x_i}{\partial \xi_{ik}} = \frac {\partial U}{\partial x_{i}}$$

and I got the force x-component of the force acting on $P_i$ I guess.

but I do not know what to conclude from this...

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#### Fred Wright

For definiteness consider a gas of electrons subject (obviously) to Coulomb potential. For N electrons you can show that:$$U_E=\frac {1}{8\pi \epsilon_0}\sum_{i=1}^{N} q_i \sum_{j=1}^{ ,N(i\neq j )}\frac{q_j}{\sqrt{\xi_{ij}^2 + \eta_{ij}^2+ \zeta_{ij}^2 }}$$
Will differentiation w.r.t xi yield units of force? I leave it to you to prove the above equation.

#### dRic2

Gold Member
Thanks for the reply, but I solved it by considering the frame of reference moving with, for example, the particle $P_k$. It this manner $\delta \xi_{ij} = \delta x_i$ (because $\delta x_j = 0$) and so it is obvious that the only work done after the displacement $\delta x_i$ comes from the force on $P_i$ due to $P_k$ (because all the other displacements vanish and thus do not contribute).

"The work function and mutual forces between particles"

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