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The work function and mutual forces between particles

  • Thread starter dRic2
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dRic2

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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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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
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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).
 

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