Let's, for the sake of argument, agree that the thermodynamic relation for equilibrium in the form of zero change of the Helmholtz potential, is applicable for particles suspended in a liquid,
$$\delta F = \delta E - T \delta S = 0, \ \ \ \ \ \ \ \ \ \ (1)$$
where $F$ is the Helmholtz potential, $E$ is the energy, $T$ is the absolute temperature and $S$ is the entropy. Then, the following integral for $\delta E$ is written
$$\delta E = -\int_0^l K \nu \delta x dx \ \ \ \ \ \ \ \ \ \ (2)$$
where $K$ is the force per unit particle, acting only along the $x$-axis, $\nu$ is the number of particles per unit volume, $\delta x$ is a variation of $x$, $\delta x$ being also a function of $x$. The liquid is bounded by the planes $x = 0$ and $x = 1$ of unit area.
Now, it can be intuitively seen how, because $K$ is a force on a unit particle, that force has to be multiplied by the number of particles residing on a plane perpendicular to the $x$-axis and that has led to the expression for $\delta E$ in eq.(2)
Then, the variation of entropy, $\delta S$, is defined as
$$\delta S = \int_0^l k \nu \frac{\partial\delta x}{\partial x} dx, \ \ \ \ \ \ \ \ \ \ (3)$$
where $k$ is Boltzmann's constant.
Now, from eq.(2) and eq.(3), using eq.(1) we get
$$-\int_0^l K \nu \delta x dx = T\int_0^l k \nu \frac{\partial\delta x}{\partial x} dx. \ \ \ \ \ \ \ \ \ \ (4)$$
or, if the integrands are continuous functions
$$-K \delta x = kT \frac{\partial \delta x}{\partial x}. \ \ \ \ \ \ \ \ \ \ (5)$$
It seems to me that l.h.s of eq.(5) is not equal to its r.h.s. -- on the l.h.s. we have work with negative sign, while on the r.h.s. we have the expression of energy for two degrees of freedom, according to the equipartition theorem, multiplied by a positive non-zero factor. What do you think about this? Am I right or am I missing something?
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