A thermodynamic inequality (from minimum work)

1. Apr 4, 2012

mSSM

EDIT: Turns out, the solution to my question is related to the determinant of a positive definite quadratic form.

This is more or less straight from Landau's Statistical Physics Part 1 (3rd edition), Chapter 21.

I don't understand how the inequality/condition (the last equation in this post) arises. I think it's a mathematical problem, but I wonder if I have ignored some physical reasoning.

The minimum work, which must be done to bring a body in an external medium from its equilibrium state to any neighbouring state:
$$\delta E - T_0 \delta S + P_0 \delta V > 0$$
where $T_0$ and $P_0$ are (constant) temperature and pressure of the external medium, and where for the temperature and the pressure of the body in equilibrium with the external medium we would have: $T=T_0$, and $P=P_0$

Regarding the energy as $E=E(S,V)$, and expanding $\delta E$ as a series:
$$\delta E = \frac{\partial E}{\partial S}\delta S + \frac{\partial E}{\partial V}\delta V + \frac{1}{2} \left[ \frac{\partial^2 E}{\partial S^2} (\delta S)^2+ 2 \frac{\partial^2 E}{\partial S\partial V} \delta S\delta V + \frac{\partial^2 E}{\partial V^2} (\delta V)^2 \right]$$

Note that $\frac{\partial E}{\partial S}=T$ and $\frac{\partial E}{\partial V}=P$. Inserting this into the first equation gives the inequality:
$$\frac{\partial^2 E}{\partial S^2} (\delta S)^2 + 2 \frac{\partial^2 E}{\partial S\partial V} \delta S\delta V+ \frac{\partial^2 E}{\partial V^2} (\delta V)^2> 0$$

If we want this inequality to be true for any values of $\delta S$ and $\delta V$, the following conditions have to hold:
$$\frac{\partial^2 E}{\partial S^2} > 0$$

$$\frac{\partial^2 E}{\partial S^2}\frac{\partial^2 E}{\partial V^2} - \left( \frac{\partial^2 E}{\partial S\partial V} \right)^2 > 0$$

Where does the last condition come from?

Last edited: Apr 4, 2012