Adiabatic compressibility always positive I am currently studying statistical physics. Intuitively it is clear to me that the adiabatic compressibility $\kappa_s = - \frac{1}{V} \left(\frac{\partial V}{\partial P}\right)_{S}$ is positive in a stable system.
Now I am trying to "prove" this. In order to do so, I tried using the internal energy: $$dE = T dS - p dV + \mu dN$$
and found: $$ \left(\frac{\partial E}{\partial V}\right)_{S,N} = -P \mathrm{\,and \, \left( \frac{\partial P}{\partial V}\right)_{S,N} = - \left(\frac{\partial^2 E}{\partial V^2}\right)_{S,N}} $$
inverting then gives me: $$ \left(\frac{\partial V}{\partial P}\right)_{S} = - \left[\left(\frac{\partial^2 E}{\partial V^2}\right)_{S,N}\right]^{-1} $$
Therefore: $$ \kappa_S =  \frac{1}{V} \left[\left(\frac{\partial^2 E}{\partial V^2}\right)_{S,N}\right]^{-1} $$
The last step would obviously be to conclude that $ \left[\left(\frac{\partial^2 E}{\partial V^2}\right)_{S,N}\right]^{-1} $ is always positive. I can not find a argument supporting this.
 A: This follows from the general convexity nature of the internal energy $E$ that can be expressed as the stability condition:
$$ \delta^2 E = \sum_{k=0}^{K} \delta Y_k \delta X_k \ge 0 \tag{1}\label{1}$$ where $X_k$ and $Y_k=\frac {\partial E}{\partial  X_k}$ are the extensive and conjugate intensives satisfying $E = \sum_{k=0}^{K} Y_k X_k$ with $k=0$:$X_0=S, \quad Y_0=T$. For a simple 2-variable system $K=1$ and $X_1=V, \quad Y_1=-p$.
Now using the definition $Y_k=\frac{\partial E}{X_k}$ and expand $\delta Y_k $ as
$$\delta Y_k = \sum_j \frac {\partial}{\partial X_j} \left( \frac {\partial E}{\partial X_k} \right) \delta X_j \tag{2}\label{2}$$
Next substitute this expansion $\eqref{2}$ in to $\eqref{1}$:
$$\delta^2 E = \sum_{k=0}^{K} \delta Y_k \delta X_k 
= \sum_{k=0}^{N} \sum_{j=0}^K   \frac {\partial^2 E}{\partial X_j \partial X_k}  \delta X_j \delta X_k \ge 0 \tag{3}\label{3}$$
In other words the matrix $\left[\frac{\partial^2 E}{\partial X_j \partial X_k}\right]$ is positive semi-definite. A necessary and sufficient condition for a matrix be positive semi-definite is that the determinants of its principal minors be non-negative.
For the 3-variable case $K=2$ this means that
$$\frac{\partial^2 E}{\partial X_0^2} \ge 0; \quad \frac{\partial^2 E}{\partial X_1^2}\ge 0; \quad \frac{\partial^2 E}{\partial X_2^2}\ge 0\\
\frac{\partial^2 E}{\partial X_0^2}\frac{\partial^2 E}{\partial X_1^2} - \left(\frac{\partial^2 E}{\partial X_0 \partial X_1}\right)^2\ge0, etc. \tag{4}\label{4}$$
Substitute $S=X_0;V=X_1;N=X_2,etc.,$ for $\delta^2 E =\delta T \delta S -\delta p \delta V +\delta \mu \delta N \ge 0\\$ and then you get, for example, $$\frac{\partial^2 E}{\partial V^2} \ge 0$$
A: My approach to this is totally different.  As  soon as I see the subscript S in the equation for $K_s$, indicating constant entropy, I immediately write $$dS=\frac{C_p}{T}dT-\left(\frac{\partial V}{\partial T}\right)_PdP=0$$So, at constant S, $$dT=\frac{T}{C_p}\left(\frac{\partial V}{\partial T}\right)_PdP\tag{1}$$The general equation for volume change in terms of  dT and dP is $$d\ln{V}=-K_TdP+\alpha dT\tag{2}$$where $K_T$ is the isothermal compressibility and $\alpha$ is the coefficient of volumetric thermal expansion.  So, combining Eqns. 1 and 2, we have for constant entropy. $$d\ln{V}=-K_TdP+\frac{VT}{C_p}\alpha^2dP=-\left[K_T-\frac{VT}{C_p}\alpha^2\right]dP$$From this it follows that $$K_S=\left[K_T-\frac{VT}{C_p}\alpha^2\right]$$We also know that$$C_p-C_v=T\left(\frac{V}{T}\right)_P\left(\frac{P}{T}\right)_V=\frac{VT}{K_TC_p}\alpha^2$$Finally, if we combine the previous two equations, we obtain:$$K_S=\frac{C_V}{C_P}K_T$$
