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In electron liquids, the compressibility $K$ is defined as $\frac{1}{K}=-V\left(\frac{\partial P}{\partial V}\right)_N=n^2\frac{\partial \mu}{\partial n}$, where $P$, $V$, $n$ and $\mu$ are pressure, volume, density and chemical potential. However, in thermodynamics we learned, I can only find the definition of isothermal and isentropic compressibility: $$\kappa_T=-\frac{1}{V}\left(\frac{\partial V}{\partial P} \right)_T$$ $$\kappa_S=-\frac{1}{V}\left(\frac{\partial V}{\partial P} \right)_S$$ which have the relation $\kappa_T/\kappa_S=\gamma$, where $\gamma$ is the Heat capacity ratio $c_P/c_V$.

My question is

  1. What is the relation between compressibility defined in electron liquid and that defined in thermodynamics?

  2. If I want to get the compressibility of electron liquid in thermodynamics' view, what should I have? the free energy? the internal energy? ...

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  • $\begingroup$ It's always true that $-V\left(\frac{\partial P}{\partial V}\right)_N=n^2\left(\frac{\partial\mu}{\partial n}\right)_N$ (you can show this by applying the triple product rule, than a Maxwell relation). However, it's ambiguous whether this compressibility is defined at constant $T$ or constant $S$. Is it possible that in the context of electron liquids, it's understood that the temperature is constant and zero? In addition, your definitions of $\kappa_T$ and $\kappa_S$ need to specify constant $N$. If you insert these missing constant parameters, the definitions in the two fields are identical. $\endgroup$
    – Chemomechanics
    Commented Mar 16, 2017 at 5:26
  • $\begingroup$ Note that the above comment assumes that $N=nV$. $\endgroup$
    – Chemomechanics
    Commented Mar 16, 2017 at 5:28

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