We know that $c^2=\frac{\partial p}{\partial ρ}$

The adiabatic compressibility is defined as: $\beta_S=-\frac{1}{V}\frac{\partial V}{\partial p}$ such that the subscript "S" stands for "adiabatic"

How can I show that $c^2=\frac{1}{\rho \beta_S}$ ?

I tried replacing $V$ by $\frac{m}{\rho}$ but I get for $\beta_S=-\rho \frac{\partial \frac{1}{\rho}}{\partial p}$


1 Answer 1


With your result

\begin{equation} \beta_S=-\rho \frac{\partial \frac{1}{\rho}}{\partial p} \end{equation}

you're nearly there. All that is missing is a little mathematical trick. Simply make the substitution $f = \frac{1}{\rho}$ and use the chain rule to differentiate $f$ by $\rho$ and you see that:

\begin{equation} \frac{\partial f}{\partial \rho} = - \frac{1}{\rho^2} \quad \Rightarrow \quad \mathrm{d} f = - \frac{1}{\rho^2} \mathrm{d} \rho \quad \Rightarrow \quad \mathrm{d} \frac{1}{\rho} = - \frac{1}{\rho^2} \mathrm{d} \rho \ . \end{equation}

Using this and the definition of $c^2$ you get to the desired result:

\begin{equation} \beta_S = -\rho \underbrace{\frac{\partial \frac{1}{\rho}}{\partial p}}_{= - \frac{\frac{1}{\rho^2} \partial \rho}{\partial p}} = \underbrace{(-\rho) (- \frac{1}{\rho^2})}_{= \frac{1}{\rho} } \underbrace{\frac{\partial \rho}{\partial p}}_{= \frac{1}{c^2}} \quad \Rightarrow \quad c^2 = \frac{1}{\rho \beta_S} \ . \end{equation}


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