How does one get from the thermodynamics definition :

$$ \chi_T = -\frac{1}{V} \left(\frac{\partial V}{\partial p}\right)_T$$

to the fluid dynamics definition :

$$ \chi_T = \frac{1}{\rho} \left(\frac{\partial \rho}{\partial p}\right)_T$$


  • 2
    $\begingroup$ Have you thought about what you could do to the equation to turn all your $V$'s into $\rho$'s? $\endgroup$
    – tpg2114
    Nov 14, 2014 at 23:41
  • 1
    $\begingroup$ I have, and I didn't find the answer to that (that's why I'm asking) $\endgroup$
    – mwa1
    Nov 15, 2014 at 0:11
  • 1
    $\begingroup$ Have you considered using an ideal gas and proving the relation is true? $\endgroup$
    – Kyle Kanos
    Nov 15, 2014 at 0:32
  • 1
    $\begingroup$ Do you know the relationship between mass, density and volume? $\endgroup$
    – tpg2114
    Nov 15, 2014 at 0:53
  • $\begingroup$ Well I know that, dimensionally speaking, density is a mass by unit volume. But I get confused because density used to be defined as $\rho = \frac{m}{V}$ and now in Fluid Mechanics I often see $\rho = \frac{dm}{d^3r}$ so I'm not sure how I'm supposed to manipulate this. $\endgroup$
    – mwa1
    Nov 15, 2014 at 1:08

1 Answer 1


If anyone is looking for the same thing, here is the solution :

$$\begin{align}m &= \rho V = \textrm{constant}\\ \Leftrightarrow~~~~~~~~~~~~~~~~ \rho~\mathrm dV + V~\mathrm d\rho &= 0\\\Leftrightarrow~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \frac{\mathrm d\rho}{\rho} &= -~ \frac{\mathrm dV}{V}\\\Leftrightarrow~~~~ \chi_T = - ~\frac{1}{V} \left(\frac{\partial V}{\partial p}\right)_T &= \frac{1}{\rho} \left(\frac{\partial \rho}{\partial p}\right)_T\end{align}$$

It's simple but not obvious if you don't know where to start...


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