when studying the equation of states in astrophysics I have stumbled upon a differential form of it, i.e.



$\chi_T=\frac{T}{P}\frac{\partial P}{\partial T}\bigg|_{\rho,X_i} $

$\chi_\rho=\frac{\rho}{P}\frac{\partial P}{\partial \rho}\bigg|_{T,X_i} $

$\rho$ - density

$X_i$ - an abundance of i's element

I would really use some hints on how to derive it with the help of the 1st law of thermodynamics.


1 Answer 1


You don't need the First Law of Thermodynamics at all. This is, in fact, just an application of basic multivariable calculus. The only assumption needed is that $P$ depends in some way on both $T$ and $\rho$, and that $P$ is differentiable with respect to $T$ and $\rho$.

So, if we have a function $P(T,\rho)$, we start by taking the differential:

$$dP=\frac{\partial P}{\partial T}\bigg|_{\rho,X_i} dT+\frac{\partial P}{\partial \rho}\bigg|_{T,X_i}d\rho$$

Then we divide by $P$:

$$\frac{dP}{P}=\frac{1}{P}\frac{\partial P}{\partial T}\bigg|_{\rho,X_i}dT+\frac{1}{P}\frac{\partial P}{\partial\rho}\bigg|_{T,X_i}d\rho$$

And then we multiply by $1=\frac{T}{T}=\frac{\rho}{\rho}$:

$$\frac{dP}{P}=\frac{T}{P}\frac{\partial P}{\partial T}\bigg|_{\rho,X_i}\frac{dT}{T}+\frac{\rho}{P}\frac{\partial P}{\partial\rho}\bigg|_{T,X_i}\frac{d\rho}{\rho}$$


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