Energy equation

The energy equation is the third MHD equation. In its most succinct form it is written:

$\frac{d}{dt}\left(\frac{p}{\rho^{\gamma}}\right)=0$

[edit] Heuristic derivation

Consider a volume with no heat transfer, that's adiabatic. Then:

$d\mathcal{E}=-PdV$

We know for our adiabatic fluid:

P V = N T

$\mathcal{E}=\frac{3}{2}NT$

So that:

$d\mathcal{E}=\frac{3}{2}NdT=-PdV$

Then:

$\frac{3}{2}\frac{dT}{T}=-\frac{dV}{V}$

We can also get from $P V = N T$ :

$\frac{dP}{P}+\frac{dV}{V}=\frac{dT}{T}$

So that:

$\frac{dP}{P}+\frac{dV}{V}=-\frac{2}{3}\frac{dV}{V}$

Or:

$\frac{dP}{P}=-\frac{5}{3}\frac{dV}{V}$

By mass conservation, $ρ V = c o n s t$ , so that $\frac{dV}{V}+\frac{d\rho}{\rho}=0$ and:

$\frac{dP}{P}=\frac{5}{3}\frac{d\rho}{\rho}$

Solving this differential equation:

P = c ρ 5 / 3

where $c$ is some constant. This means that we can take the derivative in the fluid frame (the Lagrangian derivative:

$\frac{d}{dt}\left(\frac{P}{\rho^{5/3}}\right)=0$

We can expand this derivative:

$\frac{1}{\rho^{\gamma}}\frac{dP}{dt}-\frac{\gamma P}{\rho^{\gamma+1}}\frac{d\rho}{dt}=0$

$\frac{dP}{dt}+\gamma P\nabla\cdot v=0$