# MHD Energy Equation derivation

I'm trying to go through a derivation for the energy equation in MHD in terms of partial derivatives, but I've hit a bit of a wall.

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

and need to end up with it in the following format

$$\frac{\partial p}{\partial t}+\mathbf{v} \cdot \nabla p-c_{s}^{2}\left(\frac{\partial \rho}{\partial t}+\mathbf{v} \cdot \nabla \rho\right)=0$$

where,

$$c_{s}^{2}=\frac{\gamma p_{0}}{\rho_{0}}=\frac{\gamma k_{B} T_{0}}{m}$$

by using the material time derivative

$$\frac{d}{d t} \equiv \frac{\partial}{\partial t}+\mathbf{v} \cdot \nabla$$

I really am not too sure how to get from the first equation to the second. Would appreciate some help

• If $p = p(x_{1},x_{2},\dots,t)$ then by the chain rule $$\frac{d}{dt} p = p_{t} + p_{x_{1}} (x_{1})_{t} + \dots$$ where subscripts in $t$ denote partial derivatives, then note that $(x_{i})_{t}$ represents the velocities. Now apply the product rule to the quotient $p/\rho^{\gamma}$. Dec 15, 2019 at 1:12

$$\frac{\text{D}}{\text{D}t}\left(\frac{p}{\rho^\gamma}\right) = 0,$$

where $$\text{D}/\text{D}t$$ is the material derivative. The eulerian form of the previous equation is

$$\frac{\partial}{\partial t}\left(\frac{p}{\rho^\gamma}\right) + \mathbf{v}\cdot\nabla\left(\frac{p}{\rho^\gamma}\right) = 0.$$

The change rule provides

$$\frac{1}{\rho^\gamma}\left(\frac{\partial p}{\partial t} + \mathbf{v}\cdot\nabla p\right) + p\left(\frac{\partial}{\partial t}\left(\frac{1}{\rho^\gamma}\right) + \mathbf{v}\cdot\nabla\left(\frac{1}{\rho^\gamma}\right)\right) = 0,$$

and multiplying by $$\rho^\gamma$$,

$$\frac{\partial p}{\partial t} + \mathbf{v}\cdot\nabla p + p\left(\rho^\gamma\frac{\partial}{\partial t}\left(\frac{1}{\rho^\gamma}\right) + \mathbf{v}\cdot\rho^\gamma\nabla\left(\frac{1}{\rho^\gamma}\right)\right) = 0.$$

Both time derivative and gradient can be written as

$$\rho^\gamma\frac{\partial}{\partial t}\left(\frac{1}{\rho^\gamma}\right) = -\frac{\gamma}{\rho}\frac{\partial\rho}{\partial t},$$ $$\rho^\gamma\nabla\left(\frac{1}{\rho^\gamma}\right) = -\frac{\gamma}{\rho}\nabla\rho.$$

Finally, plugging them into the last equation,

$$\frac{\partial p}{\partial t} + \mathbf{v}\cdot\nabla p - c_s^2\left(\frac{\partial\rho}{\partial t} + \mathbf{v}\cdot\nabla\rho\right) = 0.$$

where $$c_s^2 = \gamma p/\rho$$. In addition, taking the continuity equation, i.e.

$$\frac{\partial\rho}{\partial t} + \nabla\cdot\left(\rho\mathbf{v}\right) = 0,$$

or

$$\frac{\partial\rho}{\partial t} + \mathbf{v}\cdot\nabla\rho = - \rho\nabla\cdot\mathbf{v},$$

the MHD energy equation is given by

$$\frac{\partial p}{\partial t} + \mathbf{v}\cdot\nabla p + \rho c_s^2\nabla\cdot\mathbf{v} = 0,$$

or in conservative form,

$$\frac{\partial \rho p}{\partial t} + \nabla\cdot (\rho\mathbf{v}p) = -(\rho c_s)^2\nabla\cdot\mathbf{v}.$$