Eliminating mass density $\rho$ from equation of state

The equation of state (3) involves three physical quantities, namely $\rho$, $p$, and $\mathbf{u}$. It turns out that the continuity equation can be used in the equation to eliminate $\rho$. The equation of state

$$\frac{d}{d t} (p \rho^{- \gamma}) = 0,$$ (13)

can be written as

$$\frac{d p}{d t} \rho^{- \gamma} - \gamma \rho^{- (\gamma + 1)} p \frac{d \rho}{d t} = 0,$$ (14)

which simplifies to

$$\frac{d p}{d t} = \gamma \frac{p}{\rho} \frac{d \rho}{d t}$$ (15)

Expand the total derivative, giving

$$\frac{\partial p}{\partial t} +\mathbf{u} \cdot \nabla p = \gamma... ...} \left( \frac{\partial \rho}{\partial t} +\mathbf{u} \cdot \nabla \rho \right)$$ (16)

Using the mass continuity equation to eliminate $\rho$ in the above equation gives

$$\frac{\partial p}{\partial t} = - \gamma p \nabla \cdot \mathbf{u}-\mathbf{u} \cdot \nabla p.$$ (17)

This equation governs the time evolution of the pressure. A way to memorize this equation is that, if $\gamma = 1$, the equation will take the same form as a continuity equation, i.e.,

$$\frac{\partial p}{\partial t} = - \nabla \cdot (p\mathbf{u}) .$$ (18)