How Do Velocity and Pressure Relate in Linear Sound Wave Equations?

[unscientific]

Taken from my lecturer's notes, how did they make the jump from 8.5 to 8.6 and 8.7?

Even after differentiating (8.5) with time I get

$$\rho_0 \frac{\partial^2 \vec u'}{\partial t^2} + \nabla \frac{\partial p '}{\partial t} = 0$$
$$\frac{\partial^2 p'}{\partial t^2} + \rho_0 c^2 \nabla \cdot \frac{\partial \vec u'}{\partial t} = 0$$

Is there a relation between $\vec u$ and $p$ I am missing?

 

[vanhees71]

Note that
$$\vec{\nabla} \cdot \partial_t \vec{u}=\partial_t \vec{\nabla} \cdot \vec{u}.$$

 

[unscientific]

Note that
$$\vec{\nabla} \cdot \partial_t \vec{u}=\partial_t \vec{\nabla} \cdot \vec{u}.$$

Is $\nabla \cdot \vec u$ somehow related to pressure?

 

[vanhees71]

It's somewhat related to density. Using the continuity equation, which expresses mass conservation (valid in non-relativistic physics from very basic principles)
$$\partial_t \rho + \vec{\nabla} (\rho \vec{v})=0.$$

 

[unscientific]

It's somewhat related to density. Using the continuity equation, which expresses mass conservation (valid in non-relativistic physics from very basic principles)
$$\partial_t \rho + \vec{\nabla} (\rho \vec{v})=0.$$

$$\frac{\partial m}{\partial t} = - \int \rho \vec v \cdot d\vec S$$
$$\int \frac{\partial \rho}{\partial t} dV = -\int \rho \nabla \cdot \vec v dV$$

This implies that $\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec v) = 0$.

 

[boneh3ad]

Well yes that equation is satisfied by default, as it is the continuity equation.

You can relate the velocity and pressure through Euler's equation.