Sound waves in a fluid

[Delta2]

TL;DR Summary

Waves of pressure, velocity or both?

When we talk about sound waves in a fluid (air, water e.t.c.) we mean that the pressure $P(x,y,z,t)$ satisfies the wave equation, the so called velocity field of the fluid $v(x,y,z,t)$ satisfies the wave equation or both?

 

[Vanadium 50]

Velocity is the derivative of position, i.e. number density, i.e. pressure.,

 

[vanhees71]

Velocity is the derivative of position, i.e. number density, i.e. pressure.,

That doesn't make sense. The flow field, $\vec{v}(t,\vec{x})$ is the velocity of a fluid element, being momentarily at position $\vec{x}$ at time, $t$ (Eulerian description).

To answer the question in the OP: You start with Euler's equation of a free particle
$$\rho (\partial_t \vec{v} +(\vec{v} \cdot \vec{\nabla} \vec{v})=-\vec{\nabla} P$$
and the continuity equation for mass
$$\partial_t \rho + \vec{\nabla} \cdot (\rho \vec{v})=0.$$
In addition you need an equation of state, which we take as a polytrope
$$P=P_0 \left (\frac{\rho}{\rho_0} \right)^n.$$
Here $(\rho_0,P_0)$ is the mass density and pressure of the fluid at rest, and we assume that the deviations from these values and the velocity field itself are small, so that we can linearize the equations:
$$\rho_0 \partial_t \vec{v} = -\vec{\nabla} P, \qquad (1)$$
$$\partial_t \rho + \rho_0 \vec{\nabla} \cdot \vec{v}=0. \qquad (2)$$
In this approximation
$$\vec{\nabla} P=v_{\text{s}}^2 \vec{\nabla} \rho \quad \text{with} \quad v_{\text{s}}^2=\left .\frac{\mathrm{d} P}{\mathrm{d} \rho} \right|_{\rho=\rho_0}= n \frac{P_0}{\rho_0}.$$
For an adiabatic equation of state (which is consistent with the assumption of a perfect fluid made above), you have $n=C_p/C_v$. For an ideal gas that's $n=(f+2)/f$. The air, consisting mostly of molecules with two atoms, $f=5$, and thus $n=1.4$. In any case we get from (1)
$$\rho_0 \partial_t \vec{v}=-v_{\text{s}}^2 \vec{\nabla} \rho. \qquad (3)$$
Taking the divergence gives
$$\rho_0 \partial_t \vec{\nabla} \cdot \vec{v}=-v_{\text{s}}^2 \Delta \rho.$$
Taking the time-derivative of (2) finally leads to the wave equation for the density,
$$\frac{1}{v_{\text{s}}^2} \partial_t^2 \rho + \Delta \rho=0.$$
The same equation holds for $P$ since at the same linear order of the deviations from the equilibrium
$$P=P_0 + v_s^2 (\rho-\rho_0).$$
Concerning $\vec{v}$ we have to assume that it is irrotational, i.e., there's a potential for it
$$\vec{v}=-\vec{\nabla} \Phi. \qquad (4)$$
Then from (2)
$$\partial_t \rho - \rho_0 \Delta \Phi=0. \qquad (5)$$
Plugging (4) in (3) we find
$$\rho_0 \partial_t \Phi=v_{\text{s}}^2 \rho \; \Rightarrow \; \partial_t \rho=\frac{\rho_0}{v_{\text{s}}^2} \partial_t^2 \Phi$$, and finally plugging this again into (5), also the velocity potential obeys the same wave equation as $\rho$ and $P$:
$$\frac{1}{v_{\text{s}}^2} \partial_t^2 \Phi-\Delta \Phi=0.$$
So all these quantities obey the wave equation with the phase velocity $v_{\text{s}}$, which thus is the sound velocity.

 

[Delta2]

@vanhees71 So Pressure and density obey the wave equation but for velocity, the scalar potential (if it exists) of the velocity field obeys the wave equation right?

If the potential of a vector field obeys the wave equation, then can we prove that the vector field obeys the wave equation too? (Tried to prove it myself but to no result ).

 

[vanhees71]

@vanhees71 So Pressure and density obey the wave equation but for velocity, the scalar potential (if it exists) of the velocity field obeys the wave equation right?

If the potential of a vector field obeys the wave equation, then can we prove that the vector field obeys the wave equation too? (Tried to prove it myself but to no result ).

But since $\frac{1}{v_{\text{s}}^2} \partial_t^2 -\Delta$ commutes with $\vec{\nabla}$ then also $\vec{v}$ must obey the wave equation.

 

[Delta2]

But since $\frac{1}{v_{\text{s}}^2} \partial_t^2 -\Delta$ commutes with $\vec{\nabla}$ then also $\vec{v}$ must obey the wave equation.

Ok well that's wonderful but I am kind of sad that I couldn't see that myself.

BTW what are the exact mathematical conditions for the wave operator to commute with the gradient operator? Some "mixed" higher order partial derivatives of the potential of the velocity field must exist and be continuous right?

But yeah in physics we assume all the fields-potentials are infinitely differentiable in all possible ways e hehe.

 

[vanhees71]

Yes, all the partial derivatives involved must be continuous for the mixed derivatives to commute.