2
$\begingroup$

It is common to show the features and power of the Hamilton's principle by deriving the equation of vibrating string, membrane etc. using this principle. But I have never seen that used for deriving the (aero-)acoustical wave equation. Could you provide this derivation?

Edit:

I. e. equation for acoustic pressure $p$ or velocity potential $\Phi$ in d'Alembert's form:

$$ \nabla^2 \Phi - \frac{1}{c_0^2}\partial_{tt} \Phi = 0 $$

$\endgroup$
3
  • $\begingroup$ One way to answer this question would be to give you a Lagrangian description of ideal hydrodynamics (no dissipation). Would you be happy with such an answer? $\endgroup$ Commented Nov 6, 2015 at 14:21
  • $\begingroup$ Well, if you manage to incorporate the usual small amplitudes approximations leading to the wave equation then dissipation is not a problem. $\endgroup$ Commented Nov 6, 2015 at 14:26
  • 1
    $\begingroup$ Then I might suggest this paper, arxiv.org/abs/1107.0731, where they derive the constitutive relation for the stress tensor from an action. Given the constitutive relation and stress tensor conservation, one can then straightforwardly see looking at small fluctuations about an equilibrium configuration that there are two types of modes -- shear and sound. The sound modes obey the equation you wrote above. There is a fair amount of work here, which is why I'm dragging my feet and not writing an answer below. $\endgroup$ Commented Nov 6, 2015 at 20:21

1 Answer 1

3
$\begingroup$

It's actually not that complicated for a linear case. Let's derive the 1D wave equation for velocity potential $\Phi$ ($v = \partial_x \Phi$). As usual, $c_0$ denotes the speed of sound.

The kinetic energy density is obviously

$$ \mathcal{T} = \frac{1}{2}\rho_0\mathcal{v}^2 = \frac{1}{2}\rho_0\left(\frac{\partial \Phi}{\partial x}\right)^2 $$

Potential energy is derived using work done during changing a test volume using acoustic pressure $p$ (assuming adiabatic behavior). Quantitatively for its density:

$$ \mathcal{V} = \frac{1}{2}\frac{\rho_0}{c_0^2}p^2 = \frac{1}{2}\frac{\rho_0}{c_0^2}\left(\frac{\partial \Phi}{\partial t}\right)^2 $$

Hence the lagrangian:

$$ \mathcal{L} = \frac{1}{2}\rho_0\left(\frac{\partial \Phi}{\partial x}\right)^2 - \frac{1}{2}\frac{\rho_0}{c_0^2}\left(\frac{\partial \Phi}{\partial t}\right)^2 $$

The appropriate Euler-Lagrange equation for minimizing the action integral is:

$$ \frac{\partial \mathcal{L}}{\partial \Phi} = \frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial \mathcal{L}}{\partial (\partial_x \Phi)} + \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial \mathcal{L}}{\partial (\partial_t \Phi)} $$

Therefore:

$$ 0 = \partial_{xx}\Phi - \frac{1}{c_0^2}\partial_{tt}\Phi $$

Generalization to 3D case is obvious.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.