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It appears that one can derive a wave equation if the gravitational force is considered in its derivation by using $$\frac{\partial p_0}{\partial x} = -\rho_0 g.$$ Here, $\rho_0 = \rho(p_0)$ and $p_0 = p(\rho_0)$ are the density and pressure respectively in the equilibrium state. The wave equation has the following form: $$c^{-2}\frac{\partial \tilde{p}}{\partial t^2} = \nabla^2\tilde{p} + g\frac{\partial}{\partial x}\left(\frac{\tilde{p}}{c^2(x)}\right),$$ where $$c(x) = \left(\frac{\partial\rho}{\partial p}(p_0(x))\right)^{-1/2}.$$ How come that the gravitational term can be neglected if $\lambda \ll c^2/g$?

I know that $$\tilde{p} = - \rho_0 \frac{\partial \phi}{\partial t}$$ where $\phi$ is the acoustic velocity potential. So, $$\nabla^2\tilde{p} = -\rho_0\left(\frac{\partial\phi}{\partial x\partial t} + \frac{\partial\phi}{\partial y\partial t} + \frac{\partial\phi}{\partial z\partial t}\right)$$ and $$g\frac{\partial}{\partial x}\left(\frac{\tilde{p}}{c^2(x)}\right) = -g\frac{\partial}{\partial x}\left(\frac{\rho_0}{c^2(x)}\frac{\partial\phi}{\partial t}\right).$$ Now, I have no idea how to show that the gravitational term is indeed negligible if $\lambda \ll c^2/g$. Anyone has an idea?

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Straight from Euler equation you may write: $$ \rho_{0} \frac{\partial v}{\partial t}=-\mathbf{\nabla} p=-\rho_{0} g - \nabla \tilde{p}$$ So you end up comparing $\rho_{0} g$ with $\nabla \tilde{p}$. With the state equation, you may write: $$ \tilde{p}= \rho_{0}c^{2}$$ So $\nabla \tilde{p}$ is of the order of $\tilde{p} / \lambda =\rho_{0} c^{2}/ \lambda$, and the condition $\rho_{0} g \ll \nabla \tilde{p} $ yields: $$ \lambda \ll \frac{c^2}{g}$$

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  • $\begingroup$ Could you please explain what "$\nabla\tilde{p}$ is of the order $\tilde{p}/\lambda$" means? How come is that true? Where does this $\lambda$ come from? Is $\lambda$ the wavelength? Furthermore, how come that if $\rho_0g \ll \nabla\tilde{p}$, then $\lambda \ll c^2/g$? $\endgroup$
    – Vicky
    Commented Apr 17, 2020 at 11:01
  • $\begingroup$ Simply assuming a solution in the form $\tilde{p}=Ae^{j(kx-\omega t)}$ (we are talking about the solution to a wave equation, remember...) then $\nabla \tilde{p}=\frac{d\tilde{p}}{dx} =jk\tilde{p}$, with $k=2 \pi/ \lambda$. The $\ll$ symbol means you are looking for something true in orders of magnitude (powers of ten), so the $2\pi$ factor doesn't matter. You could write $1/k \ll \frac{c^{2}}{g}$, this might appear more rigorous to you but it doesn't really matter to a physicist. This type of reasoning is very often used in physics, you definitely need to become more familiar with it! $\endgroup$ Commented Apr 17, 2020 at 12:08
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We make a scaling $x\rightarrow \lambda x, c\rightarrow c_0c, t\rightarrow c_0/\lambda t,p\rightarrow \tilde{p}/\rho_0c_0^2$, then the equation for $p$ takes the form $$c^{-2}p_{tt}-\nabla^2p=\epsilon \partial_x(p/c^2)$$ where $\epsilon =g\lambda/c_0^2$. Now we need just put $\epsilon \rightarrow 0$.

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  • $\begingroup$ I was wondering what the idea behind the scaling is. How come $t \to c_0/\lambda t$? What is $c_0$? $\endgroup$
    – Vicky
    Commented Apr 17, 2020 at 11:04
  • $\begingroup$ @VicRyan $c_0$ is a typical sound speed in your problem. For numerical model with $0\le x\le L$ we can take it as $c(x)$ on the border - $c(0)$ or $c(L)$ . $\endgroup$ Commented Apr 17, 2020 at 11:20
  • $\begingroup$ I see. How about $\lambda$? Is it an arbitrary constant? or maybe an eigenvalue? Does it have a physical interpretation? $\endgroup$
    – Vicky
    Commented Apr 17, 2020 at 11:35
  • $\begingroup$ @VicRyan If we consider acoustic wave then $\lambda$ is length of wave. But if we consider package of acoustic waves then $\lambda$ is the mean length of acoustic waves. $\endgroup$ Commented Apr 17, 2020 at 13:00

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