How can the particle velocity field due to an acoustic monopole be derived in terms of pressure?

Question

I have seen in numerous literature sources (e.g. this chapter from MIT open courseware, pages 414-415) the following method of relating the pressure to particle velocity. Begin with the pressure field of an acoustic monopole (point source with uniform spherical waves): $$p(r) = \frac{A}{r}e^{j(\omega t - kr)} $$ where $r$ is radial distance from the source, $A$ is some arbitrary amplitude, $\omega$ is the angular temporal frequency of the signal, and $k$ is the angular spatial frequency. Newton's second law (as described in the literature) relates particle velocity $\textbf{u}$ to the pressure gradient: $$\rho \frac{\partial \textbf{u}}{\partial t} = -\nabla p$$ Radial symmetry in this case means that the pressure gradient is dependent only on $r$: $$\nabla p = \frac{\partial p}{\partial r} \textbf{e}_r = \left[ -\frac{A}{r^2}e^{j(\omega t - kr)} - \frac{A}{r}jke^{j(\omega t - kr)}\right] \textbf{e}_r$$ $$\nabla p = (-\frac{1}{r} - jk)p \textbf{e}_r$$ Newton's second law can therefore be written: $$\rho \frac{\partial u_r}{\partial t} = (\frac{1}{r} + jk)p $$ Assuming an oscillatory solution for $u_r$, $\dot u_r = j\omega u_r$, which gives $$ \rho j \omega u_r = (\frac{1}{r} + jk)p $$ Dividing through by $jk$, $$ \frac{\rho \omega u_r}{k} = (\frac{1}{jkr} + 1)p $$ $$ u_r = \frac{1}{\rho c} (1- j \frac{c}{2\pi fr})p$$

That's the result that I see wherever I look, but had originally taken an approach in which I considered a differential spherical element of volume $\textrm{d}V = r^2 \sin(\theta)\textrm{d}r \textrm{d}\theta \textrm{d}\phi$.

A force balance in the $\hat r$ direction on the element yields $$ \Sigma F_r = \left( p - \frac{\partial p}{\partial r} \frac{\textrm{d}r}{2} \right) \left[ \left(r- \frac{\textrm{d}r}{2} \right)^2 \sin(\theta)\textrm{d}\theta \textrm{d}\phi \right] - \left( p + \frac{\partial p}{\partial r} \frac{\textrm{d}r}{2} \right) \left[ \left(r + \frac{\textrm{d}r}{2} \right)^2 \sin(\theta)\textrm{d}\theta \textrm{d}\phi \right]$$ Dividing through by $\sin(\theta)\textrm{d}\theta\textrm{d}\phi$ and distributing terms, $$ \frac{\Sigma F_r}{\sin(\theta)\textrm{d}\theta\textrm{d}\phi} = \left( p - \frac{\partial p}{\partial r} \frac{\textrm{d}r}{2} \right) \left( r^2 - r \textrm{d}r + \frac{\textrm{d}r^2}{4} \right)- \left( p + \frac{\partial p}{\partial r} \frac{\textrm{d}r}{2} \right) \left( r^2 + r \textrm{d}r + \frac{\textrm{d}r^2}{4} \right) $$ $$ = pr^2 - pr\textrm{d}r + p \frac{\textrm{d}r^2}{4} - \frac{\partial p}{\partial r}\frac{\textrm{d}r}{2} r^2 + \frac{\partial p}{\partial r}\frac{\textrm{d}r}{2} r\textrm{d}r - \frac{\partial p}{\partial r}\frac{\textrm{d}r}{2} \frac{\textrm{d}r^2}{4} $$ $$ - pr^2 - pr\textrm{d}r - p \frac{\textrm{d}r^2}{4} - \frac{\partial p}{\partial r}\frac{\textrm{d}r}{2} r^2 - \frac{\partial p}{\partial r}\frac{\textrm{d}r}{2} r\textrm{d}r - \frac{\partial p}{\partial r}\frac{\textrm{d}r}{2} \frac{\textrm{d}r^2}{4} $$ Neglecting higher order terms, this force balance reduces to $$ \frac{\Sigma F_r}{\sin(\theta)\textrm{d}\theta\textrm{d}\phi} = -2pr\textrm{d}r - \frac{\partial p}{\partial r} r^2 \textrm{d}r $$ Dividing through by $r^2\textrm{d}r$, the sum of the forces (in the $\hat r$ direction) has been divided by the differential volume $$ \frac{\Sigma F_r}{r^2\sin(\theta)\textrm{d}r\textrm{d}\theta\textrm{d}\phi} = -\frac{2p}{r} - \frac{\partial p}{\partial r} $$ The discrepancy between the result of this approach and that of the literature is clear at this point. Moving on, recalling that $$ \frac{\partial p}{\partial r} = (-\frac{1}{r} - jk)p $$ gives the equation of motion for the fluid due to the acoustic monopole $$ \rho \frac{\partial u_r}{\partial t} = -\frac{2p}{r} + \frac{p}{r} +jkp$$ $$ \rho j \omega u_r = \left(-\frac{1}{r} + jk\right)p $$ Rearranging, we arrive at my original result: $$ u_r = \frac{1}{\rho c} (1+ j \frac{c}{2\pi fr})p$$ And compare to the solution that I find elsewhere: $$ u_r = \frac{1}{\rho c} (1- j \frac{c}{2\pi fr})p$$

It's easy to look at this inconsistency and think of it as a sign error, but I don't think that's the case. I suspect that the problem begins with the formulation of the equation of motion. All external sources that I have seen have taken as given $$\rho \frac{\partial u}{\partial t} = -\nabla p$$ but is this really correct for a spherical wave, in which the "front" and "back" surfaces have different areas on which the pressure acts? This expression looks like a plane wave solution to me.

Where in my derivation have I gone astray? Many thanks to anyone who attempts to clarify this concept for me!

Answer 1

Here it is:

The force balance on a differential fluid element around an acoustic monopole must include all forces in the $\textbf{e}_r$ direction. The symmetry of the spherical wavefront allows us to ignore any viscous effects. Forces on the element due to the pressure field include the following: $$ \left( p - \frac{\partial p}{\partial r} \frac{\textrm{d}r}{2} \right) \left[ \left(r- \frac{\textrm{d}r}{2} \right)^2 \sin(\theta)\textrm{d}\theta \textrm{d}\phi \right] \textbf{e}_r \hspace{5mm} \textrm{on the $-\textbf{e}_r$ surface} $$

$$ - \left( p + \frac{\partial p}{\partial r} \frac{\textrm{d}r}{2} \right) \left[ \left(r + \frac{\textrm{d}r}{2} \right)^2 \sin(\theta)\textrm{d}\theta \textrm{d}\phi \right] \textbf{e}_r \hspace{5mm} \textrm{on the $\textbf{e}_r$ surface} $$

$$ p \left[ r \sin{\theta} \textrm{d}r \textrm{d}\phi \right] \frac{\sin(\textrm{d}\theta)}{2} \textbf{e}_r \hspace{5mm} \textrm{on the $\textbf{e}_{\theta}$ and $-\textbf{e}_{\theta}$ surfaces} $$

$$ p \left[ r \textrm{d}r \textrm{d}\theta \right] \frac{\sin(\theta) \sin(\textrm{d}\phi)}{2} \textbf{e}_r \hspace{5mm} \textrm{on the $\textbf{e}_{\phi}$ and $-\textbf{e}_{\phi}$ surfaces} $$

Where $p$ is the pressure at the center of the element. The small angle approximations $\sin(\textrm{d}\phi) \approx \textrm{d}\phi$ and $\sin(\textrm{d}\theta) \approx \textrm{d}\theta$ simplify the force balance to

$$ \frac{\Sigma F_r}{\sin(\theta) \textrm{d}\theta \textrm{d}\phi} = \left( p - \frac{\partial p}{\partial r} \frac{\textrm{d}r}{2} \right) \left(r- \frac{\textrm{d}r}{2} \right)^2 - \left( p + \frac{\partial p}{\partial r} \frac{\textrm{d}r}{2} \right) \left(r + \frac{\textrm{d}r}{2} \right)^2 + 2pr\textrm{d}r $$

$$ = pr^2 - pr\textrm{d}r + p \frac{\textrm{d}r^2}{4} - \frac{\partial p}{\partial r}\frac{\textrm{d}r}{2} r^2 + \frac{\partial p}{\partial r}\frac{\textrm{d}r}{2} r\textrm{d}r - \frac{\partial p}{\partial r}\frac{\textrm{d}r}{2} \frac{\textrm{d}r^2}{4} $$ $$ - pr^2 - pr\textrm{d}r - p \frac{\textrm{d}r^2}{4} - \frac{\partial p}{\partial r}\frac{\textrm{d}r}{2} r^2 - \frac{\partial p}{\partial r}\frac{\textrm{d}r}{2} r\textrm{d}r - \frac{\partial p}{\partial r}\frac{\textrm{d}r}{2} \frac{\textrm{d}r^2}{4} + 2pr\textrm{d}r $$ Neglecting higher order terms, this force balance reduces to $$ \frac{\Sigma F_r}{\sin(\theta)\textrm{d}\theta\textrm{d}\phi} = - \frac{\partial p}{\partial r} r^2 \textrm{d}r $$

$$ \frac{\Sigma F_r}{r^2 \sin(\theta) \textrm{d}r \textrm{d}\theta\textrm{d}\phi} = - \frac{\partial p}{\partial r} $$ Knowing that the pressure field is $$p(r) = \frac{A}{r}e^{j(\omega t - kr)} $$ gives the equation of motion for the fluid element: $$\rho \frac{\partial u_r}{\partial t} = (\frac{1}{r} + jk)p $$ Assuming a temporally oscillatory solution for $u_r$, $\dot u_r = j\omega u_r$, which gives $$ \rho j \omega u_r = (\frac{1}{r} + jk)p $$ Dividing through by $jk$, $$ \frac{\rho \omega u_r}{k} = (\frac{1}{jkr} + 1)p $$ $$ u_r = \frac{1}{\rho c} (1- j \frac{c}{2\pi fr})p$$

In sum, my original approach failed to account for the radial force contributions of pressure on the $\textbf{e}_{\theta}$ and $\textbf{e}_{\phi}$ sides of the differential element. Careless! It should have been clear right away from the equation of motion that I derived which, for review, was $$ \rho \frac{\partial u_r}{\partial t} = -\frac{2p}{r} - \frac{\partial p}{\partial r} $$ that something was wrong. How can the particle acceleration be dependent on anything other than a pressure gradient? If this were true, then absent a pressure gradient the fluid would accelerate toward the origin under any finite pressure, which is absurd. The bottom line is that $$\rho \frac{\partial u_r}{\partial t} = -\nabla p $$ is the appropriate application of Newton's second law in this case. Hope this helps someone other than me.