A sound wave can be described by the pressure field or the sound air particle velocity (acoustic flow). Both are intrinsic to any sounds, i.e. there is no sound wave if no pressure or no particle velocity.

The module $v$ of air particle velocity of a monopole sound-source has 2 components (e.g. equation 2.108 chapter 2 of Beranek & Mellow 2012):

  • one which module is proportional to inverse distance to the sound source ($1/r$). It is the propagating component of particle velocity (associated to the active intensity), which is proportional to the pressure module in the far-field.
  • another which is proportional to the inverse square distance to the sound source ($1/r^2$). Then this component (associated to the reactive intensity) is located relatively close to the sound-source as compared to the propagating component, so it can be described as non-propagating or local component of particle velocity.

$$v(r,t)=\frac{1}{Z_{air}} \frac{1}{r} s(t-r/c)+ \frac{c}{Z_{air}} \frac{1}{r^2} ∫_t s(t-r/c)dt \tag{1}$$

with $s(t-\frac{r}{c})$ a progressive wave solution of the wave equation, moving at a velocity $c$ at time $t$ and position $r$ from the sound source, $Z_{air}$ a constant.

However, the pressure field $p$ of a monopole sound source has a single component, which module is proportional to the propagating component of the air particle velocity (~$1/r$):

$$p(r,t)=\frac{1}{r} s(t-\frac{r}{c}) \tag{2}$$

Question: how is it possible that the reactive particle velocity component is not associated to an additional pressure in the near-field of the sound source? An additional velocity field should increase the particle movement and then the force per surface, i.e. the pressure, but it does not, according to the God of maths. Why?

NB: I'm not interested in how the equation solutions (1) and (2) were found (done e.g. in above citation) but interested in their interpretation (i.e. no maths or as little as possible please),

  • $\begingroup$ Hi, could you clarify what you mean by "An additional velocity field should increase the particle movement and then the force per surface" - are you saying that an increase in particle velocity/speed = increase in pressure? $\endgroup$
    – user330563
    Commented Sep 18, 2022 at 8:07
  • $\begingroup$ @user330563 thanks for the question. I intuitively expect that the pressure would increase with particle speed, as it is the case very far from the source where both are proportionnal. Close to the source, there is an additionnal particle speed component, so why is the local pressure not affected by this additionnal particle speed (as shown by equation solutions of pressure and speed)? The equations say NO, I cant understand why intuitively in terms of particle movements, as their movements are in phase. $\endgroup$
    – Noil
    Commented Sep 18, 2022 at 9:04

2 Answers 2


Short answer:

One one level, I think the answer is "because the math says so", but I always hate this answer. I think a better answer in this case is because the particle velocity is actually a vector quantity and the acoustic pressure is a scalar quantity.


At risk of this answer becoming "because the math says so", let's look at some math. The linearized equations of mass conservation and momentum conservation may be written respectively as \begin{gather} \nabla \cdot \vec v + \frac{1}{\rho c^2}\frac{\partial p}{\partial t} = 0, \tag{C} \\ \nabla p + \rho \frac{\partial\vec v}{\partial t} = 0, \tag{M} \end{gather} where $\vec v$ is the particle velocity, $p$ is the acoustic pressure, $\rho$ is the ambient mass density, and $c$ is the sound speed. Typically, we desire to combine these two equations in to a single equation in either $\vec v$ or $p$. First, we eliminate $\vec v$ by taking the divergence of (M) and subtracting the time derivative of (C) multiplied by $\rho$, yielding the wave equation for $p$: $$\nabla^2p - \frac{1}{c^2}\frac{\partial^2p}{\partial t^2} = 0.$$ However, if we eliminate $p$ by taking the gradient of (C) and subtracting the time-derivative of (M) divided by $\rho c^2$, we obtain $$\nabla(\nabla\cdot\vec v) - \frac{1}{c^2}\frac{\partial^2\vec v}{\partial t^2} = 0.$$ Clearly, the equation for the particle velocity is different from the equation for the acoustic pressure. In one dimension, the Laplace operator and the gradient of the divergence operator become the same, but in spherical coordinates (with angular symmetry imposed), we find \begin{gather} \nabla^2p = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial p}{\partial r}\right) = \frac{\partial^2p}{\partial r^2} + \frac{2}{r}\frac{\partial p}{\partial r}, \\ \nabla\left(\nabla\cdot(\hat rv_r)\right) = \hat r\frac{\partial}{\partial r}\left( \frac{1}{r^2}\frac{\partial}{\partial r}r^2v_r \right) = \hat r\left( \frac{\partial^2 v_r}{\partial r^2} + \frac{2}{r}\frac{\partial v_r}{\partial r} - \frac{2}{r^2}v_r \right) \end{gather} Thus, the vector nature of the particle velocity leads to a different differential equation for it compared to the acoustic pressure.

Alternative perspective:

If it helps, it might be worth thinking about the acoustic intensity. The real part of $p^*\vec v$ gives the resistive component of the intensity (the energy that leaves and does not come back), while the imaginary part of $p^*\vec v$ is the reactive component (sits there and sloshes back and forth). In my mind, this energetic approach is the fundamental definition of "resistive" and "reactive". Thus, calling the real and imaginary parts of the particle velocity one or the other only makes sense in the special case where $p$ is real. You could change the initial phase of $p$, and there would be components of both the real and imaginary parts of the particle velocity contributing to both the resistive and reactive components of the intensity. But usually we take $p$ to be real (because there is no compelling reason not to!).

EDIT Unverified geometric interpretation:

I have never heard of anyone looking at this problem in the following way before, so be careful with the discussion below.

You can think of the fluid shells around the monopole as have two sources of strain (particle displacement divergence): radial compression and azimuthal compression. The radial compression comes from the typical difference-of-velocities, which is also the only type for a plane wave. The azimuthal compression comes from the fact that as a fluid shell moves toward the monopole its surface area gets smaller. This compression is independent of the pressure and becomes less significant as you move farther from the source. Furthermore, the overall effect would be to have a fluid shell oscillate about its equilibrium, and not radiate energy. I would hypothesize (again, I haven't heard of anyone else doing this analysis) that the additional term in the particle velocity comes from this azimuthal strain.

  • $\begingroup$ Thanks Michael. I edited my question again to make it as much clear as I could to say that I'm not interested in the "because the math says so" here :-). I accept the differences in the pressure and velocity solutions (which I added in the question post) and I have no problem to prove them, now I'd like to understand them. I like your "alternative perspective", this is more the direction I'd like the answers to take if possible. $\endgroup$
    – Noil
    Commented Sep 14, 2022 at 14:41
  • $\begingroup$ I don't know if this really answers your question, but the reactive field has always been taught to me as being associated with mass entrainment (this can be seen from a small $ka$ approximation). $\endgroup$
    – Michael M
    Commented Sep 14, 2022 at 15:40
  • $\begingroup$ Another thought that occurred to me is that the radial motion stretches and compresses the fluid geometrically (the surface area at $r+dr$ is greater than the surface area at $r-dr$). The farther you get from the monopole the smaller this effect is. If either of these two comments seems on the right track for you, let me know and I will try to elaborate. $\endgroup$
    – Michael M
    Commented Sep 14, 2022 at 15:42
  • 1
    $\begingroup$ I think to some extent geometry has to be part of the explanation, because you don't get this for plane waves. $\endgroup$
    – Michael M
    Commented Sep 14, 2022 at 16:05
  • 1
    $\begingroup$ I would like to enter this discussion because I find it very interesting. I think it is closely related to the fact that velocity and pressure are out of phase at the origin of the monopole. In addition to monopole sources, the effect also occurs with loudspeakers (see: acs.psu.edu/drussell/demos/burns_phd_animations/…). $\endgroup$
    – Bulbasaur
    Commented Sep 21, 2022 at 14:00

The pressure profile can be obtained entirely using energy arguments.

Consider a point source with constant frequency. At distance $r$ from the source, the power transferred from a spherical layer of air to the its larger surrounding layer is proportional to the force times the velocity: $$\frac{dW}{dt}\propto 4\pi r^2 p(r)A(r)\omega$$ Where $A$ is the amplitude of displacement at distance $r$.

In the linear regime, the amplitude and pressure are proportional, therefore: $$\frac{dW}{dt}\propto r^2 p(r)^2$$ In the stationary state, energy is not accumulated in a layer (and we are ommitting energy loss due to dissipation) and therefore: $$r^2 p(r)^2=const. \to p(r)\propto \frac{1}{r}$$


I guess, your question is related to the equation (I will try to give intuition later):

$$\rho_0 \frac{\partial v}{\partial t}=-\frac{dP}{dr}$$

For a wave, we have: $$P=P_0 +\delta P \frac{\cos (kr-\omega t)}{r} $$

Then, if we take the derivative, we will have two terms: $$\rho_0 \frac{\partial v}{\partial t}= \delta P k \frac{\sin (kr-\omega t)}{r}+\delta P \frac{\cos (kr-\omega t)}{r^2}$$

So the velocity has two parts as the OP mentioned.

What we see is that, both components of velocity are given by the same pressure profile.

To gain better insight, we work with acceleration. Acceleration is a result of force which itself is generated by pressure difference between two points. The change in the pressure comes from two different elements:

1- The fluctuations in pressure due to the oscillation of air.

2- The overall reduction in the amplitude of the pressure as we go further away from the source.

The former gives the active component of the velocity and the latter gives the reactive part, both originated from the same pressure profile.

In other words, it is the pressure change that produces velocity‌, not the other way

  • $\begingroup$ Thanks Hossein. Let me be clearer about my question. I am wondering why the local particle velocity field (~1/r^2) does not create a local pressure field in addition to the propagating pressure field associated with the propagating particle velocity field (~1/r). $\endgroup$
    – Noil
    Commented Sep 13, 2022 at 20:45
  • $\begingroup$ I edited my answer to address your question. $\endgroup$
    – Hossein
    Commented Sep 13, 2022 at 22:01
  • $\begingroup$ Thanks Hossein, I've edited/simplified my question to make clearer that I'm interested in the interpretation of the pressure profile which mismatches my intuition $\endgroup$
    – Noil
    Commented Sep 14, 2022 at 11:01
  • $\begingroup$ "2. The overall reduction in the amplitude of the pressure as we go further away from the source": do you mean the amplitude of pressure gradient? Could you elaborate on this #2 please? I'm not sure to understand. Would it be related to the geometrical interpretation given by @MichaelM? $\endgroup$
    – Noil
    Commented Sep 16, 2022 at 21:40

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