Alternatively one could ask:

Would gravity favor sound propagation towards the gravitational center if it did not affect the static density of the medium?

The answer to this could be relevant, e.g., to validate numerical simulations by assuring that all physical effects are captured by the simulation.

For visualization one could imagine the medium to be a solid with constant static density.

The question arose from my personal image of sound being a movement of the velocity of molecules. Since the molecules are affected by gravity, the movement of the velocity might be too.

I believe this question is not a duplicate of How does gravity affect sound waves? because the accepted answer therein focuses on the indirect effect through static density change.


2 Answers 2


Sound is the variation in pressure and therefore density of the air, it has nothing to do with velocity. You are probably confused by the speed of sound, which is a thermodynamical property, not an actual movement of molecules. It indicates how fast the wavefront of the pressure fluctuations travels through a medium that is in a certain thermodynamical state. These vibrations have no netto movement. For example you can hear moving objects, but the air surrounding them will be dragged along with them, making it move away from you as the object passes, yet you still hear the wind it creates.

Coming back to your question, the speed of sound is defined as $c^2 = \frac{\partial p}{\partial \rho}$, so since gravity causes the pressure and density to change as you move up and down in the atmosphere, the speed of sound will also change. However, this does not mean that it favours the gravitational direction, but that the wavefront just will not be a sphere. Since we know that over short distances the pressure and density do not vary this much (due to gravity), other influences such as temperature and wind will play a bigger role.

  • $\begingroup$ I imagine sound to be the result of velocity on a microscopic level. As you say -- macroscopically -- sound is a variation in pressure. This itself is the result of movement of molecules. Do the molecules tend to move faster/further towards the gravitational center? $\endgroup$ Jun 21, 2017 at 9:27
  • $\begingroup$ It is still a misconception that sound is the result of velocity on a mictoscopic level. Even here it's just pressure fluctuations. The movement of molecules on molecular level determines the temperature. If it also created sound, you could hear it when the air warms up or cools down, in your logic. $\endgroup$ Jun 21, 2017 at 9:49
  • $\begingroup$ Please correct me if I'm wrong, but I thought talking of pressure in a thermodynamical context always involves averaging (force/area), thus pressure can't be talked about (at least not in a straight-forward way) when tracking single/few selected molecules. $\endgroup$ Jun 21, 2017 at 9:58
  • $\begingroup$ When tracking a few molecules you cannot talk of anything, talking of pressure makes no sense, since it's an average. Talking of temperature makes no sense, since it's an average. Thinking of sound makes even less sense, since it needs a continuous medium to travel through, while "a few molecules" are not continuous, but discrete particles with nothing in between them. $\endgroup$ Jun 21, 2017 at 11:25
  • $\begingroup$ Exactly. Maybe now I can clarify my speculation: a single molecule will simply fall towards the source of gravity. "A few molecules" will also move towards it. Even if they occasionally collide, gravity will accelerate them towards its center all the time. Does the continuum limit annihilate the effect of gravity? $\endgroup$ Jun 21, 2017 at 14:40

Since here (and in the quoted question as well) the equations-based approach is missing I am gonna try to provide that. Basically, when you derive the linear acoustic wave equation, you start with linearised Euler equation and linearised continuity equation:

$$ \frac{\partial \mathbf{v}}{\partial t} = -\frac{1}{\rho_0}\nabla p + \mathbf{g} $$

$$ \frac{\partial \rho}{\partial t} + \rho_0 \nabla \cdot \mathbf{v} = 0 $$

where $\mathbf{v}$, $p$, $\rho$, $\rho_0$ and $\mathbf{g}$ denote velocity, pressure, density, equilibrium density and gravitational acceleration respectively (the latter could be generalised as an external net force acceleration).

Provided that $p = c_0^2 \rho$ we can achieve the wave equation by eliminating the velocity by:

$$ \nabla \text{Continuity} + \nabla \cdot \text{Euler} $$

And here is the answer: since $\nabla \cdot \mathbf{g} = 0$ in a homogenous gravity field, the $\mathbf{g}$ vanishes from the wave equation.

  • What if the gravity field is inhomogenous? Well, estimate wavelength on which such variation would be important. It is way in the infrasound domain...
  • Does it mean that the gravity is of no influence? Certainly not, cause it affects equilibrium pressure and density. It just doesn't appear as a direct contributor in the wave equation and doesn't provide the first order effect.
  • $\begingroup$ Thank you for this. This answers my question as far as it is possible from a standard point of view. Note the linearizations you mention are not necessary since $\mathbf{g}$ only ever appears linearly in every equation of continuum mechanics (that I have seen so far). However I am still convinced there is a higher-order effect, which has been neglected somewhere along the way, if not in the initial modeling. Boltzmann equation was such a long time ago... Fact 1: gravity affects particles. Fact 2: there is equilibrium. Conclusion: Particles must move faster downwards and longer upwards. $\endgroup$ Aug 1, 2017 at 12:08
  • $\begingroup$ @DrDoolittle Well, I can add an extended wave equation for the $\rho_0$, $p_0$ inhomogenities, if you would like me to. Or you can estimate changes in thermodynamic equilibria when the gravitational potential is present. I assume that you will end with really high order effect. $\endgroup$ Aug 1, 2017 at 19:14
  • $\begingroup$ I am not familiar with the extended wave equations. However I was arguing that sound is probably traveling faster downwards even if there were no gravity-induced changes of $\rho$,$p$. From your last comment I deduce that you -- who obviously knows a lot more in that field than I do -- would not say my thoughts are completely off, and that's satisfactory for me atm. Thanks :-) $\endgroup$ Aug 11, 2017 at 18:09

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