# If I can hear Bob, can Bob hear me? (AKA, why does the Green's function for sound propagation have to be symmetric?)

Suppose I am in location X, emitting a sound at some loudness level. Bob, standing at location Y, hears the sound with an intensity reduced by $x$ db. If Bob starts emitting sound, will I also hear it at an intensity reduced by exactly $x$ db? (In general, there may be some general configuration of doors/walls between me and Bob, not symmetric between the two of us. These barriers might absorb some sound energy, as well as reflecting, refracting and/or diffracting the sound waves).

Now, at a technical level one can make arguments that the answer is yes; for example, the propagation of sound waves at some frequency should be described by the Helmholtz equation $$(\nabla^2 + k(x)^2) \varphi = 0,$$ where $k(x)$ is the (possibly spatially-dependent) inverse wavelength of sound at location $x$. (We can also make $k(x)$ complex to include the possibility of dissipation). The intensity of sound at location $y$ emitted from position $x$ is then described by the Green's function $G(x,y)$ of the Helmholtz equation, which can be shown to be symmetric.

However, this argument seems way too technical and doesn't really answer the basic question of why the propagation of sound waves had to be described by a PDE with symmetric Green's function. Are there any exotic scenarios where this symmetry property does not hold? Does it depend on any of the approximations made in deriving the Helmholtz equation? Or is there some basic physical argument for why the symmetry property must be true?

To anticipate a possible answer, I do not believe this has anything to do with time reversal symmetry. For one thing, time reversal symmetry does not hold in the presence of absorption. And in any case, the time reverse of Bob emitting a sound is a bunch of sound waves converging on Bob, not me emitting a sound.

• Green's function solutions need to take boundary conditions into account, which could easily break the symmetry. If you and a friend sit in a room with an open window, you can hear the conversations of people outside better than they can hear yours. – J. Murray Sep 30 '17 at 4:07
• @J.Murray I don't claim to have tried this experiment, but I am doubtful. In the absence of, say, wind currents carrying sound into the room, it seems that you are still calculating the Green's function of a self-adjoint operator, and this ought to be symmetric, no? After all the boundary conditions are still symmetric between the two cases. – Dominic Else Sep 30 '17 at 4:11
• It's worth noting that background noise renders the basic notion 'if A can hear B then B can hear A' false in practice. But this is a signal-to-noise ratio thing, not something about sound propagation. – dmckee Sep 30 '17 at 4:19
• @J.Murray In my above comment, to avoid confusion, I should have said that the boundary conditions are the same in both cases instead of "symmetric between the two cases", which is somewhat ambiguous. – Dominic Else Sep 30 '17 at 4:22
• Microphone calibration and sound level measurement rely on the reciprocity relation that allows you to reverse the location of source and microphone and get the same result. I know a physicist who developed methods for measuring road noise, but this is all I recall of the details. – C. Towne Springer Sep 30 '17 at 7:22

I agree with J. Murray that boundary conditions must be taken into account one way or another. Here is why. Call $\phi_A$ the field when Alice is talking, i.e., when there is a Dirac source $\delta_A$ placed at $A$. Similarly define $\phi_B$. Then, $$(\nabla^2+k^2)\phi_A = \delta_A, \\ (\nabla^2+k^2)\phi_B = \delta_B.$$ Now multiply the first equation by $\phi_B$, second one by $\phi_A$, integrate by part (using the divergence theorem) and take the difference to get $$\int_{\partial_\Omega}(\phi_A\nabla\phi_B-\phi_B\nabla\phi_A)\cdot n\,\mathrm{d}S = \phi_A(B)-\phi_B(A).$$ This is sometimes called Maxwell-Rayleigh-Betti (or any subset thereof) reciprocal theorem. When the boundary conditions are such that the lhs is zero, we get that $\phi_A(B)=\phi_B(A)$; or in your terms, that Green's function is symmetric. This occurs for some standard boundary conditions (say $\nabla\phi\equiv 0$ at the boundary $\partial\Omega$) but is not guaranteed as far as I can tell.

Various assumption were made. Linearity for one. Also rather than time reversal symmetry, "time invariance" was assumed: somehow the medium and the boundaries are stationary. With moving boundaries, things will be different and reciprocity will fail. These assumptions are sometimes regrouped under the abbreviation "LTI" for Linear Time Invariant. You can also check the Onsager Casimir principle for more details.

• But no materials are completely impermeable to sound. So it seems to me that the only physical boundary conditions are those at infinity, and then the LHS does go to zero. – Dominic Else Sep 30 '17 at 14:46
• Actually, on second thoughts the LHS does not manifestly go to zero at infinity: $\phi_A$ and $\phi_B$ are expected to each fall off as $1/r$, and you are integrating over a surface of area $r^2$. So it does indeed look like symmetry can be violated. – Dominic Else Sep 30 '17 at 15:31
• On third thoughts, I would think that you could envelop your system in an absorbing medium far away (represented by a complex $k^2$) -- this ought not to change the Green's function nearby, but would ensure that $\phi_A$ and $\phi_B$ decay fast enough at infinity to ensure that the LHS goes to zero. So I am still not convinced that symmetry can be violated in any physical situation. – Dominic Else Sep 30 '17 at 16:37
• Empty space is impermeable to sound and many materials are also "exponentially" impermeable to sound at specific frequencies. So the question of decay at infinity seems delicate indeed and I will not venture into it. Also, changing the boundary conditions no matter how far can change the Green's function significantly everywhere: compare a giant solid sphere excited at its resonance frequency and the whole space excited at the same frequency. – Hussein Sep 30 '17 at 19:23

Boundary conditions are very important. Take the acoustics of ancient amphitheaters, the Epdidaurus for one, where at the top row you can hear a paper torn ( tried it on the excursion when we were there as students). Nobody at the stage could hear a paper torn in the upper row (even on a windless day).

The theatre is admired for its exceptional acoustics, which permit almost perfect intelligibility of unamplified spoken words from the proscenium or skēnē to all 14,000 spectators, regardless of their seating (see Ref., in Greek). Famously, tour guides have their groups scattered in the stands and show them how they can easily hear the sound of a match struck at center-stage.

This is interesting , at 1 20'

Your boundary conditions are: air completely still,no asymmetric reflective surfaces etc. etc.

• Interesting, but is possible that this failure of reciprocity was due to background noise? That is, due to the design of the theater the background noise level when standing on the stage is higher than in the stands, drowning out the torn paper sound? – Dominic Else Sep 30 '17 at 14:44
• possibly for our student experimement. But listen to the youtube clapping changes. and this analysis acoustics.org/… . these amplifications would not work the reverse way. a sound at the top row would not have the same reflections. – anna v Sep 30 '17 at 17:37

I do not know about acoustics but when RF (radio) waves are subjected to multipath reflection and the signal is received with multiple antennas (so-called antenna diversity) to overcome interference nulls it does matter which direction the wave propagates. This is a source of a common confusion because each wavelet (wavefront/ray) does indeed propagate reciprocally. The non-reciprocal behavior is a statistical one. For example, for efficient diversity reception one would like to have the antennas spaced so the received signals are uncorrelated one form another. Whether these signals are in fact uncorrelated or not will depend on the height of the antennas. For a cell-tower that separation could be several wavelengths while for a cell-phone that is situated near or on the ground may be one wavelength or less, and the reason for this is that in a city on the ground the cell-phone is surrounded by multipath reflectors but the tower high above the ground is not. Of course, in a rural environment where there are one or two reflectors even the statistics is reciprocal. Our hearing employs two receivers and our brain is a diversity receiver but I do not know if that is indeed the reason for the amazing acoustics of the ancient Greek amphitheaters @Anna_V has mentioned.