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As far as I know, the intensity of wave reflection at a boundary between two media, for normal incidence, is always governed by the ratio of the wave speeds in the two media. This holds if the wave is a scalar $\Psi$ with a requirement that $\Psi$ and $\Psi'$ both be continuous at the boundary. It also holds in somewhat different contexts such as an electromagnetic wave or a wave in the Schrodinger equation, if you reinterpret things slightly and allow for an ambiguity in the sign of the reflected amplitude.

So I was perplexed to come across this statement about mercury delay lines in old computers:

Mercury was used because the acoustic impedance of mercury is almost exactly the same as that of the piezoelectric quartz crystals; this minimized the energy loss and the echoes when the signal was transmitted from crystal to medium and back again.

The speed of sound is very different in mercury and quartz. Attempting to resolve my confusion, I dug up this article at UNSW, which defines an acoustic impedance $Z$ and a specific acoustic impedance $z$, the latter being a property of the medium alone. ($Z$ depends on both the medium and the cross-sectional area.) I also found this pdf file from Aalto University giving a quantitative treatment of reflection.

The Aalto article gives convincing physical arguments that the continuity requirement is the following: (1) the pressure has to be continuous, and (2) the normal component of the particle velocity has to be continuous. Based on this, they derive equations for reflection and transmission that look like the usual ones, but that use the ratio of $z$ rather than $c$. (They call it $z_c$.) They also say that

"In fact, also the characteristic impedance is continuous at the boundary (see this by comparing the pressures and particle velocities on both sides)"

This seems logical, and although I'm not sure whether their term "characteristic impedance" refers to what UNSW call $Z$ or what they call $z$, in fact, it seems to me that $Z_1/Z_2=z_1/z_2$, since the two only differ by a factor of the cross-sectional area $A$, which is equal for the two media at the boundary.

But wait, this implies that all media have the same $z$ at a fixed frequency, which is surely false since $z$ is a property of the medium. This is a contradiction.

(1) Can anyone help me clear up this contradiction?

(2) Are there some circumstances when acoustic reflection depends on the velocity $c$ and some in which it depends on $z$? E.g., would it be one or the other depending on whether it's a small-area port between two large volumes, or something like that? Or was I simply wrong in my expectation that it would depend on $c_1/c_2$?

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  • $\begingroup$ Could you elaborate how you arrive at the controdiction in question (1)? I do not see why you think all media should have the same $z$? $\endgroup$ – Crimson Dec 20 '17 at 13:15
  • $\begingroup$ @Crimson: That's what the second quote seems to be saying: "the characteristic impedance is continuous at the boundary (see this by comparing the pressures and particle velocities on both sides)." Their argument seems right to me, since $z=p/v$, where $p$ is the pressure, and $v$ is the particle velocity. Since $p$ and $v$ are both continuous at the boundary, it seems that $z$ must be as well. $\endgroup$ – Ben Crowell Dec 20 '17 at 15:04
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(1) The statement that the characteristic impedance is continuous at the boundary is false. In the document you refer to it is also contradicted at the same page, where is states that the characteristic impedance is different for the two different materials. There is thus a discontinuity at the boundary.

Confusion may arise because typically impedance is defined as the ratio between the complex amplitudes of the pressure and the velocity. And indeed the pressure and the velocity are continuous at the boundary. This seems to suggest that the impedance would also be continuous. However the impedance is actually defined as the ratio between the complex amplitudes of the pressure and the velocity for a traveling wave (either forward or backward. see e.g. here). If you have both forward and backward traveling waves, as in the first medium, you cannot simply obtain the impedance by dividing the pressure and the velocity. An extreme example would be a standing wave. This would, using this naive method, give zero impedance at the pressure nodes and infinite impedance at the pressure antinodes.

(2) The reflection of acoustic or electromagnetic waves always depends on the wave impedance of the two media/waveguides/cables. However for specific types of materials the ratio between $Z_1$ and $Z_2$ is the same as the ratio of the velocities of sound/light in the two media. For light this holds when both media have the same magnetic susceptibility (for instance when they are both nonmagnetic). For sound this holds when both media have the same density.

For more info see:

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  • $\begingroup$ I think there is some misunderstanding in that you think that Z1 and Z2 do not depend on material properties, while z1 and z2 do. This was not what I thought, although I may have expressed it poorly. I didn't say that Z did not depend on the medium. I said that z was a property of the medium. Z is a property both of the medium and the surrounding enclosure, since Z=ρc/A, where A is the cross-sectional area. I'll edit the question to clarify. $\endgroup$ – Ben Crowell Dec 20 '17 at 15:22
  • $\begingroup$ Your (2) makes sense, and seems to be a good answer to my (2). But I still don't see how the contradiction in my (1) is resolved. $\endgroup$ – Ben Crowell Dec 20 '17 at 15:24
  • $\begingroup$ To elaborate on Crimson's answer, physical properties of the medium change the specific acoustic impedance $z$ at specific frequencies: physics.stackexchange.com/questions/141449/… $\endgroup$ – D. Betchkal Dec 20 '17 at 22:59
  • $\begingroup$ Also helpful is this PDF by Matthew Schwartz, §7 "Impedance for other stuff" which helps show where the $A$ term becomes useful. $\endgroup$ – D. Betchkal Dec 21 '17 at 0:06
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    $\begingroup$ I think I now understand the confusion around (1). I adapted my answer. $\endgroup$ – Crimson Dec 21 '17 at 9:41

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