3.1 The impedance problem

Sound travels well in air because the air molecules can move easily. Sound travels well in water for the same reason at the molecular level, but for a different reason at the macroscopic level: water is dense and incompressible, so any compression propagates very rapidly. The relevant property is acoustic impedance, defined as the ratio of pressure to particle velocity in a propagating wave: $Z = p/v$ .

For air at room temperature, $Z_\text{air} \approx 415\,\text{Pa}\cdot\text{s/m}$ . For water — and for the perilymph of the cochlea, which is essentially salt water — $Z_\text{water} \approx 1.5 \times 10^6\,\text{Pa}\cdot\text{s/m}$ . The ratio is about 3500. This is enormous.

Consider what happens when a sound wave in air meets a water surface. The pressure has to be continuous across the interface, and so does the particle velocity. But $Z_\text{water}/Z_\text{air} = 3500$ means that for a given pressure, the water moves about 3500 times less than the air on the other side. The acoustic wave cannot push the water hard enough to keep itself going. Most of its energy reflects back.

You can compute exactly how much. The reflection coefficient at a boundary between two media is

R = \frac{Z_2 - Z_1}{Z_2 + Z_1}.

▶ Derivation: reflection at a boundary between two media

Consider an acoustic plane wave $p_i = p_0\, e^{i(\omega t - k_1 x)}$ traveling rightward in medium 1 (impedance $Z_1$ ) toward a boundary at $x = 0$ . At the boundary, some of the wave reflects ( $p_r$ ) and some transmits ( $p_t$ ) into medium 2 (impedance $Z_2$ ):

p_r = R\, p_0\, e^{i(\omega t + k_1 x)}, \quad p_t = T\, p_0\, e^{i(\omega t - k_2 x)}.

Two boundary conditions hold at $x = 0$ :

Pressure is continuous: $p_i + p_r = p_t$ at $x = 0$ , giving $1 + R = T$ .
Particle velocity is continuous: $v_i + v_r = v_t$ . Using $v = p/Z$ for a forward wave and $v = -p/Z$ for a backward wave: $(p_i - p_r)/Z_1 = p_t/Z_2$ , so $(1 - R)/Z_1 = T/Z_2$ .

From (1), $T = 1 + R$ . Substituting into (2):

\frac{1 - R}{Z_1} = \frac{1 + R}{Z_2}.

Cross-multiplying: $Z_2 (1 - R) = Z_1 (1 + R)$ , so $Z_2 - Z_1 = R(Z_1 + Z_2)$ , giving

R = \frac{Z_2 - Z_1}{Z_1 + Z_2}.

Plug in $Z_1 = Z_\text{air}$ and $Z_2 = Z_\text{water}$ : $R \approx 0.999$ . The intensity reflection coefficient is $|R|^2 \approx 0.9989$ , which means the transmitted power fraction is

T = 1 - |R|^2 \approx 0.0011,

or in decibels, about $-29.6$ dB. More than 99.9% of incoming sound power would reflect off the cochlear fluid if it were simply exposed to air. A sound at 60 dB SPL in air would arrive at the cochlea at the equivalent of 30 dB SPL — most of speech would fall below the threshold of audibility.

This is what the middle ear has to fix.