2.4 Pressure & Acoustics

Step 1: Force balance on a fluid element

Consider a thin horizontal slab of fluid at depth z with thickness dz and cross-sectional area A. The slab is in static equilibrium, so the net upward pressure force must balance the gravitational force (weight).

$$p(z + dz) \cdot A - p(z) \cdot A = \rho \, g \, A \, dz$$

Step 2: Differential form

Dividing both sides by A and taking the limit as dz approaches zero, we obtain the fundamental hydrostatic differential equation. The pressure gradient is proportional to the local fluid density and gravitational acceleration.

$$\frac{dp}{dz} = \rho(z) \, g$$

Step 3: Integration for constant density

If the fluid is incompressible (constant density $\rho$), we integrate directly from the surface (z = 0, where $p = p_0$) to depth z:

$$\int_{p_0}^{p} dp' = \int_0^z \rho \, g \, dz' \quad \Longrightarrow \quad p(z) = p_0 + \rho g z$$

Step 4: Variable-density generalization

In the real ocean, density varies with depth due to temperature, salinity, and compressibility. The integral form accounts for this by keeping $\rho(z')$ inside the integrand:

$$p(z) = p_0 + \int_0^z \rho(z') \, g \, dz'$$

Step 5: Numerical estimate

Using $\rho = 1025$ kg/m$^3$ and $g = 9.81$ m/s$^2$, the pressure increase per meter of depth is:

$$\frac{dp}{dz} = 1025 \times 9.81 = 10{,}055 \text{ Pa/m} \approx 1 \text{ dbar/m} \approx 0.1 \text{ atm/m}$$

This confirms the convenient rule: pressure in decibars is approximately equal to depth in meters.

Step 1: Definition of bulk modulus

The adiabatic bulk modulus $K_s$ measures a fluid's resistance to compression. It is defined as the ratio of an infinitesimal pressure increase to the resulting fractional decrease in volume (at constant entropy):

$$K_s = -V \frac{\partial p}{\partial V}\bigg|_s = \rho \frac{\partial p}{\partial \rho}\bigg|_s$$

Step 2: 1D wave equation in a fluid

Combining the linearized Euler equation (momentum conservation) and the continuity equation for small perturbations of pressure $p'$ and velocity $u'$ about a rest state:

$$\rho_0 \frac{\partial u'}{\partial t} = -\frac{\partial p'}{\partial x}, \qquad \frac{\partial p'}{\partial t} = -K_s \frac{\partial u'}{\partial x}$$

Step 3: Eliminate velocity to get the wave equation

Differentiate the first equation with respect to x and the second with respect to t, then combine to eliminate $u'$. This yields the classical wave equation for pressure perturbations:

$$\frac{\partial^2 p'}{\partial t^2} = \frac{K_s}{\rho_0} \frac{\partial^2 p'}{\partial x^2}$$

Step 4: Identify the wave speed

The standard wave equation has the form $\partial^2 p'/\partial t^2 = c^2 \, \partial^2 p'/\partial x^2$. By comparison, the phase speed of sound is:

$$c = \sqrt{\frac{K_s}{\rho}}$$

This is the Newton-Laplace equation. Newton originally derived $c = \sqrt{K_T/\rho}$ using the isothermal bulk modulus, which underestimated the speed of sound in air by about 16%. Laplace corrected this by recognizing that sound propagation is adiabatic (too rapid for heat exchange), requiring the adiabatic modulus $K_s = \gamma K_T$.

Step 5: Application to seawater

For seawater at the surface, $K_s \approx 2.34 \times 10^9$ Pa and $\rho \approx 1025$ kg/m$^3$:

$$c = \sqrt{\frac{2.34 \times 10^9}{1025}} \approx 1511 \text{ m/s}$$

Since both $K_s$ and $\rho$ depend on temperature, salinity, and pressure, the sound speed inherits these dependencies. The empirical Mackenzie equation captures these effects through polynomial fits to laboratory measurements.

Step 1: Snell's law for a continuously stratified medium

In a horizontally stratified ocean where the sound speed $c(z)$ varies only with depth, Snell's law requires that a conserved quantity (the ray parameter p) remains constant along each ray path. Here $\theta$ is the angle measured from horizontal:

$$p = \frac{\cos\theta(z)}{c(z)} = \frac{\cos\theta_0}{c_0} = \text{constant}$$

Step 2: Ray turning condition

A ray turns (reverses vertical direction) when $\theta = 0°$ (horizontal propagation), i.e., when $\cos\theta = 1$. This occurs at a turning depth $z_t$ where:

$$\frac{1}{c(z_t)} = \frac{\cos\theta_0}{c_0} \quad \Longrightarrow \quad c(z_t) = \frac{c_0}{\cos\theta_0}$$

Since $\cos\theta_0 < 1$ for any non-horizontal launch, the turning depth occurs where $c(z_t) > c_0$. The ray bends toward regions of lower sound speed.

Step 3: Sound speed profile with a minimum

In the typical ocean, the sound speed profile has a minimum at the SOFAR axis depth $z_{\text{SOFAR}}$. Above, temperature decrease with depth causes c to decrease. Below, pressure increase causes c to increase:

$$c(z) > c_{\min} \quad \text{for all } z \neq z_{\text{SOFAR}}$$

Step 4: Trapping condition

A ray launched near the SOFAR axis at a small angle $\theta_0$ from horizontal has a turning sound speed $c_t = c_0/\cos\theta_0$ only slightly above $c_{\min}$. Because c increases both above and below the axis, the ray encounters its turning sound speed on both sides and oscillates sinusoidally about the axis:

$$c_{\min} < c_0 / \cos\theta_0 < c_{\text{surface}}, \; c_{\text{bottom}} \quad \Longrightarrow \quad \text{ray is trapped}$$

Step 5: Maximum trapping angle

A ray remains trapped as long as its turning sound speed does not exceed the sound speed at the surface or bottom. The maximum trapping angle $\theta_{\max}$ is determined by the lesser of $c_{\text{surface}}$ and $c_{\text{bottom}}$:

$$\cos\theta_{\max} = \frac{c_{\min}}{\min(c_{\text{surface}}, c_{\text{bottom}})}$$

Rays launched at angles steeper than $\theta_{\max}$ escape the channel and interact with the surface or bottom, suffering additional losses. The SOFAR channel thus acts as a natural acoustic waveguide for rays within this angular window.

Step 6: Cylindrical spreading in the channel

Trapped rays spread only horizontally (in 2D), not vertically. Energy that would spread spherically ($\propto 1/r^2$) in an unbounded medium instead spreads cylindrically ($\propto 1/r$). The transmission loss becomes:

$$TL_{\text{SOFAR}} = 10\log_{10}(r) + \alpha r \quad \text{(dB)}$$

compared to $20\log_{10}(r)$ for spherical spreading. This 10 dB-per-decade advantage, combined with low absorption at low frequencies, allows SOFAR-channel sounds to propagate thousands of kilometers.

Step 1: Spherical spreading (geometric loss)

A point source in a homogeneous unbounded medium radiates energy uniformly in all directions. At range r, the total power P passes through a sphere of area $4\pi r^2$. The intensity (power per unit area) is:

$$I(r) = \frac{P}{4\pi r^2}$$

Step 2: Definition of transmission loss in decibels

Transmission loss (TL) is defined as the ratio of intensity at range r to the intensity at a reference distance of 1 m, expressed in decibels:

$$TL = 10 \log_{10}\!\left(\frac{I(1\text{ m})}{I(r)}\right) = 10\log_{10}\!\left(\frac{r^2}{1^2}\right) = 20\log_{10}(r)$$

Step 3: Absorption as exponential decay

In addition to geometric spreading, the medium absorbs acoustic energy. The intensity decays exponentially with distance due to viscous dissipation and chemical relaxation processes:

$$I(r) = I_0 \, e^{-2\beta r} \quad \text{where } \beta \text{ is the amplitude attenuation coefficient (Np/m)}$$

Step 4: Conversion to decibels per unit distance

Converting the exponential decay to decibel notation using $10\log_{10}(e^{-2\beta r}) = -20\beta r \log_{10}(e) = -8.686 \beta r$, we define the absorption coefficient in dB/km as $\alpha = 8686 \beta$ (with r in km):

$$TL_{\text{absorption}} = \alpha \cdot r \quad \text{(dB, with } r \text{ in km and } \alpha \text{ in dB/km)}$$

Step 5: Combined transmission loss

Since geometric spreading and absorption are independent loss mechanisms, their decibel contributions add linearly. Converting r to meters for the spreading term and km for the absorption term (or using consistent units):

$$TL = 20\log_{10}(r) + \alpha \cdot r \quad \text{(dB)}$$

Step 6: Frequency dependence of absorption

The absorption coefficient $\alpha(f)$ has contributions from chemical relaxation processes (boric acid at ~1 kHz, magnesium sulfate at ~50 kHz) and pure-water viscosity (dominant above 100 kHz). Each relaxation process contributes a term of the form:

$$\alpha_i(f) = \frac{A_i f_i f^2}{f^2 + f_i^2}$$

This is a Debye relaxation form: absorption is proportional to $f^2$ below the relaxation frequency $f_i$ and approaches a constant $A_i f_i$ above it. The total absorption is the sum of all contributions, giving the Francois-Garrison equation.