1.1 Overview of the Oceans

Step 1: Define the cumulative area distribution

The hypsometric curve describes the fraction of Earth's surface area that lies above (or below) a given elevation $h$. Let $A_{\text{total}}$ be the total surface area and $a(h)$ the area per unit elevation at height $h$. The cumulative hypsometric function is:

$$H(h) = \frac{1}{A_{\text{total}}} \int_{h}^{h_{\max}} a(h')\, dh'$$

Step 2: Normalize to fractional area

$H(h)$ ranges from 0 (at the highest peak) to 1 (at the deepest trench). Since $\int_{h_{\min}}^{h_{\max}} a(h')\,dh' = A_{\text{total}}$, we have:

$$H(h_{\min}) = 1, \quad H(h_{\max}) = 0$$

Step 3: Relate to the depth/elevation histogram

The area density function $a(h)$ is the negative derivative of the cumulative function scaled by total area:

$$a(h) = -A_{\text{total}} \frac{dH}{dh}$$

Step 4: Compute the mean depth from the hypsometric curve

The mean elevation (or depth, for the ocean) is the area-weighted integral over the hypsometric curve:

$$\bar{h} = \frac{1}{A_{\text{total}}} \int_{h_{\min}}^{h_{\max}} h \cdot a(h)\, dh = \int_0^1 h(H)\, dH$$

Step 5: Bimodal distribution interpretation

Earth's hypsometric curve is distinctly bimodal, with peaks near $+100$ m (continental platforms) and $-4500$ m (abyssal plains). This bimodality arises from the density contrast between continental crust ($\rho_c \approx 2700$ kg/m³) and oceanic crust ($\rho_o \approx 3000$ kg/m³) floating isostatically on the mantle ($\rho_m \approx 3300$ kg/m³):

$$\Delta h = h_{\text{continent}} - h_{\text{ocean}} = t_c \left(1 - \frac{\rho_c}{\rho_m}\right) - t_o \left(1 - \frac{\rho_o}{\rho_m}\right)$$

Step 6: Numerical result

For typical continental crust thickness $t_c \approx 35$ km and oceanic crust $t_o \approx 7$ km, the elevation difference is approximately:

$$\Delta h \approx 35 \times 0.182 - 7 \times 0.091 \approx 6.37 - 0.64 \approx 5.7 \text{ km}$$

This is consistent with the observed ~4.5 km difference between the two peaks of the hypsometric curve, confirming that isostasy controls Earth's large-scale topography.

Step 1: State the fundamental attenuation principle

As light propagates downward through seawater, it is attenuated by absorption and scattering. The change in irradiance $I$ over an infinitesimal depth interval $dz$ is proportional to the irradiance itself:

$$dI = -K_d \cdot I \cdot dz$$

Step 2: Write as a differential equation

Rearranging gives a first-order linear ODE where $K_d$ (m⁻¹) is the diffuse attenuation coefficient, encompassing absorption by water, dissolved organics (CDOM), phytoplankton pigments, and scattering by particles:

$$\frac{dI}{dz} = -K_d \cdot I$$

Step 3: Separate variables and integrate

Separating variables and integrating from the surface ($z = 0$, $I = I_0$) to depth $z$:

$$\int_{I_0}^{I(z)} \frac{dI'}{I'} = -K_d \int_0^z dz' \quad \Longrightarrow \quad \ln\left(\frac{I(z)}{I_0}\right) = -K_d z$$

Step 4: Exponentiate to obtain Beer's Law

Exponentiating both sides yields the Beer-Lambert law for underwater light:

$$I(z) = I_0 \, e^{-K_d z}$$

Step 5: Define the euphotic zone depth

The euphotic zone is conventionally defined as the depth where irradiance falls to 1% of the surface value. Setting $I(z_{\text{eu}}) = 0.01 \, I_0$:

$$0.01 = e^{-K_d z_{\text{eu}}} \quad \Longrightarrow \quad z_{\text{eu}} = \frac{\ln(100)}{K_d} = \frac{4.605}{K_d}$$

Step 6: Typical values and wavelength dependence

In the clearest ocean water (Sargasso Sea), $K_d \approx 0.02$ m⁻¹ at 475 nm (blue), giving $z_{\text{eu}} \approx 230$ m. In coastal waters with high CDOM and phytoplankton, $K_d \approx 0.5$ m⁻¹, giving $z_{\text{eu}} \approx 9$ m. The attenuation coefficient can be decomposed into additive contributions:

$$K_d(\lambda) = K_w(\lambda) + K_{\text{CDOM}}(\lambda) + K_{\text{phyto}}(\lambda) + K_{\text{det}}(\lambda)$$

where $K_w$ is pure water absorption, $K_{\text{CDOM}}$ is colored dissolved organic matter, $K_{\text{phyto}}$ is phytoplankton pigment absorption, and $K_{\text{det}}$ is detrital particle scattering.