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Why is the wake angle always the same?

The V-shaped wake behind a boat, a duck, or a cargo ship always opens at the same angle: or more precisely $\arcsin(1/3)\approx 19.5^\circ$. This is the Kelvin wake pattern, named after Lord Kelvin who worked out the mathematics in 1887 [CITE]. Curiously, the angle does not depend on the size of the object, nor on its speed (as long as the approximation that they are gravity waves holds). A duck, a kayak, and a tanker produce wakes with the same opening angle.

Problem is most of the derivations are not convincing. They typically invoke point disturbances created by the moving object, but these disturbances somehow have well-defined wavelength and are initially localized at the same time (plot twist: this is impossible, it's basically Heisenberg uncertainty relation). Then, they are merged together to form a wave group, and the sought opening angle appears through geometric reasoning.

My approach is not that different. But it is less handwavy – I hope so anyway. This is not to discredit the people who first think it through, I simply have access to more polished mathematics and as I understand, they only had a bunch of heurestics.

The resonance condition

Consider a point-like emitter (say, a boat) moving with constant velocity $\vec v$ through still water. Its position is $\vec x(t) = \vec v t$. The emitter interacts with water waves; it pushes water around, creating a broad spectrum of disturbances.

Here we will describe the disturbances with the usage of waves composed from sums and integrals of 2D plane waves. This is an idealized approach (such waves are characterized by vertical displacement only, and in our model can not break), but captures what happens in the world surprisingly good.

A plane wave with wavevector $\vec k$ and frequency $\omega$ has the form

\begin{equation} \psi(\vec x, t) = A \cos(\omega t - \vec k \cdot \vec x), \end{equation}

such that $\psi(x,y,t)$ can be thought of as vertical displacement of the water table at 2D horizontal position $(x,y)$ at time $t$.

What does the emitter "see" as it moves through this wave? At the emitter's position, the phase of the wave is

\begin{equation} \omega t - \vec k \cdot (\vec v t) = (\omega - \vec k \cdot \vec v) t. \end{equation}

If $\omega \neq \vec k \cdot \vec v$, this phase changes in time—the emitter moves through crests and troughs, and any effect it has on the wave averages out over time. But if

\begin{equation} \omega = \vec k \cdot \vec v, \end{equation}

the emitter always sits at the same phase of the wave. There is a resonance: the emitter can continuously feed energy into (or extract energy from) this particular wave mode. These are the waves that survive in the wake pattern.

Figure 1: A wave satisfying the resonance condition $\omega = \vec k \cdot \vec v$. The emitter (marked point) always sees the same local wave pattern.

The dispersion relation

Water waves in deep water obey the dispersion relation

\begin{equation} \omega^2 = g |\vec k|, \end{equation}

where $g$ is gravitational acceleration. This tells us that longer waves (smaller $\lvert\vec k\rvert$) travel faster. Combining this with the resonance condition $\omega = \vec k \cdot \vec v$, we get

\begin{equation}\label{eq:constraint} \vec k \cdot \vec v = \sqrt{g |\vec k|}. \end{equation}

This is the key equation. It selects which wavevectors can participate in the wake.

The constraint curve

Let's take $\vec v = (v, 0)$—the boat moves in the positive $x$-direction. Writing $\vec k = (k_x, k_y)$, the constraint becomes

\begin{equation} k_x v = \sqrt{g \sqrt{k_x^2 + k_y^2}}. \end{equation}

Squaring both sides:

\begin{equation} k_x^2 v^2 = g \sqrt{k_x^2 + k_y^2}. \end{equation}

Squaring again:

\begin{equation} k_x^4 v^4 = g^2 (k_x^2 + k_y^2). \end{equation}

Rearranging:

\begin{equation} k_y^2 = \frac{v^4}{g^2} k_x^4 - k_x^2 = k_x^2 \left( \frac{v^4 k_x^2}{g^2} - 1 \right). \end{equation}

For real solutions, we need $k_x \geq g/v^2$. The constraint defines a curve in the $(k_x, k_y)$ plane.

Figure 2: The constraint curve in wavevector space. Only wavevectors on this curve can contribute to the wake. The curve starts at $(k_x, k_y) = (g/v^2, 0)$ and extends toward larger $|k_x|$ and $|k_y|$.

The amplitude $A(\vec k)$ for wavevectors on this curve is not something we can determine from first principles without a detailed model of the source. But we can argue that only these wavevectors matter: all others average to zero. The wake is built from waves with

\begin{equation} \text{(density in } \vec k \text{-space)} \propto A(\vec k) \, \delta\!\left( \vec k \cdot \vec v - \sqrt{g|\vec k|} \right). \end{equation}

Figure 3: Selecting a wavevector. When it lies on the constraint curve, the wave crests remain stationary relative to the emitter. Off the curve, the pattern drifts.

Wave groups and group velocity

The constraint curve is one-dimensional: we can parametrize it by a single variable (say, the angle $\theta$ that $\vec k$ makes with the $x$-axis, or simply $k_x$). Different points on the curve correspond to different wavevectors, each contributing a plane wave to the wake.

Now, here is the crucial point. In typical wave physics, the group velocity is $\vec v_g = (\frac{\partial\omega}{\partial k_x}, \frac{\partial \omega}{\partial k_y})$, which for $\omega = \sqrt{g\lvert\vec k\rvert}$ points radially in $\vec k$-space. But we are not dealing with waves throughout all of $\vec k$-space: the wave are constrained to a curve, and the group velocity (of a packet of waves of similar wavevectors) must be defined along this curve.

For waves restricted to a one-dimensional subset of $\vec k$-space, the group velocity is:

Direction: tangent to the constraint curve
Magnitude: $d\omega/ds$, where $s$ is arc length along the curve

Different points on the constraint curve give wave groups propagating in different directions. As you move along the curve, the tangent direction changes, sweeping out a range of angles in real space.

Figure 4a: A 1D wave group. Superposing waves with nearby wavevectors creates a localized packet. The packet envelope moves at the group velocity, which can differ from the phase velocity.

Figure 4b: A 2D wave group from a segment of the constraint curve. The propagation direction (the group velocity direction) is tangent to the curve. Different segments produce wave groups traveling in different directions.

The wake angle

The wake boundary is the envelope of all the wave group directions. Each point on the constraint curve contributes a wave group propagating along the tangent to the curve at that point. The outermost angle—the boundary of the wake—is the maximum angle that any of these tangent directions makes with the boat's velocity.

To find this maximum, we parametrize the curve and compute the angle of the tangent as a function of position along the curve. Setting the derivative of this angle to zero gives the extremal condition. The calculation (which I'll only sketch here) proceeds as follows:

Parametrize the constraint curve, e.g., by $k_x$.
Express $k_y(k_x)$ from the constraint equation.
The tangent direction at each point is $(1, dk_y/dk_x)$.
The angle this makes with the $x$-axis is $\arctan(dk_y/dk_x)$.
Find the maximum of this angle over all valid $k_x$.

The result is

\begin{equation} \theta_{\text{max}} = \arcsin\left(\frac{1}{3}\right) \approx 19.47^\circ. \end{equation}

This is the Kelvin wake half-angle. The full wake opening is $2 \arcsin(1/3) \approx 38.94^\circ$.

Figure 5: All wave groups superposed. Each segment of the constraint curve contributes a wave group traveling in a different direction (tangent to the curve). The envelope of all these directions forms the wake boundary at $\arcsin(1/3) \approx 19.47^\circ$.

Comparison with reality

The prediction matches observation remarkably well. Satellite images of ship wakes, photographs of ducks on ponds, and careful laboratory measurements all show the same $\approx 19^\circ$ half-angle.

Figure 6: A real wake. The opening angle matches the predicted $\arcsin(1/3)$.

There are deviations. In shallow water, the dispersion relation changes (waves "feel" the bottom), and the wake angle can be different. Very fast boats can also produce narrower wakes due to finite hull-length effects.

But for most everyday situations—ducks, kayaks, sailboats, ferries—the $19.47^\circ$ angle holds. The universality comes from the dispersion relation of deep-water gravity waves and the geometry of resonance.