Kelvin wave angle
Contents
| Section | Topic |
|---|---|
| 1 | Group velocity |
| 2 | Stationarity condition |
| 3 | Ray angle |
| 4 | Maximum ray angle |
1. Group velocity
Two wave components with slightly different \(k\) and \(\omega\):
\[ \eta_1 = a\sin(\omega_1t - k_1x), \qquad \eta_2 = a\sin(\omega_2t - k_2x) \]
\[ \eta = \eta_1 + \eta_2 = a\sin(\omega_1t - k_1x) + a\sin(\omega_2t - k_2x) \]
\(\sin P + \sin Q = 2\sin\left(\dfrac{P + Q}{2}\right)\cos\left(\dfrac{P - Q}{2}\right)\), with \(P = \omega_1t - k_1x\) and \(Q = \omega_2t - k_2x\):
\[ \frac{P + Q}{2} = \frac{\omega_1 + \omega_2}{2}\,t - \frac{k_1 + k_2}{2}\,x, \qquad \frac{P - Q}{2} = \frac{\omega_1 - \omega_2}{2}\,t - \frac{k_1 - k_2}{2}\,x \]
\[ \eta = \underbrace{2a\cos\left(\frac{\omega_1 - \omega_2}{2}\,t - \frac{k_1 - k_2}{2}\,x\right)}_{\text{envelope}}\; \underbrace{\sin\left(\frac{\omega_1 + \omega_2}{2}\,t - \frac{k_1 + k_2}{2}\,x\right)}_{\text{carrier}} \]
Each part moves like a single wave: follow a point of constant phase, as in Dispersion relation, section 1.
Carrier. Constant phase between \((x_1, t_1)\) and \((x_2, t_2)\):
\[ \frac{\omega_1 + \omega_2}{2}\,t_1 - \frac{k_1 + k_2}{2}\,x_1 = \frac{\omega_1 + \omega_2}{2}\,t_2 - \frac{k_1 + k_2}{2}\,x_2 \]
\[ \frac{k_1 + k_2}{2}\,(x_2 - x_1) = \frac{\omega_1 + \omega_2}{2}\,(t_2 - t_1) \]
\[ \frac{x_2 - x_1}{t_2 - t_1} = \frac{\omega_1 + \omega_2}{k_1 + k_2} \]
The two components are close: \(k_2 \rightarrow k_1 = k\), \(\;\omega_2 \rightarrow \omega_1 = \omega\):
\[ \frac{\omega_1 + \omega_2}{k_1 + k_2} \rightarrow \frac{2\omega}{2k} = \frac{\omega}{k} \]
\[ c = \frac{\omega}{k} \quad \rightarrow \quad \text{phase velocity} \tag{1} \]
Envelope. Constant phase between \((x_1, t_1)\) and \((x_2, t_2)\):
\[ \frac{\omega_1 - \omega_2}{2}\,t_1 - \frac{k_1 - k_2}{2}\,x_1 = \frac{\omega_1 - \omega_2}{2}\,t_2 - \frac{k_1 - k_2}{2}\,x_2 \]
\[ \frac{k_1 - k_2}{2}\,(x_2 - x_1) = \frac{\omega_1 - \omega_2}{2}\,(t_2 - t_1) \]
\[ \frac{x_2 - x_1}{t_2 - t_1} = \frac{\omega_1 - \omega_2}{k_1 - k_2} \]
Both components satisfy the dispersion relation: \(\omega_1 = \omega(k_1)\), \(\;\omega_2 = \omega(k_2)\). As \(k_2 \rightarrow k_1\):
\[ \frac{\omega(k_1) - \omega(k_2)}{k_1 - k_2} \rightarrow \frac{d\omega}{dk} \qquad \text{(definition of the derivative)} \]
\[ c_g = \frac{d\omega}{dk} \quad \rightarrow \quad \text{group velocity} \tag{2} \]
\(c_g\) from \(c\). From (1), \(\omega = ck\). Product rule:
\[ c_g = \frac{d\omega}{dk} = \frac{d(ck)}{dk} = c + k\,\frac{dc}{dk} \]
Deep water:
\[ \omega = \pm\sqrt{gk} \quad \rightarrow \quad \sqrt{gk} \tag{3} \]
\[ c = \frac{\omega}{k} = \frac{\sqrt{gk}}{k} = \sqrt{\frac{g}{k}} = g^{1/2}k^{-1/2} \]
\[ \frac{dc}{dk} = -\frac{1}{2}\,g^{1/2}k^{-3/2} \]
\[ k\,\frac{dc}{dk} = -\frac{1}{2}\,g^{1/2}k^{-1/2} = -\frac{1}{2}\sqrt{\frac{g}{k}} = -\frac{1}{2}c \]
\[ c_g = c - \frac{1}{2}c = \frac{1}{2}c \]
Check, directly from (3), \(\omega = g^{1/2}k^{1/2}\):
\[ \frac{d\omega}{dk} = \overset{\text{group}}{c_g} = \frac{1}{2}\sqrt{\frac{g}{k}} = \frac{1}{2}\,\overset{\text{phase}}{c} \tag{4} \]
2. Stationarity condition
Stationarity condition:
\[ c = u\cos\theta \qquad \text{and} \qquad c_g = \frac{1}{2}c \tag{5} \]
\[ c_g = \frac{1}{2}(u\cos\theta) \]
\[ c_{g_x} = \frac{1}{2}(u\cos\theta)\cdot\cos\theta = \frac{1}{2}u\cos^2\theta \tag{6} \]
\[ c_{g_y} = \frac{1}{2}(u\cos\theta)\sin\theta = \frac{1}{2}u\sin\theta\cdot\cos\theta \tag{7} \]
3. Ray angle
Ray angle: the angle between the boat’s path and the line connecting the boat to the wave packet.
- the boat is stationary at the origin
- the wave packet is drifting away
- the wave packet apparent velocity relative to the boat is
\[ x: \; u - c_{g_x} \qquad\qquad y: \; c_{g_y} \]
\[ \tan\alpha = \frac{c_{g_y}}{u - c_{g_x}} \tag{8} \]
\[ \tan\alpha = \frac{\frac{1}{2}u\sin\theta\cdot\cos\theta}{u - \frac{1}{2}u\cos^2\theta} = \frac{\frac{1}{2}\sin\theta\cdot\cos\theta}{1 - \frac{1}{2}\cos^2\theta} \]
Multiply numerator and denominator by \(2\), then divide both by \(\cos^2\theta\):
\[ \frac{\sin\theta\cos\theta}{2 - \cos^2\theta} = \frac{\dfrac{\sin\theta\cdot\cos\theta}{\cos^2\theta}}{\dfrac{2 - \cos^2\theta}{\cos^2\theta}} = \frac{\dfrac{\sin\theta}{\cos\theta}}{\dfrac{2}{\cos^2\theta} - \dfrac{\cos^2\theta}{\cos^2\theta}} = \frac{\tan\theta}{\dfrac{2}{\cos^2\theta} - 1} \]
\[ \frac{2}{\cos^2\theta} - 1 = 1 + 2\tan^2\theta \]
\[ \tan\alpha = \frac{\tan\theta}{1 + 2\tan^2\theta} \qquad\qquad \text{Sub: } \tan\theta = t \tag{9} \]
4. Maximum ray angle
\[ \tan\alpha = \frac{t}{1 + 2t^2} \quad \rightarrow \quad \text{Find the}\ \text{max} \]
\[ f'(t) = 0 = \frac{d}{dt}\left(\frac{t}{1 + 2t^2}\right) = 0 \]
\[ \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \]
\[ \rightarrow \frac{1 + 2t^2 - t\cdot(4t)}{(1 + 2t^2)^2} = \frac{1 - 2t^2}{(1 + 2t^2)^2} = 0 \]
\[ 1 - 2t^2 = 0 \;\rightarrow\; 2t^2 = 1 \;\rightarrow\; t^2 = \frac{1}{2} \;\rightarrow\; t = \frac{1}{\sqrt{2}} \]
\[ \tan\theta = \frac{1}{\sqrt{2}} \;\rightarrow\; \theta = \arctan\left(\frac{1}{\sqrt{2}}\right) = 35.26^\circ \]
Recall:
\[ \tan\alpha = \frac{\tan\theta}{1 + 2\tan^2\theta} = \frac{1/\sqrt{2}}{1 + \frac{2}{2}} = \frac{1/\sqrt{2}}{1 + 1} = \frac{1}{2\sqrt{2}} \]
\[ \arctan\left(\frac{1}{2\sqrt{2}}\right) = \alpha \;\rightarrow\; \boxed{\;\alpha = 19^\circ\,28'\;} \tag{10} \]