# Why is the angle of the wake of a duck constant?

Why is the angle of a wake of a duck constant? And why are some conditions on the water depth necessary?

I realize that this question turns up in google searches, but I did not see a good discussion. I will be quite happy with a link.

Could anyone tell me how the two up-voted answers are related?

• Aug 3, 2017 at 11:42

The ideal Kelvin boat wake ignores surface tension, and it assumes deep water waves with an (in general) broad spectrum of frequencies $$\omega$$ with dispersion relation $$\omega^2=gk$$, where $$g\approx 9.8 \frac{m}{s^2}$$. The ideal Kelvin wake furthermore assumes that the ship sails with a constant velocity, and that the wave amplitudes of the partial waves are so small that they obey a linear superposition principle. The Kelvin wake does not describe the narrow turbulent band behind a ship, nor shock waves. The Kelvin wake consists of two types of waves: transverse and divergent waves. There are two characteristic angles

$$\alpha\approx 19^{\circ} \qquad \mathrm{and} \qquad \beta\approx 35^{\circ},$$

corresponding to

$$\tan(\alpha)= \frac{1}{2\sqrt{2}} \qquad \mathrm{and} \qquad \tan(\beta) = \frac{1}{\sqrt{2}},$$

or equivalently,

$$\sin(\alpha)= \frac{1}{3} \qquad \mathrm{and} \qquad \sin(\beta) = \frac{1}{\sqrt{3}}.$$

In polar coordinates $$(r,\theta)$$ of a co-moving coordinate system, where the position of the boat is at the origin, the transverse waves are in the region $$|\theta|\leq \beta$$, and divergent waves are in the region $$\alpha\leq |\theta|\leq \beta$$.

The angles $$\alpha$$ and $$\beta$$ are constant in at least two ways: Firstly, they don't depend on the distance $$r$$ to the ship. This is because the speed of each partial wave (with frequency $$\omega$$) is independent of the position $$(x,y)$$. Secondly, $$\alpha$$ and $$\beta$$ are, evidently, universal angles, independent of, for instance, $$g$$. This is explained in the references below.

References:

1) Howard Georgi, "The Physics of Waves", Chapter 14. (Hat tip:user1631.)

2) MIT on-line open course ware, mechanical engineering, wave propagation, lecture notes, fall 2006, Chapter 4.7.

3) Wikiwaves.

The constant wake angle is a well known phenomenon for boats and ships and is known as the Kelvin wake. As others have alluded to, it is the result of the unique relation between group and phase velocity for (deep water) gravity waves. The best detailed explanation I have seen in in the textbook "The Physics of Waves" by Howard Georgi, which is available online for free here: http://www.people.fas.harvard.edu/~hgeorgi/new.htm

• I fear you might well be overthinking it. The classic, concise, geometrical argument of Lighthill and Whitham is provided in the [Wikipedia article] (en.wikipedia.org/wiki/Wake) Wake Jan 31, 2016 at 15:29

In particle physics, we usually solve a similar problem with the Čerenkov radiation whose angle is given by $$\cos\theta = \frac{v_{\rm radiation}}{v_{\rm particle}}$$ Note that to get the radiation bounded by a specific angle, the particle has to be faster than the radiation in the given environment. However, for the duck - assuming it is moving in a deep water - the velocity of the "radiation" depends on the frequency. The speed of the duck's paddling is also variable. However, these two uncertainties combine in such a way that the angle is universally 39 degrees or so, independently of the duck's velocity.

Nevertheless, the numerical calculation of the precise numerical value of the angle is clearly difficult and depends on real-world factors. The constancy of the angle may be argued to hold by dimensional analysis and/or scaling laws. The deep water waves have phase velocity proportional to the frequency, $$c_{\rm phase} \sim f$$ This was previously denoted $v_{\rm radiation}$.

The velocity of the duck is also proportional to the frequency of its paddling; each cycle gets you that far (by a fixed distance $\delta x$), so $v_{\rm particle=duck} = \delta x\cdot f$. So if you compute the ratio of the velocities, as in the Čerenkov formula at the top, you get a constant angle because the frequency $f$ cancels.

• Looking at the sentence about dimensional analysis, would you agree that an alien duck paddling slowly in a deep mercury lake is likely to have a constant wake angle? May 5, 2011 at 12:02
• Yes, I would. Is that wrong? I just meant that the angle probably wouldn't be 39 degrees in the alien mercury lake. At any rate, I added an important clarification of the proof, namely that $c_{phase}\sim f$ for the relevant waves. May 5, 2011 at 12:04
• It is interesting that in the image the ducks following the leader take advantage of the shadow, possibly because of lower resistance to motion . Similar to the formation of geese flying. May 5, 2011 at 12:22
• @Lubos, @user9325: The angle is actually given by $\sin \alpha = 1/3,$ or $\alpha \simeq 19$ degrees. The key (or 'sufficient') result from the angle to be universal is the fact that $v_{\text{phase}} = 2v_{\text{group}}.$ If in your alien lake, the dispersion relations scale the same way as those of water, you'd find the exact same angle as in water; if they wouldn't scale like that, there wouldn't be a universal angle at all. Otherwise: the value of the angle isn't specific to water, it depends on the square root in the dispersion relation. May 5, 2011 at 12:55
• Indeed, as you caught, paddling considerations are spurious--the same wake chevron opening of 39 degrees would appear if you dragged a rubber duckie with a string, slowly. The short explanation of the universality of the angle is already in Wikipedia, (en.wikipedia.org/wiki/Wake) , and is due to Lighthill, (Whitham's book cited). Basically, all Cherenkov/sonic boom cones of varying openings configure their visible group wavefront cones to be the same universal one under discussion. Jan 31, 2016 at 20:12

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