# Waves in water always circular

I have had a question since childhood. Why do we always get circular waves (ripples) in water even when we throw irregularly shaped object in it?

• I suppose that irregularly shaped object is small compared with the ripples? – kennytm Nov 25 '10 at 6:55
• Perhaps you got my question wrong. I have edited it. – Sid Nov 25 '10 at 7:57
• Even for a long stick? – user68 Nov 25 '10 at 9:27

Actually the ripples are not circular at all. See photo below.

For example, a long stick will generate a straight water front on from its sides and circular waves from its edges. Something similar to a rectangle where the two short sides are replaced by semi-circles.

As the waves spread, the straight front will retain its length, whereas the circular sides will grow in bigger and bigger circles, hence the impression that on a large body of water the waves end up being circular - they are not, but very close.

The reason that an irregular object generates "circular" ripples is therefore this: as the waves propagate, the irregularities are maintained but spread across a larger and larger circular wave front.

A very good example of this phenomenon is the Cosmic Microwave Background (CMB) where electromagnetic waves from the Big Bang are measured after having spread for 13.7 billion years. Although the CMB is really, really smooth - because of the "circular ripple" effect, if you like, we can still measure small irregularities, which we think are due to the "irregular shape" of the Big Bang at a certain time.

• The usual name for "as the waves propagate, the irregularities are maintained but spread across a larger and larger circular wave front" is Huygen's principle en.wikipedia.org/wiki/Huygens%27_principle – j.c. Nov 25 '10 at 15:47
• Correct, I've added your link to the answer. Thank you. – Sklivvz Nov 25 '10 at 19:42

I see this question already has an accepted answer, but I'll add a few general notes for completeness.

We begin with the question of how a free surface, i.e. the interface between two fluids, responds to a pressure (ie normal stress) disturbance.

This is the Cauchy-Poisson problem. Cauchy famously solved this problem, originally a prize question posed by the French Academy of Sciences, in 1815 at the age of 26. Poisson, one of the judges, added to this in his 1816 paper, and Cauchy published his work in his 1827 memoir, with an additional several hundred pages of notes.

Now, we consider an infinitely deep, infinite basin of water with a quiescent surface at $z=0$, above which there is air, which for simplicity we assume has pressure $P_\textrm{atm}=0$. We assume the only restoring force here is gravity, since we are thinking ahead to modeling the coarse grain features of the fluid response. Note, this scenario is different than the image shown by @Sklivvz, in which capillary effects are present. These effects are very interesting, and at the end of this post I will some remarks about this phenomenon.

We begin by assuming the flow is irrotational, which means that there exists a velocity potential $\phi$, where $\nabla \phi = \textbf{u}$, with $\textbf{u}$ the fluid velocity, such that $\nabla^2\phi =0$ everywhere in the water. The conditions at the free surface $z=\eta$ are $$\eta_t+\nabla \phi \cdot \nabla \eta= \phi_z,\tag i$$ i.e. the kinematic boundary condition, constraining fluid particles on the surface to remain on the surface, and $$\phi_t+\frac{1}{2}(\nabla \phi)^2 +gz = 0,\tag{ii}$$

which is the dynamic boundary condition, ensuring continuity of pressure across the interface.

Finally, we have the condition that there is no flow at the bottom, i.e. $$\phi_z \to 0 \quad as \quad z\to -\infty\tag{iii}$$

Now, although the governing equation is linear (it's Laplace's equation), the boundary conditions are nonlinear and are evaluated at one of the variables that we are solving for, namely $\eta$. To make progress then, we assume the velocities and surface heights are weak/small, so that we can linearize these equations and evaluate them at $z=0$. The boundary conditions become

$$\eta_t = \phi_z,\tag{i'}$$ and $$\phi_t +g\eta = 0 ,\tag{ii'}$$ where again, these are evaluated at $z=0$. Now, for a particular wave with wavenumber $k$, the solution to Laplace's equation, with the condition (iii), can be found by assuming a separation of variables, which implies

$$\phi = \zeta(x,y,t) e^{kz}.$$

This means that Laplace's equation becomes

$$\nabla^2 \phi = \phi_{xx}+\phi_{yy}+k\phi_z = 0,$$

which, open substitution of $(i')$ and $(ii')$ becomes

$$\phi_{tt} - \frac{g}{k}\nabla^2_H \phi =0,$$

where $\nabla_H \equiv \partial_{xx}+\partial_{yy}$ and we recognize $\frac{g}{k}$ as the square of the deep water phase velocity, so that the above is a 2d wave equation. Note, these waves are dispersive, i.e. the phase velocity inversely depends on the wavenumber. Long waves travel faster then short waves.

Next, we assume the solutions are oscillatory in time (which can formally be shown to be the case when we assume that $\zeta$ is separable in space and time), with frequency $\omega(k)=\sqrt{gk}$, i.e. $\zeta = \psi(x,y)e^{-i\omega t}$, so that our governing equation becomes

$$\nabla^2_H \psi + k^2 \psi=0,$$

which we recognize as Helmholtz's equation. (The $k$ in the dispersion relation $\omega(k)$ for propagation along the surface is equal to $k$ in $e^{kz}$ because of the Laplace equation.)

Now, let's start with the case where we assume our solution will have a circular symmetry, and build more interesting cases from there. We transform our governing equation into polar coordinates $(r,\theta)$ to find

$$\psi_{rr} +\frac{1}{r}\psi_r +k^2 \psi = 0.$$

This is a bessel equation with solution $\psi = J_o(kr)$, where $J_o$ is a zero order Bessel function of the first kind.

This analysis was done for one wavenumber $k$, but our operators are all linear so in general the solution will be a linear superposition of these waves, i.e.

$$\eta(r,\theta,t) = \int_0^{\infty}f(k) J_o(kr)e^{-i\omega(k)t} \ dk,$$

where $f(k)$ represent the mode coefficients, which are determined by the initial conditions.

For example, if we consider the response of the fluid to a point disturbance, we have

$$\eta(r,\theta,0) =\delta(r),$$

where $\delta$ is the dirac delta function. We find $f(k)$ by the Hankel-transform, which tells us

$$f(k)= \int_0^{\infty} \delta(r) J_o(kr) \ dr =1.$$

Therefore, the governing equation is

$$\eta(r,\theta,t) = \int_0^{\infty} J_p(kr)e^{-i\omega(k)t} \ dk.$$

These integrals (related to Hankel-Transforms) are notoriously difficult to solve and progress is usually only made under asymptotic conditions, when the method of stationary phase can be applicable. For instance, under the assumption $gt^2/r \ll 1$, we find (for details see Lamb, 1932, section 239)

$$\eta(r,\theta,t) \sim \frac{\sqrt{g}t}{r^{\frac{5}{2}}}\left(\cos gt^2/4r - \sin gt^2/4r\right).$$

A sample solution is shown below.

So, we have finally developed some of the machinery necessary to talk about your question! Assuming that dropping an object into the water generates waves that are linear (this is clearly debatable for certain objects, as well as the time/length scales you're concerned with, but it is a point I will not address here), we can use superposition to model an object as a series of point excitations.

For instance, what happens when we drop a stick into the water? Well, if we model this as a super position of a bunch of point sources along the $y=0$ axis, we find the solution looks like

$$\eta(x,y,t) =\sqrt{g}t^2 \lim_{N\to \infty}\frac{1}{N}\sum_{n=1}^N \frac{1}{((x-x_n)^2+y^2)^{\frac{5}{2}}} \left(\cos \frac{gt^2}{4\sqrt{(x-x_n)^2+y^2}} - \sin \frac{gt^2}{4\sqrt{(x-x_n)^2+y^2}}\right),$$

where $x_n = n/N$, for instance. Locally, this disturbance will not produce symmetric rings, and indeed will have regions that have very minimal curvature. However, for $x\gg x_n$, this clearly will take the form of the symmetric rings given in the first example. A sample of this type of disturbance is shown below.

We can see here that the wave front is "flatter" along the y-axis.

So to conclude, a lot has to do with the scales over which you are making your observation. But it is shown, at least qualitatively, that non-symmetric patterns can be formed for axially asymmetric forcing.

I might have rambled a bit, but I hope this helps!

• Nick

Some other notes:

-This is a crude way to model swell, with a storm over the ocean acting as a normal pressure disturbance. Indeed, one can make asymptotic estimates of the wavelength as a function of distance from the disturbance, which has been corroborated for swell. Also, deep water wave dispersion implies, as surfers empirically know, when a swell arrives, the longer period waves show up first.

-If we consider a point pressure disturbance on a uniformly moving stream, one gets the famous Kelvin ship wake pattern.

• Capillary effects are very interesting in these problems. However, how to model all of the combined effects related to capillarity is nontrivial. For instance, one cannot model capillary waves without including viscous dissipation (these waves have high wavenumber, and viscous damping goes as $k^2$, hence, it's more important for these waves), as well as boundary layer effects, since capillary waves can be a strong source of vorticity. This vorticity can lead to wave front curvature, and further asymmetries which are nontrivial to model. This is an active area of research.

References:

Lamb (1932) Hydrodynamics

Whitham (1974) Linear and nonlinear waves

• +1 A handy reference for waves in water. I'd never actually appreciated that the nonlinearity people talk about in water waves is in the boundary conditions. Is this right? I guess I kind of assumed it came from the convective term in the full Navier-Stokes equations. So if you did this analysis in a finite depth, sloping bottom tank (i.e. say numerically with the full nonlinearity of the boundary conditions taken account of), do you get growing waves as they reach the "beach" that behave somewhat like real ones? – WetSavannaAnimal Dec 20 '13 at 0:15
• You can rewrite the advective term in the N-S equations as a divergent term and a rotational term so that for irrotational flow the complexity of this term, i.e. the rotational component, drops out. This is why potential flow is so developed, and also boring. Indeed, for irrotational water waves, it is the boundary conditions (which dictate all of the dynamics in the interior) that lead to the complexity. (I'll answer your question in my next comment). – Nick P Dec 20 '13 at 1:17
• The description for shallow water is analogous, except b.c. iii) changes to $\phi_z\to 0$ at $z=-h$, where $h$ is the depth of the water. To see how waves behave as they approach a beach, one would simply consider conservation of energy. For simplicity, we assume the flow is time independent, so that we have $\partial (c a^2)\partial x = 0$ which means $c a^2 = const$, where $c$ is the shallow water phase velocity $c=\sqrt{gh}$ and $a$ is the amplitude of the incident wave. This means that $a \sim h^{-\frac{1}{4}}$ so that as $h\to 0$, $a$ gets large and at some point "breaks". – Nick P Dec 20 '13 at 1:21
• Thanks: I really like your plots. Are you going to upload the "Manipulate" command and definitions to the Wolfram Demonstrations Project? I suggest this would be a good want to show. Another with a sloping bottom would be awesome! – WetSavannaAnimal Dec 20 '13 at 1:44
• @NickP I took the liberty to correct a typo in ii') and change $\eta$ to $\phi$ in the wave-equation. This is consistent with Eq.(4) en.wikipedia.org/wiki/Airy_wave_theory But I see a problem with your solution of the Helmholtz equation: you obtain standing waves only. A propagating wave, entering the Green's function for Helmholtz equation, should be a Hankel function, no just the Bessel function of the first kind. Also, I cannot agree with your decomposition of a point source into Bessel functions - $f(k)$ should be $k$, not 1. – Slaviks Jul 5 '14 at 20:48

waves always travel with a constant speed. For waves in water to travel at a constant speed they need to be circular. And hence the ripples in water are always circular.

• This has the same problem as Santosh Linkha's answer. It only explains waves coming from a point perturbation. See my comment below his answer. – Arnoques Nov 29 '11 at 22:31

I don't think that water ripples are circular in nature because of the size of the object is small compared to ripples because you can even create perfect ripple with kitchen knife.

It is circular in nature because the transmission of disturbance from a source event takes place in a wave fashion in a 2-D medium. And intensity of wave is distributed equally in all directions of medium.

As you know wave motion all particles in ripple execute a harmonic motion, they remain in phase i.e. particles at equal distance from center remain at equal height, there by creating a circle.

• This only explains why waves that come from a single point perturbation are circular. But for an irregular object the waves coming from different point will interfere and will create a pattern that's not circular. It will become asymptotically circular when the waves get much bigger than the original perturbation. – Arnoques Nov 29 '11 at 22:29

(the same question was posted on Quora and this is a copy of my answer there. Most answers there really boil down to, "It's circular. Just look at it! You get circles! That's why it's circular.")

I'd phrase it like this: The primary idea of most wave equations is that disturbances will only propagate at speed $c$. So we can look at the ideal propagated disturbance as opposed to worrying about real world water waves. (this is for the best anyways, because you could apply it to explosions or electromagnetic waves w/ no loss of generality)

So, we have some shape $S$, which is a set of points in the plane (let's take it to be the filled-in shape with its boundary). We want to look at how the set $D(t)$ changes over time, where $D(t)$ is the set of points which has a distance from some point of $S$ less than or equal to $ct$. (so $D(t)$ expands outwards in a "zone of causality")

Theorem. $\mbox{Area}(D(t))$ ~ $\pi (c t)^2$. That is:

Theorem. The area of $D(t)$ is asymptotic to $\pi (c t)^2$. That is:

Theorem. $\lim_{t\to\infty} \frac{D(t)}{\pi (c t)^2}=1$. That is:

Theorem. The zone is basically a circle. (OK, this isn't equivalent to the statement above, but the proof below can show it.)

Proof. Pick any point of the plane to be the center of your circle. Whenever I say "the circle" from now on, it means the circle with that center. At $t=0$, make one circle that encompasses the shape you start with completely. Then, if the circle has radius $r_0$, wait until time $t=\frac{2 r_0}{c}$. Now the "zone of causality" contains the circle you started with. At any time $t$, then, the border of the zone of causality will be sandwiched between the circle of radius $ct+r_0$ and the circle of radius $ct-r_0$ (which has radius $r_0$ at the time in question). Then $\pi (ct+r_0)^2\ge \mbox{Area}(D(t)) \ge \pi (ct-r_0)^2$ for all $t> \frac{2 r_0}{c}$. So $(1+\frac{r_0}{ct})^2\ge \frac{ \mbox{Area}(D(t))}{\pi (ct)^2} \ge (1-\frac{r_0}{ct})^2$. As $t\to\infty$ the left and right go to one, so the middle must too. $\square$

The statements of the form "this contains that" can be made completely rigorous using inequalities like the triangle inequality. This says nothing about the border of the zone of causality.

Equation $$η(r,θ,t)∼\frac{\sqrt g t}{r^{5/2}}(\cos gt^2/4r−\sin gt^2/4r)$$ is not physically consistent

may be

$$η(r,θ,t)∼\frac{\sqrt g t}{r^{1/2}}(\cos gt^2/4r−\sin gt^2/4r) ?$$

• may be η(r,θ,t)∼√g t/r^-1/2(cosgt2/4r−singt2/4r) ? – gabriel roson porto Jul 18 '18 at 10:22
• We do have MathJax here, I've edited this in for you but please try to keep this in mind going forwards. – Kyle Kanos Jul 18 '18 at 11:24
• Though this seems to be more of a comment on this post than an answer. – Kyle Kanos Jul 18 '18 at 11:25