The Rayleigh-Plateau instability of incompressible axisymmetric

bodies of fluid

Bruno Figeys & Sam Roelants

Supervisor: Marco Caldarelli

April 30, 2009

Abstract

It is a long known and easily verified fact that a steady jet of fluid, like the one coming out ofan ordinary kitchen faucet, eventually breaks up into spherical drops. We can thus conclude thecylindrical body of fluid is somehow unstable and decays to the more stable spherical droplets. It isexactly this instability, first researched by Plateau and by Rayleigh after him, that we will investigatein this paper. We study the stability of static and rotating cylinders of incompressible fluid undersmall perturbations in a classical framework. We also look for possible axisymmetric configurationswhich extremize the system energy, again for static and rotating bodies, and relate them in phasediagrams, showing the branches of possible stable and unstable fluid configurations.

1

Contents

1 Introduction 3

2 Boundary Conditions 4

2.1 The Young-Laplace equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 The complete set of linearized boundary conditions . . . . . . . . . . . . . . . . . . . . . . 6

3 Static axisymmetric bodies of fluid 7

3.1 Stationary configurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.1.1 Extremizing the action . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.1.2 The solutions in a phase diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3.2 Stability of a static cylinder in 3D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.3 The dispersion relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.4 Stability of a cylinder in d dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

4 Stationary rotating configurations 17

4.1 Minimizing the potential energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.2 Through the Young-Laplace equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184.4 The phase diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

5 Stability of a rotating cylinder of fluid 19

5.1 Warming up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205.2 General study of the stability of the uniform rotating cylinder of fluid . . . . . . . . . . . 215.3 The dispersion relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

6 An application to black hole physics 25

6.1 Nuclear physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

7 Conclusion 25

A Appendix I

A.1 cylindrical coordinate system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . IA.2 Generalized cylindrical coordinate system . . . . . . . . . . . . . . . . . . . . . . . . . . . I

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1 Introduction

This article is the result of an exploration of the instability of cylinders. The cylinders and all otheraxisymmetric objects discussed in this paper all have a periodicity or length L. Later on we shall encountersome examples showing how this can be achieved.

In the first section of this paper we will discuss the lesser known surface tension energy and the relatedYoung-Laplace equation. In this first part the Young-Laplace equation will also be situated in the contextit will used in in this article, i.e. as one of the frequently used boundary conditions. In the second partthe discussion we will discuss static axisymmetric bodies. We show how we obtain the equation thatdescribes these bodies and put them in a phase diagram. Next we take perhaps the most familiar ofthese shapes, the cylinder, and study its stability. The third and the fourth part are the extension of thesecond part, but this time the axisymmetric bodies are undergoing a rigid body rotation. This makes itall a lot harder to calculate, and still some open questions remain unanswered.

Despite the fact that these capillarity effects are a less widely known subject in physics it alreadyhas some history. The breakup of a steady jet of fluid into smaller droplets is a well known and quitewell understood phenomenon. It was Plateau who first ascribed this effect to the capillarity of the fluid.After analysis, he concluded that a cylindrical body of fluid of radius R is unstable to all perturbationalmodes of wavelength λ > 2πR. He thus (wrongly) assumed that after the breakup, the drops ought tobe equally spaced with an intermediate distance of 2πR. It was Lord Rayleigh who correctly arguedthat the intermediate distance is in fact determined not by the marginally unstable mode, but insteadin the maximally unstable mode. Indeed, it is the perturbational mode which grows the fastest whichwill determine the eventual spacing between the droplets, which Rayleigh found to be a fraction 0.697 ofPlateau’s initial guess. We will come across these results and many more in the course of this paper.

The Rayleigh-Plateau instability of fluids is a capillarity effect, and thus caused by the surface tensionof the fluid. The tensile properties of fluids are characterized by a parameter σ, called the surface energyor surface tension and which has dimensions of energy per unit surface or force per unit length. Wecan thus write down the potential energy of a fluid configuration due to the surface tension as U = σA,where A is of course the surface of the body of fluid.

In a simple calculation, we can verify that when considering solely surface tension as a means ofcohesion, the cylinder is an unstable configuration, in the sense that certain modes will decrease thearea of the body and thus result in a configuration of lesser energy. Suppose we start from a cylinder ofincompressible fluid of radius R0 and length L, aligned along the z-axis. Its volume V and area A arenaturally given by

V = πR20L, A = 2πR0L.

We can then perturb the radius of the cylinder to

R(z) = R0 + ǫR1 cos(kz) + ǫ2R2

We will calculate the difference is surface ∆A up to O(ǫ2). As we are dealing with a fluid of constantdensity, here, as in the rest of this paper, we need to keep the volume of fluid constant. This is why the

Figure 1: A typical example of the Rayleigh-Plateau instability on a jet of water. This image was borrowedfrom [7]

3

second order change ǫ2R2 is important to compensate for the first order change. The volume of a solidof revolution is well known to be π

∫

r2 dz, by which we easily find

Vpert = πR20L +

1

2πR2

1Lǫ2 + 2πR0R2Lǫ2 + O(ǫ3)

∆V =1

2πǫ2L

(

R21 + 4R0R2

)

.

From this, since ∆V = 0 we immediately find R2 to be

R2 = − R21

4R0

Next, we calculate the perturbed area Apert and the net change in area ∆A. The area of a solid ofrevolution is simply 2π

∫

r√

1 + r2 dz, where r denotes differentiation with respect to z. A rather quickcalculation immediately yields

Apert = 2πR0L − 2πǫ2L

(

1

4k2R2

1R0 + R2

)

+ O(ǫ3).

Upon plugging in our value for R2, we immediately find ∆A to be

∆A = 2πR2Lǫ2(

k2R20 − 1

)

.

Indeed, we find that ∆A < 0 when kR0 < 1, or equivalently, when the mode wavelength λ is greater thanthe circumference 2πR0, exactly as Plateau predicted. Intuitively, one can say that “longer cylinders aremore unstable than shorter ones”, in the sense that they are unstable under a wider range of perturbationmodes. Later in this paper we will linearize the hydrodynamic equations. It should then be clear thatwe do not need to study any other perturbations to the fluid than purely harmonic modes, since we canalways expand a more general perturbation into its Fourier series.

Thus, we have established that the cylinder is indeed an unstable configuration, which will eventuallydecay into a body of lesser surface energy, like spheres. It is then possible to study the instability of thecylinder, which is exactly what we will do in this paper. The relevance this research extends far beyondclassical hydrodynamics and has applications in an almost inconceivable range of physical domains. Afew of these domains include nuclear physics, magnetohydrodynamics and plasma physics, astrophysicsand string theory. We will come back to a few of these applications later on in the text.

2 Boundary Conditions

As is most common in fluid dynamics, throughout this paper, we will turn to the Navier-Stokes equationsin order to study the behaviour of the fluid in question. Throughout this text we will resort to a modelusing incompressible, inviscid fluids. More succinctly, we will resort to the continuity equation thatexpresses local conservation of mass and the Euler equation for ideal inviscid fluids which paraphrasesconservation of momentum. These equations are given by

∂ρ

∂t= −∇ · (ρ~v) (mass) (1)

d~v

dt= −1

ρ∇p, (momentum) (2)

where the d/dt operater denotes the convective derivative. As with any set of differential equations onewishes to solve, the appropriate boundary conditions are needed to arrive at an actual result. In thissection we look precisely for the boundary conditions to supplement our equations.

2.1 The Young-Laplace equation

Of all the hydrodynamical equations and identities that will be used throughout this work, the nextone proves to be paramount to the entire discussion. We will now discuss the Young-Laplace equation,

4

which relates the discontinuity of the pressure p at a fluid interface due to surface tension to the localshape of the surface. The Young-Laplace equation is perhaps not as famous as many of the other notableequations in fluid dynamics, but it is none the less indispensible when considering fluid capillarity.

The exact statement of the Young-Laplace equation is

∆p = σ∇ · n, (3)

with σ again being the surface energy, and n denoting the unit vector normal to the surface pointing fromregion 1 to region 2. ∆p is then the pressure in region 1 minus the pressure in region 2 on the boundary.We will not attempt to derive this equation here, but gladly defer the reader to more specialised literaturesuch as [11].

The nature of this equation is a purely geometrical one and will, more specifically, require somedifferential geometry in order to be evaluated correctly. For the reader’s convenience, we added anappendix to the end of this paper with some useful relations obtained from differential geometry.

It is also generally known from differential geometry that, if we denote by κ the mean curvature ofthe surface, ∇ · n = 2κ, so that the Young-Laplace equation may equivalently be written as

∆p = 2σκ,

or alternatively, using the pricipal radii of curvatures R1 and R2,

∆p = σ

(

1

R1+

1

R2

)

.

This is all just side information, as all of the relevant physics is contained in (3), which is the form wewill almost exclusively use. Note however, that the equation in terms of the mean curvature allows for amore transparent interpretation than the original formulation. A pressure gradient proportional to thelocal curvature of the fluid arises due to the surface tension. This pressure gradient then exerts a forcethat attempts to restore the fluid to a smoother configuration. As a consequence, we are unlikely to findfluids with sharp edges or corners, but will almost always observe configurations with smooth surfaces,such as drops, planes and cylinders.

The profile of a fluid surface may be parametrized by a function z = h(r, ϕ) in cylindrical coordinates(see A.1). It proves to be useful, however, to define the function f(r, ϕ, z) = z − h(r, ϕ), so that thesurface of the fluid is nothing but the level surface f(r, ϕ, z) = 0. In this context, we easily find the unitnormal n to the surface to be ∇f/||∇f ||, evaluated at f = 0. It is then straightforward, but perhapsa bit tedious, to obtain an expression for ∇ · n in terms of the parametrization h(r, ϕ). We will simplyprovide our result (use (A.1) and (A.2)), and to lighten the notation, we will denote differentiation withrespect to a variable by a subscript.

∇ · n = − 1

Σ3

(

hrr

(

1 +h2

ϕϕ

r2

)

+1

r2(hϕϕ + rhr)

(

1 + h2r

)

− 2

r2hϕhrϕhr

)

, (4)

where Σ = ||∇f || =√

1 + h2r + h2

ϕ/r2

In this text, we will keep only to axisymmetric bodies, so the parametrization h does not depend on ϕ.We can then simply drop all terms in this rather frightful expression containing ϕ-derivatives, i.e. hϕϕ,hϕr and hϕ.

We can, however, choose to define our surface in a different way. For example, through a functiong = r − R(z, ϕ). In this case, we can again compute ∇ · n. We find

∇ · n =1

Σ3

(

Σ2

r+

R2ϕ

r3− Rzz

(

1 +R2

ϕ

r2

)

− Rϕϕ

r2

(

1 + R2z

)

+2RϕRzRϕz

r2

)

, (5)

with Σ still defined as ||∇g||. Again, the equation simplifies significantly if we stick to axisymmetricprofiles. We will, in fact, use both parametrizations and both expressions in the future.

It is interesting to note that the Young-Laplace equation results from imposing the condition thatthe surface is in equilibrium, which is realised by balancing the normal stresses to the surface. This iscompletely equivalent to imposing that the energy of the fluid configuration be at a minimum. Hence,we will later show that minimizing the surface area (and thus the potential energy due to the surfacetension) by use of the Euler-Lagrange equation is completely equivalent to the Young-Laplace equation.

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2.2 The complete set of linearized boundary conditions

The Young-Laplace equation already gives us one condition that must hold at the fluid boundary. In thenext paragraph we will find another.

A second boundary condition

In this section, we consider a slightly different situation, in view of the stability analysis of the cylinderwe will conduct later. This time we start from a fluid surface to which we have added a small perturbation.We can then again parametrize the fluid surface as a level surface f(r, ϕ, z, t) = 0. The boundary conditionthen follows from observing that fluid particles at the boundary remain at the boundary. Expressed interms of the Lagrangian derivative, this gives us

df

dt= 0 if f = 0.

We can rewrite this as∂f

∂t+ vr

∂f

∂r+

vϕ

r

∂f

∂ϕ+ vz

∂f

∂z= 0 if f = 0. (6)

This is the second boundary condition we were looking for. Next, we will attempt to exploit the factthat the perturbations we will study are very small to significantly simplify the boundary conditions, aswell as the rest of the calculations.

Linearizing the equations

Suppose our perturbed surface is defined by a function r = R(z, ϕ, t), or equivalently by a level surfacef = r − R(z, ϕ, t) = 0. If the perturbation is small enough ( O(ε)), we can neglect any contributionsto our equations of quadratic order in ε or higher. In the context of a perturbed cylinder, we can forexample set R = R0+δR(z, ϕ, t), where R0 is the radius of the cylinder, and δR is a perturbation of orderε. This will induce a change in the velocity ~v = ~v0 + δ~v, where again ~v0 is the initial velocity field of thefluid, and δ~v is the induced perturbation, again of order ε. We can then for example linearize our secondboundary condition (6) by neglecting the products of any two or more pertubations. Remembering thatf = r − R(z, ϕ, t), we obtain

−∂R

∂t+ vr −

vϕ

r

∂R

∂ϕ− vz

∂R

∂z= 0 at r = R(z, ϕ, t).

Wich results in the linearised equation

∂δR

∂t+

v0,ϕ

r

∂δR

∂ϕ+ v0,z

∂δR

∂z= vr at r = R(z, ϕ, t) (7)

Another important simplification is found when we expand vr around R0 in a Taylor expansion. Wefind that

vr(R,ϕ, z, t) = vr(R0, ϕ, z, t) + δR∂vr

∂r+ · · · ,

meaning, as long as v0,r does not depend on r, after linearization only the leading term survives. Thus,evaluating the boundary condition at r = R in the linear approximation becomes nothing but evaluatingat r = R0.

We can also linearize the found expressions for ∇ · n. Since we are discussing perturbations on acylinder parametrized by f = r − R(z, ϕ, t) = 0, the more natural expression to use is (5). Since R0

is constant, Ri = δRi ∼ ǫ. The derivatives of R have no zeroth order contribution. We can thereforeneglect any products of derivatives of R, and keep only terms linear in Ri. One easily shows, for examplethrough a binomial expansion, that Σ = 1 + O(ǫ2). What remains is

∇ · n =1

r− δRzz −

δRϕϕ

r2.

This expression is to be evaluated at the surface r = R:

∇ · n|R =1

R− δRzz −

δRϕϕ

R2.

6

A binomial expansion tells us that R−1 = R−10 − δRR−2

0 + O(ǫ2) and R−2 = R−20 + O(ǫ2). The Young-

Laplace boundary condition is thus linearized to the following easier result:

∆p = σ

(

1

R0− δR

R20

− δRzz −δRϕϕ

R20

)

.

To sum up the work done in this section, we formulate here the two linearized boundary conditions wehave derived, which we will use several times throughout the text. We find that for small perturbationsof order ε, the fluid must satisfy to linear approximation the following two equations:

∆p = σ

(

1

R0− δR

R20

− δRzz −δRϕϕ

R20

)

(8)

vr =∂δR

∂t+

v0,ϕ

R0

∂δR

∂ϕ+ v0,z

∂δR

∂z(9)

3 Static axisymmetric bodies of fluid

With the above machinery we will first study static bodies (~v0 = 0). First we derive configurationsof fluid with an extremal potential energy. After that we will study the stability of the cylinder, in 3dimesions. As for more advanced physical topics in high-energy physics a more dimensional variant isrequired, we will perform a less complete, but still worthy, analysis without confining ourselves to thethree dimensional space. The necessary expressions derived up until now have very simple generalizationsto d dimensions by using appendix A.2.

3.1 Stationary configurations

In this section we search for the configurations of fluid with a minimal energy in the form of surfacetension. There would be an easy way to drop this surface tension and that’s by decreasing the numberof particles or, equivalently, decreasing the volume. This is of course a rather trivial way of decreasingthe surface area, and is not quite what we’re looking for. Given a constant volume, we want to knowwhich configurations have an extremal energy. When we find the shapes that extremize the energy, westill have no way of telling if it actually minimizes the energy, or if the shape is simply a saddle point.In the latter case, the configuration will be a stationary solution, but not at all a stable one.

3.1.1 Extremizing the action

The extremization problem at hand can easily be solved by the use of a Lagrange multiplier. The actionto be minimized is

I = σA − κV.

with κ the Lagrange multiplier, A the area of the body and V the volume. We won’t search for everykind of configuration, but only for axisymmetric configurations. Using the coordinates defined in (A.5),the fluid body is defined by the set of points satisfying

0 < r < R(z), 0 < z < L, (10)

0 < θi < π, for i = 1 . . . d − 3

0 < θd−2 < 2π.

We must now extremize the action for such a parametrization of a body P :

I[P] = σ

∫

∂P

rd−2√

1 + r′2dΩd−2dz − κ

∫

P

rd−2drdzdΩd−2,

where r′ denotes differentiation with respect to z. These integrals are nothing but the generalisation ofthe surface and volume of a solid of revolution to our new coordinate system. Since the integrands donot depend on the angular coordinates, the integration over Ωd−2 simply yields a constant factor, the

7

r2

rr− r+

Figure 2: The behaviour of r′2 as a function of r. The zeros correspond to maximal and minimal values of rfor the surface.

surface of a unit (d − 2)-sphere Sd−2. Since this overall factor does not matter in our optimization, wediscard it. What remains is a functional which can be written more compactly as

I[P] =

∫ L

0

Ldz

withL(r, r′, z) ≡ σrd−2

√

1 + r′2 − κ

d − 1rd−1. (11)

One way to extremize this action is by using the well known Euler-Lagrange equation

dLdr

− d

dz

(

dLdr′

)

= 0,

which yields a second order differential equation. A more clever trick however, is to use our knowledgeof Lagrangian and Hamiltionian mechanics (or, more specifically, knowledge of the Legendre transform).We compare L with the Lagrangian in classical mechanics and observe that ∂L

∂z = 0, there is no z-dependence. If z were seen as the time in classical mechanics this would mean that the Hamiltonian isa conserved quantity. Thus, we have that the Legendre transform of our Lagrangian with respect to r′

must also be conserved.

H =∂L∂r′

r′ − L = c

Plugging in our Lagrangian, we obtain the following first order differential equation:

r′2 =(d − 1)2r2(d−2)

κ2(rd−1 + c)2− 1. (12)

This equation carries several parameter freedoms (c and κ). It describes all the shapes that extremizethe surface, among which are the shapes of minimal surface area, which are the ones we are especiallyinterested in. One may argue that c < 0 results in physically unrelevant solutions, so we will not discussthem here. We will, however, briefly discuss the several other possible outcomes. Note that there isanother way to look at the things than proposed here. You could rewrite formula (12) as

1

2r′2 + V (r) = 0.

This way you could use your knowledge in classical mechanics to better understand this problem. Figures2 and 4 can then nearly be seen as the graphs of V (r), they just need to be flipped over.

8

0

L2

−L2

0 L2

z

R

Figure 3: A plot of several unduloid shapes. The limiting profiles of a sphere and a cylinder are also plotted forreference.

Spheres Should we choose c = 0, an analytical solution is available to us. The differential equationdescribing the shape of the fluid is

dr

dz= ±

√

(d − 1)2

κ2

1

r2− 1.

Rewriting this equation, we obtain

dz = ± rdr√

(d−1)2

κ2 − r2

,

which is readily integrated to

z = ±√

(d − 1)2

κ2− r2.

We have discarded the integration constant that determines the z-offset, since it is of little importanceto us. Squaring this expression, we obtain the equation of a d-dimensional sphere:

r2 + z2 =

(

d − 1

κ

)2

.

This is of course something we had at least hoped for: spherical shapes minimize (hence extremize) thesurface area for a given volume. It is therefore not very surprising that we see the jet coming out of thekitchen faucet contract into spherical drops. We can check the mean curvature of this shape R−1

1 + R−12

with R1 and R2 the principal radii of curvature of the shape. In this case, both R1 and R2 are simply2/κ, so that the mean curvature is precisely κ.

Cylinders There is another analytical solution. This is the one with constant radius, r′ = 0. In thiscase κ, c and R must be taken so that r′ = 0.

c = Rd−2

(

d − 1

κ− R

)

This is what one can see as the generalization of a cylinder to more dimensions. We will call them thecylinders in d dimensions.

Unduloids Before giving the actual numerical solutions to the differential equation, we can take a quickglance at the graph of r′ to learn quite a bit about the solutions. This will also allow for a first check ifthe found numerical solutions are at all plausible. In Figure 2 we have graphed r′2. We immediately find

9

r2

r

Figure 4: Demonstrating the behaviour of r′2 as we vary c for fixed κ. When√

c = 1/κ, the curve has only onezero, corresponding to a solution with constant radius: the cylinder.

that r′2, so r′ as well, has 2 zeros, marked by r±. Indeed, since these are radii for which r′ = 0, thesemust be extremal values for r(z). It is also obvious there can be no values of r outside the zone markedby r±, since r′2 becomes negative in this area. These solutions are already known to mathematicians,and are called unduloids. It is obvious that the cylinder and the sphere are just limiting cases of theunduloids. To understand this it is useful to understand Figure 4, in which we have plotted r′2 for a fewcombinations of c and κ to show the general behaviour. More precisely, we have chosen κ to be constantand allow c to vary. It should be clear that depending on c, there can be two, one, or zero intersectionswith the r-axis. For c very large there are no such points, consequently there cannot be any physicalstationary configuration. For the limiting case of c = ccrit there is just one such point. There is just onepossible value of r and that is r+ = r−. This solution represents the cylinder because we see that forthis value r′ = 0. For 0 < c < ccrit there are two such points, these values of c represent the unduloidswe did not know yet. For the critical value c = 0 we see that the curve is not like the ones before. Thisis because this limiting case is the sphere for which r− should be zero and r′ = −∞ at r = 0. Thecylinder is the unduloid corresponding with r+ = r−, the lowest curve in graph 4. Negative values for cdo not give physical realistic configurations, we will explain why in a later paragraph. We plotted a fewexamples of unduloids in Figure 3, along with the limiting cases of the sphere (c → 0) and the cylinder(c → κ−2, in 3D).

Constant mean curvature and the Lagrange multiplier κ Another way to derive the differentialequation found by Euler-Lagrange is by using Young-Laplace’s. With the NSE we can calculate, we willdo this later in more detail, that the pressure in the fluid must be constant everywhere in the non-rotatingcase. With Young-Laplace’s equation we can then formulate a very important statement. Looking backto the equation (3), proves that as the pressure is constant in the fluid and outside;

The mean curvature of the surface (∼ ∇ · n) must be constant.

There exists a cute way to interpret the Lagrange multiplier κ we have introduced. We begin byplugging the Lagrangian (11) in the Euler-Lagrange equation. Rewriting, one quickly obtains a secondorder differential equation:

σ

(√

1 + r′2)3

(

d − 2

r(1 + r′2) − r′′

)

= κ

We can then start from an entirely different, seemingly unrelated angle: We now know that σ∇ · n isconstant. Call this constant quantity β. We can evaluate ∇· n using the generalised formula (A.9). Sincewe are considering axisymmetric shapes, all ϕ-derivatives drop out, and we are left with a differentialequation

σ∇ · n = β =σ

(√

1 + r′2)3

(

d − 2

r(1 + r′2) − r′′

)

We find the exact same differential equation! By direct comparison we find that the Lagrange multiplierκ we introduced is up to a proportionality nothing but the mean curvature ∇ · n of the shape!

10

0

L2

−L2

0 L2

z

R

Figure 5: A plot of a solution corresponding to c < 0. The shape has a flex point on either side. There tangentlines are drawn for reference.

4.6

4.8

5

5.2

5.4

0.1 0.2 0.3 0.4 0.5

A/V

2/3

V/L3

SpheresCylindersUnduloids

Figure 6: The phase diagram of the normalized area versus the normalized volume for axisymmetric stationarysolutions. All bodies are periodic with periodicity L. The values for the unduloid branch were obtainednumerically using MatlabTM.

Another class of solutions With the previous result we can give an argument why shapes foundwith c < 0 are no realistic configurations. In Figure 5 we can see that there is a flex point. We wishto calculate the mean curvature in this point. To do this, it is easier to parametrize the surface asz − h(r) = 0. In the flex point h′ = h′′ = 0, thus ∇ · n = 0 with equation (4). The previous paragraphproves then that the mean curvature is zero everywhere and the pressure inside the fluid is the same asoutside, zero. Which means that there can be no fluid inside the defined surface. These kind of shapescan therefore not be good configurations for fluid bodies. Unlike the soap bubbles, since these adopt theform of minimal surfaces. Which are exactly those with mean curvature zero.

3.1.2 The solutions in a phase diagram

It is interesting to plot the several possible solutions in a phase diagram showing the surface of energyversus the volume, where we will stick to regular three dimensional bodies. Such a plot can be foundin Figure 6. The area A is normalized by a factor V 2/3 in order to improve the visibility of the detailsof the plot. The different branches represent different solutions, all of periodicity L. The sphere-branchand cylinder-branch are easily obtained analytically, but for the unduloid branch, one has to resort tonumerical computation of the volume and surface of several solutions. At the far left, the cylinder branch

11

(4π)2/3

14π

V/L3

A/V 2/3

CylindersUnduloids

Figure 7: A closer look of the branch point of thecylinder and unduloid branch. The branchpoint is situated exactly at the marginallyunstable cylinders with 2πR0 = L.

(36π)1/3

π6

A/V

2/3

V/L3

SpheresUnduloids

Figure 8: A closer look at the merger point. One caneasily show this point is where the spheresno longer fit in our periodicity of L

consists of the stable cylinders. This may seem strange, since we have already shown that all cylindersare stable to certain wavelengths, depending on their radius. The only reason these are stable is becauseof the periodicity we imposed on them. This makes it impossible for harmonic modes of λ > L to exist onthe cylinder. Since we know that cylinders are unstable to modes with λ < 2πR0, we find precisely thecylinders for which 2πR0 < L are indeed stable. A simple calculation allows us to situate the marginallyunstable cylinder for which 2πR0 = L on the phase diagram. We find that its volume is L3/4π, with anormalized area of (4π)2/3. The marginally unstable cylinder is situated in Figure 7. It is indeed preciselyat the bifurcation point between the cylinders and the unduloids. One may also examine the point wherethe unduloid branch and the sphere branch merge. This point is named the merger point. Intuitivelywe may anticipate there is something special about the sphere located at the merger point. Indeed, asimple calculation reveals that the merger point is precisely the sphere that has a diameter equal to theperiodicity L, which means the spheres of volume past the merger point are forbidden, since they do notfit in our periodicity. We find that the volume of a sphere of diameter L is πL3/6. The merger point islooked closer upon in Figure 8. After the next section, when we have obtained the dispersion relation forperturbations on a cylinder, we will look closer upon these branch points. It will allow us for exampleto predict where we may expect such branch points, and also give an indication of how the new solutionwill look.

3.2 Stability of a static cylinder in 3D

In this part we discuss the stability of the non-rotating cylinder in 3 dimensions. To study its stability wewill simply put a perturbation on the cylinder and study how this perturbation will evolve. We will uselinearized differential equations, this will allow us to decompose the perturbation into a Fourier series.Which offers us the possibility to calculate the time evolution of each term. The problem is thus reducedinto studying the evolution of a perturbation of the form ∼ ei(kz+mϕ).

The unperturbed cylinder

In the unperturbed case, a cylinder of incompressible fluid with length L and radius R0 is at rest. Itis shown in the previous section that this is a stationary solution. The fluid’s behaviour is determinedcompletely by the continuity equation and the Euler equation, while our boundary condition is suppliedby the Young-Laplace equation as discussed in section 2. Since ~v0 = 0 and ρ is constant the continuityequation (1) holds. The NSE reduces to ∇p0 = 0, which means that the pressure p0 must be constantin the fluid. The boundary condition tells us what the value of p0 is. If we assume that the pressureoutside is zero, then the pressure jump of the Young-Laplace equation is equal to the pressure inside the

12

cylinder.p0 = σ∇ · n

The boundary is defined by r = R0, or again by f = 0, where f(r, ϕ, z) = r − R0. Formula (5) tells usthat ∇ · n = 1

R0

. The equations for the unperturbed cylinder are

p0 = σ 1R0

~v0 = 0.

The perturbed cylinder in 3D

Now we want to explore what happens when we perturb the static cylinder through

r = R0 + δRp = p0 + δp~v = ~v0 + δ~v

. (13)

These perturbations are all of order ε. The continuity equation and NSE reduce to

∇ · δ~v = 0

ρ∂δ~v∂t + ρδ~v · ∇δ~v = −∇δp

.

Linearising the second equation kills the second term. After taking the divergence of both sides, the leftside vanishes completely. We are left with the Laplace equation for the perturbation on the pressure:

∇2δp = 0. (14)

As we mentioned before, because of the linearity of the equations, we can construct an arbitraryperturbation out of simple harmonic functions. This means we only have to study the behaviour ofperturbations ∼ ei(kz+mϕ). We do so, and assume our perturbations are of the following form:

r = R0 + εAei(kz+mϕ)+ωt

p(r, ϕ, z, t) = p0 + εP (r)ei(kz+mϕ)+ωt

~v(r, ϕ, z, t) = ~v0 + ε~V (r)ei(kz+mϕ)+ωt

.

We can plug this form for δp in our Laplace equation (A.3) and find:

∇2P (r) −(

k2 +m2

r2

)

P (r) = 0

P ′′ +1

rP ′ −

(

k2 +m2

r2

)

P = 0.

The resulting equation is a modified Bessel equation. This equation has long been extensively studied,and its solutions are generally known. We find that the solution is of the form [1]

P ∝ BIm(kr) + CKm(kr).

We have plotted a few of the modified Bessel functions in Figure 9 for reference. As we know thatthe pressure must stay finite in the center, we discard those solutions that are singular at r = 0 (beingthe Modified Bessel functions of the second kind, Kn). Thus we find the solution for the pressure to be

P (r) = BIm(kr).

Now let’s see what we can do about our boundary conditions. The boundary conditions are given by

p(R0 + δR) = ∇ · n|R0

vr(R0 + δR) =dR

dt

∣

∣

∣

∣

R0

13

0

1

2

3

4

0 1 2 3 4

I0(x)I1(x)I2(x)I3(x)

(a) modified Bessel functions of the first kind

0

1

2

3

4

0 1 2 3 4

K0(x)K1(x)K2(x)K3(x)

(b) modified Bessel functions of the second kind

Figure 9: Plots for the first few modified Bessel Functions

For the Young-Laplace equation we refer back to equation (5). Linearizing both boundary conditions,we get

p0 + δp(R0) = σ

(

1

R0+ k2δR +

m2 − 1

R20

δR

)

δvr(R0) = ωδR.

After plugging in our exponential forms of the perturbations, we get:

P (R0) = σ

(

k2 +m2 − 1

R20

)

A

Vr(R0) = ωA

The Navier-Stokes equation gives us that ρωVr(r) = −P ′(r), so that we get a set of equations

σ

R20

(

k2R20 + m2 − 1

)

A − Im(kR0)B = 0 (15)

ωA +1

ρω(kI ′m(kR0)) B = 0 (16)

If this system of linear homogeneous equations is to have non-trivial solutions for A and B, its determinantmust be zero. This results in the dispersion relation

ω2 = − σ

ρR30

(

k2R20 + m2 − 1

) kR0I′m(kR0)

Im(kR0)(17)

3.3 The dispersion relation

Finally, we have come to the point where we can study the actual stability of the cylinder. The dispersionrelation gives the rate at which any perturbational mode evolves. It should be clear that when the righthand sign turns negative, ω will be purely imaginary and the perturbation we put on the cylinderwill oscillate stably. The result of the perturbations are simply waves that will run away along thecylinder surface. We are thus especially interested in the part where ω2(k) ≥ 0 to study the rates atwhich different modes grow exponentially. In Figure 10 we have plotted the real part of the dispersionrelation with respect to the dimensionless frequency and wavenumber for an axisymmetric perturbation(m = 0). Upon inspection, we find that the modified Bessel functions Im and their derivatives arealways positive (as one can anticipate from Figure 9). Thus, the sign change is completely determinedby the factor k2R2

0 + m2 − 1, which we have dubbed the discriminant factor, for obvious reasons. It isimmediately clear that for m ≥ 1, the discriminant remains positive for all k. This result has a very

14

0

0.1

0.2

0.3

0.4

0 1R0k

√

ρR3

0

σω

0.697

Figure 10: A plot of the dispersion relation in 3 dimensions. We have chosen to plot dimensionless quantities.The maximum of the curve lies at precise kR0 = 0.697, as Rayleigh predicted. The curve changessign in kR0 = 1, as Plateau predicted.

natural interpretation. Should we expand our perturbation in angular components ∼ eimϕ, we find thatthe only components that contribute to the break up are the m = 0 modes, since the m 6= 0 are stablemodes. Thus the angular variations of the perturbation do not matter and will not contribute to thebreakup: instability is completely determined by the z-variations of the perturbation. In the m = 0 case,then, the discriminant changes sign in exactly kR0 = 1. This is indeed the stability condition Plateaudetermined, as well as the result of our little toy calculation in the introduction!

The mode for which kR0 = 1, the marginally stable mode, is one of special importance. It isimmediately clear from Figure 10 that this mode does not depend on time. In the linear approximation,this mode, also called the threshold mode, brings the cylinder into a new stationary configuration. It isof course no coincidence that the linear theory predicts a new kind of stationary solutions precisely atthe point in the phase diagram where the unduloid branch of solutions forks off. Indeed, this new kind ofstationary solutions predicted by the linear theory is precisely the zeroth and first order approximation ofthe unduloid branch in its Fourier series! This may sound like an obvious statement, but its importanceshould not be underestimated. Without any knowledge of possible stationary solutions, we were are ableto predict branch points on the phase diagram. More generally, no matter what the situation is, when ashape flips over from a stable to unstable region, the dispersion relation flips sign. And thus, there existsa time-independant threshold mode where ω(k) = 0, which indicates a new branch of solutions. We willuse this later on to predict branch points on another phase plot.

Deriving the disperion relation allows us to locate the maximum of the curve in Figure 10. We findit to be at kR0 ≈ 0.697. Again we retrieve a known result, as this is precisely Rayleigh’s correctionto Plateau’s hypothesis we have mentioned at the beginning of this text. Upon perturbing the system,we find that the mode for which kR0 grows exponentially faster than all other modes. It is thereforenatural to assume that the intermediate distance between two drops of a contracted jet corresponds tothe wavelength of this mode. We find that this wavelength is simply 2πR0/0.697 = 9.01R0. Please notethat we cannot make exact predictions using the disperion relation, or any other part of our model forthat matter, since we are only studying the phenomena up to a level where the linearization is plausible.Since the perturbations on an unstable body will grow exponentially, we quickly leave the linear regionand are forced to include non-linear terms, which complicates the calculations tremendously. We have,however, gotten quite a few results from the dispersion relation, most of which are very relevant andexperimentally measurable. Just to show how our predictions are pretty good, we have included a figuretaken from [5]. The experimental data included in Figure 11 was taken from the the work of Donnelly andGlaberson, who did a great deal of measurements concerning the capillary break up of a jet of water [6].We see that the experimental results stick to the theoretical result quite nicely. Besides what we havediscussed here, there resides a great deal more of information in the dispersion relation, which makes ita key result of this work.

15

Figure 11: The dispersion relation with the experimental data of Donelly & Glaberson (1966) plotted onto it.Taken from [5].

3.4 Stability of a cylinder in d dimensions

Unfortunately, doing the entire calculation up until now in d dimensions proved to be quite difficult.This is partly due to our lack of experience in calculations in higher dimensions, using exotic coordinatesystems with their own distinct exotic geometries. We have, however, performed the calculation in ddimensions when supposing perturbations which do not depend on any angular coordinate. This meanswe only consider m = 0 modes. We will not repeat the entire calculation, as there is hardly anydifference with the calculation done in the previous section. One needs only to look up the d dimensionalgeneralizations of the used formulas, which are all present in the appendix. We will simply state theresults here.

After performing an identical calculation, one obtains a slightly different modified Bessel equation forthe pressure:

P ′′ +d − 2

rP ′ − k2P = 0

To which the solutions are simplyP = BrαIα(kR0),

with α defined als α = −(d − 3)/2. Next we find the linearized boundary conditions:

P (R0) = σ

(

k2 − d − 2

R20

)

A

P ′(R0) = ω2ρA.

In these equations, A represents the perturbation strength. Plugging our solution for P in these boundaryconditions, and equation the determinant of the system to 0 as usual, we obtain the dispersion relation:

ω2 = − σ

ρR30

(

k2R20 − (d − 2)

) αIα(kR0) + kR0I′α(kR0)

Iα(kR0). (18)

We immediately see that in the d = 3 case, α = 0 and we retrieve the dispersion relation (17). We notethat the threshold wavenumber is kthres =

√d − 2/R0.

16

4 Stationary rotating configurations

In this section we search for axisymmetric stationary configurations of a rotating fluid in 3D. There aretwo ways to find such a configuration. One is by minimizing the (potential) energy, the other is by usingthe Young-Laplace equation on the boundary. We suppose the fluid is flowing like ~v = rΩϕ. The reasonfor this is to be found in viscosity. If the fluid were a little viscous, any configuration with shear flowscould never be a steady flow. The only rotating flow that doesn’t have such shear flows is the flow of arigid body rotation.

4.1 Minimizing the potential energy

We are again facing an optimization problem with a constraint. We want to extremize the energy, keepingthe number of particles constant. Because we are studying incompressible fluids, the density must beconstant and so must be the volume. We can introduce a Lagrange multiplier β and define the action tobe extremized as

I := Uσ + Urot − βV.

We know that Uσ = σA, but we still need to find an expression for Urot.Therefore we first derive an expression for a point mass. If we choose the potential to be zero on therotation axis then we get

U(r) = −∫ r

0

~F · d~r.

Here ~F represents the force that moves a point mass from the center to a radius r. From the point ofview of the particles in motion, they are being pushed by a fictional centrifugal force ~F = mrΩ2r. Thisresults in the following effective potential for a point mass

U(r) = −1

2mr2Ω2

To get the potential energy, we just need to integrate last equation over the entire fluid

Urot = −1

2Ω2

∫

M

r2dm

For symmetry reasons we can study only one half of the configuration, so that we can define a functionh(r) which defines the boundary as the values where f = 0 with f(r, z, ϕ) = z−h(r). The action becomes

I[P] = σ

∫

∂P

dA − 1

2Ω2ρ

∫

P

r2dV − β

∫

P

dV

I[P] = σ

∫

∂r

2πr√

1 + h′2dr −∫

∂r

(

1

2ρΩ2r2 + β

)

2πrh(r)dr

Just as in the static case, we can write this action in terms of a Lagrangian:

I =

∫

L(h, h′, r)dr,

where L is given by

L(h, h′, r) := σ2πr√

1 + h′2 −(

1

2ρΩ2r2 + β

)

2πrh.

To extremize this action we again resort to the Euler-Lagrange equation.

dLdh

− d

dr

(

dLdh′

)

= 0

Filling in this form of L and rewriting, we obtain

β +1

2ρΩ2r2 = − σ

Σ3

(

h′

r

(

1 + h′2)

+ h′′

)

17

0

L2

0 L2

z

R

(a) Shapes of a few rotating drops.

0

L2

0 L2

z

R

(b) Shapes of a few rotating non-uniformcylinders.

4.2 Through the Young-Laplace equation

The same equation can be derived by using the Young-Laplace equation on the boundary. For starterswe find the pressure jump simply as the pressure in the fluid, which we can find through the NSE.

−ρrΩ2r = −∇p

Which means that the pressure must vary as

p(r) = pc +1

2ρr2Ω2. (19)

The Young-Laplace equation tells us to equate this pressure to the curvature of the surface, givenby the divergence of the normal. If the boundary is defined as the points for which f = 0, withf(r, z, ϕ) = z − h(r), we find that the divergence is given by (4):

∇ · n = − h′

rΣ− h′′

Σ3

The Young-Laplace equation becomes;

pc +1

2ρr2Ω2 = −σ

(

h′

rΣ+

h′′

Σ3

)

This way we obtain the same differential equation as before. Through direct comparison we find thatβ = pc. So it doesn’t really matter whether we extremize the energy to find the stationary surfaces, orif we use the Young-Laplace equation, since we are basically doing the same thing. This is no differentthan the choice one has when solving a mechanical problem by balancing the forces or by minimizingthe potential energy.

4.3 Solutions

Now that we have the differential equation that describes our surfaces of stationary energy, we can tryto solve it. There is a little trick to solving equation (4.2) the divergence of the normal vector can bewritten in the following form:

pc +1

2ρr2Ω2 = −σ

r

(

rh′

√1 + h′2

)′

.

Integrating and rewriting this equation gives;

h′2 =f2

σ2r2 − f2

18

0

1

2

3

4

5

6

0 0.05 0.1

J∗

Ω∗

Figure 12: The phase plot of the dimensionless angular momentum J∗ versus the dimensionless rotationalvelocity Ω∗.

where f ≡ c + 12pcr

2 + 18ρΩ2r4 and c is the integration constant. There are several kinds of solutions

to this equation. In the non-rotating case we had as main solutions the spheres, the cylinders and theunduloids. In the rotating case, however, things get a bit more interesting. Depending on the boundaryconditions we impose on h and h′, we get different types of solutions. We will again discuss the differentsolutions briefly.

Drops It is desirable that in the Ω → 0 limit, we retrieve the solutions for the static surfaces. Thisgives us reason to believe that at least one kind of solution exists where h′(0) = 0 and h′(R) = −∞,since the we know the spheres comply to these conditions. We therefore impose these conditions andlook for numerical solutions satisfying them. These numerical solutions are plotted in Figure 12(a). Wesee that as Ω increases, a “pinching” effect becomes clear. There exists a critical Ω for which the pinchingis complete, and what remains is a toroidal shape.

Non-uniform cylinders In analogy to our previous reasoning, we would also like to find solutionsthat resemble the non-uniform cylinders, by imposing the boundary conditions h′(Rmin) = −∞ andh′(Rmax) = −∞, for some Rmin and Rmax. The uniform cylinders are then simply the solutions whereRmin = Rmax. These solutions are plotted in Figure 12(b). Again, there is a definite pinching effect asthe angular velocity increases. Just as before, there exists a critical Ω for which the pinching is complete.What remains after this pinch off is a cylindrical shape surrounded by a toroidal shape.

There are, in fact, a whole lot more solutions possible. We will not begin to discuss them all. Wewill discuss the drops and non-uniform cylinders in the phase diagram in the following section.

4.4 The phase diagram

This time we choose to plot the angular momentum versus the angular velocity of configurations. We dothis for configurations of fixed volume. The result can be found in Figure 12. Note that at first glance,this plot seems plausible. For slow rotation, the angular velocity Ω the angular momentum J = IΩ areindeed proportional, where the moment of inertia I is the proportionality. As Ω increases, however, thebody starts getting deformed, and fluid mass is displaced to a greater distance from the rotation axis, asshown in Figure 12(b). We find that I can no longer be approximated as constant, and the curve bendsover.

5 Stability of a rotating cylinder of fluid

The last major thing we will discuss is the stability of a uniform rotating cylinder in 3 dimensions. Wewill begin by repeating the calculation in the introduction of this text. This is a quick and easy way to

19

get a first glimpse of the stability behaviour of the cylinder.

5.1 Warming up

We attempt to study the instability of a perturbed cylinder by examining how the total potential energychanges as a consequence of the perturbation. The first time we did this computation, the energydepended solely on the surface area. This time, however, we must also take into consideration theeffective potential energy associated with the rotational motion. The total potential energy of the motionis therefore given by

U = Uσ + Urot.

We start from a regular cylinder which we perturb to a new shape given by

R(ϕ, z) = R0 + ǫR1 cos(kz + mϕ) + ǫ2R2.

Again, the incompressibility of the fluid forces us to take along a second order change to compensate forthe change in volume due to the first order perturbation. By imposing this incompressibility, we easilyobtain that

R2 = − R21

4R0,

which is indeed very similar to relation we found the first time. We find that the surface of our nonaxisymmetric perturbed cylinder is

A =

∫ L

0

∫ 2π

0

R√

1 + R2ϕ/R2 + R2

z dϕdz

Dropping all the higher order terms and doing the straightforward integration just like before, we findan expression for ∆Uσ:

∆Uσ = σ(k2R20 + m2 − 1)

πLR21ǫ

2

2R0.

This is indeed the result we would expect after our discussion of the stability of static cylinders. Themost notable thing is that we have in a fairly simple manner determined the discriminant factor thatdetermines the stability.

Next we compute the contribution by the rotational potential energy. We have already shown in 4.1that the potential energy of the flow associated to the rotation is precisely

Urot = −1

2ρΩ2

∫

r2 dV = −1

4ρΩ2

∫ L

0

∫ 2π

0

r4 dϕdz.

There is not much to learn from the straightforward calculations, so we immediately go to the result:

∆Urot = −πLR20R

21ǫ

2

2ρΩ2

Adding both results an arranging them in a nicer way, we get the total change of the energy:

∆U = σ

(

k2R20 + m2 − 1 − ρΩ2R3

0

σ

)

πLR21ǫ

2

2R0

This result is quite nice, since now we know exactly how the discriminant factor ought to change in therotating cylinder dispersion relation.

The new term that appears has a special significance and is called the rotational Bond number, butwe will often refer to it simply as the Bond number. The Bond number is a dimensionless quantity thatindicates the importance of a certain force in comparison to capillary effects. It is generally defined as

Σ =ρa

σ/L2,

where a is the acceleration of the force we are comparing, and L is a general length scale of the motion.Conventionally, the length scale is chosen to be R0/2. [3,10] For the moment, it’s more convenient for us

20

to define the rotational Bond number with the length scale L = R0. Since the force we compare to thecapillary force is the centrifugal repulsion, we substitute R0Ω

2 for a to obtain precisely the new term

Σ =ρΩ2R3

0

σ.

So the Bond number is a dimensionless quantity that directly indicates the rotational velocity of the flowas well as its importance compared to the capillary effects.

5.2 General study of the stability of the uniform rotating cylinder of fluid

We follow the same strategy as for the non-rotating cylinder. First we solve the NSE and the continuityequation for the rotating cylinder without a perturbation to find the zeroth order contributions. Afterthat we put a perturbation on it.

The unperturbed rotating cylinder

As already said before, we suppose the cylinder is rotating as a rigid body with ~v = rΩϕ. We havealready solved the NSE for this unperturbed case and found equation (19) that describes the pressure.We also find an expression for the central pressure pc through the Young-Laplace equation. The pressureoutside is chosen to be zero again, so that we get

pc =σ

R0− 1

2ρΩ2R2

0.

We have used the fact that the mean curvature ∇ · n is constant and equal to 1/R for a cylinder. Thisway we find that the unperturbed cylinder is determined by the following equations

p0(r, z, ϕ) = pc + 12ρΩ2r2

~v0(r, z, ϕ) = rΩϕ.

The perturbed rotating cylinder

Just as in the static case, we put a perturbation of the form (13) on the cylinder. The linearizedcontinuity equation and NSE become:

∇ · δ~v = 0

ρ∂δ~v∂t + ρ~v0 · ∇δ~v + ρδ~v · ∇~v0 = −∇δp.

We can again expand the perturbation in a Fourier series with terms ∝ ei(kz+mϕ)+ωt. This is the sameperturbation as in the static case and we will use the same notation as we did in (3.2). Using (A.4) toevaluate the terms in the NSE, we quickly get a set of equations that relate the velocity components tothe pressure:

Vr

r + V ′r + im

r Vϕ + ikVz = 0ρ (χVr − 2ΩVϕ) = −P ′

ρ (2ΩVr + χVϕ) = − imr P

ρχVz = −ikP,

(20)

where χ ≡ ω + imΩ. Upon solving the last three equations for the velocity components, and filling theseinto the first equation, we again obtain a modified Bessel equation for the pressure P (r).

P ′′ +1

rP ′ −

(

χ2 + 4Ω2

χ2k2 +

m2

r2

)

P = 0 (21)

It is interesting to note that in the non-rotating limit Ω → 0, χ → ω and we recover exactly the Besselequation we got in Section 3.2, when d is taken to be 3. The solutions are again easily found to be

P = BIm(κr),

where κ2 ≡ (χ2+4Ω2)χ2 k2.

21

Next, we impose the boundary conditions (8) and (9). The kinematic boundary condition (9) leadsus to

χA = Vr.

From equations (20), one immediately finds that

Vr = −χ

ρ

P ′

χ2 + 4Ω2− 2Ω

ρ

im

r

P

χ2 + 4Ω.

The kinematic boundary equation becomes

χA +

(

χ

ρ

κI ′m(κR0)

χ2 + 4Ω2+

2Ω

ρ

im

R0

Im(κR0)

χ2 + 4Ω2

)

B = 0 (22)

The Young-Laplace equation (8) must be handled with more care. If the pressure outside is again takento be zero, we get the linearized equation

p0(R0 + δR) + δp(R0 + δR) = σ

(

1

R0+

(

k2 +m2 − 1

R20

)

δR

)

.

The first term on the left hand side is easily calculated, since we know p0. For the second term, a quickTaylor expansion yields

δp(R0 + δR) = δp(R0) + δp′(R0)δR + · · ·So, we find that the first order contribution to δp(R0 + δR) is simply δp(R0). The final form of ourlinearized Young-Laplace equation is

p(R0) + ρΩ2R0δR + δp(R0) = σ

(

1

R0+

(

k2 +m2 − 1

R20

)

δR

)

.

Plugging in the perturbations and dividing out the exponentials, this can be rewritten as

σ

(

k2 +m2 − 1

R20

− ρΩ2R0

σ

)

A − Im(κR0)B = 0. (23)

So, the two boundary conditions (22) and (23) are

χA +

(

χ

ρ

κI ′m(κR0)

χ2 + 4Ω2+

2Ω

ρ

im

R0

Im(κR0)

χ2 + 4Ω2

)

B = 0

σ

(

k2 +m2 − 1

R20

− ρΩ2R0

σ

)

A − Im(κR0)B = 0.

If these equations are to have a non-trivial solution, the determinant must again vanish. Just like before,this provides us with the dispersion relation:

χ = − σ

ρR30

(

k2R20 + m2 − 1 − ρΩ2R3

0

σ

)

χκR0I′m(κR0) + 2imΩIm(κR0)

(χ2 + 4Ω2)Im(κR0)(24)

5.3 The dispersion relation

This equation looks very heavy, but in the end, it isn’t as bad as it looks. At first glance, we are pleasedby the fact that the discriminant factor is exactly what we anticipated from our quick computation atthe start of this section. The new term appearing is indeed precisely the Bond number describing therotational effects. A major feature that needs to be pointed out is the sign of this new term. Thenew term introduces new modes of instability. A m = 1 mode, for example, could become unstable forsufficiently large Ω. A second feature of this equation is that we can no longer express ω as an explicitfunction of k, since it appears in the argument of the Bessel functions through κ. We can still, however,plot the dispersion relation for m = 0. We do this by defining, through equation (24), an implicit functionF (ω, k). The dispersion relation graph will simply be the level contour where F (ω, k) = 0.

22

0

0.2

0.4

0.6

0.8

1

0 1 2R0k

√

ρR3

0

σω

Σ = 0

1

2

3

4

Figure 13: The dispersion relation, plotted for a m = 0 mode for different values of Ω. The graphs are labeledby their Bond number. The Σ = 0 graph corresponds to the non rotating dispersion in Figure 10

The dispersion for m = 0 modes

We have plotted the dispersion function in Figure 13 for a m = 0 mode for several values of Ω. Thedifferent curves are labeled with their corresponding Bond number. Several features of this plot are quiteinteresting to point out. First of all, the maximum of the curve shifts upward as the rotation increases.We conclude from this that faster spinning cylinders will tend to break up faster, as their exponentialgrowth rate is greater. Next, we notice that the marginally stable mode shifts upward as well. Thecylinder becomes unstable to a greater range of modes. Less formally, we may summarize these resultsby stating that “rotating cylinders are more unstable than static cylinders, and the instability increaseswith Ω.” As the maximum of the dispersion plot shifts to the right in Figure 13, λmax decreases. Thisλmax determines, as already explained, the distance between two drops when a jet of fluid collapses.Thus this distance between two distinct drops after the breakup will decrease in rotating jets.

The dispersion for m 6= 0 modes Thanks to the Bond number appearing in the discriminant factor,there is a term that can negate the m2 term in the discriminant that was keeping all m 6= 0 modesstable. We can then expect to find unstable modes for m 6= 0. Plotting the dispersion relation for thesemodes proves to be more difficult. We can, however, obtain quite a bit of information from some basicanalytical properties of the dispersion relation. For example, we can check whether or not the ω = 0and k = 0 mode is is unstable, meaning it lies on the dispersion curve. Filling in these two values intothe dispersion relation (24), we obtain an equation in Ω and m, wich gives us a condition for which theω = 0, k = 0,m 6= 0 mode indeed exists. A quick calculation teaches us this relation is precisely when

Σ = m + 1.

One may wonder what significance this result may have, as we have obtained the value of the dispersioncurve for only a single value. The point doesn’t even seem that special, because the dispersion curvesfor m = 0 also all passed through the origin of the ω, k-plane. These is, however, a major difference.In the m = 0 case, the ω = 0, k = 0 mode was hardly interesting, because it simply corresponded to

Figure 14: An example of a new branch of solutions with broken angular symmetry. This cylinder is approxi-mated to first order by the m = 3 threshold mode with k = 0. Image is taken from [10]

23

Figure 15: A plot of the dimensionless angular momentum versus the dimensionless angular velocity for dropsundergoing rigid body rotation. Image borrowed from [12].

zero perturbation. In this case however, a ω = 0, k = 0,m 6= 0 mode is a time independent perturbationwhich breaks only axial symmetry, but leaves the cylinder z-invariant. Indeed, since ω = 0 for this mode,the time dependence exp(ωt) drops out. So does the z dependence in the factor exp(ikz). This meanswe have new threshold modes. Therefore, we can expect to find new branch points on the phase diagramcorresponding to families of cylinders with broken axisymmetry. An example of such a cylinder is givenin Figure 14. Since this text is mainly a study of the rotationally invariant shapes of fluid, we will notstudy these new solutions in full detail. Now that we have the dispersion relation at our disposal, itbecomes so easy to make a few notes about these solutions, that it would be a waste not to.

We know that for every m there exists a threshold mode, determined by the condition Σ = m+1. Thiseasily allows us to determine the bifurcation points on the fase diagram. We find that the bifurcationfrequencies are

Ωthr =

√

σ(m + 1)

ρR30

.

It should be noted, though, that by no means do we claim that these are the only bifurcation points,others may exist. We have simple stated here the ones we could find, that at the same time break therotational symmetry.

It is possible to perform a linear investigation of these branch points to determine how the newbranch will start of closely to the bifurcation point. Such a study, however, is far beyond the scope ofthis text. What we can do, is take a look at a similar phase plot for rotating drops, which has long beenexplored. [12] Such a plot, given in Figure 15 resembles the one we have obtained to a high degree, andso it is a nice way to see how one might expect such bifurcations to evolve.

We indeed see a similar behaviour of the curve. Starting from the origin, we are on an axisymmetricbranch. Then we encounter a 2-lobed threshold mode, from which a new branch of 2-lobed solutionsforks. At the bifurcation point, the stability flips from stable to unstable. At every next branch point, weencounter a new branch of solutions, each time with an extra lobe, which is of course precisely what wewould expect. These branches, however, have very peculiar features. For example, there are very specificregions on the phase diagram where the drops are unstable only towards certain m modes, while they arestable with respect to all other angular modes. Going into a deeper discussion about these drops wouldbring us too far, so we simply defer the reader to more complete and technical articles such as [10,12].

24

6 An application to black hole physics

Cylinders of fluid held together by surface tension can be used as a model for black strings. Black strings,black holes with the shape of cylinders, exist in more than four dimensions. These objects have a longwavelength instability under gravitation perturbations, the Gregory-Laflamme instability, that sharesmany features with the Rayleigh-Plateau instability. This analogy, pointed out by Vitor Cardoso, OscarJ. C. Dias and Leonardo Gualtieri [4,14], is a useful tool to get some intuition of the properties of blackobjects, and extends the old ideas of the membrane paradigm. This is why the higher dimensional cylinderwhich we explored above is interesting. Just as the cylinders of fluid have a threshold wavelength abovewhich they become unstable, the black strings have one. Thus the Rayleigh-Plateau instability becomesthe hydrodynamical analog of the Gregory-Laflamme instability, and a new branch of non-uniform blackstrings emerges from the threshold mode. For large dimensions d the equation R0kthres =

√d − 2 for

fluids agrees with the Gregory-Laflamme instability. For comparison, see [4]. Just as in hydrodynamics,the angular modes of the perturbation (m 6= 0 in the discussion 3.2) are stable modes for the blackstrings, as was calculated by Kudoh [8]. There are a lot of other similarities, as for example a remarkablequalitative agreement in the dispersion relations. In 3.1.2 we plotted the solutions in a phase diagramfor 3 dimensions. As summarized by Cardoso and Dias [4] this phase diagram has some fundamentalchanges at about dimension 11. Where in lower dimensional spaces the spheres were the preferred shapesof a fixed volume fluid, as can be deducted from the phase diagram in 3.1.2, this will change for morethan 11 dimensions. Above this critical dimension the cylinders will not any more be unstable in favorof the spheres. This phenomenon also occurs in the Gregory-Laflamme instability, as discussed in thepaper of Evgeny Sorkin who first observed it [13]. A more extensive study should be done in the rotatingcase. A rotating drop could at high rotation form a rotating torus, which should be the analog of a blackring. Also these drops begin to get lobed as in Figure 15. This little summary is just a beginning of whatcan be said about this analogy. Recently, this analogy between black objects and fluid lumps has beenelevated to a precise duality [9] for particular models of gravity, and the Rayleigh-Plateau instabilityhas been proven to be the fluid dynamical dual of the Gregory-Laflamme instability [2]. This opens thepossibility to study, using fluid dynamics techniques, the complex phase structure of higher dimensionalblack holes.

7 Conclusion

Throughout this text we have learned that the Rayleigh-Plateau mechanism, while quite simple in itsoriginal scheme, has analogies and similarities with many other theories, which makes it a very relevantbranch of research, not just from a hydrodynamical point of view. We have obtained the dispersionrelations for perturbations on a cylinder, which are a great source of all kinds of information and allow formany kinds of experimental verification. Upon comparing our results with such experimental verification,we found that the linear approximation we have utilised here describes correctly the dispersion relationfor the perturbation, although it is not enough to follow the full time evolution of the instability. Wehave been able to use the linear theory to predict new branches of stationary configurations and theirproperties, which is another key result obtained through the dispersion relation. We have interpretedthe phase diagrams, and plotted the stationary solutions.

However, there still remain many open questions and many possible outsets for new research. Firstly,while this has already been done for the most part, it will prove useful to generalize the entire frameworkof the Rayleigh-Plateau instability, and all its extensions, to higher dimensional spaces, as black holeapplications are situated in such spaces. One may also look further into the m 6= 0 modes for rotatingcylinders. We did not explore in depth the dispersion relation for non-axisymmetrical perturbations.While these modes could perhaps be simply stable (excluding a discrete set of threshold modes that wefound), there is no fundamental reason to believe that this is so.

Another extension, moving a bit further from the actual Rayleigh-Plateau instability, is the researchof non-axisymmetric configurations that extremize the potential energy. The dispersion relation hasalready narrowed down several of the branch points where such configurations may start, and the lineartheory can perhaps predict their behaviour very near to the branch point, but the general behaviour ofthese solutions on the phase diagram and their stability must be researched further in order to completethe full phase diagram and obtain the cylinder analog of Figure 15.

25

A Appendix

In this appendix we will introduce some useful coordinate systems.

A.1 cylindrical coordinate system

The 3 dimensional cylindrical system is defined by following equations;

x = rcosϕ

y = rsinϕ

z = z

There are some useful formulae.

∇f =∂f

∂rr +

∂f

∂zz +

1

r

∂f

∂ϕϕ (A.1)

∇ · ~v =1

r

∂

∂r(rvr) +

∂vz

∂z+

1

r

∂vϕ

∂ϕ(A.2)

∆f =1

r

∂

∂r

(

r∂f

∂r

)

+∂2vz

∂z2+

1

r2

∂2vϕ

∂ϕ2(A.3)

~A · ∇ ~B =

[

Ar∂Br

∂r+

Aϕ

r

(

∂Br

∂ϕ− Bϕ

)

+ Az∂Br

∂z

]

r

+

[

Ar∂Bϕ

∂r+

Aϕ

r

(

Br +∂Bϕ

∂ϕ

)

+ Az∂Bϕ

∂z

]

ϕ (A.4)

+

[

Ar∂Bz

∂r+

Aϕ

r

∂Bz

∂ϕ+ Az

∂Bz

∂z

]

z

A.2 Generalized cylindrical coordinate system

This is a coordinate system which is the more dimensional variant of the cylindrical coordinate systemin 3D. Each point is described by a vector of d components: (r, z, θ1, . . . , θd−2).

x1 = r sin θ cos ϕ1

x2 = r sin θ sin ϕ1

x3 = r cos θ cos ϕ2

x4 = r cos θ sin ϕ2 cos ϕ3

... (A.5)

xd−2 = r cos θ sin ϕ2 · · · cos ϕd−3

xd−1 = r cos θ sin ϕ2 · · · sin ϕd−3

xd = z

It can be verified that ds2 = dz2 + dr2 + r2(

dθ2 + sin2 θdϕ21 + cos2 θdΩ2

d−4

)

. Where dΩd−4 is a compactnotation to write the rest of the angles, which aren’t of any interest in this context. From this we canderive the following expressions. (We don’t give the full expressions, the rest is in this context equal tozero. )

∇f =∂f

∂rr +

∂f

∂zz +

1

r

∂f

∂θ1θ1 + . . . (A.6)

∇ · ~v =1

rd−2

∂

∂r

(

rd−2vr

)

+∂vz

∂z+ . . . (A.7)

∆f =1

rd−2

∂

∂r

(

rd−2 ∂f

∂r

)

+∂2vz

∂z2+ . . . (A.8)

I

When writing a fluid profile as z = h(r, ϕ), we find the divergence of the normal to be generalized to

∇ · n =1

Σ3

[

−(

1 +R2

ϕ

r2 sin2 θ

)

Rzz −Rϕϕ

r2 sin2 θ(1 + Rzz) +

d − 2

rΣ2 +

R2ϕ

r3 sin2 θ+

2RzRϕRzϕ

r2 sin2 θ

]

(A.9)

where Σ = (1 + R2z + R2

ϕ/r2 sin2 θ)1/2.

References

[1] F. Bowman. Introduction to Bessel Functions. Dover Publications, New York, 1958.

[2] Marco M. Caldarelli, Oscar J. C. Dias, Roberto Emparan, and Dietmar Klemm. Black holes aslumps of fluid. JHEP, 0904:024, 2009.

[3] S. Chandrasekhar. The stability of a rotating liquid drop. Proc. R. Soc. Lond., 286(1404):1–26,1965.

[4] V. Cardoso & O. J. C Dias. Title: Rayleigh-plateau and gregory-laflamme instabilities of blackstrings. Phys. Rev. Lett., 96:181601, 2006.

[5] P.G. Drazin. Introduction to Hydrodynamic Stability. Cambridge University Press, Cambridge, 2002.

[6] R.J. Donnelly & W. Glaberson. Experiments on the capillary instability of a liquid jet. Proc. R.Soc. Lond. A, 290(1423):547–556, 1966.

[7] D. F. Rutland & G. J. Jameson. A non-linear effect in the capillary instability of liquid jets. Journalof Fluid Mechanics, 46:241–265, 1971.

[8] Hideaki Kudoh. Origin of black string instability. Phys. Rev. D, 73:104034, 2006.

[9] Subhaneil Lahiri and Shiraz Minwalla. Plasmarings as dual black rings, 2007.

[10] O.A. Basaran R.E Benner Jr. and L.E. Scriven. Equilibria, stability and bifurcations of the rotatingcolumns of fluid subjected to planar disturbances. Proc. R. Soc. Lond. A, 433(1887):81–99, 1991.

[11] P. Roura. Thermodynamic derivations of the mechanical equilibrium conditions for fluid surfaces:Young’s and laplace’s equations. Am. J. Phys, 73(12):1139–1146, 2005.

[12] R.A. Brown & L.E. Scriven. The shape and stability of rotating liquid drops. Proc. R. Soc. Lond.A, 371(1746):331–357, 1980.

[13] E. Sorkin. A critical dimension in the black-string phase transition. Phys. Rev. Lett., 93:031601,2004.

[14] O. J. C Dias & L. Gualtieri V. Cardoso. The return of the membrane paradigm? black holes andstrings in the water tap. Int. J. Mod. Phys. D, 17:505, 2008.

II