Capillarity and Archimedes’ Principle of Floatation

John McCuan and Ray Treinen

January 15, 2009

Abstract

We consider some of the complications that arise in attempting to generalizea version of Archimedes’ principle of floatation to account for capillary effects.The main result provides a means to relate the floating position (depth inthe liquid) of a symmetrically floating sphere in terms of readily observablegeometric quantities.

A similar result is obtained for an idealized case corresponding to a sym-metrically floating infinite cylinder. Certain possibilities are also outlined inthe event symmetry is relaxed in this latter problem.

Central to all of these results is a specialized variational formula for floatingbodies which was derived in a special case earlier [Pac. J. Math. 231 (2007)pp.167–191] and is here generalized to account for gravitational forces.

1 Introduction

We wish to consider the following version of Archimedes’ principle:

An object, when deposited into a bath of liquid, displaces a volume of liquidhaving mass equal to the effective mass of the object.

This principle takes no account of the effects of surface tension or surface energiesassociated with wetting. Indeed, simple experiments show that it is possible, undercertain circumstances, for a convex object with density greater than that of a givenliquid bath to float (only) partially submerged on the surface of the bath. Finn [Fin08]has recently given the first rigorous proof of this fact, at least in an idealized situationwhich we describe in §4 below.

1

Even in the case of a convex object with constant density ρ smaller than the densityρ0 of the liquid, how to interpret and generalize Archimedes’ principle to includesurface tension and surface energies is not entirely obvious. In order to illustrate this,let us briefly consider the notion of displaced liquid referred to in the principle above.The principle states, after all, that

ρ0Vd = ρVm, (1)

where Vm is the volume of the object and Vd is the volume of the displaced liquid.According to the account of Vetruvius the displaced volume Vd of liquid is determinedby assuming the bath is filled precisely to the rim of the vessel so that when the objectis deposited, the displaced volume is simply that which spills onto the floor, or intoa larger vessel if one wishes to avoid the mess. One presumes that the interfaceis assumed planar in this discussion, and hence for a floating object the displacedvolume is also that portion of the volume of the object below the surface of the liquidwhile floating.

With surface tension, a bath may often be filled slightly above the rim of thevessel (or far above the rim of the vessel [Mie02] if the gravity is small); the interfaceis also required to meet the floating object at a prescribed contact angle depending onmaterials and not necessarily compatible with the condition that a planar interfacemeet the object in a manner appropriate for the depth of floatation. These factorsand others may conspire to render the interface decidedly nonplanar in general andsuggest we look elsewhere for the displaced volume of liquid.

One might also use a vessel with higher sides. In fact, we shall restrict attention, inthe physical three-dimensional case, to a cylindrical vessel with circular cross sectionof radius R and having initial depth large enough to completely immerse the object,even if it floats. See Figure 1. Under the foregoing Archimedean assumptions, theobject of density ρ < ρ0 will still float in a geometric position (i.e., attitude withrespect to the interface) congruent to that obtained when the liquid spilled over therim, but the planar interface and the object will be higher; the entire level of theinterface configuration will rise, being translated upward through a vertical distanceVd/(πR

2) as if the displaced volume had been injected at the bottom of the vessel.It should be noted that the attendant volume Vr of raised liquid (i.e., the volume ofliquid above the level of the original interface) is necessarily less than the displacedvolume, being determined by the relation

Vr + Vc = Vd

where Vc is the volume of a certain cavity swept out by the submerged portion of thebottom of the object as it is vertically translated through a distance Vd/(πR

2). Thus,

2

Figure 1: The displaced liquid of Archimedean floatation

there is a somewhat delicate relation between the displaced volume, the rise in theliquid interface, and the manner in which the object floats, even in the Archimedeancase.

When surface tension and wetting energies are taken into account, the interfaceitself must also be expected to change in a global way upon introduction of the floatingobject. It is clear that making sense of the displaced volume in terms of the liquidrise in this case would be difficult at best. See Figure 2.

We are not so ambitious to propose here a general definition of the displacedvolume of liquid for an arbitrary floating object. If we assume, however, that thecontact line on the floating object is the intersection of the boundary of the objectwith a horizontal plane, then we have recourse to the alternative mentioned above.Namely, Vd is simply the volume of the object “below the surface of the liquid,” thatis to say below the horizontal plane containing the contact line. This is not onlya convenient definition, but at least in the case of a sphere leads to an interesting

Figure 2: The difficulty of finding the displaced liquid when surface tension andwetting energies are acting

3

analogue of Archimedes’ principle and, surprisingly, one that can be viewed as a directgeneralization of it.

Before stating this relation, let us recast Archimedes’ principle in the special casewe have described. Under Archimedean assumptions, a sphere of radius a floating inthe center of a cylindrical vessel should have circular contact line determined by anazimuthal angle φ. See Figure 3. The volume of the sphere below the planar interface

Figure 3: Azimuthal angles determined by a horizontal contact line (left) and differingazimuthal angles in the two-dimensional case (right)

is

Vd =1

3πa3

(

sin2 φ cos φ+ 2 + 2 cos φ)

.

Using (1), we obtain the following

Theorem 1 According to Archimedes’ principle, a homogeneous sphere of densityρ > ρ0 will sink to the bottom of a bath of density ρ0, and a homogeneous sphere ofdensity ρ < ρ0 will float at a level determined by

cos3 φ− 3 cos φ = 2

(

1 −2ρ

ρ0

)

. (2)

It is easily checked that the function F (φ) = cos3 φ − 3 cos φ is increasing from −2to 2 on [0, π] with zero derivative at the endpoints and strictly positive derivativeinterior to the interval. Thus, for each positive value 0 ≤ ρ ≤ ρ0, the condition (2)determines a unique azimuthal angle. See Figure 4.

We obtain the following result under assumptions described in detail below.

4

Theorem 2 A sphere that floats in a centrally symmetric position as described aboveunder the effects of surface tension and adhesion effects of an axially symmetric bathmust float at a level determined by the azimuthal angle φ satisfying

cos3 φ− 3 cos φ+6

κa

[

H +cos γ

a

]

sin2 φ−3 sin γ

κa2sin(2φ) = 2

(

1 −2ρ

ρ0

)

, (3)

where κ = ρg/σ is the capillary constant determined by the gravitational accelerationg and the surface tension σ, H is the mean curvature of the liquid interface at thecontact line, and γ is the contact angle of the liquid interface with the floating sphere.

The function F (φ) appearing on the left in equation (3) also takes the values −2 and2 at the endpoints φ = 0 and π respectively. However, F is decreasing at φ = 0 anddecreases to a unique local interior minimum, thus allowing for values of ρ > ρ0 anddetermining explicitly a unique maximum density ρmax = ρmax(a, γ, κ, H) for whichρ > ρmax implies no floatation is possible. From the unique interior minimum, thefunction strictly increases to a unique (interior) maximum value greater than 2 fromwhich it strictly decreases to the value 2 at the other endpoint φ = π. It will be notedfrom this description that a unique azimuthal angle φ is determined for all values of0 < ρ < ρ0, and that two values are possible for certain values of ρ ≥ ρ0 (as long asρ is not too large). We presume by continuity that the physically relevant value forheavy floating spheres is the larger one determined by (3); see the list of conjecturesand open problems at the end for further comments.

0.5 1.0 1.5 2.0 2.5 3.0

-2

-1

1

2

0.5 1.0 1.5 2.0 2.5 3.0

-2

-1

1

2

3

4

0.5 1.0 1.5 2.0 2.5 3.0

-2

-1

1

2

3

4

Figure 4: The azimuthal angles determined by Theorems 1 (left) and 2 (middle);plotted together on the right

It could be legitimately objected that the quantity H appearing in our formulais somewhat unnatural due, first of all, to the fact that there is nothing of the sortappearing in Archimedes’ principle which we wish to generalize. Upon reflection,however, it becomes clear that some globally determined quantities must appear;the situation when capillarity is taken account of is necessarily more complicated

5

than Archimedean floatation. On the other hand, we agree that the value H should,in principle, be determined by other parameters not appearing in (3), namely thevolume of liquid in the cylinder before the floating object is introduced, and thecontact angle γout between the free surface interface and the wall of the cylinder. Insummary, intuition suggests:

An accurate description of how an object floats under the influence ofcapillarity should require the inclusion of global information on the config-uration including but not limited to the volume of liquid and the contactangle between the liquid interface and relatively distant structures such thesurface of the containing vessel.

We believe this intuition is correct, but a theorem giving the exact relation amongall the relevant quantities requires an understanding of the family of solutions ofthe ordinary differential equation governing the interface which is beyond what wecurrently have. The structure of this family of solutions is notoriously complicated,and there are many basic questions even about particular special solutions which arestill open. For a survey of some of the recent results, see [Fin86, Vog82, Sie06, Sie80,Nic02, EKT04, Tur80, JP68, Tre08].

On the other hand, it simply requires a change of perspective to designate Has a locally measurable independent parameter, and the form of Theorem 2 givesquantitative content to a competing intuition which says:

The depth at which an object floats in a liquid bath (relative to the “level”of the bath) should only depend on the relative densities, the contact angleγ of the interface with the surface of the floating object, and quantitiesmeasured locally near the object.

The choice and appearance of the particular quantity H is further explained in §5 inreference to Conjecture 4. The result above may also be defended on precisely theground that it is, like Archimedes’ result, essentially algebraic and beautifully simplein that sense.

2 Variational Formulation

The general assumptions of our model are outlined in [McC07] though the derivationgiven there was aimed at the zero gravity case in which buoyancy plays no role, andthe effects of gravity were not properly considered. For the sake of making this paper

6

somewhat more self-contained we include a short review/summary of the model andamend the deficiencies in the former derivation.

Quite generally, we consider a solid structure Σ = Σs ∪ Σm consisting of a sta-tionary part Σs and a movable, or floating, part Σm. In addition, we hypothesize anequilibrium liquid interface Λ with corresponding wetted region W = Ws ∪ Wm, sothat the liquid volume V satisfies ∂V = Λ ∪W and the contact line/triple interfaceis given by ∂Λ = ∂W. Under these assumptions, we consider the variational problemassociated with

E = σ|Λ| − σβ|W| + G (4)

where G =∫

V∪ΣmG and G is a position dependent function representing field forces

such as gravity.1

Under rather general hypotheses, as described in [McC07] a family of variationsleaving Σm fixed leads to the following (standard) variational formulas

˙|V| = −

∫

Λ

2HX ·N +

∫

∂Λ

X · ~n

where H is the mean curvature defined on Λ, X is the variation vector, N is the unitnormal pointing out of the liquid volume V, and ~n is the unit conormal to N and ∂Λpointing out of Λ.

˙|W| =

∫

∂Λ

X · ~ν,

where ~ν is the unit conormal to NW and ∂W pointing out of W; note that NW

denotes the unit normal to W pointing out of V and may also be denoted by N onthe interior of W where no ambiguity arises.

G =

∫

Λ

GX ·N and ˙|V| =

∫

Λ

X ·N.

These last two formulas apparently require an interesting and somewhat delicateapplication of more general mathematical principles of fluid mechanics, and we outlinetheir derivation under more general assumptions below.

For now, we assemble E/σ − λ ˙|V| from the constituent parts above where λ is aLagrange multiplier associated with the volume constraint:

E/σ − λ ˙|V| =

∫

Λ

(−2H +G/σ − λ)X ·N +

∫

∂Λ

(X · ~n− βX · ~ν).

1We included only∫

VG in [McC07].

7

The vanishing of this quantity for all variation vectors X results in the well knowngeometric boundary value problem

2H = G/σ − λ on Λ

cos γ = β on ∂Λ(5)

since ~n = (~n ·NW)NW +cos γ ~ν. In the special case under consideration in this paper,G represents the limiting value ρ0gz taken as a limit from inside the liquid, so that

2H = κz − λ

where κ = ρ0g/σ is a capillary constant for the problem.A more general variation allowing rigid motion of Σm takes the form

X = X(p; t, h) : M × (−ǫ, ǫ) × (−δ, δ) → R3

where M = Σ ∪ V is considered as an abstract manifold; see Figure 5.

Figure 5: The variation map and its notation

It is assumed here, as indicated in the figure that h parameterizes a family of rigidmotions w = w(x; h) to which Σm is subject. Denoting derivatives with respect to hby an acute accent, we find

´|Λ| = −

∫

Λ

2HX ·N +

∫

∂Λ

X · ~n, (6)

´|W| = −

∫

Wm

2HWX ·N +

∫

∂Wm

X · ~ν, (7)

8

G =

∫

Λ

GX ·N +

∫

Wm

GX ·NW +

∫

∂Σm

GmX ·Nm. (8)

This last term requires some explanation. The quantity Gm denotes the value of thevolumetric force field potential taken as a limit from inside the movable solid structureΣm. In the special case of a floating object of density ρ, we typically take Gm = ρgz.Also in this last identity Nm denotes the unit normal to the boundary ∂Σm of themovable/floating solid structure and points out of Σm, so that Nm = −NW on theircommon domain of definition Wm. Finally, we include a brief derivation.

Up until this point, we have stated all variational formulae in their final form,that is to say with the parameters of the variation set to zero so that X represents

d

dtX(p; t)∣

∣

t=0

where X = X(p; t) : M × (−ǫ, ǫ). For this calculation, we must temporarily assumethe parameters t and h are not evaluated at zero. Notationally, this is convenientlyindicated by a tilde so that Σm = X(Σm) = X(Σm; t; h), and we will evaluate att = h = 0 at the end.

Consideration of the second term should suffice. Setting

Gm =

∫

Σm

Gm,

we have

Gm =

∫

Σm

Gm ◦Xdet DX,

where X represents the restriction of the variation to Σm and the derivative is takenin M ⊂ R

3 with respect to p. Euler’s kinematical formula tells us how a materialintegral changes with the flow of a region of fluid. We can cast our present situationinto this framework starting with the preliminary identity

∂

∂hdet DX = (divR3 v) ◦X det DX

where v(x; h) = X(X−1(x; h); h) is the spatial velocity associated with the flow X =X(p; h) and we have simply suppressed the t dependence. It might be expected (orhoped) that in our situation the motion/flow associated with the variation should beparticularly simple, at least on the solid movable object Σm, and that we might have,for example, X(p; h) ≡ w(p; h) there. However, taking into account the motion of

9

the liquid and that of the contact line of the liquid interface Λ in particular, it isclear that this would violate the continuity assumption on the variation X : M ×(−ǫ, ǫ) × (−δ, δ) → R

3. Having made this concession and subjected ourselves to theadded complication that other authors seem to have avoided, it is some consolation,as pointed out by Finn [Fin05], that the internal motion of the liquid under a variationof the free surface interface could be very complicated, and we are taking account ofsuch possibilities.

In any case, we continue to obtain

Gm =

∫

Σm

DGm · v + divR3 v

=

∫

Σm

divR3(Gmv)

=

∫

∂Σm

Gmv ·Nm,

so that

Gm∣∣

h=0

=

∫

∂Σm

GmX ·Nm.

A similar argument applies to the integral over V appearing in G and also yields

´|V| =

∫

Λ

X ·N +

∫

Wm

X ·N

where we have returned to the general assumption on evaluation, that t = h = 0.Combining this with (6-8), we have

E/σ − λ ´|V| =

∫

Λ

(−2H +G/σ − λ)X ·N +

∫

∂Λ

(X · ~n− βX · ~ν) +

β

∫

Wm

2HWX ·N +

∫

Wm

(G/σ − λ)X ·N +

∫

∂Σm

(Gm/σ)X ·Nm

=

∫

∂Wm

X · ~n− cos γ

∫

∂Wm

X · ~ν +

cos γ

∫

Wm

2HWX ·N +

∫

Wm

(G/σ − λ)X ·N +

∫

∂Σm

(Gm/σ)X ·Nm.

Next we refer to a calculation from [McC07] which uses the fact that w−1(X; h) ∈ Σm

when X = X(p; h) ∈ w(Σm; h) to show that

X − w ∈ TXΣm.

10

It follows that X may be replaced with w in the formula above. A second calculationinvolving an explicit auxiliary variation shows

∫

Wm

2HWw ·N =

∫

∂Wm

w · ~ν.

Making the indicated substitutions, we arrive at our new necessary condition forequilibrium of a floating object:

Theorem 3 If a floating object Σm subject to gravitational forces (denoted by G andGm as described above) locally minimizes energy among liquid interface configurationscompatible with a smooth family of rigid motions w = w(x; s) with w(x; 0) = idR3 andthe wetted region on the floating object is denoted by Wm, then the configuration mustsatisfy

∫

∂Wm

w · ~n+

∫

Wm

(G/σ − λ)w ·NW +

∫

∂Σm

(Gm/σ)w ·Nm = 0, (9)

where ~n is the outward pointing unit conormal along the boundary of the liquid inter-face Λ, NW is the unit normal to Σm pointing out of the liquid, Nm = −NW , and wrepresents the derivative with respect to s evaluated at s = 0.

The condition of the theorem must hold for all w ∈ R3 for free floatation, or more

generally for any collection of directions in which Σm is free to move.We next proceed to examine the consequences of (9) for the simple cases of floata-

tion suggested in the introduction.

3 Floatation in three dimensions

Here we assume a vertical circular cylindrical vessel is observed with a sphere Σm

floating symmetrically along the axis of the vessel and having symmetric circularcontact line at azimuthal angle φ = φ. Assuming the surface of the liquid is alsorotationally symmetric with respect to the same axis, the meridian of the surfacewith vertical component u and radial component r considered as functions of arclengthalong the meridian must satisfy the boundary value problem

r = cosψu = sinψ

ψ = κu− λ− sinψ/rψ = γ − φ and u = d+ a cos φ when r = r(0) = a sin φψ = π/2 − γout when r = r(ℓ) = R

(10)

11

where we have chosen coordinates so that the center of the floating sphere is (0, 0, d),and ℓ is the total length and ψ is the inclination angle of the meridian. Though thesystem of ordinary differential equations appearing in the problem above has beenstudied extensively, the structure of the family of all solutions is little understood.We conjecture that the boundary value problem as stated here has a unique embeddedsolution for each γ ∈ [0, π]. See §5 for a more precise statement. With this assertionverified, one could commence to determine the parameters d and λ (and H which willmake its appearance momentarily) in terms of natural physical parameters such asthe volume constraint |V| = V , and the contact angles γ and γout, and so on. In theabsence of such a result, we take a different path and proceed directly to the auxiliarycondition (9).

The following formulae, valid in the plane y = x2 = 0, are useful in simplifyingthe integrals in (9):

Nm[φ] = sin φe1 + cos φe3

NW [φ] = −Nm

= − sin φe1 − cos φe3

~ν[φ] = (Nm)⊥

= − cos φe1 + sin φe3

~n = cos γ~ν + sin γNW

= − cos(φ− γ)e1 + sin(φ− γ)e3

NΛ = (−~n)⊥

= sin(φ− γ)e1 + cos(φ− γ)e3.

(11)

In these formulae, the bracketed φ indicates validity in the form of the result for anarbitrary azimuthal angle on ∂Σm though the main interest is on ∂Wm; e1 and e3 arethe standard orthonormal unit vectors in R

3.Taking a vertical translation for the rigid motion of Σm so that w = e3, the three

terms of (9) are as follows:

∫

∂Wm

e3 · ~n = 2πa sin φ sin(φ− γ).

∫

Wm

(κz − λ)e3 ·N = πa2

[

(κd− λ) sin2 φ−2κa

3(1 + cos3 φ)

]

.

∫

∂Σm

κρ

ρ0

ze3 ·Nm =

4πκ

3a3 ρ

ρ0

.

12

Combining these terms and rearranging:

6 sin φ sin(φ− γ)

κa2+

3(κd− λ) sin2 φ

κa− 2 cos3 φ = 2

(

1 −2ρ

ρ0

)

. (12)

Next, we make the substitution

2H = κ(d+ a cos φ) − λ

which follows directly from (5). This leads to

6 sin φ sin(φ− γ)

κa2+

3(2H − κa cos φ) sin2 φ

κa− 2 cos3 φ = 2

(

1 −2ρ

ρ0

)

.

This last condition simplifies directly into condition (3) of Theorem 2. It remains toverify the description of the function

F (φ) = cos3 φ− 3 cos φ+6

κa

[

H +cos γ

a

]

sin2 φ−3 sin γ

κa2sin(2φ).

The values at the endpoints are immediate. We find also that

F ′[φ]

3= − cos2 φ sin φ+ sin φ+

4

κa

[

H +cos γ

a

]

sin φ cos φ−2 sin γ

κa2cos(2φ)

= sin3 φ+2

κa

[

H +cos γ

a

]

sin(2φ) −2 sin γ

κa2cos(2φ).

Thus, F ′(0) = F ′(π) = −(6/κa2) sin γ < 0. From this it is clear that F must attainan absolute min at some value less than −2 and an absolute max greater than 2. Atthese points, F ′ must vanish, and it only remains to show these are the only zeros ofF ′ on [0, π]. In fact, we see that

F ′[φ]

3= sin3 φ+ A sin(2φ−B)

for some quantities A > 0 and B independent of φ. The fact that F ′(0) < 0 tells usthat we may assume 0 < B < π. Clearly, since 0 ≤ φ ≤ π, we have sin3 φ ≥ 0 andthere can be no zero of F ′ on the interval [B/2, (B + π)/2]. For the rest, we considertwo cases.CASE I. 0 < B ≤ π/2, i.e., F ′′(0) ≥ 0.

13

In this case, both terms in the expression for F ′ are increasing on the interval0 < φ < B/2, so F ′ can have at most one zero there. (And since F ′(B) > 0 it doeshave exactly one.)

F ′ must also have a zero on [(B + π)/2, π]. Consider the largest such zero, andobserve that

F ′′[φ]

3= 3 cos φ sin2 φ+ 2A cos(2φ−B)

= (3/2) sin(2φ) sin φ+ 2A cos(2φ−B).

It must be the case that F ′′ ≤ 0 at the largest zero, but since F ′′′ is positive on[(B + π)/2, π], this means that F ′′ is negative (or F ′ is decreasing) on the intervalbetween (B + π)/2 and the largest zero. Since F ′((B + π)/2) = sin3(B + π)/2 > 0,there can be no other zeros.CASE II. π/2 ≤ B < π, i.e., F ′′(0) ≤ 0.

The reflection φ→ π − φ transforms this case into the first one with B → π −B.�

We note finally that the reader will have no trouble verifying that under theArchimedean assumptions H = 0 (a planar interface) and φ = γ (the appropriateazimuthal angle for a horizontal plane to meet the sphere at the correct contactangle) the formula in Theorem 2 reduces to the condition of Archimedes.

4 Floatation in two dimensions

Finn has recently considered a variational problem for the energy

E = −σ|Λ| − σβ|W| + G (13)

where Λ is the linear segment of intersection of an assumed planar/linear interfacewith a two-dimensional convex body and G is the specific gravitational energy we haveconsidered above. The measures appearing in the first two terms in this functionalare one-dimensional (length) and the integral is an area integral. There is no volumeconstraint in Finn’s problem, nor outer container, and it is an assumption that theinterface always lies along a fixed line. Nevertheless, he obtains the striking resultthat for some values of ρ > ρ0, σ and β there will be an equilibrium which is alocal minimum for energy in which the convex body contacts the interface, i.e., floats.While Finn gives no precise description of the geometric configuration of floatationcorresponding to our results, we can formulate and extend our results to a problem

14

dimensionally similar his. Various aspects of this two-dimensional problem seemingto be of interest, we present them now.

Physically, we envision a trough consisting of two vertical walls and a horizontalbottom. The trough is assumed to extend infinitely in the y = x2 direction and tobe filled with a sea of liquid. Into this sea is introduced a horizontal floating circularcylinder (an infinitely long log) with axis parallel to e2. Let us assume that the freesurface interface Λ also is always of cylindrical form with generator parallel to e2, sothat if the log is centrally located between the walls and the interface shares the samemidplane symmetry, then the projection of the system onto the x, z-plane resemblesthat of the system considered in the previous section (Figure 3, left), though theequation of the generating curve (and hence its shape) will be be different from thatof the meridian previously considered.

The energy of such a system can be taken to have the form of (4)

E = σ|Λ| − σβ|W| + G

where the dimensions of the measures have been lowered by one and G =∫

V∪ΣmG is

an area integral. The first order necessary conditions take the form

k = G/σ − λ on the curve Λ,

cos γ = β at the endpoints of Λ,

and

w · ~n∣∣

∂Λ

+

∫

Wm

(G/σ − λ)w ·NW +

∫

∂Σm

(Gm/σ)w ·Nm = 0, (14)

where k is the curvature of Λ, and λ arises from an area constraint on the cross sectionof liquid in the trough. In analogy to the three-dimensional case, we assume an areadensity ρ for the object, floatation in a liquid of area density ρ0, a capillary constantκ = ρ0g/σ, and that the radius of the log is a.

Before we begin an analysis of this variational problem in earnest, let us pause tonote what Archimedes’ principle would state in this lower dimensional case (becauseit will appear in a surprising way later):

Theorem 4 According to Archimedes’ principle in one lower dimension, a homoge-neous disk/log of density ρ > ρ0 will sink to the bottom of a bath of density ρ0, and ahomogeneous disk/log of density ρ < ρ0 will float at a level determined by

2φ− sin(2φ) = 2π

(

1 −ρ

ρ0

)

. (15)

15

We assume initially the contact line (i.e., the two points where Λ meets Σm) is

determined by two azimuthal angles, one φ as before and a second ¯φ measured in thecounter-clockwise direction from the vertical e3; See Figure 3, right. In addition to(11), the following identities have been found useful.

Nm[φ] = − sin ¯φe1 + cos ¯φe3

NW [φ] = −Nm

= sin ¯φe1 − cos ¯φe3

~ν[φ] = (Nm)⊥

= cos ¯φe1 + sin ¯φe3

~n = cos γ~ν + sin γNW

= cos( ¯φ− γ)e1 + sin( ¯φ− γ)e3

NΛ = (−~n)⊥

= − sin( ¯φ− γ)e1 + cos( ¯φ− γ)e3.

(16)

Taking first a horizontal motion of the floating sphere, so that w = e1, we find

e1 · ~n∣∣

∂Wm

= cos( ¯φ− γ) − cos(φ− γ)

= −2 sinB sin(A− γ),

where

A =¯φ+ φ

2and B =

¯φ− φ

2,

∫

Wm

(κz − λ)e1 ·NW = a(κd− λ)(cos ¯φ− cos φ) +

κa2

2(cos2 ¯φ− cos2 φ)

= −2a sinB sinA(κd− λ+ κa cosA cosB),

and∫

∂Σm

(κρ

ρ0

z − λ)e1 ·Nm = 0.

Since each of these terms has a factor sinB, we see from condition (14), that one

possibility is sinB = 0. If this holds, it can readily be determined that ¯φ = φ. Oncethis occurs, then since the left and right interfaces must start from the same heightand with the same inclination angle, we have a proof that the axis of the floatingcylinder must lie on the midplane between the vertical walls. This is the conclusionwe would like to make. The other alternative is that

sin(A− γ) + a sinA(κd− λ+ κa cosA cosB) = 0

16

which we rewrite as

[cos γ + a(κd− λ)] sinA +κa2

2sin(2A) cosB − sin γ cosA = 0 (17)

Leaving this open as a possibility for the moment, we turn to an independentvertical translation of Σm with w = e3.

e3 · ~n∣

∣

∂Wm

= sin( ¯φ− γ) + sin(φ− γ)

= 2 cosB sin(A− γ),

∫

Wm

(κz − λ)e3 ·NW

= a(κd− λ)(sin ¯φ+ sin φ) +κa2

4(sin(2 ¯φ) + sin(2φ)) +

κa2

2( ¯φ+ φ) − κa2π

= 2a cosB sinA(κd− λ) +κa2

2(sin(2A) cos(2B)) +

κa2

2( ¯φ+ φ) − κa2π,

and∫

∂Σm

κρ

ρ0

z e3 ·Nm = κa2π

ρ

ρ0

.

Combining these terms to form the expression in (14), we arrive at a second necessarycondition

[cos γ + a(κd− λ)] sinA cosB − sin γ cosA cosB+

κa2

2sinA cosA(1 − 2 sin2B) +

κa2

4( ¯φ+ φ)

=κa2π

2

(

1 −ρ

ρ0

)

. (18)

Multiplying the equation in (17) by cosB and subtracting the result from (18) andsimplifying, we obtain the surprising condition

¯φ+ φ− sin( ¯φ+ φ) = 2π

(

1 −ρ

ρ0

)

. (19)

This is surprising because it says that if the log floats anywhere but in the middlebetween the vertical walls of the trough, then the part of the log that rises above

17

the surface of the liquid must be prescribed exactly as Archimedes said it should.In particular, the portion that is wetted is independent of all parameters except thedensity fraction (!), a scenario which we view as highly unlikely. The fact, that wecannot rule out this possibility leads to the following curious result.

Theorem 5 In the two-dimensional floating log problem described above, either theaxis of the log lies in the vertical midplane determined by the sides of the vessel, orthe wetted/nonwetted region is determined by the generalized version of Archimedes’condition given in (19).

We will return to the last possibility in the last section of the paper. For now,we pursue a course similar to that which we were forced to pursue in the three-dimensional case by assuming symmetry of the interface with respect to the midplane.When ¯φ = φ, condition (18) associated with the vertical translation is still non-vacuous and becomes

F (φ) = 2φ+ sin(2φ) +4

κa2sin(φ− γ) +

4

κa(κd− λ) sin φ = 2π

(

1 −ρ

ρ0

)

.

Again following the three-dimensional case, we let

k = κ(d+ a cos φ) − λ

denote the curvature of the interface at the contact line on the object. Substitutionyields

Theorem 6 A log that floats in a centrally symmetric position as described above un-der the effects of surface tension and adhesion effects must float at a level determinedby the azimuthal angle φ satisfying

2φ− sin(2φ) +4

κa2sin(φ− γ) +

4k

κasin φ = 2π

(

1 −ρ

ρ0

)

. (20)

where k is the curvature of the interface at the contact line, and γ is the contact angleof the interface with the floating log.

The behavior of the function

F (φ) = 2φ− sin(2φ) +4

κa2sin(φ− γ) +

4k

κasin φ

18

is somewhat different than that in the three-dimensional case; see Figure 6. One seesfirst of all that

F (0) = −4

κa2sin γ < 0; F (π) = 2π +

4

κa2sin γ > 2π.

Thus, the endpoint values do not coincide with the extremes of the expression on theright in (20) associated with ρ = 0 and ρ = ρ0. Nevertheless, the interval between0 and 2π is clearly covered by the values of F (φ) and, in fact, each value is takenexactly once. To see this we compute

F ′(φ)

2= 1 − cos(2φ) +

2

κa2cos(φ− γ) +

2k

κacos φ

and observe first that

F ′(0)

2=

2

κa2cos(γ) +

2k

κa= −

F ′(π)

2.

It follows that F ′ is nonpositive at one of the endpoints and has the opposite signat the other. Using this, reasoning similar to that found in §3 shows F ′ can have atmost one zero on [0, π].

0.5 1.0 1.5 2.0 2.5 3.0

1

2

3

4

5

6

0.5 1.0 1.5 2.0 2.5 3.0

-2

2

4

6

8

10

0.5 1.0 1.5 2.0 2.5 3.0

-2

2

4

6

8

10

Figure 6: The azimuthal angles determined by Theorems 4 (left) and 6 (middle);plotted together on the right

Therefore, some salient features of Theorem 2 hold also in this lower dimensionalcase. For fixed k and γ, if ρ ≤ ρ0, there is a unique height at which the disk/logcan float; there is an interval ρ0 < ρ < ρmax on which there is at least one (andsometimes two) possible heights at which floatation can occur. One expects that iftwo azimuthal angles are determined by (20), the larger one is the physically relevant.

19

5 Conjectures and Questions

We consider first the three dimensional case, concerning which we make what we be-lieve is a reasonable but difficult conjecture concerning the solutions of the boundaryvalue problem (10).

Conjecture 1 Given d, λ and φ fixed (all the other parameters being obviously deter-mined by the physical problem and also given), there is at most one embedded solutionof the boundary value problem (10).

Under certain circumstances, we can sharpen this conjecture considerably. Let usassume the interface is given by the graph of a function u over the horizontal x, y-plane. In this case, the circular base of the cylinder may be partitioned into theannular region A beneath the interface and the disk B beneath the submerged portionof the sphere. Also, we may use the well known nonparametric mean curvatureequation:

div

(

Du√

1 + |Du|2

)

= κu− λ.

Integrating the equation over A, we obtain

κVA − λ|A| = 2πa sin φ sin(φ− γ) + 2πR cos γout

where VA =∫

Au is the volume of liquid present over A. On the other hand, this

volume can be computed by subtracting the volume over B from the total volume:

VA = |V| − πda2 sin2 φ+2πa3

3(1 + cos3 φ).

Similarly, |A| = πR2 − πa2 sin2 φ.Combining these relations and using the definition 2H = κ(d + a cos φ) − λ, we

find

2 sin φ(sin φ sin(φ− γ) + Ha sin φ)

=κ

πa|V| − 2

R cos γout

a−λR2

a+κa2

3(2 − cos3 φ+ 3 cos φ). (21)

On the other hand, we can rewrite condition (3), as

cos3 φ− 3 cos φ+6 sin φ

κa2(sin(φ− γ) + Ha sin φ) = 2

(

1 −2ρ

ρ0

)

,

20

so that on substituting from (21) for the last term on the left, we find

3

κa2

[

κ|V|

πa−

2R cos γout

a− λ

R2

a

]

= −4ρ

ρ0

.

It will be noted that φ has disappeared from the equation. Next it will be noted thatλ may now be determined in terms of |V|, κ, a, R, ρ/ρ0, and γout, but independentlyof γ and φ. Given this, one can evidently go back to (12), the version of (3) we hadbefore introducing H, and solve for d in terms of φ.

Conjecture 2 It should be possible as long as ρ ≤ ρ0 (and as long as the level of thebath is deep enough and |V| is large enough so that when the sphere is pushed to thebottom, there is enough liquid to cover it) to:

1. solve for λ in terms of |V|, κ, a, R, ρ/ρ0, and γout,

2. solve for d in terms of φ, γ and the other parameters, so that everything is interms of φ, and

3. find a unique value of φ and a unique embedded solution of (10) satisfying allconditions of the problem.

This conjecture is for the case ρ < ρ0. In principle similar assertions should hold insome cases for ρ ≥ ρ0. A fundamental difficulty of the conjecture is that parametricinterfaces are included; there is no known standard procedure for determining theLagrange parameter λ in such cases. Given the conjecture above, the value of d andλ and, hence, the value of H should be determined. From this, we can compute thevalue of ρmax determined by the function F (φ) appearing on the left in (3). Thereis a problem, however, in the identification of this value with the physical maximumdensity for which floatation can occur. Nevertheless, our results suggest the following:

Conjecture 3 There is a nontrivial interval of densities ρ0 < ρ < ρphys for whicha given sphere of fixed radius a and density ρ admits a solution as described in theproceeding conjecture.

It should be noted that the existence of a meridian curve satisfying the conditionsof (10) and the volume constraint etc., by no means guarantees that such a spherewill float in the position thereby described. The stability of these configurations, werethey to be determined, must still be analyzed. As a first guess we have the followingwhich comes from experimentation and is primarily of interest due to the appearanceof the quantity H :

21

Conjecture 4 (experimental) If a centrally symmetric equilibrium exists, as con-jectured above, and the sphere is lighter than the liquid (ρ < ρ0), then the resultingfloatation configuration will be stable if H < 0 and will be unstable otherwise, withthe sphere tending to the side if H > 0. If the sphere is heavier (ρ > ρ0), then thestability criteria will be reversed.

We emphasize that these conditions are derived from a limited range of parame-ters, and we expect the actual stability conditions to be (much) more complicated.Nevertheless, these observations were central in the formulation of the main resultsabove (due to the significant role played by H), and for this reason we felt it worthmentioning.

Finally, we have

Conjecture 5 For any sphere which floats in a stable configuration in the center ofa cylindrical bath with ρ ≥ ρ0, the azimuthal angle φ determined by the larger solutionof (3) is the one which will be observed physically.

We next turn to the two-dimensional case. The conjectures above, can of coursebe adapted to this case, but we leave that adaptation to the reader and proceed tosome comments concerning the possibility of having differing azimuthal angles. Themain conjecture is that (17) and (18) are incompatible conditions for ¯φ 6= φ. Werethis the case, it would say that the only possible equilibria are the symmetric ones.This also can be sharpened via an interesting calculation. Starting with (17) whichwe write as

[cos γ + a(κd− λ)] sinA− sin γ cosA + κa2 sinA cosA cosB = 0,

we consider the curvatures at the two contact points

k = κ(d+ a cos φ) − λ and ¯k = κ(d+ a cos ¯φ) − λ,

and the average kavg for which

κd− λ = kavg − κa cosA cosB.

Substituting the last into the first, we find (cos γ+kavga) sinA− sin γ cosA = 0. (Thesignificance is that B is eliminated from the equation.) Recall that A is the averageof the azimuthal angles.

22

Conjecture 6 Assuming γ, kavg, a, and ρ/ρ0 are given and fixed, there is no valueof A satisfying the system

(cos γ + akavg) sinA− sin γ cosA = 0

2A− sin(2A) = 2π(

1 − ρ

ρ0

)

.

Several of the conjectures above admit substantial numerical verification, though theremay still be outlying cases of which we are not aware.

We note finally that we have obtained global floatation configurations numerically.The stability and uniqueness of those configurations is not presently known. Belowwe give representative global configurations which have been obtained and a list ofthe relevant parameters. In all trials a = 1, κ = 1, and R = 2 and question marksrefer to Conjecture 4. In the two-dimensional case, the area of the cross section ofliquid is taken to be 10.

−3 −2 −1 0 1 2 3

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

x

heig

ht

−4 −3 −2 −1 0 1 2 3 4

0

1

2

3

4

5

6

x

heig

ht

−3 −2 −1 0 1 2 3

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

x

heig

ht

−4 −3 −2 −1 0 1 2 3 4

0

1

2

3

4

5

6

x

heig

ht

−4 −3 −2 −1 0 1 2 3 4

0

1

2

3

4

5

6

x

heig

ht

−4 −3 −2 −1 0 1 2 3 4

0

1

2

3

4

5

6

x

heig

ht

Figure 7: Two-Dimensional Case (floating logs). Table shows parameters for eachconfiguration from left to right.

ρ/ρℓ γ γout d λ φ

(1) Lightest 0 π/2 π/2 1.8850 1.0860 1.9284(2) Heavy 1 π/2 π/2 2.6504 3.4494 1.2132(3) Flat .5 π/2 π/2 1.6427 1.6427 1.5708(4) Denser 1.6 π/2 π/2 1.9934 3.9663 0.4973(5) Untable(?) .5 π/4 π/4 2.3777 2.7878 0.7328(6) Stable(?) .5 π/4 3π/4 2.7382 3.4704 1.0150

In the three dimensional cases the volume of liquid is taken to be 25.

23

−3 −2 −1 0 1 2 3

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

r

heig

ht

−3 −2 −1 0 1 2 3

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

r

heig

ht

−3 −2 −1 0 1 2 3

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

r

heig

ht

−2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5

0

0.5

1

1.5

2

2.5

3

3.5

4

r

heig

ht

−3 −2 −1 0 1 2 3

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

r

heig

ht

−3 −2 −1 0 1 2 3

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

r

heig

ht

Figure 8: Three-Dimensional Case (floating balls). Table shows parameters for eachconfiguration from left to right.

ρ/ρℓ γ γout d λ φ

(1) Lightest 0 π/2 π/2 2.1174 1.8486 1.7086(2) Heavy 1 π/2 π/2 1.8689 2.1377 1.4330(3) Flat .5 π/2 π/2 1.9902 1.9902 1.5708(4) Denser 2.1 π/2 π/2 1.4293 2.5321 0.9293(5) Untable(?) .5 π/4 π/4 1.3646 1.4679 0.7790(6) Stable(?) .5 π/4 3π/4 2.0192 2.6403 1.2570

References

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[Fin86] R. Finn. Equilibrium Capillary Surfaces. Springer-Verlag, New York, 1986.

[Fin05] R. Finn. Floating and partly immersed balls in a weightless environment.Funct. Differ. Equ., 12(1-2):167–173, 2005.

[Fin08] R. Finn. Criteria for floating I. Preprint, 2008.

[JP68] W.E. Johnson and L.M. Perko. Interior and exterior boundary value prob-lems from the theory of the capillary tube. Arch. Rational Mech. Anal.,29:125–143, 1968.

[McC07] J. McCuan. A variational formula for floating bodies. Pac. J. Math,231(1):167–191, 2007.

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[Nic02] R. Nickolov. Uniqueness of the singular solution to the capillary equation.Indiana Univ. Math. J., 51(1):127–169, 2002.

[Sie80] D. Siegel. Height estimates for capillary surfaces. Pac. J. Math., 88(2):471–515, 1980.

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