Liquid Bridges between Contacting Balls
Thomas I. Vogel
Abstract. The problem studied is that of a rotationally symmetric liq-
uid bridge between two contacting balls of equal radius, with the same
contact angle with both balls, and in the absence of gravity. The bridge
surface must be of constant mean curvature, hence a Delaunay surface.
If the contact angle is less than π2 , existence of a rotationally symmetric
bridge is shown for a large range of the relevant parameter, giving unduloidal, catenoidal, and nodoidal bridges. If the contact angle is greater
than or equal to π2 , it is shown that no stable rotationally symmetric
bridge which is symmetric across the perpendicular bisector of the line
segment between the two centers of the balls exists. Existence therefore
depends discontinuously on contact angle.
Mathematics Subject Classication (2010). Primary 76B45; Secondary
53A10.
1. Introduction
This paper continues the study of liquid bridges between solid balls in the
absence of gravity, (see Figure 1) extending and applying results from [4],
[5], and [6]. For a number of physical applications of this problem, see the
introduction to [2]. The main purpose of this paper is to investigate the
physically important special case in which the two balls are contacting, are
the same size, and the same material (leading to the same contact angle). The
fact that the theory for contacting and non-contacting balls is quite dierent
was pointed out to the author by Leonid Fel (see also [3]), and is the genesis
π
2 the behavior
is very dierent from the case that contact angle is greater than or equal
π
to
2 . In the former case, we have existence of rotationally symmetric liquid
bridges for a large range of the relevant parameter, whereas in the latter case
of this paper. We will see that if the contact angle less than
we will see a striking non-existence result. This discontinuous behavior as a
function of contact angle is reminiscent of other exotic behavior observed
for capillary surfaces. (Note 3.6 outlines another example of what might be
considered as exotic behavior, where the parameter is the separation of the
2
T. I. Vogel
Σ
B
B2
1
Figure 1. A liquid bridge between xed, solid balls
balls.) Chapters 5 and 6 of [1] contain more examples of exotic behavior
of capillary surfaces. For a discussion of the relation between the bridges
between balls problem and the more commonly studied bridges between
parallel planes problem, see the introduction to [5].
To study bridges between solid balls, we will use the framework initiated
in [6], there referred to as cutting and scaling. The procedure is to take
sections of standardized Delaunay surfaces and scale them to form bridges
making appropriate contact angles with the xed balls.
More specically, dene
x (s, A), y (s, A), and ϕ (s, A) to be solutions of
the system of ODE's
dx
ds
dy
ds
dϕ
ds
=
cos ϕ (s, A) ,
(1.1)
=
sin ϕ (s, A) ,
(1.2)
=
cos ϕ (s, A)
+ 2A,
y (s, A)
(1.3)
with initial conditions
(x (s, A) , y (s, A))
A.
The curve
choice of
x (0, A)
=
0,
y (0, A)
=
1,
ϕ (0, A)
=
0.
is the standardized Delaunay prole for that
We seek a liquid bridge making contact angle
of equal radius
r,
that the balls contact, then the distance
2r.)
γ
with two balls
whose centers are separated by a distance
R
R.
B1 , B2
(In the case
between the centers is of course
In this paper, we seek bridges which are, in addition to being rotation-
ally symmetric, also symmetric across the plane which is the perpendicular
bisector of the line segment between the two centers of the balls.
To nd such a bridge, given contact angle
standardized Delaunay curve, nd, if possible, an
f (s2 , A; γ) ≡
γ and an A specifying
s2 > 0 solving
R
x (s2 , A)
cos (ϕ (s2 , A) + γ) + sin (ϕ (s2 , A) + γ) =
y (s2 , A)
2r
(from equation (2.16) of [6]) for which
cos (ϕ (s2 , A) + γ) > 0
a
(1.4)
Contacting Balls
3
(from equation (2.8) of [6] ). Then a section of the standardized Delaunay
curve from
s1 ≡ −s2
to
s2
may be scaled to form a bridge between the two
balls with the correct contact angle.
γ ∈ 0, π2 .
2. Existence for
For Theorem 2.2 we will need to divide the proof into cases depending on the
inclination angle at the inection of an undulary. There is a simple expression
for this.
If A ∈ −1, − 12 ∪ − 12 , 0 , then the inclination angle at
an inection point satises sin ϕ = ± (2A + 1). (The inclination angle will
alternate between positive and negative values.)
Proposition 2.1.
Proof.
Given
A,
from (1.1) there is an inection at
s
if
dϕ (s, A)
cos ϕ (s, A)
=
+ 2A = 0.
ds
y
From [6], we have explicit expressions for
cos ϕ
y
=
=
y cos ϕ
and
y2 ,
so
((2A + 1) cos (2As) − 1) / (2A)
1 + (2A + 1) (1 − cos (2As)) / (2A2 )
A ((2A + 1) cos (2As) − 1)
.
2A2 + (2A + 1) (1 − cos (2As))
Set the right hand side equal to
−2A.
Solving this for
cos 2As,
we have the
equation
2
− (2A + 1) cos 2As = − (2A + 1) ,
A = − 21 (which is a straight line and uninteresting),
cos 2As = 2A + 1 at the inection.
Now look at the expression for cos ϕ from [6]. Since A < 0, it is
thus except for
that
we get
(2A + 1) cos 2As − 1
cos ϕ (s, A) = − p
.
2
4A + (4A + 2) (1 − cos 2As)
Thus, at the inection,
2
cos ϕ (s, A)
=
=
=
which is real for
A
between
−p
(2A + 1) − 1
4A2 + (4A + 2) (1 − (2A + 1))
4A2 + 4A
−√
4A2 − 8A2 − 4A
p
2 −A2 − A,
−1
and 0. Since
sin2 ϕ + 4 −A2 − A = 1,
2
sin2 ϕ = 4A2 + 4A + 1 = (2A + 1) ,
and
sin ϕ = ± (2A + 1),
as desired.
4
T. I. Vogel
For γ ∈ 0, π2 and A ∈ (−1, ∞), there exists a rotationally
symmetric liquid bridge between solid balls which are in contact. This bridge
is also symmetric across the plane between the balls.
Proof.
From (1.4), taking
there is an
ŝ > 0
Theorem 2.2.
R
to be
2r,
we must show that for such
A
and
γ
solving
f (ŝ, A; γ) ≡
x (ŝ, A)
cos (ϕ (ŝ, A) + γ) + sin (ϕ (ŝ, A) + γ) = 1
y (ŝ, A)
which also satises
g (ŝ, A; γ) ≡ cos (ϕ (ŝ, A) + γ) > 0.
We will approach this by rst looking at dierent values of
considering dierent contact angles for those
A ∈ −1, − 12 . From
between arcsin (2A + 1)
A,
and then
A's.
ϕ (s, A)
− arcsin (2A + 1), so
Case 1:
Proposition 2.1, the inclination angle
runs
(which is negative) and
that we have
γ + arcsin (2A + 1) ≤ γ + ϕ (s, A) ≤ γ − arcsin (2A + 1) .
•
γ − arcsin (2A + 1) < π2 . We must have γ + arcsin (2A + 1) >
π
− 2 , since the range of arcsin is − π2 , π2 . Thus if γ < π2 +arcsin (2A + 1),
g (s, A; γ) is positive for all s on the standardized Delaunay prole. In
Case 1a:
f (s, A; γ) = 1 has a root for positive
f (0, A; γ) = sin γ ∈ [0, 1). As s tends to innity, x (s, A)
tends to innity and y (s, A) oscillates between two positive values. Thus
x(s,A)
y(s,A) tends to innity as s → ∞. Since cos (ϕ (s, A) + γ) is positive and
bounded from zero, it follows that f (s, A; γ) → ∞ as s → ∞. Therefore,
1
by the intermediate value theorem, in the case that A ∈ −1, −
2 and
γ < π2 + arcsin (2A + 1), there exists a rotationally symmetric bridge of
this case, we need only show that
s.
Certainly,
the type described.
•
γ − arcsin (2A + 1) > π2 . Now g (s, A; γ) will be negative for
π
some s, since ϕ (s, A)+γ will cross
2 (at s = 0, ϕ is zero, so ϕ (0, A)+γ <
π
π
). Let s0 be the smallest positive solution to ϕ (s, A)+γ =
2
2 . Although
π
f (s0 , A; γ) = sin 2 = 1, this is certainly
not the root we seek, since
df g (s0 , A; γ) = 0. However, consider ds
. We have
Case 1b:
s=s0
y cos ϕ − x sin ϕ
df
=
cos (ϕ + γ)
ds
y2
x
cos ϕ
+ − sin (ϕ + γ) cos (ϕ + γ)
+ 2A
y
y
x cos ϕ
=−
+ 2A
y
y
cos (ϕ + γ) = 0, sin (ϕ + γ) = 1. The quantity
s0 . This is non-zero in this case, since ϕ does
not equal ± arcsin (2A + 1) at s0 , so that there is not an inection here.
at
s0 ,
using the fact that
in parentheses is
dϕ
ds at
Contacting Balls
In fact,
dϕ
ds
> 0
at
s0 ,
since
ϕ (0, A) = 0
for the rst time for positive
f (s, A; γ) > 1
for
s
s
at
to the left of
5
s = s0 .
s0 .
ϕ (s, A) =
df Thus
ds and
π
2
s=s0
−γ > 0
< 0,
and
f (0, A; γ) = sin γ < 1,
f (s, A; γ) = 1 has a
positive root of g (s, A; γ),
Since
it follows by the intermediate value theorem that
(0, s0 ), call it ŝ. Since s0 is the rst
g (ŝ, A; γ) > 0, and ŝ is the root we seek.
π
• Case 1c: γ − arcsin (2A + 1)
= 2 . The proof in case 1b fails, since (with
df s0 dened as before), ds
= 0, since the Delaunay curve has an
s=s0
d2 f >0
inection at s0 . However, we will show that in this case,
2
ds root in
we have
s=s0
and then apply an intermediate value theorem argument. A general
formula for the second derivative of a triple product is easy to obtain:
00
(abc) = a00 bc + ab00 c + abc00 + 2a0 b0 c + 2a0 bc0 + 2ab0 c0 .
Thus
d2 f d2 x 1
d2 1
=
cos (ϕ + γ) + x 2
cos (ϕ + γ)
ds2 ds2 y
ds
y
dx d 1
x d2
(cos (ϕ + γ)) + 2
+
cos (ϕ + γ)
y ds2
ds ds y
d 1 d
dx 1 d
(cos (ϕ + γ)) + 2x
(cos (ϕ + γ))
+2
ds y ds
ds y ds
2
dϕ
d2 ϕ
− sin (ϕ + γ)
+ cos (ϕ + γ) 2 .
ds
ds
(2.1)
dϕ
ds = 0. Substituting these
into (2.1) simplies things considerably. The only term which doesn't
At
s0 , cos (ϕ + γ) = 0, sin (ϕ + γ) = 1,
and
vanish is the third term in the rst line, so that one eventually obtains
f 00 (s0 ) =
Thus
x (s0 , A)
y (s0 , A)
3
cos (ϕ (s0 , A)) sin (ϕ (s0 , A)) > 0.
f (s) is larger than 1 for s near s0 and to the left of s0 .Thus f = 1
ŝ in (0, s0 ), where g (s) > 0, as in case 1b.
has a root
Case 2:
A = − 12 .
For
cylinder of radius 1.
A = − 12 , the standardized
Delaunay surface is simply a
π
, this cylinder can clearly be cut and
For any γ ∈ 0,
2
scaled to form a bridge between the two contacting balls.
Case 3:
A ∈ − 12 , 0 . This is similar to case 1, except that arcsin (2A + 1) > 0,
so that we now have
γ − arcsin (2A + 1) ≤ γ + ϕ (s, A) ≤ γ + arcsin (2A + 1) .
γ + arcsin (2A + 1) < π2 , case 3b is γ + arcsin (2A + 1) > π2 , and
π
case 3c is γ +arcsin (2A + 1) =
2 , but in all of these, the proofs parallel those
Case 3a is
in case 1.
6
T. I. Vogel
Case 4:
A = 0:
π
2 as s → ∞. The
π
will cross
without occurring
2
The prole is a catenary, so that
proof is the same as case 1b, since
ϕ (s, 0) + γ
ϕ (s, 0) →
at an inection.
Case 5:
A > 0: The prole is a nodary, with ϕ (s, A) increasing monotonically
s → ∞. Again, the proof is the same as case 1b.
to innity as
From [5], it is known that convex bridges which are sections of
unduloids are stable. For A > −1 suciently close to −1, the bridges whose
existence is guaranteed by Theorem 2.2 are of this type, hence stable.
Note 2.3.
3. Non-existence for
γ∈
π
,π
2
.
Although the focus of this paper is on bridges between contacting balls, in
this section we rst prove a result which applies to separated balls as well.
A ≥ 0
γ ∈
π
2 , π , no physically possible liquid
bridge of the type we are considering (axisymmetric, symmetric across the
We show that for
and for
plane which is the perpendicular bisector of the line segment between the
centers of the balls) exists between the balls, regardless of the separation.
Lemma 3.1.
On a standardized Delaunay prole with A ≥ 0, there holds
cos ϕ (s) + 2Ay (ϕ (s) , A) > 0
for all s.
Proof.
A = 0, the prole is a catenary, with ϕ (s) ∈ − π2 , π2 , and
the result follows immediately. For A > 0, from lemma 2.2 of [5], (see also
2
Note 2.2 of [6]), c ≡ y cos ϕ + Ay is constant and positive on these nodoids.
2
Thus y cos ϕ + 2Ay is certainly positive on the nodoids with A > 0, and
since y is positive, the result follows.
In the case
For A > 0, at the point where ϕ = π, we must have the x
coordinate negative.
Lemma 3.2.
Proof.
Since we are in the case
may use the inclination angle
ϕ
dx
dϕ
A>0
=
=
so that, considering
x (π) − x (0)
x
dx/ds
dϕ/ds
y cos ϕ
,
cos ϕ + 2Ay
as a function of
ˆ
=
0
ˆ
=
0
π
and the prole must be a nodoid, we
as a parameter. We have
ϕ
for the moment,
y cos ϕ
dϕ
cos ϕ + 2Ay
π/2
y cos ϕ
dϕ +
cos ϕ + 2Ay
ˆ
π
π/2
y cos ϕ
dϕ.
cos ϕ + 2Ay
Contacting Balls
In the second integral, substitute
ˆ
7
ς = π − ϕ:
ˆ π/2
y (ϕ) cos ϕ
y (π − ς) cos (π − ς)
dϕ +
dς
cos ϕ + 2Ay (ϕ)
cos (π − ς) + 2Ay (π − ς)
0
0
ˆ π/2
y (π − ϕ)
y (ϕ)
−
dϕ
=
cos ϕ
cos ϕ + 2Ay (ϕ) cos (π − ϕ) + 2Ay (π − ϕ)
0
ˆ π/2
−y (ϕ) − y (π − ϕ)
=
cos2 ϕ
dϕ.
(cos ϕ + 2Ay (ϕ)) (cos (π − ϕ) + 2Ay (π − ϕ))
0
π/2
x (π) −x (0) =
I claim that this is negative. Certainly
y (ϕ)
is positive for
ϕ ∈ (0, π),
so the
numerator of the integrand is negative. The two terms in the denominator
are positive by Lemma 3.1. Thus
x (π) = x (π) − x (0) < 0,
as desired.
For γ ∈ π2 , π , and A > 0, there are no physically possible
rotationally symmetric bridges which are also symmetric across the plane
which is the perpendicular bisector of the line segment between the centers of
the two balls, regardless of the separation of the balls.
Theorem 3.3.
Proof.
dx
dϕ > 0 at ϕ = 0, so x (ϕ) is positive for small positive ϕ.
Therefore, using Lemma 3.2, there is an angle ϕ̃ in (0, π) for which x (ϕ̃) = 0.
We have
ϕ (s1 ; A) > ϕ̃ we reach a contradiction, since the Delaunay curve will cross
y axis at inclination angles ϕ̃ and (with negative s) −ϕ̃, leading
to a non-physical self-intersection of the bridge surface. Thus ϕ (s1 ; A) ≤
ϕ̃ < π . Since we require cos (ϕ (s1 ; A) + γ) > 0, and γ ∈ π2 , π , it follows
π
3
that ϕ (s1 ; A) + γ ∈
2 π, 2π . From this we see that ϕ (s1 ; A) > 2 , for if
π
3
ϕ (s1 ; A) ≤ 2 , ϕ (s1 ; A) + γ will be less that 2 π , a contradiction. Thus the
If
itself on the
cut of the Delaunay surface in the cutting and scaling construction occurs at
a point of the nodoid where the slope is negative.
From the cutting-and-scaling construction of a bridge between balls developed in [6], the ray from the contact point to the center of the rightmost
ball is in direction
ϕ (s1 ; A) + γ − π2 .
Thus the slope of the line from the cen-
ter of the rightmost ball to the contact point is positive, and this line must
cross the nodoid. This contradicts the construction that the bridge surface is
exterior to the balls.
We now return to the main focus of this paper, bridges between contacting balls.
We consider the same equation as before:
x (s; A)
cos (ϕ (s; A) + γ) + sin (ϕ (s; A) + γ) = 1,
y (s; A)
8
T. I. Vogel
cos (ϕ (s; A) + γ) > 0. We will see that for γ ∈
A ∈ (−1, 0) there is no solution. The equation is equivalent to
x
cos (ϕ + γ) = 1 − sin (ϕ + γ)
y
cos2 (ϕ + γ)
=
,
1 + sin (ϕ + γ)
with the requirement that
π
2,π
and
which may be rewritten as
x (s; A) − y (s; A)
cos (ϕ (s; A) + γ)
= 0,
1 + sin (ϕ (s; A) + γ)
cos (ϕ (s; A) + γ) 6= 0.
π Theorem 3.4. For γ ∈
2 , π and A > −1, no liquid bridge corresponding to
that A and with contact angle γ exists between contacting balls which is both
since
rotationally symmetric and symmetric across the plane which is the perpendicular bisector of the line segment between the centers of the balls.
Proof.
For
A ≥ 0,
this is contained in Theorem 3.3. Thus, we wish to show
that
has no solution
cos (ϕ (s; A) + γ)
x (s; A) − y (s; A)
=0
1 + sin (ϕ (s; A) + γ)
π for γ ∈
2 , π , A ∈ (−1, 0), s > 0, by showing
that the left
hand side is always positive. The rst simplication is to reduce to the case
γ=
∂
∂γ
π
2 . We have
cos (ϕ + γ)
sin (ϕ + γ) + 1
=
=
=
− sin (ϕ + γ) (sin (ϕ + γ) + 1) − cos2 (ϕ + γ)
(sin (ϕ + γ) + 1)
−1 − sin (ϕ + γ)
(sin (ϕ + γ) + 1)
−1
,
sin (ϕ + γ) + 1
and on the unduloids we're considering (for
− π2 , π2 , and therefore sin (ϕ + γ) > −1,
γ . Since y is positive, we have
x (s; A) +
2
2
A ∈ (−1, 0)),
so that
we have ϕ ∈
cos(ϕ+γ)
sin(ϕ+γ)+1 decreases in
cos ϕ (s; A) + π2
y (s; A) sin ϕ (s; A)
= x (s; A) − y (s; A)
1 + cos ϕ (s; A)
1 + sin ϕ (s; A) + π2
≤ x (s; A) − y (s; A)
cos (ϕ (s; A) + γ)
.
1 + sin (ϕ (s; A) + γ)
Therefore, it suces to prove that
G (s; A) ≡ x (s; A) +
is positive for
y (s; A) sin ϕ (s; A)
1 + cos ϕ (s; A)
s > 0, A ∈ (−1, 0). G (0; A) = 0.
Contacting Balls
Looking at the derivative (suppressing
dG
ds
=
cos ϕ + sin ϕ
=
cos ϕ +
=
cos ϕ +
=
=
sin ϕ
cos ϕ + 1
sin2 ϕ
+y
cos ϕ + 1
+y
s
and
9
A
dependence),
cos ϕ (cos ϕ + 1) + sin2 ϕ
2
(cos ϕ + 1)
!
cos ϕ
1 + cos ϕ
+
2A
2
y
(1 + cos ϕ)
y cos ϕ + Ay 2 is constant on Delaunay curves, and
1 + A. Thus cos ϕ + Ay = 1+A
y . Therefore,
1
1+A
dG
=
+2
,
ds
1 + cos ϕ
y (1 + cos ϕ)
which is positive for the
and
dϕ
ds
cos ϕ
2Ay
sin2 ϕ
+
+
1 + cos ϕ 1 + cos ϕ 1 + cos ϕ
cos ϕ + cos2 ϕ + sin2 ϕ + cos ϕ + 2Ay
1 + cos ϕ
1 + 2 cos ϕ + 2Ay
.
1 + cos ϕ
Now,
s>0
!
A ∈ (−1, 0),
A's
at
we are considering. Therefore
s=0
it equals
G (s; A) > 0
for
concluding the proof.
For A < −1, there may be rotationally symmetric bridges between
contacting balls, but these are known to be unstable ([5]). Thus there do not
exist stable rotationally symmetric bridges between contacting balls with
• contact angles greater than or equal to π2 ,
• rotation symmetry,
• symmetry across the perpendicular bisector of the line segment between
the centers of the balls.
It is an open question whether the last condition may be dropped.
Note 3.5.
If the solid balls do not contact, the behavior in the case γ ≥ π2
is strikingly dierent from that predicted in Theorem 3.4. No matter how
small the separation between the balls, it is easy to put a spherical bridge
(A = −1), with center on the line segment between the centers of the solid
balls, and making contact angle γ with the solid balls. For A slightly larger
than −1, there will exist a rotationally symmetric bridge which is a convex
segment of an unduloid, hence stable ([5]).
Note 3.6.
It's natural to wonder what happens to the bridges in Note 3.6 as the
π
2 , it's easy to
verify, using elementary trigonometry, that the spherical bridges described in
separation between the solid balls tends to zero. For
γ ≥
that note have radii tending to zero as the separation of the balls decreases,
suggesting that all of the rotationally stable bridges of Note 3.6 tend to a
point.
10
T. I. Vogel
References
[1] R. Finn, Equilibrium capillary surfaces, Springer-Verlag, New York, NY, 1986.
[2] F. M. Orr, L. E. Scriven, and A. P. Rivas, Pendular rings between solids:
Meniscus properties and capillary force, J. Fluid Mech. 67 (1975), 723742.
[3] B. Y. Rubinstein and L. G. Fel, Theory of pendular rings revisited,
arXiv:1207.7096 [physics.u-dyn].
[4] T. I. Vogel, Comments on radially symmetric liquid bridges with inected proles, Dynam. Contin. Discrete Impuls. Systems Supplement (2005), 862867.
[5]
, Convex, rotationally symmetric liquid bridges between spheres, Pac. J.
Math. (2006), no. 2, 367377.
[6]
, Liquid bridges between balls: the small volume instability, J. Math.
Fluid Dynamics 15 (2013), no. 12, 397413.
Thomas I. Vogel
Department of Mathematics
Texas A&M University
College Station, TX 77843
e-mail: tvogel@math.tamu.edu