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Shape and stability of two-dimensional lipid domains with dipole-dipole Shape and stability of two-dimensional lipid domains with dipole-dipole

interactionsinteractions

Mitsumasa Iwamoto, Fei Liu, and Zhong-can Ou-Yang

Citation: J. Chem. Phys. 125125, 224701 (2006); doi: 10.1063/1.2402160

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Shape and stability of two-dimensional lipid domains

with dipole-dipole interactions

Mitsumasa Iwamotoa?

Department of Physical Electronics, Tokyo Institute of Technology, 2-12-1 O-okayama, Meguro-ku,

Tokyo 152-8552, Japan

Fei Liu

Center for Advanced Study, Tsinghua University, Beijing 100084, China

Zhong-can Ou-Yang

Institute of Theoretical Physics, The Chinese Academy of Sciences, P.O. Box 2735 Beijing 100080, China

?Received 14 June 2006; accepted 30 October 2006; published online 8 December 2006?

We study the general energy and shape of the two-dimensional solid monolayer domains with the

dipole-dipole interactions. Compared with the domain energy without tilted dipole moments ?M.

Iwamoto and Z. C. Ou-Yang, Phys. Rev. Lett. 93, 206101 ?2004??, the general dipolar energy is not

only shape and size but also boundary orientation dependent. The general shape equation derived by

this energy using variational approach predicts a circular solution and an equilibrium shape grown

from this circle. In particular, the latter is composed of two branches: a translation-induced growth

of all odd harmonic modes and a pressure-induced cooperative deformation by all even harmonic

modes. The good qualitative agreement between our prediction and the experimental observations

shows the validity of the present theory. © 2006 American Institute of Physics.

?DOI: 10.1063/1.2402160?

I. INTRODUCTION

As an extension of the order parameters in liquid crystals

?LCs?,1–4the orientational state of the floating monolayer

composed of rod polar molecules can be characterized by

three nonzero orientational order parameters Sn?n=1,2,3?.

These parameters are defined by the thermodynamic average

of the Legendre polynomials, Pn?cos ?? ?n=1, 2, and 3?, of

the orientational angle ? from the normal direction to the

surface.Among them, S1and S3do not appear in LCs and are

the specific parameters of the monolayer.5–8The two param-

eters connect to symmetry breaking and naturally relate to

the spontaneous and nonlinear polarization which have been

probed by Maxwell-displacement current ?MDC? and second

harmonic generation ?SHG?.8The presence of nonzero S1

and S3also implies that the dipole-dipole interaction as well

as the dipolar energy are inevitable when we discuss the

pattern formation of the monolayers and the domain shapes

of the monolayers. Many sophisticated microscopic tech-

niques, such as florescence microscopy9?FM? and Brewster

angle microscopy ?BAM?, were developed to detect various

monolayer domains under different physical or chemical cir-

cumstances. The simplest system should be pure single-

component phospholipid at the air-water interface.10–12

On the theoretical side, during the past two decades,

great theoretical efforts have been devoted to quantitatively

understanding the two-dimensional ?2D? domain shapes and

shape transitions in this simplest system.13–16They were al-

most based on the idea firstly given by Andelman et al.17the

shape and shape transition of the domains are determined by

a competition between the line tension at the domain bound-

ary between fluid and the domain region and the long range

dipole-dipole electrostatic force between the lipid molecules,

namely, the minimum of the energy13,14

F = ??dl + F?+ F?,

?1?

where ? is the unit line tension, F?is the energy of the

dipole-dipole interaction between the dipoles oriented nor-

mally to the water surface,

2??

F?= −??

2

t?l? · t?s?

?r?l? − r?s??dlds,

?2?

F?is the energy of the dipole-dipole interaction between the

dipoles oriented in parallel to the water surface,

2??

?r?l? − r?s??

F?=??

2

?t?l? · y ˆ0??t?s? · y ˆ0?

dlds,

?3?

r?s? describes the domain boundary curve, s is the arc length,

t=dr/ds is the unit tangential vector of the boundary, and y0

is the unit vector of the y axis; see Fig. 1. Here all dipoles are

assumed to uniformly tilt along the x direction with the same

tilted angle ?, i.e.,

? = ???,0,??? = ?0?sin ?,0,cos ??.

?4?

The different phases and their coexistence in the 2D do-

main were expected to have an analogy with the three-

dimensional ?3D? case.18Such prediction has been directly

a?Electronic mail: iwamoto@pe.titech.ac.jp

THE JOURNAL OF CHEMICAL PHYSICS 125, 224701 ?2006?

0021-9606/2006/125?22?/224701/9/$23.00© 2006 American Institute of Physics

125, 224701-1

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visualized by the FM.19,20Helfrich suggested that the shape

energy of a 3D vesicle is21

F = ?P?dV + ??dA +kc

2??2H + c0?2dA,

?5?

where ?P is the osmotic pressure, ? is the tensile stress

acting on the membrane, dA and dV are the surface and

volume elements, respectively, kcis the bending rigidity, H is

the mean curvature, and c0is the spontaneous curvature. The

last term represents the curvature-elastic energy Fc. Hence

we asked whether the 2D domain energy F can also be ex-

pressed geometrically like Eq. ?5?. If so, the previous knowl-

edge about Eq. ?5? ?Refs. 22–24? can then be mapped onto

the 2D domain case. Recently the present authors indeed

found that25when the dipole moment is perpendicular to the

water plane, the double-line integral in the energy ?Eq. ?2?? is

approximated by a sum of an additionally negative line ten-

sion and a curvature-elastic energy of the domain boundary.

Many important physical properties and experimental phe-

nomena about the 2D monolayer domain were understood

analytically and quantitatively.

In the present paper, we follow the same approximation

approach to consider the general domain energy having tilted

dipole moments. We think that such an extension is not only

for a more general formalism to treat the domain shape prob-

lem of over 20 years but also very essential to consider in

recent experiments. Radhakrishnan et al. found that an in-

plane applied electric field can destabilize the condensed

complex and induce micron-scale domain formation around

the electrode.26This observation implies that the in-plane

dipole moments of lipids might have an important effect in

the formation of the 2D monolayer domains. On the other

hand, McConnell and Moy also studied the thin shape tran-

sition of a rectangular domain induced by increasing size.

They found that a small tilted angle ? can completely oblit-

erate the sharp transition observed in the perpendicular di-

pole case.14In particular, they observed a circle to clover-

leaf-shaped

?CLS?

domain

dipalmitoylphosphatidylcholine27?DPPC? induced by com-

pression. A remarkable feature of the CLS domains is that it

is not of exact C3symmetry but a distorted triangle with C2

symmetry. Although these intriguing phenomena in the latter

experiment were attributed to the presence of the tilted di-

growthina racemic

pole moments, to the best of our knowledge, they lack quan-

titative explanation. We will show that the analytical formal-

ism developed for the tilted dipole moments in the present

work can deal with the problem.

The paper is organized as follows. In Sec. II some defi-

nitions and basic formulas are given. In Sec. III, we apply

Taylor’s expansion to approximate the general domain en-

ergy as a sum of general line tensions and curvature-elastic

energy. Then the general shape equation of the energy is

yielded by variation calculation in Sec. IV. The circular so-

lutions and their instabilities are treated in Secs. V and VI,

respectively. Section VII concludes the paper.

II. DEFINITIONS AND BASIC FORMULAS

Compared with the model given by McConnell,15we

additionally introduce a Lagrange multiplier ?P in analogy

with the Helfrich curvature-elastic energy. This multiplier is

viewed as the constraint of constant domain area and/or has

real physical meaning, i.e., ?P=?−g0, where ? is the 2D

surface pressure of the monolayer and g0is the difference in

the Gibbs free energy density between the outer ?e.g., fluid?

and inner ?solid? phases. Because the solid phase is more

stable than the fluid one, g0is always positive. Then the 2D

domain energy is

F = ?P?dA + ??dl + F?+ F?.

?6?

We describe the boundary of a domain as a closed curve

r?s? in a 2D plane. The unit tangential vector t?s? and the

outward unit normal vector m?s? of the curve at s are related

to r?s? through the Frenet formulas in the plane:28

dr?s?

ds

= t?s?,

t · t = 1,

dt?s?

ds

= ??s?m?s?,

m · m = 1,

?7?

dm?s?

ds

= − ??s?t?s?,

m · t = 0,

where ??s? is the curvature of the curve at s. Another conve-

nient representation of the unit vectors is as follows:

t?s? = ?cos ??s?,sin ??s??,

?8?

m?s? = ?sin ??s?,− cos ??s??,

where ??s? is the boundary orientational angle at s and is

related to the curvature by

??s? = −d??s?

ds

.

?9?

Given that r?s? is the boundary curve of a domain in

equilibrium, consider a slightly distorted curve,

r??s? = r?s? + ??s?m?s?,

?10?

where ??s? is a sufficiently small smooth variation function.

With the help of the formulas given in the Appendix, we

FIG. 1. Schematic illustration of the shape of a monolayer domain, whose

dipoles ? uniformly tilt along the x direction with the same tilted angle ?.

The geometric quantities describing its boundary curve are also shown, and

h is the thickness of the monolayer.

224701-2Iwamoto, Liu, and Ou-YangJ. Chem. Phys. 125, 224701 ?2006?

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calculate the variations of the boundary length L and area A

of the domain:

?L =???ds? =??− ?? +1

2?s

2+ o??3??ds

?11?

and

?A =1

2???r · mds? =??? −1

2??2+ o??3??ds.

?12?

In the derivations the following two relations are useful:

?f?s??sds = −?fs?ds

?13?

and

?f?s??ssds =?fss?ds.

?14?

Here f?s? is any smooth continuous function.

III. ENERGY FORMULA OF THE 2D DOMAIN

The key step to approximate the double-line integrals is

to rewrite the energies of the two components of the dipole

moments as

2???

2???

?r?s + x? − r?s??

F?+ F?= −??

2

t?s? · t?s + x?

?r?s + x? − r?s??dx?ds

+??

2

?t?s + x? · y ˆ0??t?s? · y ˆ0?

dx?ds,

?15?

where the range of the arc variable x?l−s is ?h,L?; h rep-

resents a nonzero monolayer thickness as a cutoff; see Fig. 1.

Substituting Eqs. ?A3?–?A5? into the above equation, the in-

ner integrals ?the brackets? are then the sum of ci?s?xi, i=

−1,0,1,..., where ci?s? are the functions of the curvature

??s?, its derivatives with different orders, and the angle ??s?.

If we consider a somewhat smooth curve and only take into

account the terms of x till the first order, after integrating we

obtain an approximation given by

2?ds?

h

h?ds +11

F?= −??

2

L

dx?1

x−11

24??s?2x + o?x2??

2L2???s?2ds

? −??

2

2

lnL

96??

?16?

and

F?=??

2

2?ds?

−?1

h

L

dx?1

xsin2??s? −1

24??s?2sin2??s??x + o?x2??

h? sin2??s?ds −

???11+ 13 cos 2??s????s?2ds.

2??s?sin 2??s?

4?s?s?sin 2??s? +11

???

2

2

lnL

1

192??

2L2

?17?

In the derivations Eq. ?9? has been invoked to remove the

derivatives of ??s?.

The validity of the above approximation is as follows: If

we consider the limit h→0 the divergences of Eqs. ?16? and

?17? depend only on the integral of the first integrand term of

1/x; it is a good approximation to neglect the terms of o?x2?.

As we pointed out in our Letter,25for the exact calculation

for a circle of radius ?0, the logarithmic divergence is

ln?8?0/e1/2h??ln?5?0/h? ?see Eq. ?2.11? in Ref. 16? and this

nearly equals ln?2??0/h? calculated by our Taylor’s expan-

sion. In fact, such an ansatz approach by expanding the line

integral was also proposed by Langer et al. ?see Eqs. ?4.11?

and ?4.12? in Ref. 16? early.

We see that F?is approximated by a sum of an addition-

ally negative line tension and a curvature-elastic energy of

the domain boundary. This indeed has a Helfrich-type form

in the 2D case. Generally, the total energy F of the domain

depends on the size ?L?, the orientation ???, and the shape

??? ?SOS?.

IV. SHAPE EQUATION

To obtain the equilibrium-shape equation, we derive the

first variation of the general energy ?Eq. ?6??, which includes

four parts:

??1?F = ?P??1??dA + ???1??ds + ??1?F?+ ??1?F?. ?18?

Equations ?11? and ?12? have given the first variations of the

boundary length L and the area of the domain A,

??1?L =???1?ds =?− ???s?ds

?19?

and

??1?A =???1?dA =???s?ds.

?20?

According to Eq. ?A12?, the first variation of F?is

2?11

2L2???1????s?2ds?

2??11

2L2???3??s?ds + 2??ss??s?ds?. ?21?

??1?F?=??

2

24L???s?2ds − lnLe

h????1?ds

+11

96??

= −??

2

24L???x?2dx − lnLe

h????s?ds

+11

96??

For F?, its first variation is

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??1?F?=??

2

2?1

L? sin2??s?ds −11

h???1??sin2??s?ds? +13

2??1

2L2?

192

48L???s?2ds −13

48L? cos 2???s?2ds????1?ds −12

2L2???1??sin2???s?2ds?

48L???x?2dx −13

2L2? sin 2???s??s?ds − ??

96??

2L2???1????s?2ds?

+??

2

2

lnL

96??

= −??

2

L? sin2??x?dx −11

48L? cos 2???x?2dx +1 + 3 cos 2??s?

2

lnL

h????s?ds

+ ??

39 cos 2??s? − 11

?3??s?ds −13

24??

2L2?

39 cos ??s? + 11

96

?ss??s?ds.

If the curve r?s? describes a boundary of an equilibrium do-

main, it must satisfy ??1?F=0 for any infinitesimal function

??s?. Combining Eqs. ?19?–?22?, we get the shape equation

of an equilibrium domain,

?P − ?? + ??3+ ??ss+ ???s= 0,

?22?

where the definitions of the coefficients are

? = ? −??

2

2

lnLe

h

+??

2?1 + 3 cos 2??

4

2L? sin2??s?ds

lnL

h

+11

48??

2L???s?2ds +??

2L??11+ 13 cos 2????s?2ds,

2

−

1

96??

? =11

96??

2L2+ ??

2L239 cos 2? − 11

192

,

?23?

? =11

48??

2L2− ??

2L211+ 13 cos 2?

96

,

? = −13

24??

2L2sin 2?,

respectively. The readers are reminded that the coefficients

are also the functions of the domain shapes. Obviously, com-

pared to the specific shape ?Eq. ?9? in Ref. 25? ???=0?, the

current equation is more complex. In addition to the more

complicated coefficients of ??s? and its derivatives, ??sis

involved.

V. THE CIRCULAR SOLUTIONS

A. The perpendicular dipole case

We rewrite the general shape equation in the absence of

??as follows:

?P − ?0? + ?0?3+ 2?0?ss= 0,

?24?

where

?0=11

96??

2L2,

?25?

?0= ? −??

2

2

lneL

h

+11

48L??

2???s?2ds.

Because the above equation is derived by defining the nor-

mal direction to be outward, ?=−1/?cfor a circle with ra-

dius ?c. According to this definition, ?=1/?tis then the cur-

vature of the inner circle of a torus with radius ?t. Hence it is

understandable that outer and inner circles can coexist if

?t??c. In the case of the planar circles ?i, i=c,t, Eq. ?24? is

simplified as

±k2?e?i

h

+ r = ln2?e?i

h

,

?26?

where k=?Ph/?e??

positive and negative signs before k correspond to the circu-

lar and torus cases, respectively. According to the values and

signs of the two constants k and r, which are also the linear

functions of the pressure ?P and the line tension ?, the so-

lutions of Eq. ?26? are divided into four cases; see Fig. 2.

2

and r=2?/??

2+11?2/12, and the

?1?

Only one circle exists for 0?r?k or 0?k?−r

?regions I and II in Fig. 2?.

Two circles of different sizes can coexist for 0?−r

?k and r?−?1+ln k? or 0?k?r and r?−?1+ln k?

?regions III, IV, and VI in the figure?.

Two circles and two tori of different sizes can coexist

for 0?r?k and r?−?1+ln k? ?region V in the figure?.

?2?

?3?

FIG. 2. Distribution of the circular and torus solutions of the equilibrium-

shape equation Eq. ?26?. The lines and the curve are k=r, k=−r, and

r=−?1+ln k?, respectively.

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Page 6

?4?

No compact circular domains exist in the other regions.

Although the distribution of these solutions seems complex,

it is only a more detailed transformation of Fig. 1 obtained in

our Letter;25their physical explanations are the same. Here

we will not repeat the discussion again.

B. General cases

Obviously, Eq. ?22? is invariant with respect to the ex-

change of ?→−? and s→−s. This means that its solution

must be symmetric with respect to the tilted direction ?the x

axis? of the dipole moments. This is the origin of the domain

shapes with C2symmetry observed not only in DPPC mono-

layer ?the CLS and Y-shaped domains in Figs. 1 and 3 of

Ref.27,respectively?

but

ristoylphosphatidic acid ?DMPA? monolayer ?Fig. 3 in Ref.

29?.

Before discussing the circular solutions of Eq. ?22?, we

rewrite it into the following form:

2cos?2???3

alsoina

L-?-dimy-

?P − ?1? + ?1?3− ??

4ln?L

h?? −13

64L2?3?

+ ??ss+ ???s= 0,

?27?

where the coefficients

?1= ? −??

2

2

lnLe

h

+??

2

4

lnL

h+11

48??

2L???s?2ds

+??

2

2L? sin2??s?ds −

???11+ 13 cos 2????s?2ds

1

96??

2L

?28?

and

?1=11

96??

2L2−

11

192??

2L2

?29?

are independent of the local variable ??s?; both of them only

depend on the global shape of the domain. The relations

between these two coefficients and ? and ? given in Eq. ?23?

are

? = ?1+3

4??

2lnL

hcos 2?

?30?

and

? = ?1+13

64L2??

2cos 2?.

?31?

Therefore, a circle is a solution of Eq. ?27? if its radius ?0

satisfies

?P + ?1?0

−1− ?1?0

−3= 0

?32?

and

3

4?0

−1ln?L

h?−13

16?2?0

−1= 0,

?33?

simultaneously. From Eq. ?33?, we have

?0=

h

2?exp?13?2

12?= 7.0? 103h.

?34?

In the same figure of the CLS domain ?Fig. 1 in Ref. 27? a

circular domain of ?0?8 ?m appears. According to Tan-

ford’s estimate for a saturated hydrocarbon chain with n car-

bon atoms, h??0.154+0.1265n? nm,30we have h?2 nm

and ?0?13.3 ?m. Thus Eq. ?34? predicts an interesting fact

that lipid chain of nanometer length can form domains with

micron size.

VI. PRESSURE-INDUCED INSTABILITY OF THE

CIRCULAR DOMAINS

McConnell15theoretically predicted and experimentally

demonstrated that the equilibrium circular domains become

unstable by either increasing their size at a fixed surface

pressure or changing the surface pressure for a fixed domain

size: the circular domains sharply transform to m-side quasi-

polygons ?the mth distorted harmonic shapes? initially, and

then continuously develop into n-branched patterns and take

on symmetries that are lower than that of the early

quasipolygons.10,31Moreover, on compressing a monolayer

of racemic DPPC, one observes that the circular domains

grow in area and form CLS.27Compared to the initial sharp

transitions, the latter transitions are very smooth. McConnell

attributed this transition difference to the coupling between

the DPPC dipolar tilt and the domain shapes and sizes.27

These experimental observations present a challenge to our

theory. In our previous work, we have studied the size-

induced instability of the circular domains.25Here we focus

only on the pressure-induced instability of the circular do-

mains.

To study the circle to CLS domain growth of the mono-

layer of racemic DPPC by compression, i.e., ?P→?P+?p,

where ?P is the equilibrium pressure at the circular solution

?0of Eq. ?22?, we assume that the growing domain is still a

solution of the equilibrium equation ?22?. We describe the

growing domain by a slightly distorted circle,

r?s? = ?0m0?s? + ??s?m0?s?,

?35?

where

m0?s? = ?sin ?0?s?,− cos ?0?s??,

?0= s/?0.

?36?

We then have the following variation form of Eq. ?22? cor-

responding to a small change of the pressure ?p:

??

?0

−??

?0

3+ ?p − ??? + ???3+ ???ss+ ????s?? = 0.

?37?

The terms of variations of ? and ? are absent here because

?sand ?ssvanish for a circle.

Considering that ?1and ?1are independent of ??s?

?Eqs. ?28? and ?29??, we have the following linear approxi-

mation for ? of the above equation after substituting Eqs.

?A12?, ?32?, and ?33?:

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Page 7

?p +a1

?0

2? + a2??s

?0

+ ?0?sss?+ a3?ss+ a4?0

2?ssss= 0, ?38?

where ?sss=d3?/ds3, ?ssss=d4?/ds4, and the definitions of

the coefficients are

a1= ?P?0+11?2

24

?2??

2− ??

2? +13?2??

2

8

cos 2?0,

a2=13?2

6

??

2sin 2?0,

a3= ?P?0+11?2

12

?2??

2− ??

2? +13?2

12

??

2cos 2?0,

a4=11?2

24

?2??

2− ??

2? −13?2

24

??

2cos 2?0,

respectively. We insert the expansion ??s?,

??s? =?

m

Cmexp?im?0?s??,

?39?

into Eq. ?38?, where i·i=−1, 0???2?, m=0,±1,...,±?,

and Cm

exp?im?0??Cm??P

*=C−mto keep the function real. We have

a1

?0

1? =?

m

?0

+11?2

24?0

2 ?,

2?2??

2− ??

2??

+ Cm−2

13?2??

16?0

2

2

+ Cm+2

13?2??

16?0

2

a2

?0

?s=?

m

exp?im?0??Cm−2

?m − 2?

?0

12?,

2

13?2??

12

2

− Cm+2

?m + 2?

?0

2

13?2??

2

a2?sss= −?

m

exp?im?0??Cm−2

?m − 2?3

?0

2

13?2??

12

2

+ Cm+2

?m + 2?3

?0

exp?im?0??Cm

2− ????− Cm−2

2

13?2??

2

12?,

2??P?0+11?2

?40?

a3?ss= −?

m

m2

?0

12

??2??

?m − 2?

?0

2

13?2??

24

2

− Cm+2

?m + 2?2

?0

2

13?2??

2

24?,

a4?0

2?ssss=?

m

exp?im?0??Cm

m4

?0

4

11?0

2?2

24

?2??

2− ???

− Cm−2

?m − 2?4

?0

?m + 2?4

?0

4

13?2?0

2??

2

48

48?.

− Cm+2

4

13?2?0

2??

2

?40?

We combine them according to Cm−2, Cm, and Cm+2and let

coefficients of exp?im?0? vanish; an important recursive re-

lation is obtained:

?m2− 1????

+ ?a + b?1 − m2??Cm? − d?m,0= 0,

2??m + 1??m + 3?Cm+2+ ?m − 1??m − 3?Cm−2?

?41?

where

a =48?P?0

13?2,

b =22

13?2??

2− ??

2?,

d =48?0

13?2?p,

2

and ?m,0=1 when m=0 and 0 otherwise. The solutions of Eq.

?41? can be derived into two cases.

For ??=0 (without dipolar tilt). Equation ?41? predicates

an mth harmonic shape transition ?i.e., Cm?0? at the critical

surface pressure,

?P =11?2??

2?m2− 1?

12?0

.

?42?

Such a pressure-induced m-side quasipolygon transition has

been observed and schematized in Ref. 31 ?Figs. 8–12

therein?. For DPPC,32?0?1 ?m, ???15 D/50 Å2at a

rough estimate, Eq. ?42? gives ?P??m2−1?m N m−1; this is

on the reasonable order as observation.31The case of m=1

means only the trivial translation of the circle. Correspond-

ingly, ?P=0 reveals that the translation happens at ?=g0.

For ???0. The dipolar tilt removes the above transition

of eigenmodes instead of the two collective deformation

branches described as follows: Branch I happens at the be-

ginning of the compression, i.e., ?p=0. From Eq. ?41? with

?m?=2n ?n=0,...,?? and C0=0, we find that all of the even

harmonic growing modes, C?2n?, must vanish. The request of

C0=0 is a consideration of the constraint of the constant

domain area at the beginning of the growth, which leads

into23

m?Cm

C0

?0

= −1

2?

?0?

2

,

?43?

i.e., C0=0 in the present linear approximation. But for the

odd modes the situation is quite different from the case of

??=0, where only C±1?0 can occur at ?P=0 ?see above?. It

is easy to find that when ?m?=1, Eq. ?41? allows not only

C±1?0 but also C±3?0. If C±3?0 happens then all of the

odd harmonic modes of C2n+1are excited. Fitting the whole

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Page 8

CLS domain ?Fig. 1 in Ref. 27?, we have the so called trans-

lation ?because C1?0? induced growth solution

? = ?0?1 +1

5cos ? +

8

34cos 3??;

?44?

see Fig. 3?a?. Here ?=?0−?/2 is the azimuth of m0. In the

above equation the higher mth harmonic modes ?m

=5,7,...? are neglected by considering them decaying rap-

idly with rates of m−2at a magic tilted angle of the dipoles

?i.e., b=0; see below for details?. The tilted dipole force can

automatically excite the growth of C3?0, which explains the

domain formation with three-arm labyrinth popularly ob-

served in the 2D lipid monolayers.10,27Branch II occurs at

?p?0 upon the further compression. First, the identity of

Eq. ?41? at ?m?=1 still holds, therefore, C±1?0 can exist, but

from the experimental observation we assumed C±3=0. From

Eq. ?41? with ?m?=2n we have

C?2?= − d*/6,

C?4?= ?a*− 3b*?d*/90,

?45?

C?6?= ?45− ?a*− 3b*??a*− 15b*??d*/3150,

where a*=a/??

simulation with the observed main domain of the CLS trido-

main structure in racemic DPPA ?Figs. 1 and 3?b? in Ref. 27?

gives such a pressure-induced solution

2, b*=b/??

2, and d*=d/??

2. A straightforward

? = ?0?1 + 0.6 cos ? + 0.06 cos 2? − 0.06 cos 4?

+ 0.02 cos 6???46?

as shown in Fig. 3?c?. Figure 3?c? reveals the two normal of

the two straight sides ?AB,CD? of the calculated main do-

main subtending exactly 120°. Both straight sides can serve

as the platforms for the growth of the two side domains ?Fig.

3?b??, both of which can be obviously described as the solu-

tions similar to Eq. ?46?. Substituting the above equation into

Eq. ?41?, we found b*=1/36 and the tilted angle

? = arctan???

???? arctan?2 = 54.7° .

?47?

To understand the surprising magic angle we calculated

the whole dipolar energy from Eqs. ?16? and ?17? for a circle

of radius ? as follows:

Fe= F?+ F?= ?2??

2− ??

2??11?2

48

−1

4ln2??

h?2??.

?48?

It becomes clear that the magic tilt makes a circular domain

grown from ?=0 to ?0with Fe=0. In physics, the shape Eq.

?22? reflects the mechanics equilibrium between two phases,

while the zero shape formation energy describes the energy

equilibrium of the solid domain grown from or melted into

the fluid phase. In fact, the shape given in Fig. 3?c? is quite

popular, e.g., in DMPA monolayer Miller and Mohwald ob-

served many such domains ?Fig. 3 in Ref. 29?. A more direct

evidence on the magic tilt may refer to the earlier x-ray dif-

fraction studies in eicosanoic acid monolayers by Durbin

et al.33They found an ?60° tilt in the nearest-neighbor ?NN?

phase. Similarly, MDC-SHG experiment on cyanobiphenyl

liquid crystal monolayers by the present authors showed us

an ?57° tilt in the condensed phase.34Given these, the CLS

tridomain structure observed in Ref. 27 can be viewed to

form by two steps: the compression and the magic dipolar tilt

first induce a translation ?C1?0? and C3?0 growth ?a dis-

torted triangle given in Fig. 3?a?? and the increasing surface

pressure ?p then excites a deformation of all even eigen-

modes and causes the domain splitting into three equilibrium

distorted hexagons characterized by Eq. ?46?. With the do-

main fission, the dipoles in the two side domains also change

their orientation as predicted by the authors of Ref. 27.

VII. CONCLUSION

In the present paper, we study the shape and stability of

the solid monolayer domains with the dipole-dipole interac-

tions. The general dipolar energy and shape equations have

been derived. Our discussion shows that the present treat-

ment by reducing the dipolar interaction, the anisotropic

double line integrals ?Eqs. ?2? and ?3??, into a SOS line ten-

sion does offer an opportunity to explain the observation of a

circle to CLS domain formation in racemic lipid monolayers.

In the earlier studies, the researchers usually assumed the

existence of an equilibrium domain with a simple shape, e.g.,

the circle14,17and torus.13They then fixed the domain size by

a variation of the energy with respect to some specific geo-

metric parameters, e.g., the radius for the circular case. Al-

though sound results have been achieved, it is not trivial to

extend this approach to the complex domain cases, such as D

form, S forms, and serpentine ones.11,12The long-lasting in-

sufficient stage of the 2D domain theory is due to the math-

FIG. 3. Calculation of equilibrium shapes grown from a

circle to ?a? a C2-symmetric triangle ?Eq. ?44?? by a

tilted dipole force along the x axis at the beginning of a

compression and then to a tridomain structure illus-

trated in ?b? with the increasing of the compression

where the main domain, the distorted hexagon, is also

an equilibrium shape predicted by theory ?Eq. ?46??. In

practical computation, ?a? is obtained by simulating a

solid domain observed in a racemic monolayer of

DPPC ?Fig. 1 in Ref. 27? and ?c? by simulating its main

domain. The latter makes us find the magic dipolar tilt

and its physical meaning, a zero shape formation energy

?Eq. ?48??.

224701-7Two-dimensional lipids with dipole-dipole interaction J. Chem. Phys. 125, 224701 ?2006?

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Page 9

ematical difficulty in calculating the double-line integrals in

Eqs. ?2? and ?3?. Therefore our approximation approach

might provide a new and reasonable way to solve this diffi-

culty. In our opinion, the general shape equation and the

other related formulas may be more useful in further inves-

tigationontheformation

phospholipid,10which is a key way toward the raft formation

in biomembranes.

of 2Dchiralcrystal of

APPENDIX: SOME USEFUL MATHEMATIC

FORMULAS

Employing Taylor’s expansion to the variable x and the

relations in Eq. ?7?, we easily have

r?s + x? = r?s? +dr

dsx +1

2!

d2r

ds2x2+1

3!

d3r

3s3x3+ o?x4?

= r?s? + t?s?x +1

2??s?m?s?x2+1

6??s?s?m?s?

− ??s?2t?s??x3+ o?x4??A1?

and

t?s + x? = t?s? +dt

dsx +1

2!

d2t

ds2x2+1

3!

d3t

dx3x3+ o?x4?t?s?

+ ??s?m?s?x +1

2??s?s?m?s? − ??s?2t?s??x2

+1

6???ss?s? − ??s?3?m?s? − 3?s?s???s?t?s??x3

+ o?x4?,

?A2?

where ?s=d??s?/ds and ?ss=d2??s?/ds2. We denote the

terms of equal or higher than the nth order in x as o?xn?. The

distance and the tangential vector inner product between s

+x and s in the curve are calculated by

?r?s + x? − r?x?? =?dr

2!

= x?1 −

dsx +1

d2r

ds2x2+1

12??s?2x2?

3!

d3r

ds3x3+ o?x4??

1

1/2

+ o?x4??A3?

and

t?s + x? · t?s? = 1 −1

2??s?2x2−3

6?s?s???s?x3+ o?x4?, ?A4?

respectively. Similarly we have

?t?s + x? · y0??t?s? · y0?

= sin2??s? − ??s?cos ??s?sin ??s?x

−1

2??s?s?cos ??s?sin ??s? + ??s?2sin2??s??x2

−1

6???ss?s? − ??s?3?cos ??s?sin ??s?

+ 3??s??s?s?sin2??s??x3+ o?x4?.

?A5?

We calculate ds?, m?, t?, and ?? for the slightly distorted

curve ?Eq. ?10??. As an illustration, we calculate

?? =dt?

ds?m?.

?A6?

Because

ds?

ds=?dr?

ds? = ?t?s? + ?s?s?m?s? − ??s???s?t?s??

= ??1 − ???2+ ?s

2?1/2,

?A7?

where ?s=d?/ds,

t? =dr?

ds?=ds

ds?

dr?

ds=ds

ds???1 − ???t + ?sm?,

?A8?

m? =ds

ds??− ?st + ?1 − ???m?,

?A9?

and m? is a unit vector perpendicular to t?, we have

dsm? =?ds

?? =ds

ds?

dt?

ds??

3

??? − ?2? + ?ss??1 − ???

+ ?s??s? + 2??s??

= ? + ?2? + ?ss+ o??2?,

?A10?

where and in the following o??n? refers to terms of equal or

higher than the nth order in ?. Hence the first order variation

of ? is

?? = ?? − ? = ?2? + ?ss+ o??2?.

?A11?

We derive other useful variations of functions with the simi-

lar processes. They are listed below:

??ds? =?− ?? +1

2?s

2+ o??3??ds,

??r · mds? = ?− ?sr · t + ?1 − ?r · m?? − ??2+ o??3??ds,

???2ds? =??3? + 2??ss+ ?4?2+3

+ 2??s?s? + 4?2?ss? + o??3??ds,

2?2?s

2+ ?ss

2

???? = ?2? + ?ss+ o??2?,

?A12?

???s? = 3??s? + ?2?s+ ?sss+ o??2?,

???ss? = 3?s

2? + 4??ss? + 5??s?s+ ?2?ss+ ?ssss+ o??2?.

??sin2?ds? =?−1

2?1 + 3 cos 2???? + o??2??ds,

???2sin2?ds? = ???3?sin2? + 2 cos 2?? − 4??ssin 2?

+ 2?sssin2??? + o??2??ds.

?A12?

We keep the second variations in the first three equations

since they will be used in the stability analysis of the domain

shapes.

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224701-9Two-dimensional lipids with dipole-dipole interaction J. Chem. Phys. 125, 224701 ?2006?

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