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The curvature elasticity of fluid membranes : A catalogue of vesicle shapes
H.J. Deuling, W. Helfrich
To cite this version:
H.J. Deuling, W. Helfrich. The curvature elasticity of fluid membranes : A catalogue of vesicle shapes. Journal de Physique, 1976, 37 (11), pp.1335-1345. �10.1051/jphys:0197600370110133500�.
�jpa-00208531�
THE CURVATURE ELASTICITY OF FLUID MEMBRANES :
A CATALOGUE OF VESICLE SHAPES
H. J. DEULING and W. HELFRICH Institut fuer Theoretische
Physik
Freie Universität
Berlin,
D 1Berlin-33,
Amimallee3, Germany
(Reçu
le 21 mai1976, accepté
le 15juin 1976)
Résumé. 2014 Les formes des membranes fermées fluides formées par la lécithine dans l’eau ont été calculées en fonction du volume délimité, de la surface de la membrane, et de la courbure spontanée.
Diverses formes symétriques par rotation sont présentées avec des indentations, des cavités et des surfaces de contact.
Abstract. 2014 Shapes of closed fluid membranes such as those formed by lecithin in water were
calculated as a function of enclosed volume, membrane area and spontaneous curvature. As the area can be taken to be constant, the only elasticity
controlling
the shapes of these vesicles is that of curvature. A large variety ofrotationally symmetric
shapes are presented, allowing for indentations,cavities and contact of the membrane with itself.
Classification .Physics Abstracts
7.130
1. Introduction. -
Lipid bilayer
membranes have been thesubject
of manyinvestigations
over the lastdecade,
becausethey
seem to beclosely
related tobiological
membranes.They
arecomposed
of a doublelayer
oflipid
molecules. Thehydrophilic
heads ofthe molecules are
pointing
towards the aqueous medium and thehydrophobic
ends of thehydrocarbon
chains are
pointing
towards the interior of the film.Lipid bilayers
can beprepared
in such a way that the membranes close to form vesicles whose diameterscan range from a few hundred
A [1]
to several mm[2].
The
shape
of such vesicles is controlledby
the elasticproperties
of thelipid bilayer.
Provided that the membrane is fluid and the vesicle does not form asphere,
theonly important elasticity
is that of curva- ture[3].
In the
following,
thelipid bilayer
membrane isalways
assumed to be a two-dimensional fluid charac- terizedby
avanishing
modulus of shearelasticity.
Area dilation of the membrane is
negligible,
except forspherical
vesicles under excess internal pressure, asmuch more energy is
required
for area dilation than for curvature.Practically
all the deformational energy of a vesicle is therefore stored in the curvature(or bending)
of the membrane. On the basis of Hooke’s law one shouldexpect
the curvature energy per unitarea of the
lipid bilayer
to beproportional
to thecurvature
squared.
Such an ansatz wouldimply
thatthe
plane bilayer
is theequilibrium
state in the absenceof external forces. A
lipid bilayer
may,however,
possess a built-in asymmetry due to either a different
lipid composition
of the two constituentmonolayers
or different environments on the two sides of the
bilayer.
Then the membrane has non-zero curvature in its state of lowest elastic energy.Allowing
for such aspontaneous curvature, we may write for the
bending
energy per unit area
[3]
where
kc
andkc
are elastic constants and cl and C2 arethe two
principal
curvatures. Thephenomenological
parameter co is the
spontaneous
curvature. If the constituentmonolayers
are free to slide over eachother,
as isgenerally assumed,
co is constant overthe entire area of a vesicle. It may be noted that for actual
shape changes
co may be atime-dependent
functional of the deformation
[4].
The concept ofcurvature elasticity, including spontaneous
curva- ture, hasrecently
beenapplied
to red blood cells[5]
toexplain
the biconcave discoidshape
oferythrocytes.
Analyzing
theexperimental
data of Evans andFung [6],
the present authors[7]
foundco = - 0.74 um-1
for the membrane of normal human
erythrocytes
under
physiological
conditions.(The
minussign
indicates a
spontaneous
curvatureopposite
to thatof the
sphere.)
In the present paper we use the concept of curvatureelasticity
to calculate theoreticalshapes of lipid bilayer
vesicles andclassify
the mostimportant
solutions of rotational symmetry.
We
also calculate thedependence
of vesicle volume on the difference in osmotic pressures inside and outside the vesicle aswell as the relation between the total curvature energy and volume for the different types of
shapes.
ToArticle published online by EDP Sciences and available at http://dx.doi.org/10.1051/jphys:0197600370110133500
designate
vesicleshapes, particularly
the more exoticones, we will often use the nomenclature
developed
for red blood cells
[8].
In all casesexamples
of theshapes
are shown in thefigures.
2.
Theory.
-Expression (1.1)
for thebending
energy per unit area of a
lipid bilayer
membrane isclosely analagous
to that for the energydensity
of aliquid crystal [9],
the first term in(1.1) corresponding
to the
splay
term ofliquid crystals
and the second termbeing analoguous
to the saddlesplay
term[10].
This second term, when
integrated
over the surface of avesicle, gives
a contribution to the total curvature energy which isindependent
of theshape according
to Gauss’ theorem.
Consequently,
theshape
of thevesicle is determined
only by
the first term in expres- sion(1.1).
For a vesicle with surface area S we definean
equivalent sphere
radius.Ro by S
= 47rR’.
Themaximal volume of the vesicle at constant surface
area S is
Vo
=(4 n/3) Ro.
When the volume V is reduced belowVo,
the vesicle becomeseasily
defor-mable and can assume a
large variety
ofshapes depend- ing
onS, V,
and co. To find theequilibrium shape
of avesicle,
we have to look for minima of the total curva-ture elastic
energy E = g c dS at constant surface
area and constant volume. We introduce
Lagrange multipliers dp
and A for the constraints V = const.and S = const.,
respectively. Ap
= Pe - pi repre- sents the osmotic pressure difference between the outer and inner medium and Arepresents
a tensilestress. The
shape
of the vesicle atequilibrium
isfound from the
equation
To
keep
theproblem simple,
we consideronly
formshaving
rotational symmetry. Theprincipal
curvaturesin this case are those
along
the meridians(ci
andalong
theparallels
of latitude(cp). , We
describe the contour of the cellby
a functionz(x),
the z-axiscoinciding with,
and xbeing
the distancefrom,
theaxis of rotation.
Denoting by §
theangle
madeby
the surface normal and the
z-axis,
we can expressthe curvatures c. and cp
by ql(x)
The
angle ql(x)
in turn is related to the contourz(x) by ,
Eliminating O(x),
we obtain a differentialequation expressing
theassumption
of rotational symmetryWe express the contour
z(x)
and volume and surfacearea in terms of x and cp
When
using expressions (2.6)
and(2.7)
we have totake into account that
z(x)
is double valued. We have tointegrate
over x from 0 to the maximal value xm and back to zero to obtain the total volume and sur-face area,
respectively.
With the aid of expres- sions(2.4), (2.6)
and(2.7)
we can rewrite the varia- tionalproblem
in thefollowing
formThe
Euler-Lagrange equation corresponding
to thisvariational
problem
readsThe two non-linear differential
equations (2.4)
and(2.9)
can be solvednumerically [5].
From the solu-tion
cp(x),
the contourz(x)
is obtainableby
a furtherintegration using expression (2.5).
Theboundary
conditions for the
equations (2.4)
and(2.9)
can bediscussed more
easily
when aslightly
different for- mulation is used. Wereplace
theindependent
variablex
by
the normalized surface area susing
the relationThe variable s ranges from s = 0 at the upper
pole
to a value
sm at
thedividing parallel
of latitude and from there to s = 1 at the lowerpole. dx/ds
ispositive
in the upper half and
negative
in the lower half. We also usef = x2
instead of x. With these substitutionswe obtain a system of three non-linear
equations
forthe three
dependent
variablescm(s), cp(s)
andf(s)
These
equations
have different classes of solutions which are characterizedby
differentboundary
condi-tions which we shall discuss in detail below.
From the solution
f (s), cp(s), cm(s)
we find thevolume V and the contour
z(s) by
furtherintegrations
For weak deformations of a
sphere,
i.e. V NVo,
there exist two solutions of
equations (2 .11 )-(2 .13) representing
anelongated ellipsoid
of revolution and an oblateellipsoid
ofrevolution, respectively.
The value of co determines which of the two
shapes
has lower elastic energy. When co is below a critical value Coc, the oblate
shape
has lower energy than theprolate
form. If however co isgreater
than Coc theelongated ellipsoid
of revolution represents the stable form. The critical spontaneous curvature Coc was found to be COc = -(39/23) Ro
1. The derivation isgiven
inappendix
B. As the volume is increased to the maximal valueVo,
the osmotic pressure differenceAp
between the inside and the outside of theellipsoids
attains the value
[3] Apc = (12 - 2coRo)kcRõ3.
In the
following
we willalways give Ap
in units ofApc
3. Numerical results and discussion. - The diffe- rential
equations
derived in section 2 have asingularity wherpver
eitherf (s)
goes to zero orf (s) c2p (s)
approa-ches 1. We can remove these
singularities by choosing
correct
boundary
conditions. At thepoles
of thevesicle,
i.e. at s = 0 and s =1,
we havef
= 0. As weapproach
thepoles, however,
the difference cm - cp also goes to zero. We can therefore take the correct limit ofequation (2.12)
and obtainWhenever
f (s)
reaches an extremum at somepoint
s = sm the
quantity f (s) c’(s)
becomesequal
to 1 andequation (2.11)
has asingularity.
We can removethis
singularity by
theboundary
condition that ats = sm the
expression
incurly
brackets in equa- tion(2.11)
also vanishes. This defines either theLagrangian multiplier
A or the curvaturecm(s)
ats=sm :
Taking
the correct limit s - s. ofequation (2.11)
we find the
equation
to be consistent with any value of y =(dcm/ds)S=Sm.
The second derivatives are all well defined at s = sm. Therefore we make aTaylor expansion
around thepoint s
= sm at which equa-tion
(2.11)
is ill-defined :Besides rotational symmetry,
ellipsoids
of revolu-tion and
disk-shaped
cells(discocytes)
also possess reflectionsymmetry
with respect to theequatorial plane. Dealing
with these forms we haveonly
toconsider the interval 0 s 0.5. At s = 0 we have
f
= 0 and cm = cp =cm.
At s = 0.5 we havef = f m,
cp
= I;. 1/2
and cm =cm.
Theparameter
y is zero because the solution issymmetric
withrespect
to thepoint s
= 0.5. We start the solution at s = 0 and ats = 0.5 and match the branches at s = 0.25 or at some other intermediate
point.
Thematching
condi-tion for the three functions
cm(s), cp(s)
andf (s) gives
three non-linear
equations
for the unknownboundary
conditions
c0m, 1m
andcm.
These three non-linearequations
which we solveby
theNewton-Raphson
method
[11]
may have several solutions as we shallsee below.
Figure
1 shows such a solution forV/Yo
= 0.75 and coRo
= - 2.0. The contour of thecell
corresponding
to this solution is foundby
a furtherintegration.
It is shown infigure
3a. This type of cell is calleddiscocyte.
The scheme outlined above serves to find solutions for agiven
value ofAp/Apc.
To cal-culate a
cell-shape
for agiven
volumeV/ Vo
we treatAp
as an additional variable which is to be determinedby equation (2.14).
When there is no reflection
symmetry
we have tosolve
theequations
in the entire interval 0 s 1.At s = 0 we
again
have cm = cp =cm
and at s = 1FIG. 1. - Plot of f (s), cm(s) and cp(s) versus s for a discocyte
with VIVo = 0.75 and co Ro = - 2.0. The curvatures are given
in units of Ro 1, f (s) is given in units of
R2
FIG. 2. - Plot of f (s), cm(s) and cp(s) versus s for a cup-shaped
cell with VIVO = 0.75 and co Ro = - 2.0.
FIG. 3. - Contours corresponding to the solutions shown in
figure la and figure 2b.
C. = cp =
c1m,
thesuperscripts
u and Ireferring
tothe upper and lower half of the cell
respectively.
cm
andcm
are unknown and still to be determinedby boundary
conditions. The functionf (s)
reaches its maximumfm
at somepoint s
= sm. At thispoint cm(s)
has some value
cm.
Theparameter y
is now different from zero. So we have to determine six unknown parameterscm, cm, cmm, fm, Y,
sm.Matching
of thefunctions
cm(s), cp(s)
andf (s)
at two intermediatepoints
in the intervals0 s sm and sm
s 1gives
us six non-linearequations
for the six unknownboundary
conditions. These non-linearequations
have more than one solution as we shall see below.
We calculate the solutions
using again
the Newton-Raphson
method.Figure
2 shows the solution forV/Vo
= 0.75 andco Ro
= - 2.0. The contour cor-responding
to this solution is shown infigure
3b.This type of vesicle is called a
cup-shaped
cell. Both solutionscorrespond
to minima in the deformational energy, as is discussed in detail inappendix
A.As
pointed
outabove,
the non-linearequations
forthe
boundary
values have several solutions. Infigure
4we show two other
symmetrical
solutionsrepresenting higher
order deformations of thesphere (only
onequadrant
isshown). Figure
5 shows ahigher
orderasymmetric
solutionassuming
abell-shaped
formwhen the volume is decreased further.
Figure
6 andfigure
7 show two otherhigher
order deformations of thesphere.
We have notinvestigated
thestability
ofthese forms.
FIG. 4. - Contours of two symmetric solutions of higher order (only one quadrant is shown, the rotational axis is along z).
When the volume is
sufficiently
reduced and the spontaneous curvature is not toosmall,
the two halves of thesymmetric discocyte (Fig. 3a)
touch at the center.They
are in contact over a circular area of radiusx o =
fJ/2
and areaf o/4 (in
units of 4nRg).
In therange 0 s
f o/4
we havecm(s)
=CP(s)
= 0. Ats =
f o/4, cm(s) changes discontinuously
to somevalue
cmo.
Thediscontinuity corresponds
to a normalforce per unit
length along
the circlebounding
the areaof contact. The force in the upper half of the cell is balanced
by
anopposite
force per unitlength
ats = 1 -
f o/4 in
the lower half. At s = 0.5 we have thesame conditions as for
discocytes.
Now we have fourunknown
boundary
valuesfo, cm, fm, c’.
The match-ing
conditionsgive
threeequations
as before and afourth
equation
is obtainedby
the condition of contact z(s
=fo/4)
= z(s
=0.5)
= 0. The fourFIG. 5. - Contours of asymmetric solutions which are charac-
terized by an almost triangular cross-section.
FIG. 6. - Contour of an asymmetric solution of higher order having pentagonal shape.
equations
are solvedby
theNewton-Raphson method.
Contours of two such torocytes are shown in
figure
8.In
calculating
the torocytes we haveneglected
theFIG. 7. - Contour of an asymmetric solution of higher order having heptagonal shape.
FIG. 8. - Two contours of torocytes (only one quadrant is shown, the rotational axis is along the z direction).
cohesive energy between the membrane
regions
whichare in contact.
For the three types of solutions discussed so far we
have calculated the
dependence
of volume V onosmotic pressure difference
J1p
for certain values ofspontaneous
curvature. Such aplot
is shown infigure
9 for coRo
= - 2.0. Theasymmetric
branchof solutions
representing cup-shaped
cells(see Fig. 3b)
meets the
symmetric
branch ofdiscocytes
at apoint
marked
by
a critical pressure differenceAp,
and acritical volume
Vt.
Noasymmetric
solutions exist forAlp > Apt
and VV,.
The occurrence of such athreshold in the volume versus pressure
diagram
is aconsequence of
breaking
reflectionsymmetry
ingoing
FIG. 9. - Plot of volume V (in units of Vo) versus pressure diffe-
rence Ap (in units of Apc) for discocytes, cup-shaped cells and torocytes calculated for co Ro = - 2.0.
FIG. 10. - Plot of volume V versus pressure difference Ap calcu-
lated for co Ro = - 4.0. A volume gap appears between the toro- cyte and the discocyte branch.
from the
discocyte
to thecup-shaped
cell.Figure
10shows the same
diagram
calculated for coRo
= - 4.0.The
discocyte
and torocyte branches are nolonger
connected. With
decreasing
pressure differenceAp,
the volume of the
discocytes
decreases at first and then increasesagain
afterhaving passed through
a mini-mum. For a
given
volume we therefore find two sym- metric solutions as is shown infigure
11 forFIG. 11. - Two discocytes of volume VIVo = 0.75. These solu- tions differ in their values for Api Apc.
V/ Vo =
0.75. The volume of the torocytes increases first withdecreasing Ap
and decreasesagain
afterpassing through
a maximum. Between this maximum volume for torocytes and the minimum volume fordiscocytes
there is a volume gap where neither toro-cytes nor
discocytes
exist(but
otherrotationally symmetric
forms may, of course,exist).
If wehad,
forexample,
adiscocyte
withV/Vo =
0.65 andco
Ro
= - 2.0 and thenby
some chemical meansdecreased the
spontaneous
curvature co, we would reach apoint
where the vesicle could nolonger
have abiconcave-discoid
shape
but wouldundergo
a suddentransition to a new
equilibrium shape
whichmight
no
longer
have rotational symmetry. We can construct such asolution,
if we compare thecup-shaped
cellsand the
discocyte
forAp
= - 0.5Ape?
which areshown in
figure
12. Both forms arenearly spherical.
FIG. 12. - Contours of a cup-shaped cell and a discocyte for
which the deviations from the spherical shape have become very localized.
The deviations from the
sphere having
become very localized at thepoles
we obtainspheres
withdefects.
We find an
energy E
= 0.043Eo
and a volumedecrease AV = - 0.047
Vo
per defect for theshape
of
figure
12.Placing
a number of such defects on asphere
in aregular
or almostregular pattern,
we obtainnon-rotationally symmetric solutions,
whichare called inverted
echinocytes.
Anotherpossible
solution could be a form where the
spikes point
to the outside of the
sphere
rather than to theinside,
which would be called an
echinocyte. However,
we cannot prove at present that the vesicle will assume
either of these forms. From the numerical results shown in
figure
10 we canonly
conclude that the dis- cocyte becomes unstable when co is madesufficiently negative.
Instead of oblate
ellipsoids
anddiscocytes
whichwe found for
Ap Apc,
we can obtainprolate ellip- soids,
if we chooseAp
>Ap,,. (This
holdsonly
forco COco For co > cos we find oblate
ellipsoids
aboveApc
andprolate ellipsoids
belowAp,.)
For theseforms we have
cm(s) cp(s).
As we reduce thevolume,
the curvature at thedividing parallel
of latitudecm
(0.5) quickly
goes to zero. If we reduce the volumefurther,
we getdumbbell-shaped vesicles,
as shown infigure
13. The calculation for dumbbells iscomplicated by
the fact that we now have an additionalsingularity.
The function
f (s)
has a minimumfo at s
= 0.5 and amaximum fm
at some value s = sm. At s = 0.5 we have the initial conditionsf
=fo,
cp =f- 1/2
=C’mm.
FIG. 13. - Contour of a dumbbell (only one quadrant is shown).
The parameter h is now found from
equation (3.2)
and the solution is started with the
expansion (3. 3),
where we have to put y =
0,
since the form is symme- tric with respect to theequatorial plane.
At s = smwe have the initial condition
f
=f.,
cp= fm l/2 .
The initial value for
cm(s)
at s = sm is now found fromequation (3. 2),
since A has been determined at s = 0.5.We start the solution
again
at s = smby using
theexpansion (3. 3),
where we now have to allow for alinear term
with y #
0. At s = 0 we have the initial condition cp = cm =cm.
The six unknown parameterscm, cm, fo, fm,
s., y have to be determinedby matching
the three functions
f (s), cp(s)
andcm(s)
at two inter-mediate
points
in the intervals0 s sm
andsm s
0.5. The calculation becomes inaccurate in thevicinity
of thepoint
where the vesicle goes from thedumbbell-shape
to theellipsoidal form,
since atthis
point
we would have to go to fourth order in theexpansion (3.3)
at s = 0.5 and at s = sm. Infigure
14we show the
dependence
of volume on pressure’ference
Ap
for coRo
= - 2.0 and coRo
= - 4.0.1 he branches for
ellipsoids
and for dumbbells do notjoin smoothly
because of theinaccuracy
of our calcu-lation mentioned above. In
figure
15 wegive
thedependence
of the totaldeformational
energy E onvolume V. The energy is
given
in units ofFIG. 14. - Plot of volume versus pressure difference for prolate ellipsoids and dumbbells.
FIG. 15. - Plot of the deformational energy E (in units of the energy of a sphere Eo) versus volume V (in units of Vo) calculated
for various forms with Co Ro = - 2.0.
which is the energy of a
sphere.
The results infigure
15calculated for co
Ro
= - 2.0 show the energy of thecup-shaped
cells to behigher
than the energy ofdiscocytes having
the same volume. Thesign
of theenergy difference is
independent
of themagnitude
ofCo.
As
pointed
outabove, cup-shaped
cells existonly
above a critical volume
Yt.
Themagnitude of Vt
de-creases with
increasing
co.Choosing
co to be zero orslightly positive,
we can obtaincup-shaped
cellswhich are very thin at the center. Such a form is shown in
figure
16 for coRo
= 0.Increasing
co and at the same timedecreasing
thevolume,
we can causethe two halves of the cell membrane to make contact
over some area at the center. Such a cell is shown in
figure
17 and referred to in thefollowing
ascodocyte
I.FIG. 16. - Contour of a deep cup-shaped cell.
FIG. 17. - Contour of a codocyte I.
We calculate this type of solution
using
thefollowing boundary
conditions. In the range of 0 s so and 1 - so s 1 the two membrane halves are in contact and the curvatures arecm(s)
=cp(s)
= const.At the
points s
= so and s = 1 - so the curvaturecp(s)
iscontinuous,
whereascm(s) changes
disconti-nuously
to a valuecm
at s = so and tocm
at s = 1 - so.The discontinuities in
cm(s)
at s = so and s = 1 - sorespectively correspond
to normal forces per unitlength along
the circlebounding
the area of contact.Since the contributions of the upper and lower halves must balance each
other,
we havein the range 0 s so and
in the
range
1 - so s 1. Theboundary
valuefo
forf (s)
iseasily
found to beAt the outer rim of the cell we have the same condi- tions as for a
cup-shaped
cell. So we now have sevenunknown
boundary
values so,C’, cm, em9 fm,
y, sm.The
matching
conditions at two intermediatepoints
in the intervals so s sm
and sm
s 1 - sogive
us sixequations.
The seventhequation
is obtainedfrom the condition of
contact z(so)
=z(l - so).
The seven non-linear
equations
are solvedby
theNewton-Raphson
method. We may view thecodocyte
Ias an
asymmetric
torocyte. Inanalogy
to the cup-shaped
cell thecodocyte
I becomes more and moresymmetric
as we reduce the volume(or
increaseAp).
In
figure
18 the volume isplotted
versus the pressure difference/p
forcodocytes
as well as for torocytes.As for the
cup-shaped cell,
we find a criticalvolume V,
below which no solution of
codocyte
I type exists.FIG. 18. - Plot of volume versus pressure difference for torocytes and codocytes I. The torocyte branch ends at a point slightly
above the threshold vt’ for codocyte I where it joins the discocyte
branch not shown in this figure.
Let us now return to the
deep cup-shaped
cell offigure
16. If we reduce the pressure differenceAp,
the volume increases
slowly
and the cell becomesmore
asymmetric. Finally
the vesicle assumes a newtype
ofshape,
calledstomatocyte,
where thedeep
indentation in
figure
16 becomes almosttotally
concave as shown in
figure
19. The numerical calcula- tions are rather difficult for these forms because the variablef (s)
now has two maxima and one minimumas
compared
toonly
one maximum for the cup-shaped
cell. The additional extrema in the functionf (s) present
two additionalsingularities
which haveFIG. 19. - Contour of a stomatocyte.
to be removed in the same way as was outlined above for the
cup-shaped
cell. We have to divide the interval0 s
1 into fiveparts
and find the solutionscm(s), cp(s), f (s)
in these five intervals. Thematching
conditions at the four intermediate
points
whichseparate
the five parts of the interval 0 s 1provide
twelve non-linearequations. Figure
20 showsthe volume
Y/ vo
as a function ofAp/Apc
for stoma-tocytes and for
cup-shaped
cells. We could not cal-culate the solutions in the range of
Ap
where the cell goes from thecup-shaped
form to the stomatocyteshape,
becausehigher
order terms in the expan- sion(3.3)
arerequired
to obtain sufficient accuracy.FIG. 20. - Plot of volume versus pressure difference for stomato- cytes and cup-shaped cells.
For co = 0 the
stomatocytes
were found to havea
slightly
lower energy thancup-shaped
cells of thesame volume. For
sufficiently negative
values of co the volume of thestomatocytes
isnearly Vo
andincreases with
decreasing Ap. Figure
21 shows astomatocyte
for coRo
= - 2.0.If we make co
sufficiently positive,
the thickness at the center of thestomatocyte
is reduced and we obtain contact as is shown infigure
22. Thistype
of solution is calledcodocyte
II. Theboundary
conditions at the circlebounding
the area of contact are identical tothose of
codocyte
I. The otherboundary
conditionsare the same as for
stomatocytes.
As the volume of thecodocyte
II isreduced,
theopening
at the top becomes narrower and the area of contact increases(Fig. 23). Figure
24gives
volume V versus pressure- differenceAp
forcodocyte
I andcodocyte II,
the curvebeing
similar to thecorresponding
function for sto- matocytes andcup-shaped
cells as shown infigure
20.4.
Concluding
remarks. -Many
of theshapes
calculated here have
recently
been observedexperi-
FIG. 21. - Contour of a nearly spherical stomatocyte.
FIG. 22. - Contour of a codocyte II.
mentally [12, 13].
The vesicles were obtainedby simple swelling
of smallpieces
of lecithin in water.Some of them were
quite large, having
diametersbetween 10 and 30 p.
Well-defined contours could be seen and
photo- graphed
under aphase
contrastmicroscope.
Thenumber of
bilayers forming
the membranevaried,
but was
apparently
one in many cases. With thespontaneous
curvature as theonly adjustable
para- meter, theoreticalshapes
could be fitted very accura-tely
to theexperimental
ones. Thesecomparisons,
which will be
published elsewhere,
leave no doubtFIG. 23. - Contour of an almost closed codocyte II.
FIG. 24. - Plot of volume versus pressure difference for codocyte I
and codocyte II.
that the
shape
of the closed membranes in their fluid state is controlledsolely by
curvatureelasticity.
They
also confirm that the two halves of themembrane,
wherethey
are in contact, do notnoticeably
attracteach
other,
as was assumed in the above calculations.An
important
difference between vesicles anddrops
which is sometimes overlooked is the absence of surface tension in the former. A closed fluid membrane
experiences virtually
no tensile stress unless water ispumped
into theenclosure,
e.g.by osmosis, beyond
the
point
where thespherical shape
is reached. Forsmaller
volumes,
themembrane,
whose area is prac-tically
constant,undergoes out-of-plane
fluctuations.They
do notdestroy
theequilibrium
contour, butwere found to be
easily
visible under themicroscope
ifthe membrane consists of
only
a small number of lecithinbilayers [12, 13].
Infact, they
can be used todetermine the value of the curvature-elastic modulus
kc. Theory [3, 4]
andexperiment [13]
indicate the order of10-12
ergfor kc. Although
theequilibrium
shape
isgoverned by
curvatureelasticity,
itgives
noinformation on
kc.
Forlarge vesicles,
the elastic modulus is too smallby
many powers of ten to affectsignificantly
internal pressure and henceshape
undergiven
osmotic conditions. As anexample,
the critical pressure calculated fromequation (B.4)
withkc
=10-12
erg andRo
=10 Jl
is of the order of10-3 dyn cm-2.
The curvature
elasticity
of fluid membranes can be used toexplain
certain red blood cellshapes [5, 7].
It may be
important
for somebiological
processes since itmight
govern the elastic behaviour ofplasma
membranes. Artificial vesicles and
bilayer elasticity
could also have other
applications.
From a funda-mental
point
ofview,
vesicles seem to offer attractivepossibilities
for thestudy
ofphase
transitions in two- dimensional systems. Forinstance,
the fluid-solid transition of lecithinbilayers
was seen under thephase
contrastmicroscope [12]
and somebilayer properties
near thetransition,
which isprobably
close to second
order,
may be obtainableby
morerefined
optical
studies.Appendix
A :Stability
of the solutionspresented
in
figures
1-3. - Toinvestigate
thestability
of thesolutions we have to calculate the deformational energy of vesicles which have the same volume and surface area as the forms in
figure
3 but differslightly
in
shape.
As isreadily
seen fromfigures
1-2 the func-tion
ep(s)
may be wellrepresented by
the first few terms of a Fourier series. We writewith five coefficients still to be determined. For a
symmetric
solution y and 6 are zeroby
definition.From
expression (A.1)
weeasily
find the value sm at whichcp(s)
has a maximum. For vesicleshaving
rotational
symmetry
the curvaturescm(s)
andcp(s)
are connected with the variable
f (s) = x2 by
thefollowing
relations which have been derived in sec-tion 2
We solve
equation (A. 2)
forf (s)
in the intervals0 .s sm
andsm s 1 starting
at s = sm with the initial conditionf (s.) =[CP(S.)] - 2
The boun-dary
conditionsf (0)
= 0 andf (1)
= 0along
withthe constraint