Biophysics: Problems IV
[4.1] A hemispherical surface is governed by the equation x2 + y 2 + z 2 = R2 , z > 0.
(a) Find h(x, y) and the normal n(x, y) for x2 + y 2 < R2
(b) Determine the mean and Gaussian curvatures.
(c) Check that your results agree with your intuition regarding properties of a sphere
[4.2] Under certain conditions, red blood cells develop dimples on their surface. Suppose that
such a dimple is modeled by a Gaussian deformation
h(x, y) = h0 exp(−[x2 + y 2 ]/2w2 )
(a) Assuming that h0 is small, calculate the mean and Gaussian curvatures as a function of
(x, y).
(b) Show that the total bending energy
1
E=
dxdy κb (σ1 + σ2 )2 + κG σ1 σ2
2 R2
is equal to
5πκb
E=
4
h0
w
2
(c) What is the total bending energy (in units of kT ) if w = 1µm, h0 = 0.1µm and κb =
10kT ?
[4.3] Obtain a relationship between the bending rigidity κb and the 2D compression modulus KA
by considering the gentle bending of a sheet into a section of cylinder with principal curvature
1/R, 0. Take the surface of zero strain (the neutral surface) to run through the middle of the
sheet, and assume that the thickness d of the sheet is unchanged after deformation.
z = d/2
z
z=0
x
z = - d/2
R
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(a) Show that the energy density of the cylindrical deformation is F = κb /2R2
(b) Using a geometrical construction show that for small deformations, the strain along the
neutral surface is given by ux = O(x3 ), uz = −x2 /R + O(x3 )
(c) Assuming that the stress perpendicular to the surface is neglible compared to internal
stresses so that σxz ≈ 0, show that the strain uxx ≈ z/R throughout the sheet
(d) Calculate the elastic energy density F = u2xx KA /2 by averaging across the membrane,
and hence show that κb = KA d2 /12
[4.4] Surface tension of a curved interface. The free energy of a binary mixture with spacedependent order parameter φ(r) is
F [φ] =
b
2
(∇φ) + f (φ) d3 r
2
where f is the local free enegegy density.
(a) Write down the Euler–Lagrange equations
(b) In the case of a curved surface interface φ(x, y, z) = ψ(x, y, z − h(x, y)) show that the
Euler–Lagrange equations reduce to the approximate form
∂f (ψ)
= b 1 + h2x + h2y ψzz
∂ψ
[Assume that spatial variations with respect to x, y are much smaller than with respect to z
and drop terms involving hxx , hyy ]. It follows that ψ can be approximated by
z−h
ψ(r) = Ψ 1 + h2x + h2y
where Ψ is the corresponding planar interfacial profile.
(c) Multiply both sides of the first equation in (b) by ψz and integrate with respect to z using
the boundary conditions ψ(z) → φ± for z → ±∞, where φ± are the bulk compositions that
minimize f (φ). Use this to show that the surface free energy is
FS = γ
1 + h2x + h2y dxdy
where
γ=b
Ψz (x)2 dz
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[4.5] Surface fluctuations. For small curvature, the surface free energy calculated in [4.4] can
be approximated by
γ
FS [h] =
2
h2x + h2y dxdy
Λ
Taking Λ to be a square of sides L, use the partition function
dh(r)e−Fs [h]/kB T
Z=
r
to show that
h(r)2 =
qmax
kB T −2
q
γL2 q=q
min
Evaluate the sum on the right-hand side in the continuum limit.
[3.6] A polyelectrolyte consists of water–soluble polymers made up of ionizable monomers. If
the polymers are reasonably stiff like DNA, then each polymer can be treated as a uniformly
charged cylindrical rod of radius r0 and charge per unit length −σ with σ > 0.
(a) Assuming that r0 λD , solve the linearized Poisson–Boltzmann equation for an infinitely
long cylinder with boundary condition
dV σ
=
dr r=r0
2πεr0
[Hint: In order to match the boundary condition, you will need to perform an asymptotic
expansion of an appropriate Bessel function]
(b) Calculate the total number of counterions per unit length of rod, and show that this
diverges as r0 → 0 when eσ > 4πεkB T . This effect leads to the condensation of counterions
on the surface of the cylinder.
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