Download Poisson Boltzmann Equation

1. Poisson-Boltzmann
1.1. Poisson equation. We consider the Laplacian operator
∂2
∂2
∂2
2
(1)
∇ = 2+ 2+ 2
∂x
∂y
∂z
which is given in spherical coordinates by
(2)
1
∂
1
∂
1
∂2
∂
∂
2
2
∇ = 2
r
+ 2
sin θ
+ 2 2
r ∂r
∂r
r sin θ ∂θ
∂θ
r sin θ ∂ϕ2
and in cylindrical coordinates by
1∂
∂
1 ∂2
∂2
(3)
r
+ 2 2 + 2.
r ∂r
∂r
r ∂θ
∂z
If ψ is the electrostatic potential and ρ is the charge
density, Poisson’s equation is
ρ
(4)
∇2 ψ = − .
εε0
where ε is the dielectric constant, taken to be 1 in
vacuo. Where there is no charge density, ψ satisfies
the Laplace equation.
The electric field is given by
(5)
E = −∇ψ.
and Poisson’s equation becomes
ρ
(6)
∇·E=
.
εε0
1.1.1. Point charge. The simplest example is the
potential of a point charge at the origin with charge
1
2
q, the Coulomb potential,
(7)
ψ=
q
.
4πεε0r
with
(8)
E=
1
r̂.
4πεε0r2
In this case
(9)
∇2 ψ = −
q
δ0
εε0
where δ is the Dirac delta function. This implies that
ψ is harmonic, ∇2ψ = 0, except at the origin.
For other charge distributions in this medium of
constant dielectric ε the potential can be computed
by integration of the Coulomb potential. If the charge
distribution is ρ(r) dV (r), the solution to Poisson’s
equation (4) is given by
Z
1
ρ(r) dV (r)
(10)
ψ(r0) =
.
4πεε0
|r0 − r|
where the integral is over all space. The integral
reduces to a surface or path integral if the charge
distribution is on a surface or a curve.
The potential ψ for some charge distributions is
easy to find by Stokes’ theorem. For example, if
a total charge q is evenly distributed on a sphere
centered at the origin the potential is seen to be the
same as the Coulomb potential outside the sphere,
and zero inside the sphere.
3
1.2. Poisson Boltzmann. The Poisson-Boltzmann
equation arises because in some cases the charge density ρ depends on the potential ψ. This makes (4)
harder to solve since ψ is on both sides of the equation. The Poisson-Boltzmann equation is often applied to salts, since both positive and negative are
present in in concentrations that vary with the potential.
The relationship of concentration of charge to potential is given by statistical mechanics. By the Boltzmann law, the probability of a particle having energy
E is proportional to e−E/kT where T is the temperature and k is Boltzmann’s constant. An ion of valence z in a field ψ has electric potential energy zeψ
where e is the elementary charge of a single electron.
By Boltzmann we can write the concentration as
c = c0e−zeψ/kT .
(11)
The constant c0 is interpreted as the bulk concentration. The charge density ρ from this ion is Fc where
F = Ne is the Faraday constant.
In the presence of many ions indexed by i, with
valences zi and concentrations ci we have
X
X
(12)
ρ=F
zici = F
zici0 e−zieψ/kT .
i
i
We make the assumption that the ion concentration
are independent. Substituting (12) in (4) gives the
Poisson-Boltzmann equation.
4
The Poisson-Boltzmann equation is a non-linear
partial differential equation. It can be simplified under the assumption that eψ/kT is very small and
so e−eψ/kT is approximately equal to 1 − eψ/kT .
This is sometimes called the Debye-Hückel approximation and the resulting Poisson-Boltzmann equation is called the linear Poisson-Boltzmann equation.
1.3. Ions and counter ions. The Poisson-Boltzmann
equation is usually applied to a 1:1 salt solution where
there are cations of valence z with a counter ion of valence −z. In this case the linear Poisson-Boltzmann
equation becomes
(13)
∇2ψ = κ2ψ
where
(14)
2z 2eFc0
.
κ =
kT ε0
2
The quantity 1/κ is called the Debye length. This
equation can be applied in the following two situations
(1) There is a central ion at the origin and the field
has spherical symmetry. This is called DebyeHückel theory.
(2) The ions are on one side of a charged plane.
1.3.1. Spherical symmetry, Debye-Hückel theory.
In case (1) above, if the central ion of charge q is at
5
the origin, the potential depends only on r in spherical coordinates. The linear Poisson-Boltzmann equation becomes
1 ∂
∂ψ
(15)
r2
= κ2ψ.
2
r ∂r
∂r
and this has a solution of the form
e−κr
(16)
ψ=A
r
for some constant A.
There are two cases based on different boundary
conditions:
(1) ∇2ψ = qδ0/ε0.
(2) There is a central ion of charge q at the origin
and a sphere of radius a free of the salt solution.
Case (1) Take A = q/ε0 and (16) approaches the
Coulomb potential as r → 0.
Case (2) Use a potential of the type
C
(17)
ψint = + D
r
inside the sphere and of type (16),
e−κr
(18)
ψext = A
r
outside the sphere. The electric field is given by
C
(19)
Eint = 2 r̂
r
(20)
Eext
e−κr
= A 2 (1 + κr) r̂.
r
6
Writing the flux over the sphere in terms of the total
charge on the central ion and using Gauss’ Theorem
gives
q
(21)
4πAe−κa(1 + κa) =
ε0
and we can solve for A. Assuming that the potential
and the electric field are continuous on the surface of
the sphere gives two equations in C and D to solve
for the potential.
We get finally
q exp(κa)
(22)
A=
4πεε0(1 + κa)
q
C=
4πεε0
−qκ
D=
.
4πεε0(1 + κa)
The potential on the interior is
1
κ
q
−
.
(23)
ψint(r) =
4πεε0 r 1 + κa
The potential on the boundary of the sphere is
q
(24)
ψext(a) = ψint(a) =
.
4πεε0a(1 + κa)
The same analysis can be applied to find ψext if
all of the charge is equally distributed over a sphere
a sphere of radius R < a. If the smaller sphere
is thought of as a protein (or amino acid) with all
charged residues evenly distributed on the surface,
this becomes the basis of the Linderstrøm-Lang model
7
for the charge on a protein. The distribution of
charge over the surface allows us to assume spherical symmetric and the analysis is the same as above.
The charge on the surface of the protein is
q
1
κ
(25)
ψint(R) =
−
4πεε0 R 1 + κa
where ε is the dielectric constant of water. This expression can be used to find the work to bring a
charge to the surface of the protein.
If the charges sum to zero, for example in the case of
a polar group, then the above analysis will not work
and the position of the charges within the protein
must be taken into account. The Tanford-Kirkwood
theory, which we discuss below, does take into account the geometry of the charge distribution.
1.3.2. Charged plane. Suppose the plane is x = 0,
The potential depends only on the distance r from
the plane and the linearized Poisson-Boltzmann becomes
d2ψ
2
(26)
=
κ
ψ
dr2
with solution
(27)
ψ = ψ0e−κr
and the potential decays exponentially.
In this case the Poisson-Boltzmann equation can
also be solved analytically without the Debye-Hückel
8
approximation. Using (12) the equation is
d2ψ 2F zc0
(28)
=
sinh(zeψ/kT ).
dr2
ε0
Replacing κr by x and zeψ/kT by y, get the reduced
Poisson-Boltzmann equation
d2y
= sinh y.
(29)
dx2
Note that x and y are dimensionless in the reduced
equation. It can be checked that a solution is
1 − γe−x
(30)
y = 2 log
1 + γe−x
where
ey0/2 − 1
(31)
γ = y /2
e 0 +1
give the proper initial condition. The solution (30)
is approximated by (27).
1.4. Inversion. One useful way to get one harmonic
function from another is inversion. The map
r → r/r2
is inversion in the unit sphere. It leaves the sphere
fixed and interchanges 0 and ∞.
If g(r) = r−1f (r/r2), we see from (2) that
(32)
∇2g(r) = r−5∇2f (r/r2).
If f satisfies the Laplace equation ∇2f = 0 (f is
harmonic) then g does also.
9
We say f is a homogeneous polynomial of degree
n if all the terms have total degree n. For such a
polynomial we have
f (tr) = tnf (r).
for any real number t. By (32), if f is a harmonic
and homogenous polynomial of degree n then g(r) =
r−2n−1f (r) is harmonic and homogeneous of degree
−(n + 1).
1.5. Dipoles. The potential function for a collection of point charges can be obtained by adding the
functions for the individual charges. If the charges
are qi at the points ri then the potential function is
X
qi
(33)
ψ(r) =
.
4πε
|r
−
r
|
0
i
i
A dipole is a configuration consisting of a positive
and a negative charge of equal magnitude. If there
are charges q at r2 and −q at r1, the potential is
q
1
1
(34)
ψ(r) =
−
.
4πε0 |r − r2| |r − r1|
For studying the potential of a dipole, it is useful
to have the Taylor expansion
X
1
p
(35)
=
η k Pk (x)
1 + η 2 − 2xη
k
10
where Pk are the Laguerre polynomials
(36)
P0(x) = 1
P1(x) = x
1
2
P2(x) =
3x − 1
2
1
2
P3(x) =
5x − 3x
2
...
We have
1
1
= p
(37)
|r − r1| r 1 + η 2 − 2 cos θη
where η = r1/r and cos θ = r̂ · r̂1 is the cosine of
the angle between the vectors r and r1. The series
converges for |η| < 1.
Using (35) we get for r large,
1 X r1 k
1
=
(38)
Pk (cos θ).
|r − r1| r
r
k
Using (38) the dipole potential (34) can be approximated for r large by just the one term
1 r̂
1 r̂
·
q(r
−
r
)
=
·D
(39) ψ(r) ≈
2
1
4πε0 r2
4πε0 r2
where D = q(r2 −r1) is called the dipole vector. The
magnitude of ψ is of order 1/r2.
Using (39) the electric field for a dipole can also be
approximated
1 1
[3(r̂ · D)r̂ − D] .
(40)
E = −∇ψ ≈
4πε0 r3
11
The magnitude is of order 1/r3.
1.5.1. Other charge distributions. For a number of
point charges qj at points rj the potential ψ is the
sum of the Coulomb potentials,
X qj
(41)
ψ(r) =
.
|r
−
r|
j
j
Similarly, for a continuous charge distribution ρ,
ZZZ
ρ(r1)
dr1.
(42)
ψ(r) =
|r1 − r|
Using (38) ψ can be expanded in terms of Legendre
polynomials
Z Z Z
∞
X
(43) ψ(r) =
r−n−1
r1nPn(rˆ1 · r̂) dr1
n=0
for r outside a sphere about the origin containing all
of the charge
Z Z Z
∞
X
(44) ψ(r) =
rn
r1−n−1Pn(rˆ1 · r̂) dr1
n=0
for r inside a sphere about the origin containing none
of the charge. Each term in brackets is a spherical
harmonic and can be further expanded in the form
n
X
(45)
CnmPnm(cos ϑ)eimϕ
m=−n
where (r, ϑ, ϕ) are spherical coordinates in some fixed
coordinate frame and Pnm are the associated Legendre
functions.
12
The most general expansion of a function ψ satisfying Laplace’s equation is
(46)
∞ X
n
∞ X
n
X
X
−n−1 m
imϕ
Cnmr
Pn (cos ϑ)e +
GnmrnPnm(cos ϑ)eimϕ
n=0 m=−n
n=0 m=−n
where Cnm and Gnm are suitable constants. To get
the most general function replace the constants by
power series in r, Cnm(r) and Gnm(r).
1.6. General solution for Poisson-Boltzmann.
We saw that the solution of the linear Poisson-Boltzmann
equation in the spherically symmetric case is of the
form Ar−1 exp(−κr). A general solution in spherical coordinates using a series expansion can be found.
Take the reduced Poisson-Boltzmann equation
∇2 y = y
(47)
and substitute y = e−r u to get
(48)
∇2u = 2r−1 (r · ∇u + u) .
Solution of the form
(49)
K(r)Q(r)
where Q is harmonic and homogeneous of degree
−n − 1 can be found by substituting in (48). Taking
derivatives of (49) get
(50)
∇u = K 0Q r̂ + K∇Q
r · ∇u = [rK 0 − (n + 1)K] Q
∇2u = K 00 − 2nr−1K 0 Q.
13
Substituting in (48) gives the differential equation
rK 00(r) − 2K 0(r)(n + r) + 2nK(r) = 0.
P
s
Substituting K(r) = ∞
s=0 Ks r into (51), get a recursion relation
(51)
(52)
Ks+1 =
2(n − s)
Ks .
(s + 1)(2n − s)
It follows that K(r) is a constant times
n
X
(53)
Knsxs.
s=0
where
(54)
Kns
2sn!(2n − s)!
=
s!(2n)!(n − s)!
and k is a constant. The polynomials Kn(r) =
P
n
s
s=0 Kns r for n = 0, 1, 2 are given by
n
Kn(r)
0
1
1
1+r
2 1 + r + r2/3
Table 1.
Thus the functions e−r Kn(r)r−2n−1Pn(r) where Pn
is a homogeneous polynomial of degree n satisfy the
14
reduced Poisson-Boltzmann equation (47), for example
(55)
e−r r−1A
e−r (1 + r)r−3(Ax + By + Cz)
2
r
e−r 1 + r +
r−5(Ax2 + By 2 − (A + B)z 2 + Cxy + Dxz + Eyz)
3
More generally, adding solutions of this type, any
series of the form
∞
X
(56)
e−r
Kn(r)r−2n−1Pn(r)
n=0
where Pn is a homogeneous polynomial of degree n,
gives a solution of (47) vanishing infinity. Any such
polynomial Pn can be written as the sum of 2n + 1
spherical harmonics
rnPnm(cos ϑ)eimϕ,
−n ≤ m ≤ n.
1.7. Tanford-Kirkwood theory. The protein is
modeled as a sphere of radius R containing point
charges and with the solvent outside a larger circle
of radius a.
Description of potentials: For convenience write
(57)
Ynm(ϑ, ϕ) = Pnm(cos ϑ)eimϕ.
These are the spherical harmonics without normalization constants.
15
notation
region
description
potential dielectric
R1
r<R
inside protein
V1 = ψ1 + ψ
εi
R2
R < r < a outside protein, ion free
V2
ε
R3
a<r
ionic solution
V3
ε
Table 2. Regions and potentials in the Tanford-Kirkwood theory
(1) ψ1 is sum of Coulomb potentials from charges
in R1.
1 X qk
.
ψ1(r) =
4πεiε0
|r − rk |
k
Using (44) there is an expansion of the form
∞ X
n
X
ψ1(r) =
Enmr−n−1Ynm(ϑ, ϕ)
n=0 m=−n
for r near R. There is an explicit expression for
Enm in terms of the qk and the rk giving the
charges and their locations.
(2) ψ is potential in R1 arising from charges in R3.
It satisfies Laplace’s equation and has expansion
∞ X
n
X
ψ(r) =
BnmrnYnm(ϑ, ϕ)
n=0 m=−n
(3) V1 is the total potential in R1
∞ X
n
X
V1(r) = ψ1(r)+ψ(r) =
(Enmr−n−1+Bnmrn)Ynm(ϑ, ϕ)
n=0 m=−n
(4) V2 is potential in R2 arising from charges in R1
and R3. It satisfies Laplace’s equation and has
16
expansion
V2(r) =
∞ X
n
X
(Cnmr−n−1 + Gnmrn)Ynm(ϑ, ϕ)
n=0 m=−n
(5) V3 is the potential in the ionic region R3. It
satisfies ∇2V3 = κ2V3 and can be expanded in
the form
V3(r) =
∞ X
n
X
Anme−κr Kn(κr)r−n−1Ynm(ϑ, ϕ)
n=0 m=−n
The coefficients Anm, Bnm, Cnm, Gnm are computed
from the boundary conditions below which insure
that the potentials and the fluxes are continuous on
the spheres r = R and r = a.
surface boundary conditions
r=R
V1 = V2
εir · ∇V1 = εr · ∇V2
r=a
V2 = V3
r · ∇V2 = r · ∇V3
Table 3. Boundary conditions for potentials
To compute the boundary conditions, note that
r · ∇(rnYnm) = n(rnYnm)
r · ∇(r−n−1Ynm) = (−n − 1)(r−n−1Ynm)
17
so
r · ∇V1(r) =
r · ∇V2(r) =
r · ∇V3(r) =
∞ X
n
X
n=0 m=−n
∞ X
n
X
n=0 m=−n
n
∞ X
X
[−(n + 1)r−n−1Enm + nrnBnm]Ymn
[−(n + 1)r−n−1Cnm + nrnGnm)]Ymn
Anme−κr [−(κr + n + 1)Kn(κr) + κrK 0(κr)]r−
n=0 m=−n
Now equating the coefficients of Ymn, the boundary
conditions become
Enm + BnmR2n+1 = Cnm + GnmR2n+1
εi[−(n + 1)Enm + nR2n+1Bnm] = ε[−(n + 1)Cnm + nR2
Cnm + Gnma2n+1 = Anme−κaKn(κa)
Anme−κa[−(κa + n + 1)Kn(κa) + κaK 0(κa)] = −(n + 1)Cnm + na2n+
This, assuming that E is known, this gives four linear
equations in B, C, A, G. The matrix of coefficients
is
(58)

Rn+1
1
0
−Rn+1
 εinRn+1εi ε(n + 1)
0
−nRn+1


0
1
−Kn(κa)
a2n+1
0
n + 1 −e−κa(2n + 1)Kn+1(κa) −na2n+1




18
and the right side of the equation is given by the
column matrix 

−Emn
 εi(n + 1)Emn 
.

(59)


0
0
Solve for the coefficients Bmn to get the potential
ψ given by
∞ X
n
X
ψ(r) =
BnmrnYnm(ϑ, ϕ).
n=0 m=−n
The energy of the interactions between the charges
qk and the ions is then
X
qk ψ(rk ).
k