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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