1. Poisson-Boltzmann 1.1. Poisson equation. We consider the Laplacian. which is given in spherical coordinates by (2)

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1 1. Poisson-Boltzmann 1.1. Poisson equation. We consider the Laplacian operator (1) 2 = 2 x y z 2 which is given in spherical coordinates by (2) 2 = 1 ( r 2 ) + 1 r 2 r r r 2 sin θ θ and in cylindrical coordinates by ( 1 (3) r ) + 1r 2 r r r 2 θ z 2. ( sin θ ) + 1 θ r 2 sin 2 θ If ψ is the electrostatic potential and ρ is the charge density, Poisson s equation is (4) 2 ψ = ρ εε 0. 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 (6) E = ρ εε Point charge. The simplest example is the potential of a point charge at the origin with charge 1 2 ϕ 2

2 2 q, the Coulomb potential, q (7) ψ = 4πεε 0 r. with (8) E = In this case 1 4πεε 0 r 2ˆr. (9) 2 ψ = q εε 0 δ 0 where δ is the Dirac delta function. This implies that ψ is harmonic, 2 ψ = 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 (10) ψ(r 0 ) = 1 4πεε 0 ρ(r) dv (r). r 0 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 (11) c = c 0 e zeψ/kt. The constant c 0 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 z i and concentrations c i we have 3 (12) ρ = F i z i c i = F i z i c i0 e z ieψ/kt. We make the assumption that the ion concentration are independent. Substituting (12) in (4) gives the Poisson-Boltzmann equation.

4 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 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) 2 ψ = κ 2 ψ where (14) κ 2 = 2z2 efc 0 kt ε 0. 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 Debye- Hückel theory. (2) The ions are on one side of a charged plane 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 (15) 1 r 2 r ( r 2 ψ ) r = κ 2 ψ. and this has a solution of the form (16) ψ = A e κr r for some constant A. There are two cases based on different boundary conditions: (1) 2 ψ = qδ 0 /ε 0. (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/ε 0 and (16) approaches the Coulomb potential as r 0. Case (2) Use a potential of the type (17) ψ int = C r + D inside the sphere and of type (16), (18) ψ ext = A e κr r outside the sphere. The electric field is given by (19) E int = C r 2 ˆr (20) E ext = A e κr r 2 (1 + κr) ˆr. 5

6 6 Writing the flux over the sphere in terms of the total charge on the central ion and using Gauss Theorem gives (21) 4πAe κa (1 + κa) = q ε 0 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 (22) A = q exp(κa) 4πεε 0 (1 + κa) C = q 4πεε 0 qκ D = 4πεε 0 (1 + κa). The potential on the interior is (23) ψ int (r) = q ( 1 4πεε 0 r κ ). 1 + κa The potential on the boundary of the sphere is q (24) ψ ext (a) = ψ int (a) = 4πεε 0 a(1 + κa). The same analysis can be applied to find ψ ext 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 (25) ψ int (R) = q [ 1 4πεε 0 R κ ] 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 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 (26) with solution d 2 ψ dr 2 = κ2 ψ (27) ψ = ψ 0 e κr and the potential decays exponentially. In this case the Poisson-Boltzmann equation can also be solved analytically without the Debye-Hückel

8 8 approximation. Using (12) the equation is d 2 ψ (28) dr = 2F zc 0 sinh(zeψ/kt ). 2 ε 0 Replacing κr by x and zeψ/kt by y, get the reduced Poisson-Boltzmann equation d 2 y (29) = sinh y. 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 (31) γ = ey 0/2 1 e y 0/2 + 1 give the proper initial condition. The solution (30) is approximated by (27) Inversion. One useful way to get one harmonic function from another is inversion. The map r r/r 2 is inversion in the unit sphere. It leaves the sphere fixed and interchanges 0 and. If g(r) = r 1 f(r/r 2 ), we see from (2) that (32) 2 g(r) = r 5 2 f(r/r 2 ). If f satisfies the Laplace equation 2 f = 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) = t n f(r). for any real number t. By (32), if f is a harmonic and homogenous polynomial of degree n then g(r) = r 2n 1 f(r) is harmonic and homogeneous of degree (n + 1) 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 q i at the points r i then the potential function is 9 (33) ψ(r) = i q i 4πε 0 r r i. A dipole is a configuration consisting of a positive and a negative charge of equal magnitude. If there are charges q at r 2 and q at r 1, the potential is (34) ψ(r) = q 4πε 0 ( 1 r r 2 1 r r 1 ). For studying the potential of a dipole, it is useful to have the Taylor expansion (35) η2 2xη = k η k P k (x)

10 10 where P k are the Laguerre polynomials (36) P 0 (x) = 1 P 1 (x) = x P 2 (x) = 1 ( 3x 2 1 ) 2 P 3 (x) = 1 ( 5x 2 3x ) 2. We have 1 (37) r r 1 = 1 r 1 + η 2 2 cos θη where η = r 1 /r and cos θ = ˆr ˆr 1 is the cosine of the angle between the vectors r and r 1. The series converges for η < 1. Using (35) we get for r large, 1 (38) r r 1 = 1 ( r1 ) k Pk (cos θ). r r k Using (38) the dipole potential (34) can be approximated for r large by just the one term (39) ψ(r) 1 ˆr 4πε 0 r q(r 2 2 r 1 ) = 1 ˆr 4πε 0 r D 2 where D = q(r 2 r 1 ) is called the dipole vector. The magnitude of ψ is of order 1/r 2. Using (39) the electric field for a dipole can also be approximated (40) E = ψ 1 1 [3(ˆr D)ˆr D]. 4πε 0 r3

11 The magnitude is of order 1/r Other charge distributions. For a number of point charges q j at points r j the potential ψ is the sum of the Coulomb potentials, (41) ψ(r) = q j r j j r. Similarly, for a continuous charge distribution ρ, ρ(r 1 ) (42) ψ(r) = r 1 r dr 1. Using (38) ψ can be expanded in terms of Legendre polynomials [ ] (43) ψ(r) = r n 1 r1p n n ( ˆr 1 ˆr) dr 1 n=0 for r outside a sphere about the origin containing all of the charge [ ] (44) ψ(r) = r n r1 n 1 P n ( ˆr 1 ˆr) dr 1 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 (45) C nm Pn m (cos ϑ)e imϕ m= n where (r, ϑ, ϕ) are spherical coordinates in some fixed coordinate frame and Pn m are the associated Legendre functions. 11

12 12 The most general expansion of a function ψ satisfying Laplace s equation is (46) n n C nm r n 1 Pn m (cos ϑ)e imϕ + G nm r n Pn m (cos ϑ)e imϕ where C nm and G nm are suitable constants. To get the most general function replace the constants by power series in r, C nm (r) and G nm (r) 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 (47) 2 y = y and substitute y = e r u to get (48) 2 u = 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 Q ˆr + K Q r u = [rk (n + 1)K] Q 2 u = [ K 2nr 1 K ] Q.

13 Substituting in (48) gives the differential equation 13 (51) rk (r) 2K (r)(n + r) + 2nK(r) = 0. Substituting K(r) = s=0 K sr s into (51), get a recursion relation (52) K s+1 = 2(n s) (s + 1)(2n s) K s. It follows that K(r) is a constant times (53) where n K ns x s. s=0 (54) K ns = 2s n!(2n s)! s!(2n)!(n s)! and k is a constant. The polynomials K n (r) = n s=0 K nsr s for n = 0, 1, 2 are given by n K n (r) r r + r 2 /3 Table 1. Thus the functions e r K n (r)r 2n 1 P n (r) where P n is a homogeneous polynomial of degree n satisfy the

14 14 reduced Poisson-Boltzmann equation (47), for example (55) e r r 1 A e r (1 + r)r 3 (Ax + By + Cz) ( ) e r 1 + r + r2 r 5 (Ax 2 + By 2 (A + B)z 2 + Cxy + Dxz + Eyz) 3 More generally, adding solutions of this type, any series of the form (56) e r K n (r)r 2n 1 P n (r) n=0 where P n is a homogeneous polynomial of degree n, gives a solution of (47) vanishing infinity. Any such polynomial P n can be written as the sum of 2n + 1 spherical harmonics n m n. r n P m n (cos ϑ)e imϕ, 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) Y m n (ϑ, ϕ) = P m n (cos ϑ)e imϕ. These are the spherical harmonics without normalization constants.

15 notation region description potential dielectric R 1 r < R inside protein V 1 = ψ 1 + ψ ε i R 2 R < r < a outside protein, ion free V 2 ε R 3 a < r ionic solution V 3 ε Table 2. Regions and potentials in the Tanford-Kirkwood theory 15 (1) ψ 1 is sum of Coulomb potentials from charges in R 1. ψ 1 (r) = 1 q k 4πε i ε 0 r r k. Using (44) there is an expansion of the form n ψ 1 (r) = E nm r n 1 Yn m (ϑ, ϕ) for r near R. There is an explicit expression for E nm in terms of the q k and the r k giving the charges and their locations. (2) ψ is potential in R 1 arising from charges in R 3. It satisfies Laplace s equation and has expansion ψ(r) = n k B nm r n Y m n (ϑ, ϕ) (3) V 1 is the total potential in R 1 n V 1 (r) = ψ 1 (r)+ψ(r) = (E nm r n 1 +B nm r n )Yn m (ϑ, ϕ) (4) V 2 is potential in R 2 arising from charges in R 1 and R 3. It satisfies Laplace s equation and has

16 16 expansion V 2 (r) = n (C nm r n 1 + G nm r n )Yn m (ϑ, ϕ) (5) V 3 is the potential in the ionic region R 3. It satisfies 2 V 3 = κ 2 V 3 and can be expanded in the form V 3 (r) = n A nm e κr K n (κr)r n 1 Y m n (ϑ, ϕ) The coefficients A nm, B nm, C nm, G nm 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 V 1 = V 2 ε i r V 1 = εr V 2 r = a V 2 = V 3 r V 2 = r V 3 Table 3. Boundary conditions for potentials To compute the boundary conditions, note that r (r n Y m n ) = n(r n Y m n ) r (r n 1 Y m n ) = ( n 1)(r n 1 Y m n )

17 so 17 r V 1 (r) = r V 2 (r) = r V 3 (r) = n [ (n + 1)r n 1 E nm + nr n B nm ]Ym n n n [ (n + 1)r n 1 C nm + nr n G nm )]Y n m A nm e κr [ (κr + n + 1)K n (κr) + κrk (κr)]r Now equating the coefficients of Y n m, the boundary conditions become E nm + B nm R 2n+1 = C nm + G nm R 2n+1 ε i [ (n + 1)E nm + nr 2n+1 B nm ] = ε[ (n + 1)C nm + nr 2 C nm + G nm a 2n+1 = A nm e κa K n (κa) A nm e κa [ (κa + n + 1)K n (κa) + κak (κa)] = (n + 1)C nm + na 2n+ This, assuming that E is known, this gives four linear equations in B, C, A, G. The matrix of coefficients is (58) R n R n+1 ε i nr n+1 ε i ε(n + 1) 0 nr n K n (κa) a 2n+1 0 n + 1 e κa (2n + 1)K n+1 (κa) na 2n+1

18 18 and the right side of the equation is given by the column matrix E mn (59) ε i (n + 1)E mn 0. 0 Solve for the coefficients B mn to get the potential ψ given by ψ(r) = n B nm r n Y m n (ϑ, ϕ). The energy of the interactions between the charges q k and the ions is then k q k ψ(r k ).

Theoretische Physik 2: Elektrodynamik (Prof. A-S. Smith) Home assignment 11

Theoretische Physik 2: Elektrodynamik (Prof. A-S. Smith) Home assignment 11 WiSe 22..23 Prof. Dr. A-S. Smith Dipl.-Phys. Matthias Saba am Lehrstuhl für Theoretische Physik I Department für Physik Friedrich-Alexander-Universität Erlangen-Nürnberg Problem. Theoretische Physik 2: