Classical Field Theory: Electrostatics-Magnetostatics April 27, 20101 1J.D.Jackson, "Classical Electrodynamics", 2nd Edition, Section 1-5 Classical Field Theory: Electrostatics-Magnetostatics Electrostatics The behavior of an electrostatic field can be described by two differential equations: r~ · E~ = 4πρ (1) (Gauss law) and r~ × E~ = 0 (2) the latter equation being equivalent to the statement that E~ is the gradient of a scalar function, the scalar potentialΦ: E~ = −r~ Φ (3) Eqns (1) and (3) can be combined into one differential equation for a single scalar functionΦ( ~x): r2Φ = −4πρ (4) This equation is called Poisson equation. In the regions of space where there is no charge density, the scalar potential satisfies the Laplace equation: r2Φ = 0 (5) Classical Field Theory: Electrostatics-Magnetostatics For a general distribution ρ(~x 0), the potential is expected to be the sum over all increments of charge d 3x 0ρ(~x 0), i.e., Z ρ(~x 0) Φ(~x) = d 3x 0 (6) j~x − ~x 0j This potential should satisfy Poisson's equation. But does it? If we operate with r2 on both sides of (6) we get (on ~x not on ~x 0) Z 1 r2Φ(~x) = ρ(~x 0)d 3x 0r2 (7) j~x − ~x 0j But r2(1=j~x − ~x 0j) = 0 as long as ~x 6= ~x 0 !(Why ?) The singular nature of r2(1=j~x − ~x 0j) = r2(1=r) can be best expressed in terms of the Dirac δ-function. Since r2(1=r) = 0 for r 6= 0 and its volume integral is −4π (Why?) we can write 1 r2 = −4πδ(j~x − ~x 0j) (8) j~x − ~x 0j Classical Field Theory: Electrostatics-Magnetostatics By definition, if the integration volume contains the point ~x = ~x 0 Z δ3(~x − ~x 0)d 3x = 1 otherwise is zero. This way we recover Poisson' s equation 2 0 r Φ(~x) = −4πρ(~x )~x0=~x (9) Thus, we have not only shown that the potential from Coulomb's law satisfies Poisson's eqn, but we have established (through the solution of Poisson's eqn) the important result that : the potential from a distributed source is the superposition of the individual potentials from the constituent parcels of charge. We may consider situations in which ρ is comprised of N discrete charges 0 qi , positioned at ~xi so that N 0 X 3 0 0 ρ(~x ) = qi δ (~x − ~xi ) (10) i=1 In this case the solution for the potential is a combination of terms proportional to1 =j~x − ~x 0j. Classical Field Theory: Electrostatics-Magnetostatics Green Theorem If in the electrostatic problem involved localized discrete or continuous distributions of charge with no boundary surfaces, the general solution (6) would be the most convenient and straight forward solution to any problem. To handle the boundary conditions it is necessary to develop some new mathematical tools, namely, the identities or theorems due to George Green (1824). These follow as simple applications of the divergence theorem Z I r~ · A~ d 3x = A~ · ~n da (11) V S which applies to any well-behaved vector field A~ defined in the volume V bounded by the closed surface S. Let A~ = Φr~ ,(Φ and arbitrary scalar fields). Then r~ · (Φr~ ) = Φr2 + r~ Φ · r~ (12) @ Φr~ · ~n = Φ (13) @n where @=@n is the normal derivative at the surface S. Classical Field Theory: Electrostatics-Magnetostatics When (12) and (13) substituted into the divergence theorem (8) produces the so-called Green's 1st identity Z I @ Φr2 + r~ Φ · r~ d 3x = Φ da : (14) V S @n If we rewrite (14) withΦ and interchanged, and then subtract it from (14) we obtain Green's 2nd identity or Green's Theorem: Z I @ @Φ Φr2 − r2Φ d 3x = Φ − da (15) V S @n @n Now we can apply Poisson's equation (8) for discrete charge, substituting for = 1=j~x − ~x 0j Z 4πρ(~x 0) 3 ~ ~ 0 ~ 0 3 −4πδ (x − x )Φ(x ) + 0 d x V j~x − ~x j I @ 1 1 @Φ 0 = Φ 0 0 − 0 0 da (16) S @n j~x − ~x j j~x − ~x j @n Classical Field Theory: Electrostatics-Magnetostatics Integrating the Dirac delta function over all values of ~x 0 within V and for ~x within the volume V yields a nonzero result Z ρ(~x 0) 1 I 1 @Φ @ 1 ~ 3 0 0 Φ(x) = 0 d x + 0 0 − Φ 0 0 da V j~x − ~x j 4π S j~x − ~x j @n @n j~x − ~x j (17) • The (blue) correction term goes to zero as the surface S goes to infinity (because S falls of faster than1 =j~x − ~x 0j) • If the integration volume is free of charges, then the first term of equation (17) becomes zero, and the potential is determined only by the values of the potential and the values of its derivatives at the boundary of the integration region (the surface S). Classical Field Theory: Electrostatics-Magnetostatics • Physical experience leads us to believe that specification of the potential on a closed surface defines a unique potential problem. This is called Dirichlet problem or Dirichlet boundary conditions. • Similarly it is plausible that specification of the electric field (normal derivative of the potential) everywhere on the surface (corresponding to a given surface-charge density) also defines a unique problem. The specification of the normal derivative is known as the Newmann boundary condition. • As it turns out either one of the two conditions results in unique solution. It should be clear that a solution to the Poisson eqn with both Φ and @Φ=@n specified arbitrarily on a closed boundary (Cauchy boundary conditions) does not exist since there are unique solutions for Dirichlet and Newmann condition separately. Classical Field Theory: Electrostatics-Magnetostatics Green Functions The solution of the Poisson or Laplace eqn in a finite volume V with either Dirichlet or Neumann boundary conditions on the boundary surface S can be obtained by means of Green's theorem (15) and the so-called Green functions. In obtaining the result (17) we have chosen = 1=j~x − ~x 0j satisfying 1 r2 = −4πδ(j~x − ~x 0j) (18) j~x − ~x 0j The function1 =j~x − ~x 0j is only one of a class of functions depending on the variables ~x and ~x 0 and called Green functions. In general r02G (~x;~x 0) = −4πδ(j~x − ~x 0j) (19) where 1 G (~x;~x 0) = + F (~x;~x 0) (20) j~x − ~x 0j and F satisfying the Laplace equation inside the volume V r02F (~x;~x 0) = 0 (21) Classical Field Theory: Electrostatics-Magnetostatics If we substitute G(~x;~x 0) in eqn (17) we get Z 1 I @Φ @ ~ ~ 0 ~ ~ 0 3 0 ~ ~ 0 ~ 0 ~ ~ 0 0 Φ(x) = ρ(x )G(x; x )d x + G(x; x ) 0 − Φ(x ) 0 G(x; x ) da V 4π S @n @n (22) The freedom in the definition of G means that we can make the surface integral depend only on the chosen type of BC. • For the Dirichlet BC we demand: 0 0 GD (~x;~x ) = 0 for ~x 2 S (23) Then the 1st term on the surface integral of (22) vanishes Z 1 I @ ~ ~ 0 ~ ~ 0 3 0 ~ 0 ~ ~ 0 0 Φ(x) = ρ(x )GD (x; x )d x − Φ(x ) 0 GD (x; x )da (24) V 4π S @n • For Neumann BC the simplest choice of BC on G is @G N (~x;~x 0) = 0 for ~x 0 2 S (25) @n0 but application o the Gauss theorem on (19) shows that (how?) I @GN 0 0 da = −4π 6= 0 S @n which is incosistent with r02G(~x;~x 0) = −4πδ3(~x − ~x 0). Classical Field Theory: Electrostatics-Magnetostatics This will mean that the outflux of G cannot be zero when there is a source enclosed by S. Then the simplest boundary condition that we can use is @G 4π N (~x;~x 0) = − for ~x 0 2 S (26) @n0 S S is the total area of the boundary surface. Then the solution will be: Z 1 I @Φ ~ ~ 0 ~ ~ 0 3 0 ~ ~ 0 0 Φ(x) = ρ(x )GN (x; x )d x + hΦiS + 0 GN (x; x )da (27) V 4π S @n where hΦiS is the average value of the potential over the whole surface I 1 0 0 hΦiS ≡ Φ(~x )da (28) S S In most cases S is extremely large (or even infinite), in which case hΦiS ! 0. The physical meaning of F (~x;~x 0) : it is a solution of the Laplace eqn inside V and so represents the potential of charges external to the volume V chosen as to satisfy the homogeneous BC of zero potential on the surface S. Classical Field Theory: Electrostatics-Magnetostatics Green Function • It is important to understand that no matter how the source is distributed, finding the Green function is completely independent or ρ(~x 0). • G(~x;~x 0) depends exclusively on the geometry of the problem, is a \template", potential and not the actual potential for a given physical problem. • In other words G(~x;~x 0) is the potential due to a unit charge, positioned arbitrarily within the surface S consistent either with GD = 0 or @GN =@n = −4π=S on the surface. • The \true potential" is a convolution of this template with the given ρ(~x 0). Classical Field Theory: Electrostatics-Magnetostatics Laplace Equation in Spherical Coordinates In spherical coordinates( r; θ; φ) the Laplace equation can be written in the form 1 @2 1 @ @Φ 1 @2Φ (rΦ) + sin θ + = 0 (29) r 2 @r 2 r 2 sin θ @θ @θ r 2 sin2 θ @φ2 If we assume 1 Φ = U(r)P(θ)Q(φ) (30) r Then by substituting in (29) and multiplying with r 2 sin2 θ=UPQ we obtain 1 d 2U 1 d dP 1 d 2Q r 2 sin2 θ + sin θ + = 0 (31) U dr 2 r 2 sin θP dθ dθ Q dφ2 We see that the therms depending on φ have been isolated and we can set 1 d 2Q = −m2 (32) Q dφ2 with solution Q = e±imφ (33) Classical Field Theory: Electrostatics-Magnetostatics Similarly the remaining terms can be separated as: 1 d dP m2 sin θ + l(l + 1) − P = 0 (34) r 2 sin θ dθ dθ sin2 θ d 2U l(l + 1) − U = 0 (35) dr 2 r 2 The radial equation will have a solution U = Ar l+1 + Br −l (36) while l is still undetermined.