Classical Field Theory: Electrostatics-Magnetostatics

Transcription

1 Classical Field Theory: Electrostatics-Magnetostatics April 27, J.D.Jackson, Classical Electrodynamics, 2nd Edition, Section 1-5

2 Electrostatics The behavior of an electrostatic field can be described by two differential equations: E = 4πρ (1) (Gauss law) and E = 0 (2) the latter equation being equivalent to the statement that E is the gradient of a scalar function, the scalar potential Φ: E = Φ (3) Eqns (1) and (3) can be combined into one differential equation for a single scalar function Φ( x): 2 Φ = 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: 2 Φ = 0 (5)

3 For a general distribution ρ( x ), the potential is expected to be the sum over all increments of charge d 3 x ρ( x ), i.e., Φ( x) = ρ( x ) x x d 3 x (6) This potential should satisfy Poisson s equation. But does it? If we operate with 2 on both sides of (6) we get (on x not on x ) ( ) 1 2 Φ( x) = ρ( x )d 3 x 2 x x (7) But 2 (1/ x x ) = 0 as long as x x! (Why?) The singular nature of 2 (1/ x x ) = 2 (1/r) can be best expressed in terms of the Dirac δ-function. Since 2 (1/r) = 0 for r 0 and its volume integral is 4π (Why?) we can write ( ) 1 2 x x = 4πδ( x x ) (8)

4 By definition, if the integration volume contains the point x = x δ 3 ( x x )d 3 x = 1 otherwise is zero. This way we recover Poisson s equation 2 Φ( x) = 4πρ( x ) x = 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 q i, positioned at x i so that ρ( x ) = N i=1 q i δ 3 ( x x i ) (10) In this case the solution for the potential is a combination of terms proportional to 1/ x x.

5 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 A d 3 x = A n da (11) V which applies to any well-behaved vector field A defined in the volume V bounded by the closed surface S. Let A = Φ ψ, (Φ and ψ arbitrary scalar fields). Then S (Φ ψ) = Φ 2 ψ + Φ ψ (12) Φ ψ n = Φ ψ n where / n is the normal derivative at the surface S. (13)

6 When (12) and (13) substituted into the divergence theorem (8) produces the so-called Green s 1st identity ( Φ 2 ψ + Φ ψ ) d 3 x = Φ ψ da. (14) n V If we rewrite (14) with Φ and ψ interchanged, and then subtract it from (14) we obtain Green s 2nd identity or Green s Theorem: ( Φ 2 ψ ψ 2 Φ ) [ d 3 x = Φ ψ n ψ Φ ] da (15) n V Now we can apply Poisson s equation (8) for discrete charge, substituting for ψ = 1/ x x [ 4πδ 3 ( x x )Φ( x ) + 4πρ( x ] ) V x x d 3 x [ = Φ ( ) ] 1 1 Φ n x x x x n da (16) S S S

7 Integrating the Dirac delta function over all values of x within V and for x within the volume V yields a nonzero result ρ( x ) Φ( x) = V x x d 3 x + 1 [ 1 Φ 4π S x x n Φ ( )] 1 n x x da (17) The (blue) correction term goes to zero as the surface S goes to infinity (because S falls of faster than 1/ x x ) 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).

8 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.

9 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 ψ = 1/ x x satisfying 2 ( 1 x x ) = 4πδ( x x ) (18) The function 1/ x x is only one of a class of functions depending on the variables x and x and called Green functions. In general 2 G ( x, x ) = 4πδ( x x ) (19) where G ( x, x ) = 1 x x + F ( x, x ) (20) and F satisfying the Laplace equation inside the volume V 2 F ( x, x ) = 0 (21)

10 If we substitute G( x, x ) in eqn (17) we get Φ( x) = ρ( x )G( x, x )d 3 x + 1 4π V S [ G( x, x ) Φ n Φ( x ) n G( x, x ) (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: V G D ( x, x ) = 0 for x S (23) Then the 1st term on the surface integral of (22) vanishes Φ( x) = ρ( x )G D ( x, x )d 3 x 1 Φ( x ) 4π n G D( x, x )da (24) For Neumann BC the simplest choice of BC on G is G N n ( x, x ) = 0 for x S (25) but application o the Gauss theorem on (19) shows that (how?) G N n da = 4π 0 S which is incosistent with 2 G( x, x ) = 4πδ 3 ( x x ). S ] da

11 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 N n ( x, x ) = 4π for x S (26) S S is the total area of the boundary surface. Then the solution will be: Φ( x) = V ρ( x )G N ( x, x )d 3 x + Φ S + 1 4π S Φ n G N( x, x )da (27) where Φ S is the average value of the potential over the whole surface Φ S 1 Φ( x )da (28) S In most cases S is extremely large (or even infinite), in which case Φ S 0. The physical meaning of F ( x, x ) : 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. S

12 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 ). G( x, x ) 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 ) is the potential due to a unit charge, positioned arbitrarily within the surface S consistent either with G D = 0 or G N / n = 4π/S on the surface. The true potential is a convolution of this template with the given ρ( x ).

13 Laplace Equation in Spherical Coordinates In spherical coordinates (r, θ, φ) the Laplace equation can be written in the form 1 2 r 2 r 2 (rφ) + 1 ( r 2 sin θ Φ ) 1 2 Φ + sin θ θ θ r 2 sin 2 θ φ 2 = 0 (29) If we assume Φ = 1 U(r)P(θ)Q(φ) (30) r Then by substituting in (29) and multiplying with r 2 sin 2 θ/upq we obtain [ 1 r 2 sin 2 d 2 ( U θ U dr d r 2 sin θ dp )] + 1 d 2 Q sin θp dθ dθ Q dφ 2 = 0 (31) We see that the therms depending on φ have been isolated and we can set 1 d 2 Q Q dφ 2 = m2 (32) with solution Q = e ±imφ (33)

14 Similarly the remaining terms can be separated as: ( 1 d r 2 sin θ dp ) ] + [l(l + 1) m2 sin θ dθ dθ sin 2 P θ = 0 (34) d 2 U l(l + 1) dr 2 r 2 U = 0 (35) The radial equation will have a solution while l is still undetermined. U = Ar l+1 + Br l (36)

15 Legendre Equation and Legendre Polynomials The equation (34) for P(θ) can be expressed in terms of x = cos θ [ d (1 x 2 ) dp ] ] + [l(l + 1) m2 dx dx 1 x 2 P = 0 (37) This is the generalized Legendre equation and its solutions are the associated Legendre functions. We will consider the solution of (54) for m 2 = 0 d dx [ (1 x 2 ) dp dx ] + l(l + 1)P = 0 (38) The solution should be single valued, finite, and continuous on the interval 1 x 1 in order that it represents a physical potential. The solution can be found in the form of a power series P(x) = x k j=0 where k is a parameter to be determined. a j x j (39)

16 By substitution in (54) we get the recurrence relation (how?) a j+2 = while for j = 0, 1 we find that (k + j)(k + j + 1) l(l + 1) a j (40) (k + j + 1)(k + j + 2) if a 0 0 then k(k 1) = 0 (41) if a 1 0 then k(k + 1) = 0 (42) These two relations are equivalent and it is sufficient to choose either a 0 or a 1 different from zero but not both (why?). We also see that the series expansion is either only odd or only on even powers of x. By choosing either k = 0 or k = 1 it is possible to prove the following properties: The series converges for x 2 < 1, independent of the value of l The series diverges for x = ±1, unless it terminates. Since we want solution that is finite at x = ±1, as well as for x 2 < 1, we demand that the series terminates. Since k and j are positive integers or zero, the recurrence relation (40) will terminate only if l is zero or a positive integer.

17 If l is even (odd), then only the k = 0 (k = 1) series terminates. The polynomials in each case have x l as their highest power. By convention these polynomials are normalized to have the value unity for x = +1 and are called the Legendre polynomials of order l. The first few are: P 0 (x) = 1 P 1 (x) = x P 2 (x) = 1 2 (3x 2 1) (43) P 3 (x) = 1 2 (5x 3 3x) P 4 (x) = 1 8 (35x 4 30x 2 + 3) The Legendre polynomials can be taken from Rodrigue s formula: d l P l (x) = 1 2 l l! dx l (x 2 1) l (44)

18 The Legendre polynomials form a complete orthonormal set of functions on the interval 1 x 1 (prove it) 1 1 P l (x)p l (x) = 2 2l + 1 δ l l (45) Since the Legendre polynomials form a complete set of orthonormal functions, any function f (x) on the interval 1 x 1 can be expanded in terms of them i.e. f (x) = A l P l (x) (46) where (how?) A l = 2l l=0 1 1 f (x)p l (x)dx (47) Thus for problems with azimuthal symmetry i.e. m = 0 the general solution is: [ Φ(r, θ) = A l r l + B l r (l+1)] P l (cos θ) (48) l=0 where the coefficients A l [it is not the same as in eqn (47)] and B l can be determined from the boundary conditions.

19 Example Boundary Value Problems with Azimuthal Symmetry Let s specify as V (θ) the potential on the surface of a sphere of radius R, and try to find the potential inside the sphere. If there are no charges at the origin (r = 0) the potential must be finite there. Consequently B l = 0 for all l. Then on the surface of the sphere V (r = R, θ) = A l R l P l (cos θ) (49) l=0 and the coefficients A l will be taken via eqn (47) A l = 2l + 1 2R l π 0 V (θ)p l (cos θ) sin θdθ (50)

20 If, for example V (θ) = ±V on the two hemispheres then the coefficients can be derived easily and the potential inside the sphere is (how?): [ 3 ( r ) Φ(r, θ) = V P 1 (cos θ) 7 ( r ) 3 P3 (cos θ) + 11 ( r ) ] 5 P5 (cos θ)... 2 R 8 R 16 R (51) To find the potential outside the sphere we merely replace (r/r) l by (R/r) l+1 and the resulting potential will be (how?): Φ(r, θ) = 3 ( ) [ 2 R V P 1 (cos θ) 7 ( ) ] 2 R P 3 (cos θ) +... (52) 2 r 12 r

21 Legendre Polynomials An important expansion is that of the potential at x due to a unit point charge at x 1 x x = l=0 r< l r> l+1 P l (cos γ) (53) where r < (r > ) is the smaller (larger) of x and x and γ is the angle between x and x. Show that the potential is : 1 x x = ( A l r l + B l r (l+1)) P l (cos γ) l=0 1 x x = 1 r > l=0 ( r< r > ) l on the x-axis on the z-axis

22 Associated Legendre Functions and Spherical Harmonics For problems without azimuthal (axial) symmetry, we need the generalization of P l (cos θ), namely, the solutions of [ d (1 x 2 ) dp ] ] + [l(l + 1) m2 dx dx 1 x 2 P = 0 (54) for arbitrary l and m. It can be shown that in order to have finite solutions on the interval 1 x 1 the parameter l must be zero or a positive integer and that the integer m can take on only the values l, (l 1),..., 0,...,(l 1), l (why?). The solution having these properties is called an associated Legendre function Pl m (x). For positive m it is defined as Pl m (x) = ( 1) m (1 x 2 ) m/2 d m dx m P l(x) (55) If Rodrigues formula is used an expression valid for both positive and negative m is obtained: P m l (x) = ( 1)m 2 l (1 x 2 ) m/2 d m l! dx m P l(x) (56)

23 There is a simple relation between P m l (x) and P m l (x) : P m m (l m)! l (x) = ( 1) (l + m)! Pm l (x) (57) For fixed m the functions Pl m (x) form an orthonormal set in the index l on the interval 1 x 1. The orthogonality relation is 1 1 Pl m (x)pm l (x)dx = 2 (l + m)! 2l + 1 (l m)! δ l l (58) We have found that Q m (φ) = e imφ, this function forms a complete set of orthogonal functions in the index m on the interval 0 φ 2π. The product Pl m Q m forms also a complete orthonormal set on the surface of the unit sphere in the two indices l, m. From the normalization condition (58) we can conclude that the suitably normalized functions, denoted by Y lm (θ, φ), are : 2l + 1 (l m)! Y lm (θ, φ) = 4π (l + m)! Pm l (cos θ)e imφ (59) and also Y l, m (θ, φ) = ( 1) m Y lm(θ, φ) (60)

24 The normalization and orthogonality conditions are: 2π 0 dφ π 0 sin θdθy l m (θ, φ)y lm(θ, φ) = δ l lδ m m (61) An arbitrary function g(θ, φ) can be expanded in spherical harmonics g(θ, φ) = where the coefficients are A lm = l l=0 m= l A lm Y lm (θ, φ) (62) dωy lm(θ, φ)g(θ, φ). (63) The general solution for a boundary-value problem in spherical coordinates can be written in terms of spherical harmonics and powers of r in a generalization of (48) : Φ(r, θ, φ) = l l=0 m= l [ A lm r l + B lm r (l+1)] Y lm (θ, φ) (64) If the potential is specified on a spherical surface, the coefficients can be determined by evaluating (64) on the surface and using (63).

25 Spherical Harmonics Y lm (θ, φ) 1 l = 0 Y 00 = 4π 3 l = 1 Y 11 = sin θeiφ 8π 3 Y 10 = 4π cos θ l = 2 Y 22 = π sin2 θe 2iφ 15 Y 21 = sin θ cos θ eiφ 8π ( 5 3 Y 20 = 4π 2 cos2 θ 1 ) 2

26 Spherical Harmonics Y lm (θ, φ) l = 3 Y 33 = π sin3 θ e 3iφ Y 32 = π sin2 θ cos θe 2iφ Y 31 = π sin θ(5 cos2 θ 1) e iφ Y 30 = π sin θ ( 5 cos 3 θ 3 cos θ ) (65)

27 Figure: Schematic representation of Ylm on the unit sphere. Ylm is equal to 0 along m great circles passing through the poles, and along l m circles of equal latitude. The function changes sign each time it crosses one of these lines.

28 Addition Theorem for Spherical Harmonics The spherical harmonics are related to Legendre polynomials P l by a relation known as the addition theorem. The addition theorem expresses a Legendre polynomial of order l in the angle γ in terms of products of the spherical harmonics of the angles θ, φ and θ, φ : P l (cos γ) = 4π l(l + 1) l m= l Y lm(θ, φ )Y lm (θ, φ) (66) where γ is the angle between the vectors x and x, x x = x x cos γ and cos γ = cos θ cos θ + sin θ sin θ cos(φ φ ). P l (cos γ) is a function of the angles θ, φ with the angles θ, φ as parameters and it maybe expanded in a series (63) : P l (cos γ) = l A l m (θ, φ )Y l m (θ, φ) (67) m= l l =0

29 The addition theorem offer the possibility to extend the expansion valid for a point charge (axially symmetric distribution) to an arbitrary charge distribution. By substituting (66) into (53) we obtain 1 x x = 4π l l=0 m= l 1 2l + 1 r< l r> l+1 Y lm(θ, φ )Y lm (θ, φ) (68) This equation gives the potential in a completely factorized form in the coordinates x and x. This is useful in any integration over charge densities, when one is the variable of integration and the other the observation point.

30 Multipole Expansion A localized distribution of charge is described by the charge density ρ( x ), which is nonvanishing only inside a sphere a around some origin. The potential outside the sphere can be written as an expansion in spherical harmonics Φ( x) = l 4π 2l + 1 q 1 lm r l+1 Y lm(θ, φ) (69) l=0 m= l This type of expansion is called multipole expansion; The l = 0 term is called monopole term, The l = 1 are called dipole terms etc. The problem to be solved is the determination of the constants q lm in terms of the properties of the charge distribution ρ( x ). The solution is very easily obtained from the integral for the potential Φ( x) = ρ( x ) x x d 3 x (70)

31 Using the expansion (68) for 1/ x x i.e. Φ( x) = l l=0 m= l 4π 1 2l + 1 r l+1 Y lm(θ, φ) Y lm(θ, φ )ρ( x )r l d 3 x (71) Consequently the coefficients in (69) are : q lm = Ylm(θ, φ )ρ( x )r l d 3 x (72) and called the multipole moments of the charge distribution ρ( x )

32 Monopole moment l = 0 Here, the only component is q 00 = ρ( x )r 0 Y00(θ, φ )d 3 x = 1 4π ρ( x )d 3 x = q 4π (73) Observed from a large distance r, any charge distribution acts approximately as if the total charge q (monopole moment) would be concentrated at one point since the dominant term in (68) Φ( x) = 4πq00 1 r Y 00 + = q r +... (74) NOTE: The moments with m 0 are connected (for real charge density) too the moments with m < 0 through q l, m = ( 1) m q lm (75)

33 Dipole moment l = 1 In Cartesian coordinates the dipole moment is given by p = x ρ( x )d 3 x (76) In Spherical representation one obtains q11 = ρ( x )r Y11(θ, φ )d 3 x 3 = ρ( x )r (sin θ cos φ i sin θ sin φ ) d 3 x 8π in Cartesian represantation 3 3 = ρ( x )(x iy )d 3 x = 8π 8π (p x ip y ) (77) also q 10 = = ρ( x )r Y 10(θ, φ )d 3 x = 3 4π z ρ( x )d 3 x = 3 ρ( x )r cos θ d 3 x 4π 3 4π p z (78)

34 Quadrupole moment l = 2 q22 = ρ( x)r 2 Y22(θ, φ )d 3 x = 1 15 ρ( x) [r sin θ (cos φ i sin φ )] 2 d 3 x 4 2π = 1 15 ρ( x)(x iy ) 2 d 3 x 4 2π because (x iy ) 2 = 1 [ (3x 2 r 2 ) 6ix y (3y 2 r 2 ) ] 3 = π (Q 11 2iQ 12 Q 22 ) (79) where Q ij is the traceless (why?) quadrupole moment tensor: (3x Q ij = i x j r 2 ) δ ij ρ( x )d 3 x (80)

35 Quadrupole moment l = 2 Analogously 15 q21 = ρ( x )r 2 Y21(θ, φ )d 3 x = ρ( x )z (x iy )d 3 x 8π = π (Q 13 iq 23 ) (81) and q 20 = = 1 2 ρ( x )r 2 Y20(θ, φ )d 3 x = π Q π From eqn (60) we can get the moments with m < 0 through ρ( x )(3z r 2 )d 3 x (82) q l, m = ( 1) m q lm (83)

36 Multipole Expansion By direct Taylor expansion of 1/ x x the expansion of Φ( x) in rectangular coordinates is: Φ( x) = q p x + r r x i x j Q ij 2 r (84) The electric field components for a given multipole can be expressed most easily in terms of spherical coordinates. From E = Φ and (69) for fixed (l, m) we get(how?): i,j 4π(l + 1) E r = 2l + 1 q 1 lm r l+2 Y lm E θ = 4π 2l + 1 q 1 lm r l+2 θ Y lm (85) E φ = 4π 2l + 1 q 1 1 lm r l+2 sin θ φ Y lm For a dipole p along the z-axis they reduce to: E r = 2p cos θ r 3, E θ = p sin θ r 3, E φ = 0 (86)

37 Multipole Expansion The field intensity at a point x due to a dipole p at a point x 0 (r = x x 0 ) is E( x) = 3 n ( p n) p x x 0 3 with n = r r (87) because Φ( x) = p x/r 3 and ( ) E( x) = Φ ( p x) 1 3 n ( p n) p = r 3 ( p x) r 3 = x x 0 3 (88)

38 Energy of a Charge Distribution in an External Field The multipole expansion of the potential of a charge distribution can also be used to describe the interaction of the charge distribution with an external field. The electrostatic energy of the charge distribution ρ( x) placed in an external field Φ( x) is given by W = ρ( x)φ( x)d 3 x (89) V The external field (if its is slowly varying over the region where ρ( x) is non-negligible) may be expanded in a Taylor series: Φ( x) = Φ(0) + x Φ(0) Φ x i x j (0) +... (90) 2 x i x j Since E = Φ for the external field Φ( x) = Φ(0) x E(0) 1 2 i=1 j=1 3 3 i=1 j=1 x i x j E j x i (0) +... (91)

39 Since E = 0 for the external field we can substract 1 6 r 2 E(0) from the last term to obtain finally the expansion: Φ( x) = Φ(0) x E(0) i=1 j=1 ( 3xi x j r 2 δ ij ) E j x i (0) +... (92) When this inserted into (89) the energy takes the form: W = qφ(0) p E(0) i=1 j=1 Q ij E j x i (0) +... (93) Notice, that: the total charge interacts with the potential, the dipole moment with the electric field, the quadrupole with the electric field gradient, etc

40 Examples EXAMPLE 1 : Show that for a uniform charged sphere all multipole moments vanish except q 00. If the sphere has a radius R 0 and constant charge density ρ( x) = ρ then R0 qlm = ρ r l Y lm(θ, φ )r 2 dr dω = ρ Rl+3 0 Ω 0 l + 3 Ylm(θ, φ )dω Ω but since Y 00 = 1/ 4π we have from the orthogonality relation qlm = ρ Rl+3 0 4π Y l + 3 lmy 00 dω = ρ Rl+3 0 4πδl0 δ m0 Ω l + 3

41 EXAMPLE 2: Perform multipole decomposition of a uniform charge distribution whose surface is a weakly deformed sphere: ) 2 R = R 0 (1 + a2my 2m (θ, φ), a 2m << 1 m= 2 The multipole moments are R(θ,φ ) qlm = ρr l Ylmr 2 dr dω Ω 0 = ρ Rl+3 0 l + 3 Ylm(θ, φ ) Ω ( m= 2 a 2mY 2m(θ, φ ) ) l+3 dω

42 ( qlm ρ Rl+3 0 Y l + 3 lm(θ, φ ) 1 + (l + 3) Ω 2 m= 2 a 2mY 2m(θ, φ ) Apart from the monopole moment q 00 of the previous example, we find qlm = ρ Rl+3 0 l + 3 (l+3) 2 µ= 2 a 2µ Ω Y ) dω lmy 2µ dω = ρr l+3 a2µδ mµ δ l2 (l > 0) Thus the nonvanishing multipole moments are : q 2m = ρr5 0 a 2m. 0 µ

43 Magnetostatics There is a radical difference between magnetostatics and electrostatics : there are no free magnetic charges The basic entity in magnetic studies is the magnetic dipole. In the presence of magnetic materials the dipole tends to align itself in a certain direction. That is by definition the direction of the magnetic flux density, denoted by B (also called magnetic induction). The magnitude of the flux density can be defined by the magnetic torque N exerted on the magnetic dipole: N = µ B (94) where µ is the magnetic moment of the dipole. In electrostatics the conservation of charge demands that the charge density at any point in space be related to the current density in that neighborhood by the continuity equation ρ/ t + J = 0 (95) Steady-state magnetic phenomena are characterized by no change in the net charge density anywhere in space, consequently in magnetostatics J = 0 (96)

44 Biot & Savart Law If d l is an element of length (pointing in the direction of current flow) of a wire which carries a current I and x is the coordinate vector from the element of length to an observation point P, then the elementary flux density d B at the point P is given in magnitude and direction by d B = ki d l x x 3 (97) NOTE: (97) is an inverse square law, just as is Coulomb s law of electrostatics. However, the vector character is very different. If we consider that current is charge in motion and replace Id l by q v where q is the charge and v the velocity. The flux density for such a charge in motion would be v x B = kq x 3 (98) This expression is time-dependent and valid only for charges with small velocities compared to the speed of light.

45 Diff. Equations for Magnetostatics & Ampere s Law The basic law (97) for the magnetic induction can be written down in general form for a current density J( x): B( x) = 1 J( x ) ( x x ) c x x 3 d 3 x (99) This expression for B( x) is the magnetic analog of electric field in terms of the charge density: E( x) = 1 ρ( x ) ( x x ) c x x 3 d 3 x (100) In order to obtain DE equivalent to (99) we use the relation ( x x ( ) ( ) ) 1 x x 3 = x x as 1 = r r r 3 (101) and (99) transforms into B( x) = 1 c J( x ) x x d 3 x (102)

46

47 From (102) follows immediately: B = 0 (103) This is the 1st equation of magnetostatics and corresponds to E = 0 in electrostatics. By analogy with electrostatics we now calculate the curl of B B = 1 c J( x ) x x d 3 x (104) which for steady-state phenomena ( J = 0) reduces to (how?) 2 B = 4π c J (105) This is the 2nd equation of magnetostatics and corresponds to E = 4πρ in electrostatics. 2 Remember : A = A 2 A

48 The integral equivalent of (105) is called Ampere s law It can be obtained by applying Stokes s theorem to the integral of the normal component of (105) over the open surface S bounded by a closed curve C. Thus B n da = 4π J nda (106) c S is transformed into B d l = 4π c C S S J nda (107) Since the surface integral of the current density is the total current I passing through the closed curve C, Ampére s law can be written in the form: B d l = 4π c I (108) C

49 Vector Potential The basic differential laws in magnetostatics are B = 4π c J and B = 0 (109) The problems is how to solve them. If the current density is zero in the region of interest, B = 0 permits the expression of the vector magnetic induction B as the gradient of a magnetic scalar potential B = Φ M, then (109) reduces to the Laplace equation for Φ M. If B = 0 everywhere, B must be the curl of some vector field A( x), called the vector potential and from (102) the general form of A is: A( x) = 1 c B( x) = A( x) (110) J( x ) x x d 3 x + Ψ( x) (111)

50 The added gradient of an arbitrary scalar function Ψ shows that, for a given magnetic induction B, the vector potential can be freely transformed according to A A + Ψ (112) This transformation is called a gauge transformation. Such transformations are possible because (110) specifies only the curl of A. If (110) is substituted into the first equation in (109), we find ( ) A = 4π c ( ) J or A 2 A 4π = c J (113) If we exploit the freedom implied by (112) we can make the convenient choice of gauge (Coulomb gauge) A = 0. Then each component of the vector potential satisfies the Poisson equation 2 A = 4π c J (114) The solution for A in unbounded space is (111) with Ψ =constant: A( x) = 1 J( x ) c x x d 3 x (115)

51 Further Reading Green Function Greiner W. Classical Electrodynamics, Chapter 1, 2, 3, 4 Jackson J.K Classical Electrodynamics, 2nd Edition, Sections , , ,