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Classical Field Theory:ElectrostaticsMagnetostatics
April 27, 20101
1J.D.Jackson, Classical Electrodynamics, 2nd Edition, Section 15Classical Field Theory: ElectrostaticsMagnetostatics

Electrostatics
The behavior of an electrostatic field can be described by two differentialequations:
~ ~E = 4 (1)(Gauss law) and
~ ~E = 0 (2)the latter equation being equivalent to the statement that ~E is thegradient of a scalar function, the scalar potential :
~E = ~ (3)
Eqns (1) and (3) can be combined into one differential equation for asingle scalar function (~x):
2 = 4 (4)
This equation is called Poisson equation.In the regions of space where there is no charge density, the scalarpotential satisfies the Laplace equation:
2 = 0 (5)
Classical Field Theory: ElectrostaticsMagnetostatics

For a general distribution (~x ), the potential is expected to be the sumover all increments of charge d3x (~x ), i.e.,
(~x) =
(~x )
~x ~x d3x (6)
This potential should satisfy Poissons equation. But does it? If weoperate with 2 on both sides of (6) we get (on ~x not on ~x )
2(~x) =(~x )d3x 2
(1
~x ~x 
)(7)
But 2(1/~x ~x ) = 0 as long as ~x 6= ~x ! (Why ?)The singular nature of 2(1/~x ~x ) = 2(1/r) can be best expressedin terms of the Dirac function.Since 2(1/r) = 0 for r 6= 0 and its volume integral is 4 (Why?) wecan write
2(
1
~x ~x 
)= 4(~x ~x ) (8)
Classical Field Theory: ElectrostaticsMagnetostatics

By definition, if the integration volume contains the point ~x = ~x 3(~x ~x )d3x = 1
otherwise is zero. This way we recover Poisson s equation
2(~x) = 4(~x )~x=~x (9)
Thus, we have not only shown that the potential from Coulombs lawsatisfies Poissons eqn, but we have established (through the solution ofPoissons eqn) the important result that :the potential from a distributed source is the superposition of theindividual potentials from the constituent parcels of charge.We may consider situations in which is comprised of N discrete chargesqi , positioned at ~x
i so that
(~x ) =N
i=1
qi3 (~x ~x i ) (10)
In this case the solution for the potential is a combination of terms
proportional to 1/~x ~x .Classical Field Theory: ElectrostaticsMagnetostatics

Green Theorem
If in the electrostatic problem involved localized discrete or continuousdistributions of charge with no boundary surfaces, the general solution(6) would be the most convenient and straight forward solution to anyproblem.To handle the boundary conditions it is necessary to develop some newmathematical tools, namely, the identities or theorems due to GeorgeGreen (1824). These follow as simple applications of the divergencetheorem
V
~ ~A d3x =
S
~A ~n da (11)
which applies to any wellbehaved vector field ~A defined in the volume Vbounded by the closed surface S .Let ~A = ~, ( and arbitrary scalar fields). Then
~ (~) = 2 + ~ ~ (12)
~ ~n = n
(13)
where /n is the normal derivative at the surface S .
Classical Field Theory: ElectrostaticsMagnetostatics

When (12) and (13) substituted into the divergence theorem (8)produces the socalled Greens 1st identity
V
(2 + ~ ~
)d3x =
S
nda . (14)
If we rewrite (14) with and interchanged, and then subtract it from(14) we obtain Greens 2nd identity or Greens Theorem:
V
(2 2
)d3x =
S
[
n
n
]da (15)
Now we can apply Poissons equation (8) for discrete charge, substitutingfor = 1/~x ~x 
V
[43(~x ~x )(~x ) + 4(
~x )
~x ~x 
]d3x
=
S
[
n
(1
~x ~x 
) 1~x ~x 
n
]da (16)
Classical Field Theory: ElectrostaticsMagnetostatics

Integrating the Dirac delta function over all values of ~x within V and for~x within the volume V yields a nonzero result
(~x) =
V
(~x )
~x ~x d3x +
1
4
S
[1
~x ~x 
n
n
(1
~x ~x 
)]da
(17) The (blue) correction term goes to zero as the surface S goes toinfinity (because S falls of faster than 1/~x ~x ) If the integration volume is free of charges, then the first term ofequation (17) becomes zero, and the potential is determined only by thevalues of the potential and the values of its derivatives at the boundaryof the integration region (the surface S).
Classical Field Theory: ElectrostaticsMagnetostatics

Physical experience leads us to believe that specification of thepotential on a closed surface defines a unique potential problem. This iscalled Dirichlet problem or Dirichlet boundary conditions. Similarly it is plausible that specification of the electric field (normalderivative of the potential) everywhere on the surface (corresponding to agiven surfacecharge density) also defines a unique problem. Thespecification of the normal derivative is known as the Newmannboundary condition. As it turns out either one of the two conditions results in uniquesolution. It should be clear that a solution to the Poisson eqn with both and /n specified arbitrarily on a closed boundary (Cauchyboundary conditions) does not exist since there are unique solutions forDirichlet and Newmann condition separately.
Classical Field Theory: ElectrostaticsMagnetostatics

Green Functions
The solution of the Poisson or Laplace eqn in a finite volume V witheither Dirichlet or Neumann boundary conditions on the boundary surfaceS can be obtained by means of Greens theorem (15) and the socalledGreen functions.In obtaining the result (17) we have chosen = 1/~x ~x  satisfying
2(
1
~x ~x 
)= 4(~x ~x ) (18)
The function 1/~x ~x  is only one of a class of functions depending onthe variables ~x and ~x and called Green functions.In general
2G (~x ,~x ) = 4(~x ~x ) (19)where
G (~x ,~x ) =1
~x ~x + F (~x ,~x ) (20)
and F satisfying the Laplace equation inside the volume V
2F (~x ,~x ) = 0 (21)
Classical Field Theory: ElectrostaticsMagnetostatics

If we substitute G (~x ,~x ) in eqn (17) we get
(~x) =
V
(~x )G (~x ,~x )d3x +1
4
S
[G (~x ,~x )
n (~x )
nG (~x ,~x )
]da
(22)The freedom in the definition of G means that we can make the surfaceintegral depend only on the chosen type of BC. For the Dirichlet BC we demand:
GD(~x ,~x) = 0 for ~x S (23)
Then the 1st term on the surface integral of (22) vanishes
(~x) =
V
(~x )GD(~x ,~x)d3x 1
4
S
(~x )
nGD(~x ,~x
)da (24)
For Neumann BC the simplest choice of BC on G isGNn
(~x ,~x ) = 0 for ~x S (25)
but application o the Gauss theorem on (19) shows that (how?)S
GNn
da = 4 6= 0
which is incosistent with 2G (~x ,~x ) = 43(~x ~x ).Classical Field Theory: ElectrostaticsMagnetostatics

This will mean that the outflux of G cannot be zero when there is asource enclosed by S . Then the simplest boundary condition that we canuse is
GNn
(~x ,~x ) = 4S
for ~x S (26)
S is the total area of the boundary surface. Then the solution will be:
(~x) =
V
(~x )GN(~x ,~x)d3x + S +
1
4
S
nGN(~x ,~x
)da (27)
where S is the average value of the potential over the whole surface
S 1
S
S
(~x )da (28)
In most cases S is extremely large (or even infinite), in which caseS 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 .
Classical Field Theory: ElectrostaticsMagnetostatics

Green Function
It is important to understand that no matter how the source isdistributed, finding the Green function is completely independent or (~x ). G (~x ,~x ) depends exclusively on the geometry of the problem, is atemplate, potential and not the actual potential for a given physicalproblem. In other words G (~x ,~x ) is the potential due to a unit charge, positionedarbitrarily within the surface S consistent either with GD = 0 orGN/n = 4/S on the surface. The true potential is a convolution of this template with the given(~x ).
Classical Field Theory: ElectrostaticsMagnetostatics

Laplace Equation in Spherical Coordinates
In spherical coordinates (r , , ) the Laplace equation can be written inthe form
1
r22
r2(r) +
1
r2 sin
(sin
)+
1
r2 sin2
2
2= 0 (29)
If we assume
=1
rU(r)P()Q() (30)
Then by substituting in (29) and multiplying with r2 sin2 /UPQ weobtain
r2 sin2
[1
U
d2U
dr2+
1
r2 sin P
d
d
(sin
dP
d
)]+
1
Q
d2Q
d2= 0 (31)
We see that the therms depending on have been isolated and we canset
1
Q
d2Q
d2= m2 (32)
with solutionQ = eim (33)
Classical Field Theory: ElectrostaticsMagnetostatics

Similarly the remaining terms can be separated as:
1
r2 sin
d
d
(sin
dP
d
)+
[l(l + 1) m
2
sin2
]P = 0 (34)
d2U
dr2 l(l + 1)
r2U = 0 (35)
The radial equation will have a solution
U = Ar l+1 + Brl (36)
while l is still undetermined.
Classical Field Theory: ElectrostaticsMagnetostatics

Legendre Equation and Legendre Polynomials
The equation (34) for P() can be expressed in terms of x = cos
d
dx
[(1 x2)dP
dx
]+
[l(l + 1) m
2
1 x2
]P = 0 (37)
This is the generalized Legendre equation and its solutions are theassociated Legendre functions.We will consider the solution of (54) for m2 = 0
d
dx
[(1 x2)dP
dx
]+ l(l + 1)P = 0 (38)
The solution should be single valued, finite, and continuous on theinterval 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) = xkj=0
ajxj (39)
where k is a parameter to be determined.
Classical Field Theory: ElectrostaticsMagnetostatics

By substitution in (54) we get the recurrence relation (how?)
aj+2 =(k + j)(k + j + 1) l(l + 1)
(k + j + 1)(k + j + 2)aj (40)
while for j = 0, 1 we find that
if a0 6= 0 then k(k 1) = 0 (41)if a1 6= 0 then k(k + 1) = 0 (42)
These two relations are equivalent and it is sufficient to choose either a0or a1 different from zero but not both (why?). We also see that the seriesexpansion is either only odd or only on even powers of x . By choosingeither k = 0 or k = 1 it is possible to prove the following properties:
I The series converges for x2 < 1, independent of the value of l
I The series diverges for x = 1, unless it terminates.
Since we want solution that is finite at x = 1, as well as forx2 < 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.
Classical Field Theory: ElectrostaticsMagnetostatics

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. Byconvention these polynomials are normalized to have the value unity forx = +1 and are called the Legendre polynomials of order l .The first few are:
P0(x) = 1
P1(x) = x
P2(x) =1
2(3x2 1) (43)
P3(x) =1
2(5x3 3x)
P4(x) =1
8(35x4 30x2 + 3)
The Legendre polynomials can be taken from Rodrigues formula:
Pl(x) =1
2l l!
d l
dx l(x2 1)l (44)
Classical Field Theory: ElectrostaticsMagnetostatics

The Legendre polynomials form a complete orthonormal set of functionson the interval 1 x 1 (prove it) 1
1Pl(x)Pl(x) =
2
2l + 1ll (45)
Since the Legendre polynomials form a complete set of orthonormalfunctions, any function f (x) on the interval 1 x 1 can be expandedin terms of them i.e.
f (x) =l=0
AlPl(x) (46)
where (how?)
Al =2l + 1
2
11
f (x)Pl(x)dx (47)
Thus for problems with azimuthal symmetry i.e. m = 0 the generalsolution is:
(r , ) =l=0
[Al r
l + Bl r(l+1)
]Pl(cos ) (48)
where the coefficients Al [it is not the same as in eqn (47)] and Bl can
be determined from the boundary conditions.
Classical Field Theory: ElectrostaticsMagnetostatics

Example
Boundary Value Problems with Azimuthal Symmetry
Lets specify as V () 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 finitethere. Consequently Bl = 0 for all l . Then on the surface of the sphere
V (r = R, ) =l=0
AlRlPl(cos ) (49)
and the coefficients Al will be taken via eqn(47)
Al =2l + 1
2R l
0
V ()Pl(cos ) sin d (50)
Classical Field Theory: ElectrostaticsMagnetostatics

If, for example V () = V on the two hemispheres then the coefficientscan be derived easily and the potential inside the sphere is (how?):
(r , ) = V
[3
2
( rR
)P1(cos )
7
8
( rR
)3P3(cos ) +
11
16
( rR
)5P5(cos ) . . .
](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?):
(r , ) =3
2
(R
r
)2V
[P1(cos )
7
12
(R
r
)2P3(cos ) + . . .
](52)
Classical Field Theory: ElectrostaticsMagnetostatics

Legendre Polynomials
An important expansion is that of thepotential at ~x due to a unit point chargeat ~x
1
~x ~x =l=0
r l
Pl(cos ) (53)
where r) is the smaller (larger) of ~x and ~x  and is the angle between ~x and ~x .
Show that the potential is :
1
~x ~x =l=0
(Al r
l + Bl r(l+1)
)Pl(cos ) on the zaxis
1
~x ~x =
1
r>
l=0
(r
)lon the xaxis
Classical Field Theory: ElectrostaticsMagnetostatics

Associated Legendre Functions and Spherical Harmonics
For problems without azimuthal (axial) symmetry, we need thegeneralization of Pl(cos ), namely, the solutions of
d
dx
[(1 x2)dP
dx
]+
[l(l + 1) m
2
1 x2
]P = 0 (54)
for arbitrary l and m.It can be shown that in order to have finite solutions on the interval1 x 1 the parameter l must be zero or a positive integer and thatthe 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 Legendrefunction Pml (x) . For positive m it is defined as
Pml (x) = (1)m(1 x2)m/2dm
dxmPl(x) (55)
If Rodrigues formula is used an expression valid for both positive andnegative m is obtained:
Pml (x) =(1)m
2l l!(1 x2)m/2 d
m
dxmPl(x) (56)
Classical Field Theory: ElectrostaticsMagnetostatics

There is a simple relation between Pml (x) and Pml (x) :
Pml (x) = (1)m (l m)!
(l + m)!Pml (x) (57)
For fixed m the functions Pml (x) form an orthonormal set in the index lon the interval 1 x 1. The orthogonality relation is 1
1Pml (x)P
ml (x)dx =
2
2l + 1
(l + m)!
(l m)!ll (58)
We have found that Qm() = eim, this function forms a complete set of
orthogonal functions in the index m on the interval 0 2.The product Pml Qm forms also a complete orthonormal set on the surfaceof the unit sphere in the two indices l ,m.From the normalization condition (58) we can conclude that the suitablynormalized functions, denoted by Ylm(, ), are :
Ylm(, ) =
2l + 1
4
(l m)!(l + m)!
Pml (cos )eim (59)
and alsoYl,m(, ) = (1)mY lm(, ) (60)
Classical Field Theory: ElectrostaticsMagnetostatics

The normalization and orthogonality conditions are: 20
d
0
sin dY lm(, )Ylm(, ) = llmm (61)
An arbitrary function g(, ) can be expanded in spherical harmonics
g(, ) =l=0
lm=l
AlmYlm(, ) (62)
where the coefficients are
Alm =
dY lm(, )g(, ) . (63)
The general solution for a boundaryvalue problem in sphericalcoordinates can be written in terms of spherical harmonics and powers ofr in a generalization of (48) :
(r , , ) =l=0
lm=l
[Almr
l + Blmr(l+1)
]Ylm(, ) (64)
If the potential is specified on a spherical surface, the coefficients can bedetermined by evaluating (64) on the surface and using (63).
Classical Field Theory: ElectrostaticsMagnetostatics

Spherical Harmonics Ylm(, )
l = 0 Y00 =
1
4
l = 1 Y11 =
3
8sin e i
Y10 =
3
4cos
l = 2 Y22 =1
4
15
2sin2 e2i
Y21 =
15
8sin cos e i
Y20 =
5
4
(3
2cos2 1
2
)
Classical Field Theory: ElectrostaticsMagnetostatics

Spherical Harmonics Ylm(, )
l = 3 Y33 = 1
4
35
4sin3 e3i
Y32 =1
4
105
2sin2 cos e2i
Y31 = 1
4
21
4sin (5 cos2 1) e i
Y30 =1
2
7
4sin
(5 cos3 3 cos
)(65)
Classical Field Theory: ElectrostaticsMagnetostatics

Figure: Schematic representation ofYlm on the unit sphere. Ylm is equalto 0 along m great circles passingthrough the poles, and along l mcircles of equal latitude. Thefunction changes sign each time itcrosses one of these lines.
Classical Field Theory: ElectrostaticsMagnetostatics

Addition Theorem for Spherical Harmonics
The spherical harmonics are related toLegendre polynomials Pl by a relation known asthe addition theorem.
The addition theorem expresses a Legendrepolynomial of order l in the angle in terms ofproducts of the spherical harmonics of theangles , and , :
Pl(cos ) =4
l(l + 1)
lm=l
Y lm(, )Ylm(, ) (66)
where is the angle between the vectors ~x and ~x , ~x ~x = x x cos and cos = cos cos + sin sin cos( ).Pl(cos ) is a function of the angles , with the angles
, asparameters and it maybe expanded in a series (63) :
Pl(cos ) =
l=0
lm=l
Alm(, )Ylm(, ) (67)
Classical Field Theory: ElectrostaticsMagnetostatics

The addition theorem offer the possibility to extend the expansion validfor a point charge (axially symmetric distribution) to an arbitrary chargedistribution.By substituting (66) into (53) we obtain
1
~x ~x = 4
l=0
lm=l
1
2l + 1
r l
Y lm(, )Ylm(, ) (68)
This equation gives the potential in a completely factorized form in thecoordinates ~x and ~x . This is useful in any integration over chargedensities, when one is the variable of integration and the other theobservation point.
Classical Field Theory: ElectrostaticsMagnetostatics

Multipole Expansion
A localized distribution of charge is described bythe charge density (~x ), which is nonvanishingonly inside a sphere a around some origin.The potential outside the sphere can be writtenas an expansion in spherical harmonics
(~x) =l=0
lm=l
4
2l + 1qlm
1
r l+1Ylm(, ) (69)
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 qlm interms of the properties of the charge distribution (~x ).The solution is very easily obtained from the integral for the potential
(~x) =
(~x )
~x ~x d3x (70)
Classical Field Theory: ElectrostaticsMagnetostatics

Using the expansion (68) for 1/~x ~x  i.e.
(~x) =l=0
lm=l
4
2l + 1
1
r l+1Ylm(, )
Y lm(
, )(~x )r ld3x (71)
Consequently the coefficients in (69) are :
qlm =
Y lm(
, )(~x )r ld3x (72)
and called the multipole moments of the charge distribution (~x )
Classical Field Theory: ElectrostaticsMagnetostatics

Monopole moment l = 0
Here, the only component is
q00 =
(~x )r 0Y 00(
, )d3x =14
(~x )d3x =
q4
(73)
Observed from a large distance r , any charge distribution actsapproximately as if the total charge q (monopole moment) would beconcentrated at one point since the dominant term in (68)
(~x) = 4q001
rY00 + =
q
r+ . . . (74)
NOTE:The moments with m 0 are connected (for real charge density) too themoments with m < 0 through
ql,m = (1)mqlm (75)
Classical Field Theory: ElectrostaticsMagnetostatics

Dipole moment l = 1
In Cartesian coordinates the dipole moment is given by
~p =
~x (~x )d3x (76)
In Spherical representation one obtains
q11 =
(~x )r Y 11(
, )d3x
=
3
8
(~x )r (sin cos i sin sin) d3x
in Cartesian represantation
=
3
8
(~x )(x iy )d3x =
3
8(px ipy ) (77)
also
q10 =
(~x )r Y 10(
, )d3x =
3
4
(~x )r cos d3x
=
3
4
z (~x )d3x =
3
4pz (78)
Classical Field Theory: ElectrostaticsMagnetostatics

Quadrupole moment l = 2
q22 =
(~x)r 2Y 22(
, )d3x
=1
4
15
2
(~x) [r sin (cos i sin)]2 d3x
=1
4
15
2
(~x)(x iy )2d3x
because (x iy )2 = 13
[(3x 2 r 2) 6ix y (3y 2 r 2)
]=
1
12
15
2(Q11 2iQ12 Q22) (79)
where Qij is the traceless (why?) quadrupole moment tensor:
Qij =
(3x i x
j r2ij
)(~x )d3x (80)
Classical Field Theory: ElectrostaticsMagnetostatics

Quadrupole moment l = 2
Analogously
q21 =
(~x )r 2Y 21(
, )d3x =
15
8
(~x )z (x iy )d3x
= 13
15
8(Q13 iQ23) (81)
and
q20 =
(~x )r 2Y 20(
, )d3x =1
2
5
4
(~x )(3z r 2)d3x
=1
2
5
4Q33 (82)
From eqn (60) we can get the moments with m < 0 through
ql,m = (1)mqlm (83)
Classical Field Theory: ElectrostaticsMagnetostatics

Multipole Expansion
By direct Taylor expansion of 1/~x ~x  the expansion of (~x) inrectangular coordinates is:
(~x) =q
r+~p ~xr3
+1
2
i,j
Qijxixjr5
+ . . . (84)
The electric field components for a given multipole can be expressedmost easily in terms of spherical coordinates. From ~E = ~ and (69)for fixed (l ,m) we get(how?):
Er =4(l + 1)
2l + 1qlm
1
r l+2Ylm
E = 4
2l + 1qlm
1
r l+2
Ylm (85)
E = 4
2l + 1qlm
1
r l+21
sin
Ylm
For a dipole ~p along the zaxis they reduce to:
Er =2p cos
r3, E =
p sin
r3, E = 0 (86)
Classical Field Theory: ElectrostaticsMagnetostatics

Multipole Expansion
The field intensity at a point ~x due to a dipole ~p at a point ~x0(r = ~x ~x0) is
~E (~x) =3~n (~p ~n) ~p~x ~x03
with ~n =~r
r(87)
because (~x) = ~p ~x/r3 and
~E (~x) = ~ = (~p ~x)r3
(~p ~x)(
1
r3
)=
3~n (~p ~n) ~p~x ~x03
(88)
Classical Field Theory: ElectrostaticsMagnetostatics

Energy of a Charge Distribution in an External Field
The multipole expansion of the potential of acharge distribution can also be used todescribe the interaction of the chargedistribution with an external field.The electrostatic energy of the chargedistribution (~x) placed in an external field(~x) is given by
W =
V
(~x)(~x)d3x (89)
The external field (if its is slowly varying over the region where (~x) isnonnegligible) may be expanded in a Taylor series:
(~x) = (0) + ~x ~(0) + 12
3i=1
3j=1
xixj2
xixj(0) + . . . (90)
Since ~E = ~ for the external field
(~x) = (0) ~x ~E (0) 12
3i=1
3j=1
xixjEjxi
(0) + . . . (91)
Classical Field Theory: ElectrostaticsMagnetostatics

Since ~ ~E = 0 for the external field we can substract
1
6r2~ ~E (0)
from the last term to obtain finally the expansion:
(~x) = (0) ~x ~E (0) 16
3i=1
3j=1
(3xixj r2ij
) Ejxi
(0) + . . . (92)
When this inserted into (89) the energy takes the form:
W = q(0) ~p ~E (0) 16
3i=1
3j=1
QijEjxi
(0) + . . . (93)
Notice, that:
I the total charge interacts with the potential,
I the dipole moment with the electric field,
I the quadrupole with the electric field gradient, etc
Classical Field Theory: ElectrostaticsMagnetostatics

Examples
EXAMPLE 1 : Show that for a uniform charged sphere all multipolemoments vanish except q00.
If the sphere has a radius R0 and constant charge density (~x) = then
qlm =
R00
r lY lm(, )r 2dr d =
R l+30l + 3
Y lm(, )d
but since Y00 = 1/
4 we have from the orthogonality relation
qlm = R l+30l + 3
4
Y lmY00d =
R l+30l + 3
4l0m0
Classical Field Theory: ElectrostaticsMagnetostatics

EXAMPLE 2: Perform multipole decomposition of a uniform chargedistribution whose surface is a weakly deformed sphere:
R = R0
(1 +
2m=2
a2mY2m(, )
), a2m

qlm R l+30l + 3
Y lm(, )
(1 + (l + 3)
2m=2
a2mY2m(
, )
)d
Apart from the monopole moment q00 of the previous example, we find
qlm = R l+30l + 3
(l+3)2
=2a2
Y lmY2d = R l+30
a2ml2 (l > 0)
Thus the nonvanishing multipole moments are : q2m = R50 a2m.
Classical Field Theory: ElectrostaticsMagnetostatics

Magnetostatics
There is a radical difference between magnetostatics andelectrostatics : 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 itselfin a certain direction. That is by definition the direction of the magneticflux density , denoted by ~B (also called magnetic induction). The magnitude of the flux density can be defined by the magnetictorque ~N exerted on the magnetic dipole:
~N = ~ ~B (94)where ~ is the magnetic moment of the dipole. In electrostatics the conservation of charge demands that the chargedensity at any point in space be related to the current density in thatneighborhood by the continuity equation
/t + ~ ~J = 0 (95) Steadystate magnetic phenomena are characterized by no change inthe net charge density anywhere in space, consequently in magnetostatics
~ ~J = 0 (96)Classical Field Theory: ElectrostaticsMagnetostatics

Biot & Savart Law
If d ~ is an element of length (pointing in thedirection of current flow) of a wire which carriesa current I and ~x is the coordinate vector fromthe element of length to an observation pointP, then the elementary flux density d~B at thepoint P is given in magnitude and direction by
d~B = kId ~ ~x~x 3
(97)
NOTE: (97) is an inverse square law, just as is Coulombs law ofelectrostatics. However, the vector character is very different.
If we consider that current is charge in motion and replace Id ~ by q~vwhere q is the charge and ~v the velocity. The flux density for such acharge in motion would be
~B = kq~v ~x~x 3
(98)
This expression is timedependent and valid only for charges with small
velocities compared to the speed of light.Classical Field Theory: ElectrostaticsMagnetostatics

Diff. Equations for Magnetostatics & Amperes 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
c
~J(~x ) (
~x ~x )~x ~x 3
d3x (99)
This expression for ~B(~x) is the magnetic analog of electric field in termsof the charge density:
~E (~x) =1
c
(~x )
(~x ~x )~x ~x 3
d3x (100)
In order to obtain DE equivalent to (99) we use the relation
(~x ~x )~x ~x 3
= ~(
1
~x ~x 
)as ~
(1
r
)=
~r
r3(101)
and (99) transforms into
~B(~x) =1
c~
~J(~x )~x ~x 
d3x (102)
Classical Field Theory: ElectrostaticsMagnetostatics

Classical Field Theory: ElectrostaticsMagnetostatics

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 = 1c~ ~
~J(~x )~x ~x 
d3x (104)
which for steadystate phenomena (~ ~J = 0) reduces to (how?) 2
~ ~B = 4c~J (105)
This is the 2nd equation of magnetostatics and corresponds to~ ~E = 4 in electrostatics.
2Remember : ~
~ ~A
= ~
~ ~A2~A
Classical Field Theory: ElectrostaticsMagnetostatics

The integral equivalent of (105) is called Amperes law
It can be obtained by applying Stokesstheorem to the integral of the normalcomponent of (105) over the open surface Sbounded by a closed curve C . Thus
S
~ ~B ~n da = 4c
S
~J ~nda (106)
is transformed intoC
~B d ~= 4c
S
~J ~nda (107)
Since the surface integral of the current density is the total current Ipassing through the closed curve C , Amperes law can be written in theform:
C
~B d ~= 4c
I (108)
Classical Field Theory: ElectrostaticsMagnetostatics

Vector Potential
The basic differential laws in magnetostatics are
~ ~B = 4c~J and ~ ~B = 0 (109)
The problems is how to solve them. If the current density is zero in the region of interest, ~ ~B = 0permits the expression of the vector magnetic induction ~B as the gradientof a magnetic scalar potential ~B = ~M , then (109) reduces to theLaplace equation for M . If ~ ~B = 0 everywhere, ~B must be the curl of some vector field~A(~x), called the vector potential
~B(~x) = ~ ~A(~x) (110)
and from (102) the general form of ~A is:
~A(~x) =1
c
~J(~x )~x ~x 
d3x + ~(~x) (111)
Classical Field Theory: ElectrostaticsMagnetostatics

The added gradient of an arbitrary scalar function shows that, for agiven magnetic induction ~B, the vector potential can be freelytransformed according to
~A ~A + ~ (112)This transformation is called a gauge transformation. Suchtransformations are possible because (110) specifies only the curl of ~A.
If (110) is substituted into the first equation in (109), we find
~(~ ~A
)=
4
c~J or ~
(~ ~A
)2~A = 4
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 ofthe vector potential satisfies the Poisson equation
2~A = 4c~J (114)
The solution for ~A in unbounded space is (111) with =constant:
~A(~x) =1
c
~J(~x )~x ~x 
d3x (115)
Classical Field Theory: ElectrostaticsMagnetostatics

Further Reading
I Green Function
I Greiner W. Classical Electrodynamics, Chapter 1, 2, 3, 4I Jackson J.K Classical Electrodynamics, 2nd Edition,
Sections 1.71.10, 3.13.6, 4.14.2, 5.15.4
Classical Field Theory: ElectrostaticsMagnetostatics