Multipole Expansion of Electrostatic Potential

Multipole expansion of Electrostatic Potential

We have seen earlier that when the charge distribution has sufficient symmetry or when the potential is required on a symmetry axis of the distribution, one can obtain potential due to it at a point outside the distribution by solving Laplace’s equation with appropriate boundary conditions. In general, the problem of finding potential due to an arbitrary charge distribution is difficult and one must take recourse to approximations. One such procedure is what is known as the multipole expansion of the potential at a point far removed from the charge distribution. In this approximation, the potential is expressed as a sum of contributions due to charge monopole, a dipole, a quadrupole etc. The 1 essential idea of the distribution is to expand as power series in 1/r where ~r is |~r − ~r0| the position vector of the point at which the potential is required and ~r0 is the position vector of a charge element in the charge distribution. 1.0 Expansion in Spherical coordinates We will first consider the expression for the potential in spherical coordinates.

r − r’ P

r’ r

Consider a charge distribution shown in the ﬁgure above. The potential at the point P (~r) due to the charge distribution is given by

0 1 ρ(~r ) 3 0 V (~r) = 0 d r (1) 4π0 |~r − ~r |

Far from the charge distribution, we may obtain an expansion of the potential as a power series in 1/r, where r is the distance of the point P from the origin. denoting the angle between ~r and ~r0 by θ, we have

1 1 r02 r0 1/2 = 1 + − 2 cos θ (2) |~r − ~r0| r r2 r

The expansion is obtained by expressing the above as a power series using a binomial 2 c D.K. Ghosh, IIT Bombay expansion of the above,

r02 r0 1/2 r02 r0 1 + − 2 cos θ = 1 − − cos θ + r2 r 2r2 r 1 1 3 r02 r0 2 (− )(− ) − 2 cos θ + 2! 2 2 r2 r 1 1 3 5 r02 r0 3 (− )(− )(− ) − 2 cos θ + ... (3) 3! 2 2 2 r2 r

We will collect the ﬁrst four terms of the above. The ﬁrst term is clearly 1. Thus if the 1 1 series is truncated here, the potential simply becomes R ρ(~r0)d3r0, which is just the 4π0 r potential one would obtain if all the charges are assumed to be concentrated at the origin. This is also called the monopole term in the expansion of the potential. r0 r0 The term proportional to is cos θ. This leads to a potential term r r

Z 1 1 0 0 3 0 V = 2 r cos θρ(~r )d r (4) 4π0 r Z 1 1 0 0 3 0 = 2 (ˆr · ~r )ρ(~r )d r 4π0 r 1 rˆ · ~p = 2 (5) 4π0 r where in the last but one step, we have used the fact that θ is the angle between the vectors ~r and ~r0. The equation (5) represents the dipole term in the potential, which is obtained by generalizing the deﬁnition of an elementary dipole to the case of a charge distribution. Consider an elementary dipole which consists of two charges ±q, separated by a distance d. The potential due to it at a point P which is at a distance r from the centre of the dipole is given by +q r+ θ P d/2 r O d/2 r_ _ q

1 q q V (r) = − 4π0 r+ r− c D. K. Ghosh, IIT Bombay 3

d2 d d2 where r2 = r2 + ∓ rd cos θ = r2 1 ∓ cos θ + . For r d, we have ± 4 r 4r2

1 1 d 1/2 ' 1 ∓ cos θ r± r r 1 d ' 1 ± cos θ r 2r Thus 1 1 d − ≈ 2 cos θ r+ r− r which gives the dipole potential to be given by 1 qd cos θ 1 p cos θ Vdip = 2 = 2 (6) 4π0 r 4π0 r where ~p = qdpˆ, the direction of dipole moment being from the negative charge to the positive charge. This shows that the potential decreases more rapidly than that due to a monopole. This deﬁnition for two charges is generalized to the case of a collection of discrete charges, X ~p = qi~ri (7) i Or, in case of continuous charge distribution, Z ~p = ~rρ(~r)d3r (8)

Using the relationship (7) we can see that the dipole moment is independent of origin if the total charge of the distribution is zero, for if we let ~ri → ~ri + ~a, the expression on P P the right of this equation becomes i qiri + ~a i qi. For the potential given by (6), the electric ﬁeld components are ∂V 2p cos θ Er = − = 3 (9) ∂r 4π0r 1 ∂V p sin θ Eθ = − = 3 (10) r ∂θ 4π0r 1 ∂V E = − = 0 (11) ϕ r sin θ ∂ϕ r02 The next two terms in the expansion (3) gives terms proportional respectively to and r2 r03 and are r3 r02 3 r02 r02 1 − + cos2 θ = (3 cos2 θ − 1) (12) 2r2 2 r2 r2 2 3 r03 1 15 8r03 r03 (−4 cos θ) + (− )(− cos3 θ) = (5 cos3 θ − 3 cos α) (13) 8 r3 6 8 r3 2r3 4 c D.K. Ghosh, IIT Bombay

Equation (22) gives the quadrupole term in the expansion of the potential, which is given by Z 2 1 1 02 0 (3 cos θ − 1) 3 0 Vquad = 3 r ρ(~r ) d r (14) 4π0 r 2 Similarly (13) gives the octupole term in the expansion of the potential

Z 3 1 1 03 0 (5 cos θ − 3 cos θ) 3 0 Voctu = 4 r ρ(~r ) d r (15) 4π0 r 2 This expressions are generalization of the potential due to a quadrupole and an octupole, respectively, made of discrete charges shown in the ﬁgure.

y +q _ q _ +q q _ q _ q +q d/2 +q x +2q _ O q

_ d/2 _ q _ +q _ q q +q

(a) A planar quadrupole (b) a linear quadrupole An Octupole

One can continue the expansion by including higher and higher order terms. Equations (4),(14), (15), along with the monopole term in the expansion of the potential can be written as Z Z 1 1 0 3 0 1 0 0 3 0 V (~r) = ρ(~r )P0(cos θ)d r + 2 r ρ(~r )P1(cos θ)d r 4π0 r r 1 Z 1 Z + r02ρ(~r0)P (cos θ)d3r0 + r03ρ(~r0)P (cos θ)d3r0 (16) r3 2 r4 3

In equation (16), the angle dependent integrands are

P0(cos θ) = 1

P1(cos θ) = cos θ 1 P (cos θ) = (3 cos2 θ − 1) 2 2 1 P (cos θ) = (5 cos3 θ − 3 cos θ) 3 2 Legendre polynomials of orders 0 to 2. In general, the expansion of the potential can be written down as ∞ 1 X 1 Z V (~r) = r0nρ(~r0)P (cos θ)d3r0 (17) 4π rn+1 n 0 n=0 Note that this expansion assumes that the point P lies on the z-axis so that the angle θ is usual polar angle. One can obtain an unrestricted expansion in the spherical basis, which leads to spherical harmonics instead of the Legendre polynomial. This would bring out c D. K. Ghosh, IIT Bombay 5 the tensor nature of the quadrupole moment. However, we will not touch upon this in our course. Instead, we will consider the expansion in rectangular coordinates. Multipole expansion in Cartesian Coordinates One can expand the potential expression (1) in Cartesian coordinates. In this method, we 1 expand the function in a Taylor series in three variables x, y and z around origin. |~r − ~r0| Using results proved in multi-variable calculus, we have

f(x, y, z) = α0 + αx(x − x0) + αy(y − y0) + αz(z − z0) 2 2 2 + αxx(x − x0) + αyy(y − y0) + αzz(z − z0)

+ αxy(x − x0)(y − y0) + αyz(y − y0)(z − z0) + αzx(z − z0)(x − x0) + ... (18) where the coeﬃcients of the expansion are given by

α0 = f(x0, y0, z0) ∂f ∂f ∂f α = | ; α = | ; α = | x ∂x x0,y0,z0 y ∂y x0,y0,z0 z ∂z x0,y0,z0 1 ∂2f 1 ∂2f 1 ∂2f α = | ; α = | ; α = | xx 2! ∂x2 x0,y0,z0 yy 2! ∂y2 x0,y0,z0 zz 2! ∂z2 x0,y0,z0 1 ∂2f 1 ∂2f 1 ∂2f α = | ; α = | ; α = | (19) xy 2! ∂x∂y x0,y0,z0 yz 2! ∂y∂z x0,y0,z0 zx 2! ∂z∂x x0,y0,z0 where it is assumed that the order of diﬀerentiation in the last term is unimportant. The function that we need to expand is 1 1 = |~r − ~r0| p(x − x0)2 + (y − y0)2 + (z − z0)2

0 and since we are expanding about the origin with r r , we have x0 = 0 = y0 = z0. The diﬀerentiations are with respect to the primed variables. We have ∂ 1 1 −2(x − x0) = − ∂x0 p(x − x0)2 + (y − y0)2 + (z − z0)2 2 [(x − x0)2 + (y − y0)2 + (z − z0)2]3/2 (x − x0) = [(x − x0)2 + (y − y0)2 + (z − z0)2]3/2 Substituting x0 = y0 = z0 = 0, we get x α = x r3 By symmetry, it follows that y z α = ; α = y r3 z r3 Let us calculate the second derivatives. We have ∂2 1 ∂ (x − x0) = ∂x02 p(x − x0)2 + (y − y0)2 + (z − z0)2 ∂x0 [(x − x0)2 + (y − y0)2 + (z − z0)2]3/2 6 c D.K. Ghosh, IIT Bombay

Diﬀerentiating we get

−3 −2(x − x0)2 1 ( ) − 2 [(x − x0)2 + (y − y0)2 + (z − z0)2]5/2 [(x − x0)2 + (y − y0)2 + (z − z0)2]3/2

3(x − x0)2 − [(x − x0)2 + (y − y0)2 + (z − z0)2] = [(x − x0)2 + (y − y0)2 + (z − z0)2]3/2 Once again, substituting x0 = y0 = z0 = 0 we have

3x2 − r2 α = xx 2r5 and likewise 3y2 − r2 3z2 − r2 α = ; α = r5 yy 2r5 zz 2 For the other set of three second derivatives, we have

∂2 1 ∂ (x − x0) = ∂y0∂x0 p(x − x0)2 + (y − y0)2 + (z − z0)2 ∂y0 [(x − x0)2 + (y − y0)2 + (z − z0)2]3/2

The right hand side evaluates to

3(x − x0)(y − y0) [(x − x0)2 + (y − y0)2 + (z − z0)2]5/2 which gives 3xy α = α = xy yx 2r5 Analogously 3yz 3zx α = α = ; α = α = yz zy 2r5 zx xz 2r5 In the present case, the terms in equation (18) are to be obtained by letting x, y, z → 0 0 0 x , y , z and setting x0 = y0 = z0 = 0. Thus the expression for the potential (1), becomes, keeping terms only up to quadrupole moment terms

" Z 3 Z 1 1 X xi V (x, y, z) = ρ(r0)d3r0 + x0 ρ(r0)d3r0 4π r r3 i 0 i=1 3 2 2 Z 3 3 Z # X 3x − r X X 3xixj + i x02ρ(r0)d3r0 + x0 x0 ρ(r0)d3r0 (20) 2r5 i 2r5 i j i=1 i=1 j=1,j6=i

The last two terms inside the bracket of the above expression can be rearranged as follows:

3 3 Z 2 Z X X xixj r 3x0 x0 ρ(r0)d3r0 − r02ρ(r0)d3r0 2r5 i j 2r5 i=1 j=1 c D. K. Ghosh, IIT Bombay 7

P 02 02 2 P 2 P where we have used i xi = r . We may rewrite r = i xi = i,j δi,jxixj. This makes the above to be equal to

3 3 1 X X x x (3x0 x0 − δ r02) 2r5 i j i j i,j i=1 j=1

With these simpliﬁcations, the expression for the potential becomes

" Z 3 Z 1 1 X xi V (x, y, z) = ρ(r0)d3r0 + x0 ρ(r0)d3r0 4π r r3 i 0 i=1 3 3 # 1 X X Z + x x (3x0 x0 − δ r02)ρ(r0)d3r (21) 2r5 i j i j i,j i i=1 j=1,

The quadrupole moment tensor is a tensor of rank 2 having 9 components, which are deﬁned by the integral in the last term of (21) Z 0 0 0 02 3 0 Qij = ρ(r )(3xixj − δi,jri )d r (22)

It is a traceless tensor which can be seen easily

Q11 + Q22 + Q33 = 0

Conventionally, when the charge distribution has azimuthal symmetry, one has Q11 = Q22 one deﬁnes the quadrupole moment to be given by Q33, which is then also equal to −2Q11 = −2Q22 in view of the traceless nature of the quadrupole moment. It is easy to see that in spherical coordinates Z Z 0 2 2 0 2 Q33 = ρ(r )(3z − r ) = ρ(r )(3 cos θ − 1) as obtained before. Examples : Example 1 : Consider the planar quadrupole shown in Figure 2(a). The charge density is discrete and is given by ρ(r0) = +qδ(x0 − d/2)δ(y0 − d/2)δ(z0) − qδ(x0 + d/2)δ(y0 − d/2)δ(z0) + qδ(x0 + d/2)δ(y0 + d/2)δ(z0) − qδ(x0 − d/2)δ(y0 + d/2)δ(z0). In this case, only 2 Q12 and Q21 are non-zero and each is given by Q = 3d q. The potential is the given by 1 1 3qd2 xy V = 5 [Q12xy + Q21yx] = 5 4π0 2r 4π0 r Example 2 : (From Griﬃths Problem 3.27) : A sphere of radius R , centered at the origin, carries a charge density R ρ(r, θ) = k (R − 2r) sin θ, r2 8 c D.K. Ghosh, IIT Bombay where k is constant. Find the approximate potential outside the sphere along z-axis, far from the sphere. We will expand the potential in multipoles. The monopole terms is zero , as can be seen by calculating the total charge,

Z R Z π R − 2r 2 Q =kR dr 2 sin θ2πr sin θdθ 0 0 r Z R Z π 2 2 R π = 2πkR (R − 2r)dr sin θdθ = 2πkR(Rr − r )|0 × = 0 0 0 2 We next calculate the dipole term, which is obtained by multiplying the integrand above by r cos θ. In this case, the angle integration is

Z π 3 2 sin θ π sin θ cos θdθ = |0 = 0 0 3 Thus the dipole term also does not contribute to the potential. We next calculate the quadrupole term, given by Q = R ρ(r)r2(3 cos2 θ−1)d3r. This can be calculated as follows:

Z R Z π 2 2 2 Qzz = 2πkR r (R − 2r)dr sin θ(3 cos θ − 1)dθ 0 0 3 4 Z π Rr r R 2 2 = 2πkR( − 2 )|0 × (3 cos θ − 1) sin θdθ 3 4 0 R4 π kπ2R5 = 2πkR × = 6 8 24 The potential due to the quadrupole is

5 1 Qzz kπR V = 3 = 3 4π0 2r 1920r