Pablo Laguna Georgia Institute of Technology,...

ElectromagnetismChapter 4: Electric Multipoles

Pablo LagunaGeorgia Institute of Technology, USA

Fall 2012

Notes based on textbook: Modern Electrodynamics by A. Zangwill

Laguna Electromagnetism

Electrostatic Multipole Expansion

ϕ(r) =1

4πε0

∫d3r′ ρ(r′)|r− r′|

Electrostatic multipole expansion: approximating to ϕ(r) when r lies farfrom any point within a localized charge distribution.

Localized source: a source distribution a finite volume of space ofcharacteristic size R.

Therefore r R.

The ratio R/r is a small parameter that can be used as an expansionparameter.

Thus1

|r− r′| =1r− r′ · ∇1

r+

12

(r′ · ∇)2 1r− · · ·


Substitution of

1|r− r′| =

1r− r′ · ∇1

r+

12

(r′ · ∇)2 1r− · · ·

into

ϕ(r) =1

4πε0

∫d3r′ ρ(r′)|r− r′|

yields

ϕ(r) =1

4πε0

[∫d3r′ρ(r′)

]1r−[∫

d3r′ρ(r′)r ′i

]∇i

1r

+

[12

∫d3r′ρ(r′)r ′i r ′j

]∇i∇j

1r− · · ·

.


The expression

ϕ(r) =1

4πε0

[∫d3r′ρ(r′)

]1r−[∫

d3r′ρ(r′)r ′i

]∇i

1r

+

[12


]∇i∇j

1r− · · ·

.

can be rewritten as

ϕ(r) =1

4πε0

Qr

+p · rr 3 + Qij

3ri rj − r 2δij

r 5 + · · ·

where

Q =

∫d3r ρ(r) monopole moment

pi =

∫d3r ρ(r)ri dipole moment

Qij =12

∫d3r ρ(r)ri rj quadrupole moment


Electric Dipole

For an electrostatic ally neutral object (Q = 0) the dominant component ofthe electric potential is

ϕ(r) =1

4πε0

p · rr 3 r R

The numerical value of p is independent of the choice of origin when Q = 0.

Proof: Consider the dipole p′ obtained when one shifts the origin by a vectord so r = r′ + d. Then, because d3r = d3r

′and ρ(r′) = ρ(r),

p′ =

∫d3r′ρ(r′)r′ =

∫d3r ρ(r)(r− d) = p−Q d.

Hence, p′ = p, if Q = 0.

Notice that the electric dipole moment is not uniquely defined for any systemwith a net charge.


Since E = −∇ϕ, given

ϕ(r) =1

4πε0

p · rr 3

the dipole electric field reads

E(r) =1

4πε0

3r(r · p)− pr 3

Notice the O(r−3) dependence.

If the dipole is not located at the origin, but at r0, the electric dipole filed isgiven by

E(r) =1

4πε0

3n(n · p)− p|r− r0|3

wheren =

r− r0

|r− r0|


Electric Dipole Field Lines

Consider a dipole aligned with the z-axis; that is, p = p z. SIncez = r cos θ − θ sin θ then

E(r , θ) =p

4πε0r 3

[2 cos θ r + sin θ θ

].

The electric field lines are found from:

drEr

=rdθEθ⇒ 1

rdrdθ

= 2 cot θ ⇒ r = k sin2 θ.


Exercise

Let E(r) be the electric field produced by a charge density ρ(r) which liesentirely inside a spherical volume V of radius R. Show that the electric dipolemoment of the distribution is given by

∫r<R

E(r) d3r = − 13ε0

p

If the charge density ρ(r) is all exterior to the sphere V , then

∫r<R

E(r) d3r =4πR3

3E(0)


Recall that the electric field of a dipole is given by

E(r) =1

4πε0

3n(n · p)− p|r− r0|3

and that on the other hand∫r<R

E(r) d3r = − 13ε0

p

In order for these two expressions to be compatible, an extra term isneeded in E(r). Specifically

E(r) =1

4πε0

3n(n · p)− p|r− r0|3

− 13ε0

pδ(r− r0)

Notice that the extra term does not contribute to the far field.

The extra term can also be viewed as the contribution from a point dipole


Electric Field of a Point DipoleRecall that the electrostatic potential of a dipole is

ϕ =−1

4π ε0p · ∇1

r

From E = −∇ϕ

E =1

4π ε0∇[p · ∇1

r

]=

14π ε0

(p · ∇)∇1r

but∂

∂r i

∂

∂r j

1r

=3 ri rj − r 2 δij

r 5 − 4π3δij δ(r)

thus

pi ∂

∂r i

∂

∂r j

1r

=pi (3 ri rj − r 2 δij )

r 5 − 4π3

pj δ(r)

=3 pi ni nj − pj

r 3 − 4π3

pjδ(r)

so

E =1

4πε0

3n(n · p)− pr 3 − 1

3ε0pδ(r)


Charge Density of a Point Dipole

From ∇ · E = ρ/ε0 and ϕ = −14π ε0

p · ∇ 1r

ρ = ε0∇ · E = −ε0∇2ϕ(r)

= −ε0∇2[−1

4π ε0p · ∇1

r

]=

p4π· ∇∇2 1

r

Thusρ = −p · ∇δ(r)


Dipole Force

Consider the force exerted on a neutral charge distribution ρ(r′) by anexternal electric field E(r′).Approximate E(r′) by

E(r′) = E(r) +[(r′ − r) · ∇

]E(r) + · · ·

where r is a reference point located somewhere inside ρ(r′)Therefore, since Q = 0 by assumption, the force on ρ(r) is

F =

∫d3r′ρ(r′)E(r′) =

∫d3r′ρ(r′)(r′ · ∇)E(r)

ThusF =

∫d3r′ρ(r′)(r′ · ∇)E(r) = (p · ∇)E(r)

If p is constant, with the help of

∇(p · E) = p× (∇× E) + E× (∇× p) + (p · ∇)E + (E · ∇)p

one gets thatF = ∇(p · E).

which emphasizes that an electric field gradient is needed to generate aforce.


Dipole Torque

The Coulomb torque exerted on ρ(r) by a field E is given by:

N =

∫d3r′r′ × E(r′)ρ(r′) =

∫d3r′δ(r′ − r)(p · ∇′)(r′ × E)

= (p · ∇)(r× E) = p× E + r× F

where we have used that for a point dipole

ρ(r) = −p · ∇δ(r− r0).


The Dipole Potential Energy

The potential energy of interaction between ρ(r) and E(r) can be calculatedusing ρ(r) = −p · ∇δ(r− r0)

VE (r) =

∫d3r′ϕ(r′)ρ(r′) = −

∫d3r′ϕ(r′)p · ∇′δ(r′ − r)

Integration by parts yields

VE (r) = −p · E(r)

Notice thatF = −∇VE = ∇(p · E).


Dipole-Dipole Interaction

Given the electric field of a dipole

E(r) =1

4πε0

3r(r · p)− pr 3

It costs no energy to bring the first dipole.

The work done by us to bring p2 into position is exactly the interactionenergy with the dipole electric field of p1:

W12 = −p2 · E1(r2) =1

4πε0

p1 · p2

|r2 − r1|3− 3p1 · (r2 − r1)p2 · (r2 − r1)

|r2 − r1|5

Repeating for N prticles

UE = W =1

4πε0

12

N∑i=1

N∑j 6=i

pi · pj

|ri − rj |3− 3pi · (ri − rj )pj · (ri − rj )

|ri − rj |5

In a compact form

UE = −12

N∑i=1

pi · E(ri ).


Electric Dipole Layers

A distribution of point electric dipoles confined to a surface may becharacterized by a dipole moment per unit area τ (r).

If the layer is made up of point dipoles p = q b, with b the characteristicsize of the point dipole and q its effective charge, then

τ =pS

=qbS

= σb.

where S is the area occupied by each point dipole and σ the surfacecharge density of the layer.

Therefore, a crude representation of the layer is

b b

Figure: Side view of a dipole layer: two oppositely charged surfaces withareal charge densities ±σ separated by a distance b. In themacroscopic limit, b→ 0 and σ →∞ with τ = σb = constant.


Potential Electric Dipole Layers

In general

τ =dpdS

Recall the electrostatic potential dϕ of a point dipole dp

dϕ =−1

4π ε0dp · ∇ 1

|r− rS |

where rS is a point on the surface layer.

Therefore,

ϕ =

∫S

dϕ = − 14πε0

∫S

dp · ∇ 1|r− rS |

= − 14πε0

∫S

dS τ · ∇ 1|r− rS |

Recall that τ is parallel to p and thus to the surface’s normal n.Therefore

ϕ = − 14πε0

∫S

dS τ · ∇ 1|r− rS |

= − 14πε0

∫S

τ n · ∇(

1|r− rS |

)dS


One can show that

n · ∇(

1|r− rS |

)dS = −cos θ dS

|r− rS |2= −dΩ

where dΩ is the solid angle subtended at the observation point by dS

Therefore

ϕ = − 14πε0

∫S

τ n · ∇(

1|r− rS |

)dS =

14πε0

∫S

τ dΩ

If we let the observation point to be infinitesimally close to the dipolelayer, one can show that there is a jump discontinuity

ϕ+ − ϕ− = ∆ϕ = τ/ε0


Consider the following dipole layer consisting of two parallel plane surfaces,top positively charged and bottom negatively charged:

The electric field has magnitude is E = σ/ε0, therefore

∆ϕ = E d =σ dε0

Recall that in the limit d → 0, one has that σ d → τ thus

∆ϕ =τ

ε0

Note the similarity with equation

(E2 − E1) · n21 = σ/ε0


The Electric Quadrupole

The third (electric quadrupole) term dominates the multipole expansionwhen ( Q = p = 0).

The quadrupole potential is given by

ϕ(r) =1

4πε0Qij

3ri rj − δij r 2

r 5

whereQij = 1

2

∫d3r ρ(r)ri rj

Notice that the tensor Qij is symmetric ( Qij = Qji ), thus it only has 6independent components.

One can show that Qij is uniquely defined tensor and does not dependon the choice of the origin of coordinates if Q = p = 0.

Also, it is not difficult to show that

ρ(r) = Qij∇i∇jδ(r)


Traceless Quadrupole Tensor

Recall

ϕ(r) =1

4πε0

12


3r i r j − r 2δij

r 5

which can be rewritten as

ϕ(r) =1

4πε0

12

∫d3r′ρ(r′)

(3r ′i r ′j − δij r ′2

) ri rj

r 5

Define traceless quadrupole tensor as

Θij =12

∫d3r′ρ(r′)(3r ′i r ′j − r ′2δij ) = 3Qij −Qkkδij

soϕ(r) =

14πε0

Θijri rj

r 5

Notice that Θii = δij Θij = 0. Therefore Θij only has 5 independent

components.


Force and Torque on a Quadrupole

The force and torque exerted on a point quadrupole in an external electricfield are respectively

F = Q ij∇i∇jE

andN = 2(Q · ∇)× E + r× F.

The corresponding electrostatic interaction energy is

VE (r) = −Q ij∇iEj (r) = − 13 Θij∇iEj (r)

where in the last equality it is assumed that the source charge for the externalfield E is assumed to be far from the quadrupole.

The expressions above are valid for Q = p = 0 if the electric field does notvary too rapidly over the volume of the distribution.


Legendre Polynomials

Recall1

|r− r′| =1r− r′ · ∇1

r+

12

(r′ · ∇)2 1r− · · ·

An alternative would be to instead

1|r− r′| =

1√r 2 − 2r · r′ + r ′2

=1r

[1− 2(r · r′) r ′

r+

r ′2

r 2

]−1/2

=1r

1√1− 2 x t + t2

where t ≡ r ′/r and x ≡ r · r′ = cos θ

If |x | ≤ 1 and 0 < t < 1, one has that

1√1− 2xt + t2

≡∞∑`=0

t`P`(x)

where P`(x) are the Legendre polynomials


Orthogonality

1∫−1

dxP`(x)P`′ (x) =2

2`+ 1δ``′ ,

Completeness∞∑`=0

2`+ 12

P`(x)P`(x ′) = δ(x − x ′),

Parity: P`(−x) = (−1)` P`(x)

Differential Equation:

(1− x2)P′′n (x)− 2 x P′n(x) + n(n + 1) Pn(x) = 0

Examples

P0(x) = 1

P1(x) = x

P2(x) =12

(3x2 − 1)

P2(x) =12

(5x3 − 3x)


Spherical Harmonics

Consider the partial differential equation

1sin θ

∂

∂θ

(sin θ

∂Y`m∂θ

)+

1sin2 θ

∂2Y`m∂φ2 = −`(`+ 1)Y`m

The solutions to this eigenvalue equation are the spherical harmonics,Y`m(θ, φ).

Orthogonality:

2π∫0

dφ

π∫0

dθ sin θY`m(θ, φ)Y ∗`′m′(θ, φ) = δ``′δmm′

Completeness:

∞∑`=0

∑m=−`

Y`m(θ, φ)Y ∗`m(θ′, φ′) = δ(cos θ − cos θ′)δ(φ− φ′)


The first few spherical harmonics are

Y00(θ, φ) =1√4π,

Y10(θ, φ) =

√3

4πcos θ

Y1±1(θ, φ) = ∓√

38π

sin θ exp(±iφ)

Y20(θ, φ) =

√5

16π

3 cos2 θ − 1

Y2±1(θ, φ) = ∓√

158π

sin θ cos θ exp(±iφ)

Y2±2(θ, φ) =

√15

32πsin2 θ exp(±2iφ)

Notice that Y`m ∝ exp(i m φ)


Inverse DistanceIn terms of the Legendre polynomials, we saw that

1|r− r′| =

1r

∞∑`=0

(r ′

r

)`P`(r · r′) r ′ < r .

But from the spherical harmonic addition theorem,

P`(r · r′) =4π

2`+ 1

+∑m=−`

Y ∗`m(θ′, φ′)Y`m(θ, φ).

we have that

1|r− r′| =

1r

∞∑`=0

4π2`+ 1

(r ′

r

)` ∑m=−`

Y ∗`m(Ω′)Y`m(Ω) r ′ < r .

Notice that an exchange of primed and unprimed produces a formula validwhen r ′ > r . Thus

1|r− r′| =

1r>

∞∑`=0

4π2`+ 1

(r<r>

)` ∑m=−`

Y ∗`m(Ω<)Y`m(Ω>)

where r< = min(r , r ′) and r> = max(r , r ′)Laguna Electromagnetism

Electrostatic Potential Expansion

From

ϕ(r) =1

4πε0

∫d3r ′

ρ(r′)|r− r′| .

and1

|r− r′| =1r>

∞∑`=0

4π2`+ 1

(r<r>

)` ∑m=−`

Y ∗`m(Ω<)Y`m(Ω>)

In the exterior

ϕ(r) =1

4πε0

∞∑`=0

∑m=−`

A`mY`m(Ω)

r `+1 r > R

whereA`m =

4π2`+ 1

∫d3r ′ ρ(r′) r ′` Y ∗`m(Ω′)


Electrostatic Potential Expansion

From

ϕ(r) =1

4πε0

∫d3r ′

ρ(r′)|r− r′| .

and1

|r− r′| =1r>

∞∑`=0

4π2`+ 1

(r<r>

)` ∑m=−`

Y ∗`m(Ω<)Y`m(Ω>)

In the interior

ϕ(r) =1

4πε0

∞∑`=0

∑m=−`

B`m r ` Y ∗`m(Ω) r < R.

where

B`m =4π

2`+ 1

∫d3r ′

ρ(r′)r ′`+1 Y`m(Ω′)


Azimuthal Symmetry

If ρ(r , θ, φ) = ρ(r , θ) then m = 0 and then

Y`0(θ, φ) =

√2`+ 1

4πP`(cos θ)

and

12π

2π∫0

dφY`m(θ, φ) =

√2`+ 1

4πP`(cos θ)δm, 0.

Also A`m = A`δm0√

4π/(2`+ 1) and B`m = B`δm0√

4π/(2`+ 1) with

A` =

∫d3r ′ r ′`ρ(r ′, θ′)P`(cos θ′)

andB` =

∫d3r ′

1r ′`+1 ρ(r ′, θ′)P`(cos θ′)


So

ϕ(r) =1

4πε0

∞∑`=0

∑m=−`

A`mY`m(Ω)

r `+1 r > R

ϕ(r) =1

4πε0

∞∑`=0

∑m=−`

B`m r ` Y ∗`m(Ω) r < R.

becomes

ϕ(r , θ) =1

4πε0

∞∑`=0

A`r `+1 P`(cos θ) r > R,

ϕ(r , θ) =1

4πε0

∞∑`=0

B` r `P`(cos θ) r < R.


Spherical Harmonics Summary

ϕ(r , θ, φ) =1

4πε0

∞∑`=0

∑m=−`

A`mY`m(Ω)

r `+1 r > R

ϕ(r , θ, φ) =1

4πε0

∞∑`=0

∑m=−`

B`m r ` Y ∗`m(Ω) r < R.

with

A`m =4π

2`+ 1

∫ρ(r ′, θ′, φ′) r ′`+2 Y ∗`m(Ω′)dr ′dΩ′

B`m =4π

2`+ 1

∫ρ(r ′, θ′, φ′)

r ′`−1 Y`m(Ω′)dr ′dΩ′

Also Orthogonality:

2π∫0

dφ

π∫0

dθ sin θY`m(θ, φ)Y ∗`′m′(θ, φ) = δ``′δmm′


Legendre Polynomials Summary

ϕ(r , θ) =1

4πε0

∞∑`=0

A`r `+1 P`(cos θ) r > R,

ϕ(r , θ) =1

4πε0

∞∑`=0

B` r `P`(cos θ) r < R.

with

A` =

∫r ′`+2ρ(r ′, θ′)P`(cos θ′)dr ′dΩ′

B` =

∫ρ(r ′, θ′)

r ′`−1 P`(cos θ′)dr ′dΩ′

Also Orthogonality

1∫−1

dxP`(x)P`′ (x) =2

2`+ 1δ``′ ,


Examples

Example 1:Show that for a uniform charge sphere all multipoles momentvanish except A00

Example 2: Perform a multiple decomposition of a uniform charge distributionwhose surface is deformed as

R(θ, φ) = R0

[1 +

2∑m=−2

a∗2m Y2m(θ, φ)

]|a2m| << 1

Example 3: Find the potential of an isolated sphere of radius a at r > a if thepotential at its surface is given by ϕ(a, θ, φ) = V0 cos3 θ

Example 4: Find the potential of the sphere with a hole in problem 3.12 atr > R using spherical harmonics.


Examples

Example 5: Calculate the potential inside a sphere of radius a such that itspotential in the surface is given by ϕ(a, θ) = V if 0 ≤ θ ≤ π/2 andϕ(a, θ) = −V if π/2 ≤ θ ≤ π.

Example 6: Two concentric spheres have radii a, b (b > a) and each isdivided into two hemispheres by the same horizontal plane. The upperhemisphere of the inner sphere and the lower hemisphere of the outer sphereare maintained at potential V . The other hemispheres are at zero potential.Determine the potential in the region a ≤ r ≤ b as a series in Legendrepolynomials. Include terms at least up to l = 4. Check your solution againstknown results in the limiting cases b →∞ , and a→ 0.


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