Método de investigación

4.1 Dipole Moment Practice

In both cases, the total charge of the system is zero. Therefore, the dipole moment is unique and independent of the choice of origin. We choose the natural center of each as the origin.

(a) For a linear charge density λ = λ(φ) around a ring, the volume charge density is

ρ(r) = λ(φ)δ(θ− π/2) the ring is

p =

Only the ˆx integral is non-zero above. This result of the integration around the ring is

p = ¹₂λ₀R²x.ˆ

(b) For a surface charge density σ = σ(θ) on a spherical shell, the volume charge density is ρ(r) = σ(θ)δ(r− R).

4.2 Smolochowski’s Model of a Metal Surface

(a) The form of n+(z) “smears out” the charge due to the positive nuclei but recognizes that there is a well-defined “last layer” of nuclei at z = 0. The form of n₋(x) models the fact that the electron wave functions “spill out” into the vacuum beyond the last row of nuclei.

n₊

z n₋

(b) There is only a z-component to the dipole moment p by symmetry. So, the dipole moment per unit area is

pz =

_∞

−∞dz z (n+− n₋) = ¹₂n¯ d dκ


0

−∞dz e^{κ z}+

_∞

0

dz e^−κz 

=−n¯ κ².

(c) The total charge density is ρ(z) = n+(z)− n₋(z) =−sgn(z)¹₂n exp(−κ|z|) . It must be¯ that E = E(z)ˆz by symmetry. Therefore, because sgn(z) = d|z|/dz, Gauss’ law gives

dE(z)

dz = ρ(z)

0

=− n¯ 2
0

d|z|

dz exp(−κ|z|).

This can be integrated by inspection to

E(z) = n¯

2
0κexp(−κ|z|).

The integration constant is zero because only a charged sheet produces an electric field at infinity. The electric field is finite at z = 0 so the potential must be continuous there. Hence, if we let ϕ(0) = 0,

ϕ(z) =−

_z

0

dz^E(z^) =− n¯ 2
0κ

_z

0

dz^exp(−κ|z|) = n¯

2
0κ²sgn(z){exp(−κ|z|) − 1} .

z ϕ(z)

(d) This potential gives ϕ(∞)−ϕ(−∞) = −¯n

κ²
₀ = p_z/
₀. The right side is the change in potential which occurs across a double layer. Rather than a sudden jump, the change is spread out over the entire length of the system.

(e) The total energy per unit area is

UE =¹₂
0

_∞

−∞dz E²(z) = ¯n² 8
0κ²


0

−∞dz e^{2κ z}+

_∞

0

dz e^−2κz 

= n¯² 8
0κ³.

4.3 The Charge Density of a Point Electric Dipole (a) Begin with two point charges arranged as shown below.

P

O r − sb − r₀ sb

r₀ r

r − r₀ +q/s

−q/s

The charge density of this system is ρ(r) = q

s[δ(r− r0− sb) − δ(r − r0)] .

For the point electric dipole, we are interested in the limit as s→ 0. Therefore, we expand the argument of the delta function of the positive charge to get

ρ_D(r) = lim

s→0

q

s[δ(r− r0)− sb · ∇δ(r − r0) +· · · − δ(r − r0)] .

All the higher terms in the expansion are proportional to s, s², etc. and thus go to zero in the limit. Therefore, with p = qb, we get the advertised result,

ρ_D(r) =−p · ∇δ(r − r0).

(b) The suggested charged density is correct because the electrostatic potential it produces is

ϕ(r) = 1 4π
0

d³r^ ρD(r^)

|r − r^|

= − p

4π
0 ·

d³r^∇^δ(r^− r0)

|r − r^|

= p

4π
₀ · ∇0

d³r^δ(r^− r0)

|r − r^|

= p

4π
0 · ∇0

1

|r − r0|

= − p

4π
0 · ∇ 1

|r − r0|.

4.4 Stress Tensor Proof of No Self-Force

Let S be any surface in vacuum which completely encloses the distribution ρ(r) in question.

The net force on ρ(r) is

F =
0

S

dS



(ˆn· E)E −1

2n(Eˆ · E)

 .

Nothing changes if we expand S all the way out to infinity. If ρ(r) has a net charge, the asymptotic electric field varies as 1/r². Therefore, dSE² ∝ 1/r² as r→ ∞ and the surface integral is zero. We get the same result if ρ(r) does not have a net charge because the field goes to zero even faster as r→ ∞.

4.5 Point Charge Motion in an Electric Dipole Field The electrostatic potential and electric field of the dipole are

ϕ(r, θ) = p cos θ

4π
0r² E = p 4π
0r³

2 cos θ ˆr + sin θ ˆθ

 .

The initial condition is υ = 0 when r = R =

x²₀+ y₀²and θ = π/2. Therefore, conservation of energy guarantees that

1

2mυ²+qp cos θ 4π
0R² = 0.

On the other hand, the motion will be circular if the radial force equals the centripetal acceleration, that is, if

mυ²

R =−qEr=−2qp cos θ 4π
0R³ .

This equation is identical to the energy conservation equation so the motion is indeed semi-circular. A moment’s reflection shows that the particle moves periodically back and forth along the arc shown below.

p

q

Source: R.S. Jones, American Journal of Physics 63, 1042 (1995).

4.6 The Energy to Assemble a Point Dipole

We know that dW = ϕ(r)dq is the work required to move charge dq quasistatically from infinity to the point r. Therefore, the work required to bring charge dq to r and charge−dq to r + δ is

dW = ϕ(r)dq− ϕ(r + δ)dq.

dq –dq

r δ

We will take the limit δ → 0 presently so it is appropriate to perform a Taylor expansion to get

dW = ϕ(r)dq− [ϕ(r) + δ · ∇ϕ(r)]dq = −δq · ∇ϕ(r).

From the figure, it is consistent to define dp =−δdq in the limit when dq → ∞ and δ → 0 such that their product remains finite. Therefore, because E = −∇ϕ, we get the desired result,

dW =−E(r) · dp.

Source: A.M. Portis, Electromagnetic Fields (Wiley, New York, 1978).

4.7 Dipoles at the Vertices of Platonic Solids The electric field of a point dipole is

E(r) = 1 4π
0

3ˆn(ˆn· p) − p

|r − r0|³ −4π

3 p δ(r− r0)

 ,

where ˆn = (r− r0)/|r − r0|. The delta function has no effect since we are interested in the field E(0) at the center of each polyhedron. Also, ˆn = ˆr₀ at this observation point.

(a) The positions r0 of the dipoles for the octahedron on the far left can be taken to be

±aˆx, ±aˆy, and ±aˆz. Therefore, r0 = a and ˆn takes the values±aˆx, ±aˆy, and ±aˆz when we sum over dipoles. Hence, the total field at the origin is

E(0) = 1 4π

1

a³ [−6p + 3ˆxpx+ 3(−ˆx)(−px) + 3ˆyp_y+ 3(−ˆy)(−py) + 3ˆzp_z+ 3(−ˆz)(−pz)]

= 0.

(b) The positions r0 of the dipoles for the tetrahedron in the middle are a(ˆx + ˆy + ˆz), a(−ˆx − ˆy + ˆz), a(−ˆx + ˆy − ˆz), and a(ˆx − ˆy − ˆz). Therefore, r0 =√

3a and ˆn takes the values (ˆx + ˆy + ˆz)/√

3, (−ˆx − ˆy + ˆz)/√

3, (−ˆx + ˆy − ˆz)/√

3, and (ˆx− ˆy − ˆz)√ 3.

Hence, the total field at the origin is

E(0) = 1 4π
0

1

3a³ [−4p + (ˆx + ˆy + ˆz)(px+ py + pz) + (−ˆx − ˆy + ˆz)(−px− py + pz)]

+ 1

4π
₀ 1

3a³ [(−ˆx + ˆy − ˆz)(−px+ py − pz) + (ˆx− ˆy − ˆz)(px− py− pz)]

= 0.

(c) The eight dipoles at the corners of the cube are the superposition of two tetrahedra with dipoles at their corners rotated by 90^◦ with respect to one another. From part (b), each tetrahedron contributes zero to the electric field at the center. Hence, E(0) = 0 for this case also.

Source: A.M. Portis, Electromagnetic Fields (Wiley, New York, 1978).

4.8 Two Coplanar Dipoles

Choose a polar coordinate system with p ˆz. The field produced by p in this system is E = p

4π
₀r³



2 cos θˆr + sin θ ˆθ

 .

p

θ θ

zˆ

rˆ

ˆ

′ p′′

θ

At equilibrium, the potential energy V =−p^· E is a minimum. In the pictured coordinate system,

p^= p^cos θ^ˆr + p^sin θ^θ.ˆ Therefore,

V =−(2 cos θ cos θ^+ sin θ sin θ^), and the minimum energy occurs at

∂V

∂θ^ =−(sin θ cos θ^− 2 cos θ sin θ^) = 0.

This gives the final result as

tan θ = 2 tan θ^.

4.9 Potential of a Double Layer

(a) Begin with our fundamental formula for the potential due to a double layer:

ϕ(r) =− 1 4π
₀

S

dSτ (r_S)· ∇ 1

|r − rS|.

Now dSτ = dSτ ˆn = dSτ . Therefore, working out the gradient, ϕ(r) =− 1

4π
₀

S

dS· ∇ 1

|r − rS|τ (r_S) = 1 4π
₀

S

dS· r− rS

|r − rS|³τ (r_S).

On the other hand, the solid angle is defined as Ω(r) =

S

dΩ =

S

dS· rS− r

|rS − r|³. Combining the preceding equations completes the demonstration.

(b) Let r_L(r_R) be a point infinitesimally close to the surface point r_S in region L(R). The surface appears to have infinite extent when viewed at very close range, so

ϕR(rS)− ϕL(rS) =− 1

4π
₀τ (rS) [ΩR − ΩL] = τ (rS)

₀ .

The square brackets contribute 4π because Ω_L = 2π and Ω_R = −2π are the solid angles subtended at r_L and r_R by an infinite plane.

4.10 A Spherical Double Layer

The outward normal to the sphere is ˆr. Therefore, using the divergence, the potential at any point in space due to a surface dipole density τ = τ ˆr is

ϕ(r) = − 1 4π
₀

S

dS^τ· ∇ 1

|r − r^|

= − τ

4π
0

S

dS^ˆr· ∇ 1

|r − r^|

= − τ

4π
0

dS^· ∇ 1

|r − r^|

= − τ

4π
0

V

d³r^∇^· ∇ 1

|r − r^|

= + τ

4π
0

V

d³r^∇² 1

|r − r^|

= − τ

4π
₀

V

d³r^δ(r− r^).

Because r is the observation point and V is the volume enclosed by the spherical shell, the last integral above gives

ϕ(r) =

 −τ/
0 r < R,

0 r > R.

Notice that the matching condition is satisfied:

ϕ(r > R)− ϕ(r < R) = τ/
0.

4.11 The Distant Potential of Two Charged Rings

In cylindrical coordinates (s, φ, z), the charge density of the inner ring is ρ(s, φ, z) = Q

2πsδ(s− a)δ(z).

x

Since x = s cos φ, the x-component of the electric dipole moment vector is

p_x =

The y-component vanishes similarly because y = s sin φ. The z-component vanishes because of the factor δ(z). Hence, p = 0.

The components of the primitive Cartesian quadrupole tensor are

Qij= 1

All the off-diagonal elements are zero because of the φ-integration. The diagonal elements are

Qxx = Qy y= Qa²

8π Qz z = 0.

We conclude that the distant electric field of this ring is

ϕa(r) = Q

The potential of the outer ring is similar except with opposite charge. The monopole terms cancel and the quadrupole terms add. Therefore, the asymptotic potential is a pure quadrupole:

ϕ(r) = Q(a²− b²) 8π
₀

x²+ y²− 2z² r⁵ .

4.12 The Potential Far from Two Neutral Disks

Each disk has no charge and no dipole moment. The latter is true because the charge density depends only on the radial distance from the disk center. Therefore, with respect to its own symmetry axis, each disk produces a quadrupole potential

ϕ = 1

where Q is a quadrupole moment, θ is the polar angle from its symmetry axis, and r = x²+ y²+ z². We now restrict ourselves to the x-y plane, where r = s, and write θ for the

polar angle for the horizontal disk and θ^ for the polar angle of the tipped disk. This gives the total potential,

ϕ_tot(x, y) = 1

This formula will have the desired form (independent of all angles) if cos²θ + cos²θ^ = 1, which will be true if θ^= θ + π/2. Hence, α = π/2.

Source: J.A. Greenwood, British Journal of Applied Physics 17, 1621 (1966).

4.13 Interaction Energy of Adsorbed Molecules

(a) The interaction energy between a point dipole p₁ at r₁ and a point dipole p₂ at r₂ is

U₁₂= 1

where UN N comes from the four nearest neighbors at a distance a, UN N N comes from the four next-nearest neighbors at a distance√

2a, and the factor ¹₂ corrects for double-counting in the total. By direct evaluation, we get

U_{N N} = 2

as required. The energy is independent of the angle α!

Source: V.M. Rozenbaum and V.M. Ogenko, Soviet Physics Solid State 26, 877 (1984).

(b) A point charge representation of each N2 molecule is +− +. A qualitative argument to find the preferred orientation focuses on maximizing the Coulomb attraction between molecules on the four nearest-neighbor sites. This suggests that the most favorable arrangement is the following.

+ − + + − + + − +

+ − + +

+− +

+−

+ +− +

+−

+ +−

Source: L. Mederos, E. Chac´on, and P. Tarazona, Physical Review B 42, 8571 (1990).

4.14 Practice with Cartesian Multipole Moments

(a) The total charge Q = 0. There is no dipole moment because the charge is distributed symmetrically about the origin. The components of the quadrupole moment tensor are

Qij =¹₂

d³r rirjρ(r).

Since

ρ(x, y) = qδ(z){δ(x − a)δ(y − a) + δ(x + a)δ(y + a)}

− qδ(z) {δ(x − a)δ(y + a) + δ(x − a)δ(y + a)} ,

all four terms contribute equally to both Q_xy and Q_{y x}. In detail,

Q = 2qa²

⎡

⎣ 0 1 0 1 0 0 0 0 0

⎤

⎦ .

(b) The total charge is q = 2λ. The dipole moment is zero because the charge is symmet-rical around the origin. All Qij = 0 except

Q_{z z} =¹₂λ



−

dz z² = ¹₃λ³.

(c) The charge density is ρ(x, y, z) = λδ(z)δ(r− R) in cylindrical coordinates. The total charge is q = 2πRλ trivially. The dipole moment is zero because the charge is symmet-rically distributed. The charge lies entirely in the x-y plane so Q_xz = Q_{y z} = Q_{z z} = 0.

This leaves only

Qxy = ¹₂λ

2π 0

dθ sin θ cos θ

_+∞

0

dr r³δ(r− R) = 0

Qxx = Q_{y y} =¹₂λ

_2π

0

dθ cos²θ

_∞

0

drr³δ(r− R) = ¹₂λπR³.

4.15 The Many Faces of a Quadrupole

(a) The components of the primitive electric quadrupole moment tensor are

Qij =¹₂

d³r rirjρ(r).

Since

ρ(x, y) = qδ(z){δ(x − a)δ(y − a) + δ(x + a)δ(y + a)}

− qδ(z) {δ(x − a)δ(y + a) + δ(x − a)δ(y + a)} ,

all four terms contribute equally to both Q_xy and Q_{y x}. Therefore,

Q = 2qa²

⎡

⎣ 0 1 0 1 0 0 0 0 0

⎤

⎦ .

The potential produced by this quadrupole is

ϕ(r) = Q_ij3rirj− δijr²

r⁵ = 12qa² xy (x²+ y²)^5/2. (b) The primitive quadrupole tensor is Q = 2qa²(ˆxˆy + ˆy ˆx).

(c) Writing the Cartesian unit vectors in terms of the spherical polar unit vectors gives ˆ

x = sin θ cos φˆr + cos θ cos φˆθ− sin φ ˆφ ˆ

y = sin θ sin φˆr + cos θ sin φˆθ + cos φ ˆφ ˆ

z = cos θˆr− sin θˆθ.

Substituting these into part (b) and simplifying yields the nine matrix elements of Q in spherical polar coordinates:

Q_{r r} = 4qa²sin²θ cos φ sin φ

Q_{θ θ} = 4qa²cos²θ cos φ sin φ Qφφ =−4qa²sin φ cos φ Qr θ = Qθ r = 4qa²sin θ cos θ sin φ cos φ Q_{r φ} = Q_φr = 2qa²sin θ(cos²φ− sin²φ) Q_{θ φ} = Q_φθ = 2qa²cos θ(cos²φ− sin²φ).

(d) Since r = rˆr in spherical polar coordinates, all terms involving rθ and rφ will vanish.

Thus

ϕ_Q(r) = Q_ij3r_ir_j − δijr²

r⁵ = Q_{r r}2r² r⁵ − Qθ θ

r² r⁵ − Qφφ

r² r⁵. (e) Substituting the appropriate results from part (c) into part (d) gives

ϕ_Q(r) = 4qa²r²

r⁵ sin φ cos φ(2 sin²θ− cos²θ + 1).

Since x = r sin θ cos φ and y = r sin θ sin φ, the electric potential is ϕQ(r) = 12qa² xy

(x²+ y²)^5/2, the same as in part (a).

(f) The definition of the components of the quadrupole tensor in part (a) is valid in Carte-sian coordinates only.

Source: Prof. R. Grigoriev, Georgia Institute of Technology (private communication).

4.16 Properties of a Point Electric Quadrupole

(a)

ϕ(r) = 1 4π
₀

d³r^ ρ(r^)

|r − r^| = 1 4π
₀

d³r^ 1

|r − r^|Qij∂_i^∂_j^δ(r^− r0)

= 1

4π
0

d³r^δ(r^− r0)Qij∂^_i∂^_j 1

|r − r^| = 1 4π
0

Qij∂i∂j

d³r^δ(r^− r0) 1

|r − r^|

= 1

4π
₀Qij∂i∂j

1

|r − r0|.

This is a quadrupole potential so ρ(r) is correct as stated.

(b) F =

d³rρ(r)E(r) = Qij

d³r E(r)∂i∂jδ(r− r0) = Qij∂i∂jE(r0).

(c) The torque is τ =

d³r r× ρ(r)E(r) so τi =
ij kQm 

d³r rjEk∂m∂δ(r− r0)

=
_{ij k}Q_{m }

d³r δ(r− r0)∂_m∂_(r_jE_k)

=
_{ij k}Q_{m }

d³r δ(r− r0){δj ∂_mE_k + δ_{m j}∂_E_k+ r_j∂_m∂_E_k}

=
ij kQm j∂mEk(r0) +
ij kQj ∂Ek(r0) + (r× F)i,

where F is given by part (b). Finally, Qm j = Qj m so the total torque can be written as

N = 2(Q· ∇) × E + r × F, where (Q· ∇)i= Qij∇j.

(d) VE =

d³r ϕ(r)ρ(r) = Qij

d³r δ(r− r0)∂i∂jϕ(r) =−Qij∂iEj(r0).

4.17 Interaction Energy of Nitrogen Molecules

The leading contribution to the interaction energy may be calculated by treating each molecule as a point quadrupole. The potential produced by molecule A is

ϕ_A(r) = 1

2Q^A_ij∇i∇j

1

|r − rA|. The charge density associated with molecule B is

ρB(r) = Q^B_{k m}∇k∇mδ(r− rB).

Therefore, the interaction between the two is

VE =

d³rρB(r)ϕA(r)

= 1

4π
₀Q^A_ijQ^B_{k }∇i∇j∇k∇

1

|rA− rB|

= 1

8π
0

Q^A_ijQ^B_{k m}

d³r∇k∇mδ(r− rB)∇i∇j

1

|r − rA|

= 1

8π
0

Q^A_ijQ^B_{k m}

d³r δ(r− rB)∇k∇m∇i∇j

1

|r − rA|

= 1

8π
0

Q^A_ijQ^B_{k m}∇^B_k∇^B_m∇^B_i ∇^B_j 1

|rB − rA|.

This shows that the interaction energy varies as R⁻⁵ where R =|rA− rB|.

4.18 A Black Box of Charge

Place a point charge at the center of the box. This gives an  = 0 multipole and no others.

Now take a point charge q and surround it by a spherical shell of uniformly distributed surface charge which integrates to −q. This point-plus-shell (PPS) object produces no electric field outside of itself. Therefore, placing any number of these PPS objects in the box away from the exact center produces a non-spherically symmetric charge distribution (with respect to the center of the box) with the desired property.

Source: Prof. Scott Tremaine, Institute for Advanced Study (private communication).

4.19 Foldy’s Formula

(a) As indicated in the figure below, we choose an arbitrary origin O and locate the center of the neutron charge distribution at r. In that case, the electrostatic interaction energy is

VE(r) =

d³s ρN(s)ϕ(r + s).

O r

s ρN

(b) When ϕ(r) varies slowly over the size of ρN(s), a Taylor series expansion is appropriate:

ϕ(r + s) = ϕ(r) + s_k∂_kϕ(r) + 1

2s_js_kϕ(r) +· · · . Substituting this above with ρ_N(s) = ρ_N(s) gives

V_E(r) =



d³s ρ_N(s)

 ϕ(r) +



d³s s_kρ(s)



s· ∇ϕ(r) +1

2



d³s s_js_kρ_N(s)



∂_j∂_kϕ(r) +· · · .

The first bracketed integral is zero because the neutron has no charge. The second bracketed integral is zero because ρN is spherically symmetric. For the same reason, only the j = k terms survive in the third bracketed integral. Hence,

VE(r) = 1 2



d³s s²_kρN(s)



∇²kϕ(r) = 1 2



d³ss² 3ρN(s)



∇²ϕ(r).

The last equality follows because the spherical symmetry of ρN(s) implies that the integrals with s²_x, s²_y, and s²_z are all equal to 1/3 of the integral with s² = s²_x+ s²_y+ s²_z. Finally, Poisson’s equation for the electrostatic potential of the electron is

₀∇²ϕ =−eδ(r − r0).

Therefore, integrating VE(r) over all of space gives the desired formula.

Source: L.L. Foldy, Reviews of Modern Physics 30, 471 (1958).

4.20 Practice with Spherical Multipoles

(a) The volume charge density of the shell is

ρ = σ₀δ(r− R) sin θ cos φ = σ0

2π

3 δ(r− R) [Y1,−1(θ, φ)− Y1,1(θ, φ)] . If B =

2π/3, the orthonormality of the spherical harmonics gives the exterior mul-tipole moments as

Am = 4π 2 + 1

d³r ρ(r)r^Y_m^∗ (θ, φ)

= 4πσ0

2 + 1B

∞

0

dr r^{2+ }δ(r− R)

dΩ{Y1,−1(Ω)− Y1,1(Ω)} Y_m^∗ (Ω)

= 4πσ0

3 R³Bδ,1{δm ,−1− δm ,1} . (b) The potential outside the sphere is

ϕ(r > R, θ, φ) = 1 4π
0

∞

= 0

 m =−

Am

Ym(θ, φ) r^{+ 1}

= 1

4π
0

4πσ₀ 3

R³

r² B{Y1,−1(θ, φ)− Y1,1(θ, φ)}

= σ₀ 3
0

R³

r² sin θ cos φ

= σ0R³ 3
₀

x r³.

(c) The charge density is real so ρ(r) = ρ^∗(r) and we can compute the interior spherical multipole moments from

B_m = 4π 2 + 1

d³rρ^∗(r)

r^{+ 1} Y_m(θ, φ)

= 4πσ0

2 + 1B

∞

0

dr r^1−δ(r− R)

dΩ

Y_1,−1^∗ (Ω)− Y_1,1^∗ (Ω) Y_m^∗ (Ω)

= 4πσ₀

3 Bδ_,1{δm ,−1− δm ,1} .

(d) The potential inside the sphere is

ϕ(r < R, θ, φ) = 1 4π
0

∞

= 0

 m =−

B_mr^Y_m^∗ (θ, φ)

= 1

4π
0

4πσ0

3 rB

Y_1,−1^∗ (θ, φ)− Y_1,1^∗ (θ, φ)

= σ₀ 3
0

r sin θ cos φ

= σ₀ 3
0

x.

(e) The potential is continuous at r = R as it should be. The tangential components of the electric field are similarly continuous at r = R. As for the normal component of the electric field, direct calculation shows that this matching condition is also satisfied:

∂ϕ<

∂r −∂ϕ>

∂r



r = R

=

σ0

3
₀ −−2σ0R³ 3
₀r³



r = R

sin θ cos φ

= σ0

₀ sin θ cos φ

= σ(θ, φ)

0

.

(f) A general electric dipole potential is

ϕ(r) = 1 4π
0

p· r r³ .

Comparing this with results of (b) shows that the shell carries an electric dipole moment p = ¹₃QRˆx where Q = 4πR²σ0 is the total charge of the shell.

4.21 Proof by Interior Multipole Expansion

Choose the origin at the center of a charge-free spherical sub-volume. All the source charge must be outside the sphere so an interior multipole expansion for points inside the sphere is

ϕ(r, θ, ϕ) = 1 4π
0



m

B_mr^Y_m^∗ (θ, ϕ)

where

Bm = 4π 2 + 1

d³rρ(r)

r^{+ 1}Ym(θ, φ).

Using Y00 = 1/ = √

4π and the orthonormality of the spherical harmonics, the desired average is

ϕ_S = 1

Rewriting this slightly gives the desired result:

ϕ =

4.22 The Potential outside a Charged Disk

(a) The exterior moments are

A_m = 4π volume charge density of the disk. This gives

Am = 4πσ Substituting this into the exterior multipole expansion,

ϕ(r) = 1

(b) The potential on the z-axis of the disk is

On the other hand, 1 + t²− 1 =

and P_(1) = 1. Therefore, (2) is indeed the same as the multipole expansion (1) evaluated at θ = 0.

4.23 Exterior Multipoles for Specified Potential on a Sphere

(a) The general form of an exterior, spherical multipole expansion is given by

ϕ(r) = On the surface of the sphere,

ϕ(R, Ω) =

The orthonormality of the spherical harmonics gives the expansion coefficients as A_m = R^{+ 1}

(b) By examining figure (b), it is clear that we need the potential to change signs every time φ is an integer multiple of π/2. Thus, m =±2, which in turn implies that  ≥ 2. For the asymptotic form of the potential we need only keep the lowest value of  necessary.

Examining figure (a), we can see that the potential must change signs every time θ is

an integer multiple of π/2 as well. It is clear that Y₂₂ ∝ sin²θ does not satisfy this requirement. However, Y₃₂ ∝ sin²θ cos θ ∝ cos θ − cos 3θ does. Thus, the potential must be a linear combination of Y_3±2 and must equal±V at r = R. That is,

ϕ(r) = V

R r

4

2 2π

105(Y32+ Y₃₋₂) = V

R r

4

sin²θ cos θ cos 2φ, r→ ∞.

4.24 A Hexagon of Point Charges

(a) Choose the origin of coordinates r = 0 at the center of the hexagon. The charges q_α (α = 1, . . . , 6) are positioned at r = r_α. Using the geometry of a hexagon, we label the (x, y) position of each charge beginning with the topmost and proceeding clockwise:

q −q q −q q −q

(0, a) (x, a/2) (x,−a/2) (0,−a) (−x, −a/2) (−x, a/2) Then, by direct computation,

Q =

α

qα= q− q + q − q + q − q = 0.

The dipole component p_z = 0 because z = 0 for all the charges. Otherwise,

p_x =

α

q_αx_α = 0− qx + qx + 0 − qx + qx = 0

p_y =

α

q_αy_α = qa− qa/2 − qa/2 + qa − qa/2 − qa/2 = 0.

The quadrupole matrix is symmetric and Q_{z z} = Q_xz = Q_{y z} = 0 because z = 0 for all the charges. Otherwise,

Qxy ∝

α

qαxαyα = 0− qxa/2 − qxa/2 + 0 + qxa/2 + qxa/2 = 0

Qxx∝

α

qαx²_α= 0− qx²+ qx²+ 0 + qx²− qx² = 0

Q_{y y} ∝

α

q_αy_α² = qa²− qa²/4 + qa²/4− qa²/4− qa²+ qa²/4− qa²/4 = 0.

However, the next (octupole) moment has non-zero components. An example is O_{y y y} ∝

α

q_αx³_α = qa³− qa³/8− qa³/8 + qa³− qa³/8− qa³/8 = 3qa³/2.

(b) The potential of a monopole, dipole, and quadrupole vary as ϕ ∝ r⁻¹, ϕ∝ r⁻², and ϕ∝ r⁻³. Therefore, the next term in the expansion must behave as

ϕ(r)∝ r⁻⁴.

4.25 Analyze This Potential

(a) The charge distribution must have (i) zero net charge; (ii) no dipole moment; and (iii) a quadrupole potential with only the single term Bx²/r⁵. Consider the traceless multipole expansion,

ϕ(r) = 1 4π
0

Q r +p· r

r³ + Θij

rirj

r⁵ +· · ·

 ,

and focus on the quadrupole term. The desired term is Θ_xxx²/r⁵, so we must have Θxx= 0. However, the traceless condition is

Θxx+ Θy y+ Θz z = 0.

Therefore, we must have Θ_{y y} = 0 or Θz z = 0 or both. In other words, quadrupole terms like y²/r⁵ or z²/r⁵ or both must also be present if x²/r⁵is present. We conclude that no charge distribution can produce the stated potential.

(b) We can also use the primitive Cartesian multipole expansion,

ϕ(r) = 1 4π
0

Q r +p· r

r³ + Qij

3rirj − r²δij

r⁵ +· · ·

 .

We eliminate the terms with no factor of x²/r⁵ by requiring that Q_xy = Q_xz = Q_{y z} = 0. The desired factor x²/r⁵ appears in the remaining diagonal terms,

ϕ_quad= 1 4π
0

Q_xx(2x²− y²− z²) + Q_{y y}(2y²− x²− z²) + Q_{z z}(2z²− x²− y²)

r⁵ .

Rearranging this gives

ϕquad= 1 4π
0

x²(2Qxx− Qy y− Qz z) + y²(2Qy y− Qz z− Qxx) + z²(2Qz z− Qxx− Qy y)

r⁵ .

We want to eliminate the y²/r⁵ and z²/r⁵ terms. This means that 2Qy y = Qxx+ Qz z and 2Qz z − Qy y− Qz z.

The only solution to these equations in Qxx = Qy y = Qz z. However, this makes the coefficient of the x²/r⁵ term zero also. Therefore, once again, we conclude that there is no charge distribution with a quadrupole potential of the form Bx³/r⁵.

(c) If we permit charge to extend to any point in space, a charge distribution which produces the stated potential can always be found from the Poisson equation,

ρ(r) =
0∇²ϕ(r).

In document Derecho de impugnación en las sentencias por contravenciones de tránsito sin pena privativa de libertad (página 37-40)