The Liquid Drop Model of Nuclear Fission

Problems

Application 4.4 The Liquid Drop Model of Nuclear Fission

4.6 Spherical and Azimuthal Multipoles 113

The corresponding potential is

ϕ(r, θ)= σ0

30

⎧⎨

⎩

rcosθ r < R, R³

r² cosθ r > R. (4.99) Note that the two expansions agree atr=R. This reflects the continuity of the electrostatic potential as the observation point passes through a layer of charge. The fact that only the=1 term survives tells us that the electric field outside the shell is exactly dipolar. The interior “dipole” potential is proportional torcosθ =z. This gives a constant electric fieldEˆzinside the shell becauseE= −∇ϕ.

4.6.4 Preview: the Connection to Potential Theory

The multipole calculation outlined in Application 4.3 exploited the fact that all the source charge lay on the surface of a spherical shell. Another approach to this problem, calledpotential theory, uses the same information to infer thatϕ(r) satisfies Laplace’s equation both inside the shell and outside the shell:

∇²ϕ(r)=0 r=R. (4.100)

From this point of view, (4.95) and (4.96) are linear combinations of elementary solutions of Laplace’s equation with azimuthal symmetry. The coefficientsAandB are chosen so the potentials inside and outside the sphere satisfy the required matching conditions (Section 3.3.2) atr=R. If the source charge on the shell is not azimuthally symmetric, similar remarks apply to a potential given by the simultaneous use of (4.86) and (4.89). We will explore potential theory in detail in Chapters 7 and 8.

Example 4.5 Find the electrostatic potential produced by a uniformly charged line bent into an origin-centered circular ring in thex-yplane. The charge per unit length of the line isλ=Q/2π R.

Solution: The potentialϕ(r, θ) is determined completely by an interior and exterior azimuthal multipole expansion with respect to an origin-centered sphere of radiusR. In spherical coordinates, the volume charge density of the ring is

ρ(r)=(λ r)δ(r−R)δ(cosθ).

Note that the right side of this equation is dimensionally correct and satisfies the necessary condition Q=

d³rρ(r)=2π Rλ. Using (4.93) and (4.94), the expansion coefficients areA=QRP(0) andB =QR⁻⁽⁺¹⁾P(0). Therefore,

ϕ(r, θ)= Q 4π ₀r

∞ =0

R r

P(0)P(cosθ) r≥R, ϕ(r, θ)= Q

4π 0R ∞ =0

"r R

#

P(0)P(cosθ) r≤R.

As it must be, the potential is continuous as the observation point passes through the ring.

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114 ELECTRIC MULTIPOLES: APPROXIMATE ELECTROSTATICS FOR LOCALIZED CHARGE

area. Electrostatics favors fission because the distortion lowers the Coulomb self-energy of the system.

The latter dominates when the nuclear charge is large enough. To see this, we will show the following.

Claim A: Up to quadrupole order, the most general azimuthally symmetric distortion of a uniform sphere with chargeqand radiusR0which preserves both the volume and the position of the center of mass has a surface described by

R(θ)=R0

1−α²

5 +αP2(cosθ) +

α1. (4.101)

Claim B: If U0 is the self-energy of the undistorted nucleus, the change in electrostatic energy induced by the shape distortion (4.101) is

UE = −¹₅U0α². (4.102)

Proof of Claim A: The most general azimuthally symmetric form for the nuclear shape (up to quadrupole order) is

R(θ)=R₀{1+α₀P₀(cosθ)+α₁P₁(cosθ)+α₂P₂(cosθ)}. (4.103) Theα1 term can be discarded because it corresponds to a rigid displacement of the center of mass of the undistorted sphere along thez-axis. To see this, note first thatP1(cosθ)=cosθ=z on the unit sphere. Then, to first order inα11, the presumptive shape equationr=1+α1zimplies that r²=1+2α1z. On the other handr²=x²+y²+z². Equating these two expressions forr²implies thatx²+y²+(z−α1)²=1 (to first order inα1). This is the equation of the displaced unit sphere shown in Figure 4.17(a).

z z

( )a ( )b

Figure 4.17: Small-amplitude distortions of a unit sphere: (a)R(θ)=α1P1(cosθ);

(b)R(θ)=α₂P₂(cosθ).

To preserve the sphere volume, we useP0(cosθ)=1 and demand that 4π

3 R₀³=2π π 0

dθsinθ R(θ)

0

drr²= 2π 3 R₀³

1

−1

dx[1+α₀+α₂P₂(x)]³. (4.104) The integral on the far right is performed using the orthogonality relation (4.75) for the Legendre polynomials. By omitting terms that are higher order than quadratic in the alpha parameters, we find

4π

3 R₀³= 4π

3 R³₀+4π R³₀.

α0+α₀²+¹₅α²₂/

. (4.105)

To second order in the alpha parameters, we must putα0= −α₂²/5 to make (4.105) an equality. This proves Claim A if we setα=α2in (4.101).

Proof of Claim B: Letϕ0(r) be the electrostatic potential of the undistorted nucleus and use a step function to write the undistorted nuclear charge density asρ0(r)=(R₀−r). The same quantities

4.6 Spherical and Azimuthal Multipoles 115

for the shape-distorted nucleus areϕ(r) andρ(r, θ)=(R(θ)−r). With this notation, the change in total Coulomb energy (3.78) upon distortion is

UE= ¹₂

d³r[ρ(r)ϕ(r)−ρ0(r)ϕ0(r)]. (4.106) To evaluate the first integral in (4.106), we focus onρ(r) and expand(R(θ)−r) to second order in αand writeδ(R0−r) for the radial derivative ofδ(R0−r). This gives

ρ(r, θ)ρ0(r)+αP2(cosθ)R0δ(R0−r)

−α²₁

5R0δ(R0−r)+¹₂R₀²P₂²(cosθ)δ(R0−r)

+ · · · (4.107) It is convenient to write (4.107) asρρ0+ρ1+ρ2where the subscript indicates the power of the small parameterαin each term. The corresponding potential isϕϕ0+ϕ1+ϕ2. This means that the productρ(r)ϕ(r) contains nine terms. One of these terms isρ0ϕ0. Three other terms are of orderα³ and can be neglected. The remaining five terms can be reduced to three because Green’s reciprocity relation (Section 3.5.2) implies that

d³rρ0(r)ϕ1(r)=

d³rρ1(r)ϕ0(r) (4.108)

and

d³rρ0(r)ϕ2(r)=

d³rρ2(r)ϕ0(r). (4.109)

We conclude that (to second order inα) the distortion-induced change in Coulomb energy (4.106) is UE

d³r{ρ0ϕ1+¹₂ρ1ϕ1+ρ2ϕ0} =U0+U1+U2. (4.110) To findϕ0, it is simplest to evaluate the formula derived in Example 3.4 at the end of of Section 3.4.1:

ϕ0(r)= 1 0r

_r

0

dss²ρ0(s)+ 1 0

_∞

r

dssρ0(s). (4.111)

The charge densityρ0(r) equalswhenr < R0and zero otherwise. Therefore,

ϕ0(r)=

⎧⎪

⎪⎨

⎪⎪

⎩

(3R²₀−r²) 60

r < R, R³₀

3₀r r > R0.

(4.112)

To findϕ1, we note from (4.107) thatρ1(r) has the formρ1=σ δ(R0−r) with effective surface charge densityσ=αϕP2(cosθ)R0. This means that the matching condition (3.19) must be imposed atr=R0:

∂ϕ1

∂r

r=R₀⁻

− ∂ϕ1

∂r

r=R⁺₀

= σ

0 = αR0P2(cosθ) 0

. (4.113)

The potential itself has an exterior multipole expansion (4.95) forr > R0and an interior multipole expansion (4.96) forr < R0. However in light of (4.113), it should be sufficient to keep only the=2 term in each. Therefore, imposing the continuity ofϕ1atr=R0, we write

ϕ1(r, θ)=

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩ 1 4π 0

C2

r R0

2

P2cos(θ) r < R0, 1

4π ₀C₂ R0

r 3

P₂cos(θ) r > R₀.

(4.114)

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116 ELECTRIC MULTIPOLES: APPROXIMATE ELECTROSTATICS FOR LOCALIZED CHARGE

Substituting (4.114) into (4.113) determines the coefficientC2. The final result is

ϕ1(r, θ)=

⎧⎪

⎪⎨

⎪⎪

⎩

αr²P2(cosθ) 50

r < R0, αR₀⁵P2(cosθ)

50r³ r > R0.

(4.115)

Withρ0,ρ1,ρ2,ϕ0, andϕ1in hand, it is straightforward to evaluate the three energy integrals in (4.110) which determine UE using the Legendre orthogonality relation (4.75) and (for the ρ2ϕ0

piece) the fact thatϕ₀(r) is continuous atr=R0. The results are U0=0 U1= 2π

250

²R⁵₀α² U2= − 2π 150

²R₀⁵α². (4.116) Since=Q/(4π R₀³/3), the sum of the three energies in (4.116) gives the total change in electrostatic energy produced by a quadrupole shape distortion of a spherical nucleus with constant charge density as

UE = − 3

5 Q² 4π 0R0

+α²

5 . (4.117)

This proves Claim B because, using (3.77), the total energy of the undistorted nucleus is U0= 1

2

d³rρ0(r)ϕ0(r)= 3 5

Q² 4π 0R0

. (4.118)

4.7 Primitive and Traceless Multipole Moments

There is a deep connection between the spherical multipole expansion derived in Section 4.6 and a traceless version of the Cartesian multiple expansion derived in Section 4.1.1, namely that the number of independent traceless Cartesian multipole moments T_{ij ...m}⁽⁾ is equal to the number of spherical multipole momentsAmBeginning at=2, the number in question is smaller (often much smaller) than the number of independent primitive Cartesian multipole moments,C_{ij ...m}⁽⁾ . Our proof of these statements begins with an explicit formula for the entire primitive Cartesian multiple expansion. A convenient form is

ϕ(r)= 1 4π 0

∞

=0

C_{ij ...m}⁽⁾ N_{ij ...m}⁽⁾ (r), (4.119) where the primitive electric moment of orderis

C_{ij ...m}⁽⁾ = 1

!

d³r ρ(r)rirj· · ·rm

3 45 6

terms

(4.120) and

N_{ij ...m}⁽⁾ (r)=(−1)∇i∇j· · · ∇m

3 45 6

terms

1

r. (4.121)

A complete traceless Cartesian multipole expansion is obtained by (i) substituting the expansion (4.78) of|r−r|⁻¹into Coulomb’s formula (4.85) for the electrostatic potential; and (ii) writing out the Legendre polynomials (4.77) explicitly using ˆr·ˆr=rir_i/rrandr²=rirjδij. The first few terms

4.7 Primitive and Traceless Multipole Moments 117

1 0 1

.

²

. .

0 1 2

. . .

¹

Number of balls in A

Number of balls in B

Figure 4.18: (+1)×(+1) matrix for calculation of the number of independent components of a completely symmetric-rank tensor.

generated by this procedure are ϕ(r)= 1

4π 0

Q r + p·r

r³ +ij

rirj

r⁵ + · · · +

. (4.122)

The monopole (=0) and dipole (=1) terms in (4.122) are the same as those in the primitive multiple expansion (4.7). However, the=2 term is not the quadrupole potential (4.52) expressed in terms of the primitive momentsQij. Instead, it is the quadrupole potential (4.64) expressed in terms of the traceless momentsij. The complete series (4.122) is

ϕ(r)= 1 4π 0

∞ =0

T_{ij ...m}⁽⁾

terms

5 63 4 rirj· · ·rm

r²⁺¹ , (4.123)

where the scalarsT_{ij ...m}⁽⁾ are the components of anth-ordertracelessCartesian multipole moment:

T_ij⁽⁾_···m= (−1)

!

d³yρ(y)y²⁺¹ ∂

∂yi

∂

∂yj · · · ∂

∂ym

1

|y|. (4.124)

The meaning of “traceless” is that we getzerowhen the trace operation is performed over any pair of indices:

δijT_{ij ...m}⁽⁾ =δimT_{ij ...m}⁽⁾ = · · · =δmjT_{ij ...m}⁽⁾ =0. (4.125) The representation of the electrostatic potential by this traceless Cartesian expansion (4.123) is com- pletely equivalent to the primitive Cartesian expansion (4.119). However, as we will now show, the traceless representation is much more efficient.

4.7.1 Counting Multipole Moments

The primitive Cartesian momentsC_ij⁽⁾_···mdefined in (4.120) are symmetric with respect to the inter- change of any two indices. This means that the number of independent components of the tensor C⁽⁾—call this numberM—is equal to the number of ways that each ofindices takes one of the three values (x, y, z) regardless of their order. This is equivalent to counting the number of ways to place indistinguishable balls in three distinguishable urns,A,B, andC. Figure 4.18 represents this problem in matrix form. Each row indicates the number of balls put into urnA. Within any row, each column

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118 ELECTRIC MULTIPOLES: APPROXIMATE ELECTROSTATICS FOR LOCALIZED CHARGE

indicates the number of balls put into urnB. The remaining balls are put into urnC. Each shaded box of the matrix thus represents one possible placement of the balls in the three urns. This means that the number of shaded boxes in Figure 4.18 is the numberMwe seek. Hence,

M=1+2+ · · · +(+1)= ¹₂(+2)(+1). (4.126) The traceless tensorT_{ij ...m}⁽⁾ defined by (4.123) is also symmetric with respect to index interchange.

This tensor would have M independent components also, except that each of the trace conditions (4.125) reduces this number by one. Moreover, each trace represents a choice of two indices to be made equal out of a total ofindices. This means that the number of traces is the binomial coefficient

2

= !

2!(−2)! =¹₂(−1). (4.127)

Therefore, the number of independent components ofT_{ij ...m}⁽⁾ is

M= ¹₂(+2)(+1)−¹₂(−1)=2+1. (4.128) Comparing (4.126) to (4.128) shows thatMMwhen1. Beginning with the quadrupole term, the traceless moments represent the potential with increasing efficiency.

The number 2+1 can been seen in another way using Maxwell’s construction of multipole moments. The geometrical constructions shown in Figures 4.5 and 4.10 do not immediately generalize to higherprimitivemoments because there is no unique way to choose and orient two quadrupoles, two octupoles, etc. so they are “equal and opposite” as we did with two charges and two dipole vectors.

However, fortracelessmoments, the dipole and quadrupole potential formulae (4.13) and (4.54)can be generalized. It turns out¹⁵that the potential of a “point 2-pole” located at the origin can be written in the form

ϕ(r)= (−) 4π ₀

terms

5 63 4 (a· ∇)(b· ∇)· · ·(p· ∇)1

r. (4.129)

This representation is correct only if|a| = |b| = · · · = |p|. Thus, the potential (4.129) is characterized by two Cartesian components for each vector and one (common) magnitude, i.e., 2+1 independent parameters.

It is very satisfying to find that the spherical expansion (4.86) and the traceless Cartesian expansions (4.123) and (4.129) all involve a multiple moment with 2+1 components for each spatial factor r⁻⁽⁺¹⁾. This means we can generalize the remarks made at the end of Section 4.6.1 for the dipole and quadrupole and assert that eachA mcan be written as a linear combination of the 2+1 components of T⁽⁾ and vice versa. In the language of group theory, both representations of the potential are irreducible, i.e., maximally efficient in our language.