1

Bohr & Wheeler Fission Theory Calculation 4 March 2009

(1) Introduction The goal here is to reproduce the calculation of the limiting Z2/A against spontaneous fission

€

Z 2

A

lim

= 2 aS

aC

. (1.1)

as first done by Bohr & Wheeler in their famous paper The Mechanism of Nuclear Fission [Phys. Rev. 56, 426-450 (1939)]. aS and aC are the surface and Coulomb energy factors, respectively ~ 18 MeV and 0.72 MeV. The development given here draws heavily on M. S. Plesset’s analysis in American Journal of Physics 9(1), 1-10 (1941). Plesset incorporates more terms in the analysis than are necessary to determine the limit and his notation is a little confusing, so what is given here is more minimalist than what he presented. The deformed nucleus is presumed to be modeled as a sphere of initial radius Ro which is distorted:

€

r θ( ) = Ro 1 + α0 + α2P2 cosθ( )[ ], (1.2)

where P2 is a Legendre polynomial. The idea is that both perturbing coefficients α0 and α2 are small. Two coefficients are included on the rationale that the volume of the nucleus is assumed to be conserved (incompressible) even as it distorts, so that one can then solve for, say, α0 in terms of α2. It turns out that the lowest-order contributions to the surface and Coulomb energies are both of order

€

α22. Thus, I carry no terms to any higher orders.

θ

r

2

There are several steps in the calculation: The volume integral – section 2 below. The surface area integral - section 3. The Coulomb electrostatic potential-energy integral – section 4. Results of the area and Coulomb integrals are combined to give the total energy, upon insisting on volume conservation. Detailed working of various integrals appear in appendices A-H.

3

(2) The Volume Integral The element of volume in spherical coordinates is

€

r2 sinθ dr dθ dφ . Hence

€

V = r2 sinθ dr dθ dφφ= 0

2π

∫r= 0

r(θ )

∫θ = 0

π

∫ . (2.1)

Note the order of integration with r and θ. The upper limit of r is a function of θ :

€

r θ( ) = Ro 1 + α0 + α2P2[ ] . (2.2) The r integral must be done first, then the θ integral. The φ integral gives 2π directly. Integrating over r:

€

V = 2π3

r3 θ( ) sinθ dθθ = 0

π

∫ . (2.3)

This integral proceeds as

€

V = 2π Ro3

3

1+α0 +α2P2[ ]3 sinθ dθθ = 0

π

∫ . (2.4)

We will have numerous integrals of this form. In all such cases, it is convenient to make a change of variable x = cosθ, which gives

€

V = 2π Ro3

3

1+α0 +α2P2[ ]3 dx−1

1

∫ . (2.5)

It is also handy to have the normalization condition for Legendre polynomials:

€

Pi Pj dx = 2δij

i + j +1−1

1

∫ . (2.6)

Expanding out (2.5):

€

V = 2π Ro3

3

1+α0( )3 dx

−1

1

∫ + 3 1+α0( )2α2 P2dx + 3 1+α0( )α22 P2

2dx + α23 P2

3dx−1

1

∫−1

1

∫−1

1

∫

.

4

The first integral gives 2, the second integral vanishes by (2.6), the third integral gives 2/5 by (2.6), and the last integral is dropped as I retain terms only to order

€

α22. Hence

€

V = 4π Ro3

3

1+α0( )3 + 3

51+α0( )α2

2

. (2.7)

If volume is to be conserved, then the contents of the brace bracket in (2.7) must equal unity:

€

1 + 3α0 + 3α02 + α0

3 + 35α22 + 3

5α0α2

2 = 1.

At this point I follow Plesset’s argument. The idea here is to solve for

€

α0 in terms of α2. Both are small. Begin by dropping the presumably very small cross term

€

α0α22 as well as the

€

α03 term;

what remains is

€

α02 + α0 + 1

5α22 = 0,

a quadratic in α0. Solving gives

€

α0 = −1 ± 1 − 4α2

2 52

~ −1 ± 1 − 2α2

2 5 + ...( )2

where I invoked a binomial expansion of the square root. To decide on which root to choose, if

€

α2 → 0 and volume is to be conserved, we must have

€

α0 → 0 as well. The upper sign is the correct choice, leaving

€

α0 ~ − 15α22 . (2.8)

This result will prove valuable in computing the area and Coulomb energies. .

5

(3) The Surface Area Integral The deformed nucleus does not have a spherical profile, so we have to use the general arc-length for spherical coordinates:

€

ds2 = dr2 + r2dθ 2 + r2 sin2θ dφ 2 . (3.1) Consider a line of “longitude” at fixed φ:

€

ds2 = dr2 + r2dθ 2 = dθ 2 r2 + drdθ

2

= r2dθ 2 1 + 1

r2drdθ

2

. (3.2)

So, the area of a ribbon of surface at polar angle θ is

€

dA = 2π r sinθ ds = 2π r2 sinθ 1 + 1r2

drdθ

2

dθ . (3.3)

Put

€

x = 1r2

drdθ

2

. Presuming x to be small, that is, that the nucleus is not greatly distorted, so

that dr/dθ ~ 0, then

€

1+ x( )1/ 2 ~ 1 + 12

x − 18

x 2 + ... . (3.4)

θ

r

ds

6

Now, the nucleus is described by

€

r θ( ) = Ro 1 + α0 + α2P2[ ] . This means that

€

dr / dθ( ) = α2 Ro dP2 / dθ( ), so, to order

€

α22, we can stop at the x/2 term in (3.4):

€

dA = 2π r2 sinθ 1 + 121r2

drdθ

2

+ ...

dθ

€

= 2π sinθ r2 + 12

drdθ

2

+ ...

dθ . (3.5)

So the surface area of the deformed nucleus, to this level of approximation, has two contributions:

€

A = 2π r2 sinθ dθ + 12

drdθ

2

sinθ dθ + ...0

π

∫0

π

∫

. (3.6)

These integrals are done in Appendices A and B:

€

r2 sinθ dθ0

π

∫ = 2Ro2 1+α0( )2 + 1

5α22

. (3.7)

€

12

drdθ

0

π

∫2

sinθ dθ = 65

Ro2α2

2. (3.8)

These give, in (3.6),

€

A ~ 4π Ro2 1+α0( )2 + 4

5α2

2 + ...

. (3.9)

Substituting into this the result of volume conservation,

€

α0 ~ −α22 / 5, [(2.8) from above] gives,

to terms of order

€

α22,

€

A ~ 4π Ro2 1 + 2

5α2

2 + ...

.

If Ω is the factor which converts surface area to equivalent energy, and invoking the usual approximation

€

Ro ~ aoA1/ 3, then we have the surface energy US as

€

US ~ aS A2 / 3( ) 1 + 25α2

2 + ...

(3.10)

7

where

€

aS = 4π Ωao2 , known to be ~ 18 MeV from fits to the semi-empirical mass formula. The

areal energy increases upon perturbation of the nucleus from its initially spherical shape.

8

(4) The Coulomb Potential Integral The charge distribution gets divided into elements of volume. The Coulomb self-potential is given by

€

UC = 12

ρ2

4πεo

dV1 dV2

r12V∫

V∫ , (4.1)

where ρ is the electrical charge density and r12 is the distance between the two elements of volume. There is one integral for volume elements labeled “1” and one for those labeled “2”. The volume elements are both of the form

€

r2 sinθ dr dθ dφ , so (4.1) is actually a sextuple integral:

€

UC = 12

ρ2

4πεo

1r12

r22

φ1

∫φ2

∫θ1

∫θ 2

∫ r12 sinθ2 sinθ1 dr1 dr2

r1

∫r2

∫ dθ1 dθ2dφ1 dφ2 . (4.2)

As in the volume and area integrals, the limits on the r’s are functions of θ since

€

r θ( ) = Ro 1 + α0 + α2P2[ ] . Again the r integrals must be done first, then the θ integrals. Great Care must be taken to keep track of the “1” and “2” integrals. For the r12 in the denominator, use the identity:

r1

r2

r12

θ12

9

€

1r12

=

r2k

r1k +1

Pk cosθ12( ), r2 < r1

k= 0

∑r1

k

r2k +1

Pk cosθ12( ), r2 > r1

k= 0

∑

(4.3)

where

€

θ12 is the angle between the directions from the origin to volume elements 1 and 2. Using dτ’s to represent volume elements in the usual way, Eq. (4.2) can be expressed more compactly as

€

UC = ρ2

8πεo

dτ 2r122( )

∫

1( )∫ dτ1, (4.4)

where the (1) and (2) are to remind that each integral is actually a triple integral over the coordinates for volume elements 1 and 2. Do the inner integral first; label it U2. Break it into two regimes, one for r2 = 0 to r1 (for which r2 < r1) and then from r1 to

€

r2 θ2( ) (for which r2 > r1), and use (4.3):

€

U2 = dτ 2r122( )

∫ = r2k

r1k +1

Pk dτ 2

k

∑0

r1

∫ + r1k

r2k +1

Pk dτ 2

k

∑r1

r2 θ 2( )

∫ . (4.5)

Note carefully that Pk still means

€

Pk cosθ12( ) . These integrals are over the “2” coordinates; the factors of r1 can be extracted but must remain within the sums. Write the volume elements as

€

dτ = r2dr dΩ:

€

U2 = 1r1

k +1 r2k +2Pkdr2 dΩ2

0

r1

∫θ ,φ∫

k

∑ + r1k r2

1−k

r1

r2 θ 2( )

∫θ ,φ∫ Pkdr2 dΩ2

k

∑ . (4.6)

The first integral over r2 is trivial and gives

€

r1k+3 k + 3( ); the second requires a bit of care for the

case of k = 2, where a logarithmic form arises:

€

U2 = r12 Pk

k + 3( )θ ,φ∫

k

∑ dΩ2 + r1k

2 − k( )r22−k θ( )Pk

θ ,φ∫

k≠2

∑ dΩ2

− r12 1

2 − k( )Pk

θ ,φ∫

k≠2

∑ dΩ2 + r12 ln

r2 θ( )r1

P2

θ ,φ∫ dΩ2.

(4.7)

10

Now, from the addition theorem for spherical harmonics, the

€

Pk cosθ12( ) can be written in terms of Associated Legendre polynomials whose arguments are the cosines of the individual direction angles of the “1” and “2” volume elements:

€

Pk cosθ12( ) = k −m( )!k + m( )!

Pkm cosθ1( )

m=−k

k

∑ Pkm cosθ2( )exp ιm φ1 −φ2( )[ ]. (4.8)

Imagine (4.8) substituted into (4.7). In the integrals over φ1 and φ2, only m = 0 will give non-zero contributions. The Associated Legendre polynomials thus reduce to regular Legendre polynomials. Designate them as Pk(1) and Pk(2) to keep track of which belongs to coordinates “1” and “2”; thus, for example, Pk(1) designates the k’th-order Legendre polynomial for coordinate set “1”.

€

U2 = r12 Pk 1( )

k + 3( )Pk 2( )

θ ,φ∫

k

∑ dΩ2 + r1

kPk 1( )

2 − k( )r22−k θ( )Pk 2( )

θ ,φ∫

k≠2

∑ dΩ2

− r12 Pk 1( )

2 − k( )Pk 2( )

θ ,φ∫

k≠2

∑ dΩ2 + r12P2 1( ) ln

r2 θ( )r1

P2 2( )

θ ,φ∫ dΩ2.

(4.9)

The first and third integrals in (4.9) will vanish except when k = 0; on integrating over φ2 and θ2 they respectively give

€

4π r12 / 3 and

€

−2π r12 , for a total of

€

−2π r12 / 3:

€

U2 = − 2π3

r12 +

r1kPk 1( )

2 − k( )r22−k θ( )Pk 2( )

θ ,φ∫

k≠2

∑ dΩ2 + r12P2 1( ) ln

r2 θ( )r1

P2 2( )

θ ,φ∫ dΩ2. (4.10)

The second and third terms on the right side are done in Appendices C and D. The second term emerges as (to order

€

α22, as usual)

€

r1kPk 1( )

2 − k( )r22−k θ( )Pk 2( )

θ ,φ∫

k≠2

∑ dΩ2 = 2π Ro2 1+α0( )2 + πα2

2 Ro2 Pk 1( )

r1Ro

k

k≠2

∑ 1− k( )1+α0( )k k,2,2{ }(4.11)

where

€

k,2,2{ } designates the integral of a product of three Legendre polynomials over θ :

€

k,2,2{ } = PkP22

−1

1

∫ dx . (4.12)

The third term in (4.10) comes out as

11

€

r12Pk 1( ) ln

r2 θ( )r1

Pk 2( )

θ ,φ∫ dΩ2 = 4π

5P2 1( )

r12α2

1+α0( ) −

π P2 1( ) r12α2

2

1+α0( )22,2,2{ }. (4.13)

Look at the last terms on the right sides of each of (4.11) and (4.13). The one that is missing from (4.11), that for k = 2, is precisely that which appears in (4.13). Gathering (4.11) and (4.13) into (4.10) then gives:

€

U2 = − 2π3

r12 + 2π Ro

2 1+α0( )2

+ 4π5

P2 1( )r12α2

1+α0( ) + πα2

2 Ro2 Pk 1( )

r1Ro

k

k

∑ 1− k( )1+α0( )k k,2,2{ }.

(4.14)

This expression goes back into (4.4). Write the volume element for coordinates “1” out explicitly:

€

UC = ρ2

8π εo

−2π3

r14dr1dΩ1

0

r1 θ( )

∫θ ,φ∫ + 2π Ro

2 1+α0( )2

r12dr1dΩ1

0

r1 θ( )

∫θ ,φ∫

+ 4π5

α2

1+α0( )P2 1( ) r1

4dr1dΩ1

0

r1 θ( )

∫θ ,φ∫ + πα2

2 Ro2 1− k( ) k,2,2{ }

Rok 1+α0( )k

k

∑ r1k +2Pk 1( ) dr1dΩ1

0

r1 θ( )

∫θ ,φ∫

.

(4.15)

The four terms on the right side are done in Appendices E, F, G, and H. The results are (to order

€

α22):

€

−2π3

r14dr1dΩ1

0

r1 θ( )

∫θ ,φ∫ = − 4π

2Ro5

152 1+α0( )5 + 4 1+α0( )3α2

2[ ] . (4.16)

€

2π Ro2 1+α0( )2 r1

2dr1dΩ1

0

r1 θ( )

∫θ ,φ∫ = 4π

2

3Ro5 2 1+αo( )5 + 6

51+αo( )3α2

2

. (4.17)

€

4π5

α2

1+α0( )P2 1( ) r1

4dr1dΩ1

0

r1 θ( )

∫θ ,φ∫ = 16π

2

25Ro5 1+α0( )3α2

2 . (4.18)

12

€

πα22 Ro

2 1− k( ) k,2,2{ }Ro

k 1+α0( )k

k

∑ Pk 1( ) r1k +2dr1dΩ1

0

r1 θ( )

∫θ ,φ∫ = 8

15π 2α2

2 Ro5 1+α0( )3 . (4.19)

(4.16) – (4.19) into (4.15) gives

€

UC = ρ2

8π εo

−4π 2Ro

5

152 1+α0( )5 + 4 1+α0( )3α2

2[ ] +

43π 2 Ro

5 2 1+α0( )5 + 651+α0( )3α2

2

+ 16π2

25Ro5 1+α0( )3α2

2 + 815π 2α2

2 Ro5 1+α0( )3

.

Rearrange

€

UC = ρ2

8π εo

π 2Ro5 1+α0( )5 − 8

15 + 8

3

+ 1+α0( )3α2

2 −1615

+ 2415

+ 1625

+ 815

,

or

€

UC = ρ2

8π εo

π 2Ro5 3215

1+α0( )5 + 12875

1+α0( )3α22

,

The charge density is given by

€

ρ = 3Z e4π Ro

3 .

so

€

UC = 9Z 2 e2

16π 2 Ro6

π 2Ro

5

8π εo

3215

1+α0( )5 + 12875

1+α0( )3α22

,

or

€

UC = 9Z 2 e2

128 π εoRo

3215

1+α0( )5 + 12875

1+α0( )3α22

.

Take the 128 into the brace bracket, and take a factor of 3 from the numerator of the prefactor into the brace bracket:

13

€

UC = 3Z 2 e2

π εoRo

1201+α0( )5 + 1

251+α0( )3α2

2

.

Extract a factor of 20 into the denominator of the prefactor:

€

UC = 3Z 2 e2

20 π εoRo

1+α0( )5 + 4

51+α0( )3α2

2

.

Now use the empirical nuclear size relationship

€

Ro = aoA1/ 3, ao ~ 1.2 fm:

€

UC = 3 e2

20 π εoao

Z 2

A1/ 3

1+α0( )5 + 4

51+α0( )3α2

2

.

The first prefactor is the Coulomb energy parameter aC ~ 0.72 MeV:

€

UC = aCZ 2

A1/ 3

1+α0( )5 + 4

51+α0( )3α2

2

. (4.20)

Now substitute the volume-conservation condition

€

α0 ~ −α22 / 5, [(2.8) from above] into (4.20):

€

UC = aCZ 2

A1/ 3

1−

15α22

5

+ 451− 15α22

3

α22

~ aCZ 2

A1/ 3

1 − 1

5α22 + ...

. (4.21)

The Coulomb self-energy decreases upon perturbation of the nucleus from its initially spherical shape.

14

(5) Limiting Z2/A Calculation From (3.10) and (4.21) we have the area and Coulomb energies of the distorted nucleus as

€

US ~ aS A2 / 3( ) 1 + 25α2

2 + ...

(3.10)

€

UC ~ aCZ 2

A1/ 3

1 − 1

5α22 + ...

. (4.21)

Fission will proceeds spontaneously if the total energy of the deformed nucleus is less than that of the initial spherical undeformed (α2 = 0) shape:

€

US + UC( )deformed − US + UC( )undeformed < 0 . (5.1) With (3.10) and (4.21) this corresponds to

€

Z 2

A > 2 aS

aC

, (5.2)

as shown by Bohr & Wheeler.

15

Appendix A:

€

r2 θ( ) sinθ dθ0

π

∫

Here we have, with

€

r θ( ) = Ro 1 + α0 + α2P2[ ]; set x = cosθ :

€

A = Ro2 1+α0( )2 dx

−1

1

∫ + 2 1+α0( )α2 P2 dx + α22 P2

2 dx−1

1

∫−1

1

∫

. (A.1)

The orthogonality relation is

€

Pi Pj dx = 2δij

i + j +1−1

1

∫ . (A.2)

The first integral gives 2, the middle integral vanishes, and the third integral gives 2/5:

€

A = 2Ro2 1+α0( )2 + 1

5α22

. (A.3)

16

Appendix B:

€

12

drdθ

2

sinθ dθ0

π

∫

Here we need to deal with the derivatives dr/dθ. Now,

€

r θ( ) = Ro 1 + α0 + α2P2[ ] . (B.1) But, the P’s are functions of cosθ. Let x = cosθ as usual:

€

dPdθ

= dPdx

dxdθ

= − sinθ dPdx

. (B.2)

So

€

drdθ

= − Ro sinθ α2dP2dx

. (B.3)

Hence

€

B = 12

drdθ

2

sinθ dθ0

π

∫ = α22 Ro

2

2dP2dx

2

sin3θ dθ0

π

∫ = α22 Ro

2

2dP2dx

2

1− x 2( ) dx−1

1

∫ (B.4)

Now, P2 = (3x2-1)/2, so dP/dx = 3x, hence

€

B = 92α22 Ro

2 x 2 1− x 2( ) dx−1

1

∫ = 65α22 Ro

2. (B.5)

17

Appendix C:

€

r1kPk 1( )

2 − k( )r22−k θ( )Pk 2( )

θ ,φ∫

k≠2

∑ dΩ2

Begin by substituting

€

r θ( ) = Ro 1 + α0 + α2P2[ ]:

€

C = r1

kPk 1( )Ro2−k

2 − k( )1 + α0 + α2P2 k( )[ ]

2−kPk 2( )

θ ,φ∫

k≠2

∑ dΩ2 . (C.1)

Within the square bracket, factor out

€

1+α0( ) :

€

C = r1

kPk 1( )Ro2−k 1+α0( )2−k

2 − k( )1 +

α2P2 k( )

1+α0( )

2−k

Pk 2( )θ ,φ∫

k≠2

∑ dΩ2 . (C.2)

For the square bracket, do a binomial expansion to order

€

α22; also break up the Ro term:

€

C = Ro2 r1

Ro

k Pk 1( ) 1+α0( )2−k

2 − k( )1 +

2 − k( )1+α0( )

α2P2 2( ) + 2 − k( ) 1− k( )2 1+α0( )2

α22P2 2( )

2 + ...

Pk 2( )

θ ,φ∫

k≠2

∑ dΩ2 .

(C.3) Consider the terms in the integrand. For the first, only k = 0 will give a non-zero contribution, and the integral of P0(2) over all (θ,φ) is 4π. The second would give a non-zero contribution for k = 2 [the integral of Pk(2) P2(2)], but k cannot have this value in the sum, so this term contributes nothing. Hence, to order

€

α22,

€

C = 2π Ro2 1+α0( )2 + α2

2Ro2 r1

Ro

k 1− k( )Pk 1( )

2 1+α0( )k P2 2( )2 Pk 2( )

θ ,φ∫

k≠2

∑ dΩ2 . (C.4)

Integrating over φ gives 2π ; the factor of 2 cancels that in the denominator of the sum. For the θ integral, transform to x = cosθ, and use the

€

k,2,2{ } notation of the main text:

€

C = 2π Ro2 1+α0( )2 + πα2

2Ro2 r1

Ro

k 1− k( )Pk 1( )

1+α0( )k

k≠2

∑ k,2,2{ }. (C.5)

18

Appendix D:

€

r12P2 1( ) ln

r2 θ( )r1

P2 2( )

θ ,φ∫ dΩ2

Expand the logarithm:

€

D = r12P2 1( ) ln r2 θ( )[ ]P2 2( )

θ ,φ∫ dΩ2 − ln r1( ) P2 2( )

θ ,φ∫ dΩ2

. (D.1)

The second integral vanishes. Put

€

r θ( ) = Ro 1 + α0 + α2P2[ ] in for r2(θ) in the first integral, and factor out

€

1+α0( ) as was done in Appendix C above:

€

D = r12P2 1( ) ln Ro 1+α0( ) 1 +

α2P2 2( )

1+α0( )

P2 2( )

θ ,φ∫ dΩ2. (D.2)

Again expand the logarithm; the integral with ln [

€

Ro 1+α0( )] vanishes like the second integral in (D.1) above, leaving

€

D = r12P1 1( ) ln 1 +

α2P2 2( )

1+α0( )

P2 2( )

θ ,φ∫ dΩ2 . (D.3)

Do a binomial expansion on the logarithm, again keeping terms to order

€

α22:

€

ln 1+ x( ) ~ x − 12

x 2 + ... (D.4)

hence

€

D = r12P2 1( )α2

1+α0( )P2 2( )2

θ ,φ∫ dΩ2 −

r12P2 1( )α2

2

2 1+α0( )2P2 2( )3

θ ,φ∫ dΩ2. (D.5)

The first integral in (D.5) evaluates to 4π/5, and the second to 2π

€

2,2,2{ } as in Appendix C above. Hence

€

D = 4π5

r12P2 1( )α2

1+α0( ) −

π r12P2 1( )α2

2

1+α0( )22,2,2{ } . (D.6)

19

Appendix E:

€

−2π3

r14dr1dΩ1

0

r1 θ( )

∫θ ,φ∫

Integrate over r1. Integrating over φ gives a factor of 2π . Also invoke the usual transformation x = cosθ :

€

E = − 2π3

r14dr1dΩ1

0

r1 θ( )

∫θ ,φ∫ = − 4π

2

15r15 θ( ) dx

−1

1

∫ . (E.1)

Put

€

r θ( ) = Ro 1 + α0 + α2P2[ ] . In expanding this to the fifth power, keep only terms to order

€

α22:

€

E = − 4π2Ro

5

151+α0( )5 + 5 1+α0( )4α2P2 2( ) + 10 1+α0( )3α2

2P2 2( )2 + ...[ ] dx

−1

1

∫ . (E.2)

The middle term in the integrand makes no contribution as the integral of P2(2) over x = (-1, 1) is zero. The other two terms are standard:

€

E = − 4π2Ro

5

152 1+α0( )5 + 4 1+α0( )3α2

2[ ]. (E.3)

20

Appendix F:

€

2π Ro2 1+α0( )2

€

r12dr1dΩ1

0

r1 θ( )

∫θ ,φ∫

This proceeds very much like Appendix E above. Integrate over r1, invoke a factor of 2π from integrating over φ, and invoke the usual transformation x = cosθ :

€

F = 2π Ro2 1+α0( )2 r1

2dr1dΩ1

0

r1 θ( )

∫θ ,φ∫ = 4π

2

3Ro2 1+α0( )2 r1

3 θ( )dx−1

1

∫ . (F.1)

Put

€

r θ( ) = Ro 1 + α0 + α2P2[ ] . In expanding this to the third power, keep only terms to order

€

α22:

€

F = 4π2

3Ro5 1+α0( )2 1+α0( )3 + 3 1+α0( )2α2P2 2( ) + 3 1+α0( )α2

2P2 2( )2 + ...[ ]dx

−1

1

∫ . (F.2)

The middle term in the integrand makes no contribution as the integral of P2(2) over x = (-1, 1) is zero. The other two terms are standard:

€

F = 4π2

3Ro5 2 1+α0( )5 + 6

51+α0( )3α2

2

. (F.3)

21

Appendix G:

€

4π5

α2

1+α0( )P2 1( ) r1

4dr1dΩ1

0

r1 θ( )

∫θ ,φ∫

This is similar to E and F above. Integrate over r1, invoke a factor of 2π from integrating over φ, and invoke the usual transformation x = cosθ :

€

G = 8π2

25α2

1+α0( )P2 1( ) r1

5 θ( ) dx−1

1

∫ . (G.1)

Put

€

r θ( ) = Ro 1 + α0 + α2P2[ ] . In expanding this to the fifth power, keep only terms to order

€

α22:

€

G = 8π2

25α2Ro

5

1+α0( )P2 1( ) 1+α0( )5 + 5 1+α0( )4α2P2 1( ) + 10 1+α0( )3α2

2P2 1( )2 + ...[ ] dx

−1

1

∫ . (G.2)

Note the presence of the factor of P2(1) in front of the square bracket in the integrand. This results in the first term within the square bracket making no contribution as the integral of P2(1) over x = (-1, 1) is zero. We can ignore the last term in the square bracket as it will lead to a term of order

€

α23 when combined with the factor of α2 outside the integral. Only the second term gives a

surviving contribution:

€

G = 8π2

5Ro5 1+αo( )3α2

2 P2 1( )2 dx

−1

1

∫ = 16π2

25Ro5 1+α0( )3α2

2 . (G.3)

22

Appendix H:

€

πα22 Ro

2 1− k( ) k,2,2{ }Rok 1+α0( )kk

∑ Pk 1( ) r1k+2dr1dΩ1

0

r1 θ( )

∫θ ,φ∫

Integrate over r1, invoke a factor of 2π from integrating over φ, and invoke the usual transformation x = cosθ :

€

H = 2π 2α22 Ro

2 1− k( ) k,2,2{ }Ro

k 1+α0( )k k + 3( )k

∑ r1k +3 θ( )

−1

1

∫ Pk 1( ) dx . (H.1)

We want to keep terms to order

€

α22, but there already is a factor of

€

α22 in front of the integral.

Hence, in expanding out

€

r1k+3 θ( ) , we need only keep the zero-th order term, that is,

€

r1k +3 θ( ) ~ Ro

k +3 1+α0( )k +3 , so we have

€

H = 2π 2α22 Ro

5 1− k( ) k,2,2{ } 1+α0( )3

k + 3( )k

∑−1

1

∫ Pk 1( ) dx . (H.2)

The only term for which the integral gives a non-zero value is k = 0; the value of the integral is 2:

€

H = 43π 2α2

2 Ro5 1+α0( )3 0,2,2{ }. (H.3)

The value of {0,2,2} is 2/5:

€

H = 815π 2α2

2 Ro5 1+α0( )3. (H.4)