204
6 Appendices
By (6.32), the first here integral gives 2 since you can imagine extracting one
factor of P 0 = 1 out in front of the integral to leave two such factors inside. This is
an important point: Since P 0 = 1, we can always extract a factor of P 0 from within
an integral, or, equivalently, multiply any integrand we come across by P 0 as is
convenient; this can be a handy trick to see if an integral will vanish by virtue of
(6.31). The second integral in (6.34) vanishes via precisely this trick, and the third
integral gives 2/5 by (6.32). The last integral gets dropped as we retain terms only to
order α
2
2 . To this order, the volume evaluates as
V =
4 π R
3
O
3
(1 + α 0 )
3
+
3
5
(1 + α 0 ) α
2
2 + · · ·
.
(6.35)
If volume is to be conserved, then the contents of the brace bracket in (6.35) must
equal unity. If α 0 and α 2 are presumed small, then the α 0 α
2
2 and α
3
0 terms can be
dropped; what remains is a quadratic equation in α 0 whose solution is
α 0 ∼ −
1
5
α
2
2 .
(6.36)
This result will prove valuable in computing the area and Coulomb energies.
6.5.3 The Area Integral
Figure 6.3 shows a ribbon of surface area at spherical-coordinate polar angle θ and
angular width dθ that goes all the way around the nucleus. The area of the ribbon will
be its arc length times its circumference 2πr (sinθ). However, the deformed nucleus
does not have a spherical profile, so the arc length is not simply r dθ. Rather, we have
to compute it by using the general expression for arc-length in spherical coordinates
for a trajectory running along a line of constant “longitude” φ:
ds
2
= dr
2
+ r
2 dθ
2
.
(6.37)
Since r is a function of θ , we can write this as
ds
2
= dr
2
+ r
2 dθ
2
= r
2 dθ
2
1 +
1
r 2
dr
dθ
2
.
(6.38)
6 Appendices
By (6.32), the first here integral gives 2 since you can imagine extracting one
factor of P 0 = 1 out in front of the integral to leave two such factors inside. This is
an important point: Since P 0 = 1, we can always extract a factor of P 0 from within
an integral, or, equivalently, multiply any integrand we come across by P 0 as is
convenient; this can be a handy trick to see if an integral will vanish by virtue of
(6.31). The second integral in (6.34) vanishes via precisely this trick, and the third
integral gives 2/5 by (6.32). The last integral gets dropped as we retain terms only to
order α
2
2 . To this order, the volume evaluates as
V =
4 π R
3
O
3
(1 + α 0 )
3
+
3
5
(1 + α 0 ) α
2
2 + · · ·
.
(6.35)
If volume is to be conserved, then the contents of the brace bracket in (6.35) must
equal unity. If α 0 and α 2 are presumed small, then the α 0 α
2
2 and α
3
0 terms can be
dropped; what remains is a quadratic equation in α 0 whose solution is
α 0 ∼ −
1
5
α
2
2 .
(6.36)
This result will prove valuable in computing the area and Coulomb energies.
6.5.3 The Area Integral
Figure 6.3 shows a ribbon of surface area at spherical-coordinate polar angle θ and
angular width dθ that goes all the way around the nucleus. The area of the ribbon will
be its arc length times its circumference 2πr (sinθ). However, the deformed nucleus
does not have a spherical profile, so the arc length is not simply r dθ. Rather, we have
to compute it by using the general expression for arc-length in spherical coordinates
for a trajectory running along a line of constant “longitude” φ:
ds
2
= dr
2
+ r
2 dθ
2
.
(6.37)
Since r is a function of θ , we can write this as
ds
2
= dr
2
+ r
2 dθ
2
= r
2 dθ
2
1 +
1
r 2
dr
dθ
2
.
(6.38)