1.11 A Numerical Model of the Fission Process
43
V neck = 2π R
3
π/2−γ
0
cos β(2 sin γ − cos β)
2 dβ.
(1.98)
This integral is tedious but straightforward, and gives
V neck =
2π R
3
3
13 sin
2
γ cos γ + 2 cos γ − 3 sin γ
π − 2γ + sin(2γ )
. (1.99)
On adding (1.94), (1.95), and (1.99), the volume of the partially-fissioned nucleus
can be written as
V =
4
3
π R
3 f V (γ ),
(1.100)
where
f V (γ ) = 1 + 7 sin
2
γ cos γ + 2 cos γ −
3
2
sin γ
π − 2γ + sin(2γ )
. (1.101)
The original nucleus was of volume 4π R
3
O
3, so demanding conservation of
volume during the fission process gives the radius R at any time in terms of R O and
γ as
R(γ ) = f
−1/3
V
(γ )R O .
(1.102)
Calculation of the surface area of the deformed nucleus proceeds similarly. The
element of surface area in spherical coordinates is d S = r
2 sin θ dθ dφ. The surface
area of the two end caps is
S caps = 2R
2
π−γ
θ=0
2π
φ=0
sin θ dθ dφ = 4π R
2
(1 + cos γ ).
(1.103)
The cones contribute no surface area as they are embedded within the caps. As
for the neck, look again to Fig. 1.16. An element of arc length along the edge of the
neck for angle dβ will be Rdβ. The area of the edge of the narrow disk must then be
2π x R dβ. Hence, for the entire neck, we have, with (1.96) for x,
S neck = 4π R
2
π/2−γ
0
(2 sin γ − cos β)dβ.
(1.104)
This evaluates to