246
T. R. Routray et al.
l
0 = ρ 0
t 0
2
(1 − x 0 ) +
W +
B
2
− H −
M
2
f (r )d
3 r
, (17.43a)
ul
0 = ρ 0
t 0
2
(2 + x 0 ) +
W +
B
2
f (r )d
3 r
,
(17.43b)
l
γ =
t 3
12
ρ
γ +1
0
(1 − x 3 ),
(17.43c)
ul
γ =
t 3
12
ρ
γ +1
0
(2 + x 3 ),
(17.43d)
l
ex = ρ 0
M +
H
2
− B −
W
2
f (r )d
3 r,
(17.43e)
ul
ex = ρ 0
M +
H
2
f (r )d
3 r.
(17.43f)
The neutron (proton) single particle potential in (17.16), (17.17) now becomes
u
n( p)
(k, ρ, Y P ) =
l
0
ρ n( p)
ρ 0
+
ul
0
ρ p(n)
ρ 0
+
l
γ
ρ
γ +1
0
ρ n( p) +
ul
γ
ρ
γ +1
0
ρ p(n)
ρ
1 + bρ
γ
+
l
ex
ρ n( p)
ρ 0
I (k, k n( p) ) +
ul
ex
ρ p(n)
ρ 0
I (k, k p(n) )
+
l
γ (ρ
2
n + ρ
2
p )
2ρ
γ +1
0
+
ul
γ ρ n ρ p
ρ
γ +1
0
γρ
γ −1
(1 + bρ) γ +1 ,
(17.44)
where
I (k, k n( p) ) =
j 0 (kr)
3 j 1 (k n( p) r )
k n( p) r
f (r )d
3 r
f (r )d 3 r
.
(17.45)
In SNM, the energy density H (ρ) and single particle potential u(k, ρ) can be obtained
from (17.40) and (17.44), respectively, by substituting ρ n = ρ p =
ρ
2
and k n = k p =
k f , and the resulting expressions are
H (ρ) =
3
2 k
2
f ρ
10m
+
0
2
ρ
2
ρ 0
+
γ
2ρ
γ +1
0
ρ
2
ρ
1 + bρ
γ
+
ex
2ρ 0
ρ
2 J (k f ), (17.46)
and
u(k, ρ) =
0
ρ 0
ρ +
γ
ρ
γ +1
0
1 + bρ +
γ
2
ρ
1 + bρ
(γ +1)
+
ex
ρ 0
ρ I (k, k f ), (17.47)