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Inflation in super gravity
be solved. Let this minimum be at tP = a. From (2.158), supersymmetry
conservation requires
aw
ai+t/>W =0.
(8.5)
From (8.1), V = 0 requires, in addition, that W = O. Thus, we require that
aw = w =0 at (8.6)
atP
In general, we require
00
00
LJ..n(ln =0
and
LnJ..n(ln-1 = O.
(8.7)
n=O
11=0
In the special case (8.4), these lead to
J..I
a = - -
and
4l.oJ..2 = Af.
(8.8)
2A2
Then,
W(tP) = IL 2 J..2(tP - a)2.
(8.9)
Then, from (8.1)
e.
the effective potential corresponding to (8.9) is (exercise 2)
V = 2 IL4J..~[tP6 - 4atP S + (00 2 + I)3 +
+ (a 4 - 00 2 + 4) Assuming that slow roll occurs from close to the origin, we need
V' (0) = O.
(8.11 )
Then
a«(l2 - I) = o.
(8.12)
For the minimum not to be at the origin (since tP must roll from a maximum), we
do not want a = O. Thus, a must be ± I and we take
(I = 1
(8.13)
without loss of generality. Then (8.9) becomes
W(tP) = IL 3 J..2(t/> - 1)2
(8.14)
(still in units where Mp = 1) and (8.10) is
V(tP) = e~IL4A~(l - tP 2 - 4s + ,;6).
(8.15)