Nucleation of true vacuum
61
which has degenerate minima at ~ = ±JJ./ v'A with Vo = O. Then (2.178) leads
(exercise 5) to
tP - 4>0 = :r tanh [ I (p - PO)]
(2.180)
where 4>0 is the value of tP at some reference value po of p. Choosing po to be
the value at which ~ takes the average of its values in the true and false vacua,
namely 4>0 = 0, then
tP = ~ tanh [!!:.(P - PO)].
(2.181)
v'f.
2
Assuming that PO » J1. -I, the length scale on which tP varies, then
J1.
as p -+ 0
(2.182)
tP -+ - v'A
and, in any case,
JJ.
as p -+ 00.
(2.183)
tP -+ .fi
It will be seen later that this is correct for the bounce solution that minimizes
SE. If (after lifting the degeneracy of the two vacua using the O(f) term in V)
~ = tP- = -J1./v'A is the true vacuum and tP = tP+ = JJ./.fi is the false vacuum,
then the bounce solution describes a bubble of true vacuum embedded in the false
vacuum with wall thickness of order J1. -I, where the rapid variation of tP occurs,
separating the two regions. Under the assumption that PO » J1.- I , the radius of
the bubble is large compared with the thickness of the wall, which explains the
'thin-wall' approximation terminology.
The next steps are to calculate B and to justify the various assumptions made.
In the thin-wall approximation.
~(p) = - ~
for p «po
(2.184)
v'f.
_ J1.
[J1.
- v' A tanh i(p - PO)]
for p ~ PO
(2.185)
_ J1.
for p» po.
(2.186)
-..If.
The contribution to SE from outside the wall (p » po) is zero because here
!(dtP/dp)2 + V(~) ~ O. The contribution from inside the wall (p « po) is
obtained by first noting that here
1 (d~)2
2" dp + V(tP) ~ -£.
(2.187)
As a consequence, the contribution to SE from inside the wall is
1l'2
SE ~ -T£pg.
(2.188)