1.2 Classical Nucleation Theory
17
Now we may consider full solutions. We may treat the system size, V, and time,
t, independently:
dP(V, t) = (∂ P/∂ V ) · dV + (∂ P/∂t) · dt
(1.2.25)
Then, Eq. (1.2.18) for unit time and Eq. (1.2.24) for unit system size yield the
general form
P(V, t) = 1 − [1 − P(dV, dt)]
(V /dV ).(t/dt)
(1.2.26)
We selected the pre-factor of the nucleation rate to be the inverse of the expected
induction time, τ, at which it is equally likely that the system has nucleated or not,
or equivalently, the survival probability becomes 50% [8]. A pertinent question is:
how can one find τ 2 that renders P(V 2 , t 2 ) = F(V 2 , τ 2 ) = 0.5 when the three other
variables of V 1 , V 2 , and τ 1 are known.
We may consider two systems, Systems 1 and 2, that have the system sizes of V 1
and V 2 , respectively, and apply Eq. (1.2.26) to each of the two systems. Then
P(V 2 , t 2 ) = 1 − [1 − P(dV, dt)]
(V 2 /dV )·(t 2 /dt)
(1.2.27)
for the first system and
P(V 1 , t 1 ) = 1 − [1 − P(dV, dt)]
(V 1 /dV )·(t 1 /dt)
(1.2.28)
for the second system. Taking the natural logarithm of both sides of Eq. (1.2.27)
yields
ln[1 − P(V 2 , t 2 )] = (V 2 /dV ) · (t 2 /dt) · ln[1 − P(dV, dt)]
(1.2.29)
Likewise, taking the natural logarithm of both sides of Eq. (1.2.28) yields
ln[1 − P(V 1 , t 1 )] = (V 1 /dV ) · (t 1 /dt) · ln[1 − P(dV, dt)]
(1.2.30)
Dividing Eq. (1.2.29) by Eq. (1.2.30) then yields
ln[1 − P(V 2 , t 2 )]/ ln[1 − P(V 1 , t 1 )] = (V 2 /V 1 ) · (t 2 /t 1 )
(1.2.31)
We defined the most probable or expected induction time of System 1, τ 1 , so that
P(V 1 , τ 1 ) becomes P(V 1 , τ 1 ) = F(V 1 , τ 1 ) = 0.5. Therefore
ln[1 − P(V 2 , t 2 )] = −(V 2 /V 1 ) · (t 2 /τ 1 ) · ln 2 = −Ct 2
(1.2.32)
where C = (V 2 /V 1 )·(ln 2/τ 1 ). Solving Eq. (1.2.32) yields
P(V 2 , t 2 ) = 1 − exp(−Ct 2 )
(1.2.33)