
Now we can look at some typical cases
and see what this formula means. However, first we have to find the right
values for l 


For this we have to take the given
values of the initial cooling rate, which we call l', and see what l
values correspond to these cooling rates. 


The initial cooling rate l' is the derivative of the T(t)
function at t = t_{0} = 0, we thus have 


d
dt 
(T_{0} · exp –
l · t 
÷
÷

t = 0 
= 
l' =

– l
·T_{0} · exp –l ·
t 
÷
÷

t = 0 
= – l ·T_{0} 




and obtain 





The "–" sign cancels,
because our l' must carry a minus sign, too,
if it is to be a cooling and not a heating rate. 

Replacing l by l'/T_{0} yields the final formula:



L = 
æ
ç
è 
2D_{0} ·
kT_{0}^{2}
l' · E 
ö
÷
ø 
1/2 
· exp 

E_{}
2kT_{0} 




We have to evaluate this formula for cooling
rates l' given as (–) 1
^{o}K/s, 10 ^{o}K/s, 50 ^{o}K/s,
10^{4} ^{o}K/s, and activation energies of E =
1.0 eV, 2.0 eV, 5 eV. For D_{0} we take
D_{0} = 10^{–5}
cm^{2}s^{–1}. 


The result (including the [1 –
kT_{0}/E] term is shown below 




What can we learn from the formula
and the curves? 


 The cooling rate is not all that important. Differences in the cooling rate
of a factor of 50 produce only an order of magnitude effect or less
since L is only proportional to (1/l)^{1/2}.
 The starting temperature T_{0} is slightly more
important than the activation energy E; both have the same weight
in the exponential, but T_{0} appears directly in the
preexponential while E enters only as square root.
 The preexponential factor D_{0} of the
diffusion coefficient is exactly as important as l' and E in the preexponential factor
of the equation for L



What can we do with the numbers? Quite simple:
 L gives you the average of the largest distance between some
point defect agglomerates, e.g. precipitates, because point defects farther
away than L from some nuclei cannot reach it and must form their
own agglomerate.
 The average number of point defects in an agglomerate divided by
L^{3} gives a lower limit for the point defect
concentration, because at least as many point defects as we find in an
agglomerate must have been in the volume L^{3}.



