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Now you must solve a simple looking
integral. There are several ways of doing that |
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- Find a good math book with lots of integrals and take the solution from
there (the "Bronstein", however, won't do)
- Do a sensible approximation and solve it yourself in a simple way
- Go all the way and solve it completely - if you can.
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Here we go the second route. |
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We use a Taylor expansion for
1/u around u0 because that's where
u is felt most critically - for large values of u
everything tends to be zero anyway. In full generality we have |
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If we keep it really simple, we could just use
the first term, having 1/u »
1/u0; but we will go one step beyond this and take |
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This gives us |
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The second term of the Taylor
expansion brought in the factor [1
kT0/E] and since kT0 «
E in all normal cases, it is indeed not very important. If we neglect it,
we may simply give the desired solution as |
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L = |
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2D0 · kT0
λ · E |
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1/2 |
· exp |
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E
2kT0 |
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Now we can look at some typical cases
and see what this formula means. However, first we have to find the right
values for λ |
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For this we have to take the given values of the
initial colling rate which we call λ'and see
what λ values correspond to these cooling
rates. |
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The initial cooling rate λ' is the derivative of the T(t)
function at t = t0 = 0, we thus have |
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d
dt |
(T0 · exp λ · t |
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t = 0 |
= |
λ' = |
λ ·T0
· exp λ · t |
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t = 0 |
= λ
·T0 |
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and obtain |
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The "" sign cancels,
because our λ' must carry a minus sign, too,
if it is to be a cooling and not a heating rate. |
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Replacing λ by λ'/T0 yields the final formula:
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L = |
( |
2D0 · kT02
λ' · E |
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1/2 |
· exp |
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E
2kT0 |
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We have to evaluate this formula for cooling
rates λ' given as () 1
oK/s, 10 oK/s, 50 oK/s,
104 oK/s, and activation energies of E =
1.0 eV, 2.0 eV, 5 eV. For D0 we take
D0 = 105
cm2s1. |
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The result (including the [1
kT0/E] term is shown below |
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What can we learn from the formula
and the curves? |
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- 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/λ)1/2.
- The starting temperature T0 is slightly more
important than the activation energy E; both have the same weight
in the exponential, but T0 appears directly in the
pre-exponential while E enters only as square root.
- The pre-exponential factor D0 of the
diffusion coefficient is exactly as important as λ' and E in the pre-exponential factor
of the equation for L
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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
L3 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 L3.
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© H. Föll