#format jsmath
== Diffusion ==

Diffusion turns curvatures in distributions into changes in time.  The effect on a point on a curve is to take a little away from the point, and add a little from what is on both sides of it.   This can be expressed as


$$ { { \partial F(x,t) } \over { \partial t } } ~ = ~ d { { \partial^2 F(x,t) } \over { \partial x^2 } } $$

Where $F(x,t)$ is a function in space and time and $ d $ is the diffusion constant.

As a difference equation, this can be reformulated as:

$$ {{F(x,t+\delta t) - F(x,t)}\over{\delta t}} ~ = ~ D {{ F(x + \delta x, t) +F(x - \delta x, t) - 2*F(x,t)}\over{\delta x^2}} $$

So, what is D?  Let's figure it out for Rutherford scattering pitch angle diffusion.

Assume a band of N thinsat atoms occupying a magnetic equatorial band ( B = 90° ) between magnetic radius $L$ and $L + d L $.   Each atom adds a scattering area of:

$$ { { \partial \sigma } \over { \partial \alpha } } ~ = ~ { { - \pi C^2 ~ \cos ( \theta ) } \over { 16 ~ \sin^3 ( \theta ) } } $$

For small $ \theta $:

$$ { { \partial \sigma } \over { \partial \alpha } } ~ \approx ~ { { - \pi C^2  } \over { 16 {\theta}^3 } } $$

This is in all directions.  For diffusion that changes the pitch angle $ \alpha $, we need to look at the component in the direction of the field lines only.   That will be an annular ring, radius $ \theta $, width $ d \theta $, solid area $2 \pi \theta d \theta $. 

Let $ d \alpha = \cos( \gamma ) d \theta $ .  The contribution to the diffusion is proportional to the integral of $ {\alpha}^2 $ over the circle:

$$ D \propto \int_0^{2 \pi} {\alpha}^2 d \gamma ~ = ~ \int_0^{2 \pi} \left( \cos( \gamma ) \theta \right)^2 d \gamma ~ = ~  {\theta }^2 \int_0^{2 \pi} \cos( \gamma )^2 d \gamma ~ = ~ { 1 \over 2 } {\theta }^2 $$

The area amount $ d \sigma $ is multiplied by $ d N $ and divided by the area $ 2 \pi L d L $ to make $ d D $.

$$ D ~ \approx ~ \left( { d N / 2 \pi L d L } \right) ~\int_{\theta_0}^{\theta_1} { \left( { { 1 \over 2 } { \theta }^2 } \right) \left( { { -\pi C^2  } \over { 16 {\theta}^3 } } \right) d \theta } $$

$$ D ~ \approx ~ \left( { d N } \over { d L } \right)  \left( { C^2 } \over { 64 L } \right) \ln{ \left( { {\theta_0} \over {\theta_1} } \right) } $$


$ \theta_0 $ is a very small angle that is a function of the crystal lattice spacing, about 140 picomenters (WAG) - if the distance $ s $ approaches that, then the nucleus is shielded by the electrons and the deflecting charge is zero.  

$ \theta_1 $ must be less than the small angle approximation.

'''This needs reworking''', but the point is that we can derive a diffusion constant $ D $, given the energy and the density of the server sky aluminum atoms over the !McIlwain L band.

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