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

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.

MoreLater

Diffusion (last edited 2013-08-28 14:24:36 by KeithLofstrom)