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