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'''Not right ...''' '''Not right ... we need to integrate from the smallest (but not zero) angle, set by atomic radius perhaps ???'''

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 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_0^{\infty} { \left( { { 1 \over 2 } { \theta }^2 } \right) \left( { { -\pi C^2 } \over { 16 {\theta}^3 } } \right) d \theta }

Not right ... we need to integrate from the smallest (but not zero) angle, set by atomic radius perhaps ???

MoreLater

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