Diff for "Diffusion"

Differences between revisions 3 and 4

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 ...

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-  ⇤ ← Revision 3 as of 2013-08-28 03:15:24 → 
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  Editor: KeithLofstrom
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+   ← Revision 4 as of 2013-08-28 04:21:38 → ⇥
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  Editor: KeithLofstrom
  Comment:
-Deletions are marked like this.
+Additions are marked like this.
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-Assume a band of N thinsat atoms occupying a magnetic equatorial band ( B = 0 ) between magnetic radius $L$ and $L + \delta L $.   Each
atom adds a scattering of
+Assume a band of N thinsat atoms occupying a magnetic equatorial band ( B = 90&deg; ) between magnetic radius $L$ and $L + d L $.   Each atom adds a scattering area of:
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-$$ { \partial Area } \over { \partial \theta } =  { { - \pi C^2 ~ \cos ( \Delta\alpha / 2 ) } \over { 16 ~ \sin^3 ( \Delta\alpha / 2 ) } }  $$
+$$ { { \partial \sigma } \over { \partial \alpha } } ~ = ~ { { - \pi C^2 ~ \cos ( \theta ) } \over { 16 ~ \sin^3 ( \theta ) } } $$

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 ...'''