Diff for "Diffusion"

Differences between revisions 5 and 6

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_{\theta_0}^{\theta1} { \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) ( ~\int_{\theta_0}^{\theta1} { { d \theta } \over \theta }

\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 thart, 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

-  ⇤ ← Revision 5 as of 2013-08-28 04:23:18 → 
  Size: 2236
  Editor: KeithLofstrom
  Comment:
+   ← Revision 6 as of 2013-08-28 14:07:00 → ⇥
  Size: 2769
  Editor: KeithLofstrom
  Comment:
-Deletions are marked like this.
+Additions are marked like this.
 Line 33:
-$$ D ~ \approx ~ \left( { d N / 2 \pi L d L } \right) ~\int_0^{\infty} { \left( { { 1 \over 2 } { \theta }^2 } \right) \left( { { -\pi C^2  } \over { 16 {\theta}^3 } } \right) d \theta } $$
+$$ D ~ \approx ~ \left( { d N / 2 \pi L d L } \right) ~\int_{\theta_0}^{\theta1} { \left( { { 1 \over 2 } { \theta }^2 } \right) \left( { { -\pi C^2  } \over { 16 {\theta}^3 } } \right) d \theta } $$
 Line 35:
-'''Not right ... we need to integrate from the smallest (but not zero) angle, set by atomic radius perhaps ???'''
+$$ D ~ \approx ~ \left( { d N } \over { d L } \right)  \left( { C^2 } \over { 64 L } \right) ( ~\int_{\theta_0}^{\theta1} { { d \theta } \over \theta } $$
 Line 38:
+$ \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 thart, then the nucleus is shielded by the electrons and the deflecting charge is zero.
-Line 39:
+Line 40:
+$ \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.