Using Greens Theorem To Derive A New Form Of The Larmor Energy

[Aethelwulf]

Using Greens Theorem to Derive a new form of the Larmor Energy

 

The Larmor energy is written as a Hamiltonian

 

[math]\Delta H = \frac{2 \mu}{\hbar Mc^2 e} r^{-1}\frac{\partial V®}{\partial r}(L \cdot S)[/math]

 

The part we will concentrate on is

 

[math]\frac{\partial V®}{\partial r}[/math]

 

And we will use Greens theorem to derive an equivalence with this expression. We begin with the determinant

 

[math]\nabla \times F = \begin{vmatrix}\hat{n}_1 & \hat{n}_2 & \hat{n}_3 \\ \partial_x & \partial_y & \partial_z \\F_x & F_y & 0 \end{vmatrix}[/math]

 

You can write this as

 

[math]\nabla \times F = \frac{\partial F_y}{\partial z}\hat{n}_1 - \frac{\partial F_x}{\partial z}\hat{n}_2 + (\frac{\partial F_y}{\partial z} - \frac{\partial F_x}{\partial z})\hat{n}_3[/math]

 

The first set of terms cancel out

 

[math]\nabla \times F = (\frac{\partial F_y}{\partial z} - \frac{\partial F_x}{\partial z})\hat{n}_3[/math]

 

A unit vector squared just comes to unity, so if you multiply a unit vector of both sides we get

 

[math]\nabla \times F \cdot \hat{n} = (\frac{\partial F_y}{\partial z} - \frac{\partial F_x}{\partial z})[/math]

 

Now, a force equation can be given as

 

[math]F = \frac{\partial V®}{\partial r} \hat{n}[/math]

 

Notice, apart from the unit vector, this is an identical term found in the Larmor energy. Again, if one multiplies the unit vector on both sides we get

 

[math]F \cdot \hat{n} = \frac{\partial V®}{\partial r}[/math]

 

A quick check over the original Larmor energy

 

[math]\Delta H = \frac{2 \mu}{\hbar Mc^2 e} r^{-1}\frac{\partial V®}{\partial r}(L \cdot S)[/math]

 

Shows that the Larmor energy can be written as

 

[math]\Delta H = \frac{2 \mu}{\hbar Mc^2 e} r^{-1} F \cdot \hat{n} (L \cdot S)[/math]

 

 

http://en.wikipedia.org/wiki/Spin%E2%80%93orbit_interaction

Edited January 26, 2013 by Aethelwulf