A Geometric Interpretation 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(r)}{\partial r}(L \cdot S)[/math]

The part we will concentrate on is

[math]\frac{\partial V(r)}{\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 x}\hat{n}_1 - \frac{\partial F_x}{\partial y}\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(r)}{\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(r)}{\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(r)}{\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]

Here

[math] L \cdot S = |L| |S| \cos \theta [/math]

appears like an equation describing an angle between two vector quantities, the momentum and it's spin coupling.

we replaced like terms with the inverse curl operator

[math] F \cdot \hat{n} =(\nabla \times)^{-1} \mathbf{A}[/math]

[math](\frac{\partial F_y}{\partial z} - \frac{\partial F_x}{\partial z}) = \mathbf{A}[/math]

so that

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

is related to the quantity

[math](\nabla \times)^{-1} \mathbf{A}[/math]

(which if my memory serves right) shouldn't be much of a surprise because the inverse of the curl involves the potential of a system, so that the Larmor energy can be written as

[math]\Delta H = \frac{2 \mu}{\hbar Mc^2 e} r^{-1} (\nabla \times)^{-1} \mathbf{A} |L||S| cos \theta[/math]

The geometric interpretation of this equation isabout the orbit itself, a closed curve displacement. Specifically, the perimeter of the closed curve in which the electron is moving in, in vector notation.