TrappedLight Posted September 19, 2013 Posted September 19, 2013 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. I don't know why some of these equations are not showing up :/
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