Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Classical Electrodynamic gauge interaction.

Posted by peeterjoot on October 22, 2010

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Motivation.

In [1] chapter 6, we have a statement that in classical mechanics the electromagnetic interaction is due to a transformation of the following form

\begin{aligned}\mathbf{p} &\rightarrow \mathbf{p} - \frac{e}{c} \mathbf{A} \\ E &\rightarrow E - e \phi\end{aligned} \hspace{\stretch{1}}(1.1)

Let’s verify that this does produce the classical interaction law. Putting a more familiar label on this, we should see that we obtain the Lorentz force law from a transformation of the Hamiltonian.

Hamiltonian equations.

Recall that the Hamiltonian was defined in terms of conjugate momentum components p_k as

\begin{aligned}H(x_k, p_k) = \dot{x}_k p_k - \mathcal{L}(x_k, \dot{x}_k),\end{aligned} \hspace{\stretch{1}}(2.3)

we can take x_k partials to obtain the first of the Hamiltonian system of equations for the motion

\begin{aligned}\frac{\partial {H}}{\partial {x_k}} &= - \frac{\partial {\mathcal{L}}}{\partial {x_k}}  \\ &= - \frac{d}{dt} \frac{\partial {\mathcal{L}}}{\partial {\dot{x}_k}} \end{aligned}

With p_k \equiv {\partial {\mathcal{L}}}/{\partial {\dot{x}_k}}, and taking p_k partials too, we have the system of equations

\begin{subequations}

\begin{aligned} \frac{\partial {H}}{\partial {x_k}} &= - \frac{d p_k}{dt}\end{aligned} \hspace{\stretch{1}}(2.4a)

\begin{aligned} \frac{\partial {H}}{\partial {p_k}} &= \dot{x}_k\end{aligned} \hspace{\stretch{1}}(2.4b)

\end{subequations}

Classical interaction

Starting with the free particle Hamiltonian

\begin{aligned}H = \frac{\mathbf{p}}{2m},\end{aligned} \hspace{\stretch{1}}(3.5)

we make the transformation required to both the energy and momentum terms

\begin{aligned}H - e\phi = \frac{\left(\mathbf{p} - \frac{e}{c} \mathbf{A}\right)^2 }{2m} = \frac{1}{{2m}} \mathbf{p}^2 - \frac{e}{m c} \mathbf{p} \cdot \mathbf{A} + \frac{1}{{2m}} \left(\frac{e}{c}\right)^2 \mathbf{A}^2 \end{aligned} \hspace{\stretch{1}}(3.6)

From 2.4b we find

\begin{aligned}\frac{d x_k}{dt} = \frac{\partial {H}}{\partial {p_k}} = \frac{1}{{m}} \left( p_k - \frac{e}{c} A_k \right),\end{aligned} \hspace{\stretch{1}}(3.7)

or

\begin{aligned}p_k = m \frac{d x_k}{dt} + \frac{e}{c} A_k.\end{aligned} \hspace{\stretch{1}}(3.8)

Taking derivatives and employing 2.4a we have

\begin{aligned}\frac{d p_k}{dt} &= m \frac{d^2 x_k}{dt^2} + \frac{e}{c} \frac{d A_k}{dt}  \\ &= -\frac{\partial {H}}{\partial {x_k}} \\ &=\frac{1}{{m}} \frac{e}{c} p_n \frac{\partial {A_n}}{\partial {x_k}} - e \frac{\partial {\phi}}{\partial {x_k}}- \frac{1}{{m}} \left(\frac{e}{c}\right)^2 A_k \frac{\partial {A_k}}{\partial {x_k}} \\ &=\frac{1}{{m}} \frac{e}{c} \left(m \frac{d x_n}{dt} + \frac{e}{c} A_n\right)\frac{\partial {A_n}}{\partial {x_k}} - e \frac{\partial {\phi}}{\partial {x_k}}- \frac{1}{{m}} \left(\frac{e}{c}\right)^2 A_k \frac{\partial {A_k}}{\partial {x_k}} \\ &=\frac{e}{c} \frac{d x_n}{dt}\frac{\partial {A_n}}{\partial {x_k}} - e \frac{\partial {\phi}}{\partial {x_k}}\end{aligned}

Rearranging and utilizing the convective derivative expansion d/dt = (d x_a/dt) {\partial {}}/{\partial {x_a}} (ie: chain rule), we have

\begin{aligned}m \frac{d^2 x_k}{dt^2} &=\frac{e}{c} \frac{d x_n}{dt}\left( \frac{\partial {A_n}}{\partial {x_k}}- \frac{\partial {A_k}}{\partial {x_n}} \right) - e \frac{\partial {\phi}}{\partial {x_k}}\end{aligned} \hspace{\stretch{1}}(3.9)

We guess and expect that the first term of 3.9 is e (\mathbf{v}/c \times \mathbf{B})_k. Let’s verify this

\begin{aligned}(\mathbf{v} \times \mathbf{B})_k&= \dot{x}_m B_d \epsilon_{k m d} \\ &= \dot{x}_m ( \epsilon_{d a b} \partial_a A_b ) \epsilon_{k m d} \\ &= \dot{x}_m \partial_a A_b \epsilon_{d a b} \epsilon_{d k m}\end{aligned}

Since \epsilon_{d a b} \epsilon_{d k m} = \delta_{a k} \delta_{b m} - \delta_{a m} \delta_{b k} we have

\begin{aligned}(\mathbf{v} \times \mathbf{B})_k&= \dot{x}_m \partial_a A_b \epsilon_{d a b} \epsilon_{d k m} \\ &=\dot{x}_m \partial_a A_b \delta_{a k} \delta_{b m} -\dot{x}_m \partial_a A_b \delta_{a m} \delta_{b k} \\ &= \dot{x}_m ( \partial_k A_m - \partial_m A_k )\end{aligned}

Except for a difference in dummy summation variables, this matches what we had in 3.9. Thus we are able to put that into the traditional Lorentz force vector form

\begin{aligned}m \frac{d^2 \mathbf{x}}{dt^2} &= e \frac{\mathbf{v}}{c} \times \mathbf{B} + e \mathbf{E}.\end{aligned} \hspace{\stretch{1}}(3.10)

It’s good to see that we get the classical interaction from this transformation before moving on to the trickier seeming QM interaction.

References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

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