## 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

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 as

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

With , and taking partials too, we have the system of equations

\begin{subequations}

\end{subequations}

# Classical interaction

Starting with the free particle Hamiltonian

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

From 2.4b we find

or

Taking derivatives and employing 2.4a we have

Rearranging and utilizing the convective derivative expansion (ie: chain rule), we have

We guess and expect that the first term of 3.9 is . Let’s verify this

Since we have

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

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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