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For the Lagrangian $$L = \frac{1}{2}mu^2 - q(\phi - \frac{\vec{A}}{c}\cdot \vec{u})$$ of a non-relativistic point particle in an electromagnetic potential, what gauge is used for the electromagnetic potential $\vec{A}$ and $\phi$?

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up vote 8 down vote accepted

There is no specific gauge being used in this electromagnetic Langrangian interacting with classical particles; in other words, the Lagrangian you wrote is still gauge-invariant so you may use any gauge.

To show that it's gauge invariant, perform a gauge transformation $$ \phi \to \phi + \frac{\partial}{\partial t} \lambda,\quad \vec A\to \vec A + c \nabla \lambda $$ where $\lambda = \lambda(\vec x,t) = \lambda(\vec x(t),t)$. With this change of the potentials, your Lagrangian will change to $$ L\to L - q \frac{\partial}{\partial t} \lambda - q\vec u \cdot\nabla \lambda $$ The final $L$ may be rewritten if one knows that $\vec u = \partial \vec x / \partial t$: $$ L\to L - q \frac{\partial}{\partial t} \lambda - \frac{q\partial \vec x}{\partial t} \cdot\nabla \lambda $$ But via something like the chain rule, that's nothing else than the total time derivative of $\lambda$: $$ L \to L - q \frac{d}{dt}\lambda $$ which is seen because $\lambda$ is a function of $\vec x, t$ but we must use the right value of the argument $\vec x$, and because $\vec x$ itself depends on $t$, the total time derivative is given by the partial time derivative plus the extra spatial term.

Now, the action is $$ S = \int dt\, L $$ and if you add a "total time derivative" to the Lagrangian, just like we did above, this extra term may be integrated and gives you the difference of $q\lambda$ at $t=+\infty$ and $t=-\infty$. Assuming that at both of these extreme moments, the particle had or will have positions $\vec x$ that are far from the spacetime region where the gauge transformation was nontrivial i.e. where $\lambda$ was nonzero, the integral of the total derivative vanishes. So we have $$ S \to S,$$ the action is gauge-invariant. It's common in physics that the argument why the boundary things - and total derivatives - don't matter is simplified. There are other reasons to ignore the total derivatives "immediately" and "generally".

Generalizations: quantum physics and QFT

In quantum physics, particles are described as fields, so one would couple the particles to the electromagnetic field a bit differently, e.g. via $\bar\psi \gamma^\mu A_\mu \psi$, and the phase of the field $\psi$ would have to be changed by something like $\exp(i\lambda)$ to make the gauge invariance work.

But it would still be true that the Lagrangian as naturally written is gauge-invariant. If one couldn't write basic Lagrangians (e.g. for a single charged particle) that are gauge-invariant, gauge invariance wouldn't exist (it wouldn't really be possible in an interesting theory) and we wouldn't ever be talking about it. If you want the Lagrangian to pick a specific gauge, you have to include some extra gauge-fixing terms to the Lagrangian such as $K'(\partial_\mu A^\mu)^2$.

But it would be silly to do such a thing without a good reason: in most cases, gauge invariance is a damn useful symmetry (more precisely: a redundancy) that allows you to simplify some calculations. In quantum field theory, a good reason to gauge-fix the Lagrangian is the desire to have a well-defined photon propagator. The propagator is the inverse of the differential operator in the equations of motion; without gauge-fixing, the operator has a reduced rank and its inverse would be singular.

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Good answer. It should probably be stressed that the $A_{\mu}$ field in the Lagrangian $L$ mentioned in the question is not dynamically active, but just an electromagnetic background, which is required to satisfy Maxwell's eqs. (The dynamically active variable is the position of the point charge.) To make $A_{\mu}$ dynamically active as well, we should first introduce the standard $F_{\mu\nu}^2$ term, and secondly, gauge-fixing terms. – Qmechanic May 10 '11 at 22:44
Thanks, Qmechanic, but nope. $A_\mu$ may be treated as a background but even with this Lagrangian, it may be perfectly dynamical as well, so that the charged fields influence the electromagnetic field and vice versa. It's exactly how physics was supposed to work throughout the 19th century. The Lagrangian is that of ordinary electrodynamics so why should the key field be non-dynamical? Of course, there should also be $(E^2-B^2)/2$ in the Lagrangian which is gauge-invariant, too. Again, there is no need to gauge-fix it - don't get confused. Gauge symmetry is a virtue not vice. – Luboš Motl May 11 '11 at 4:38

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