What gauge is used in the Lagrangian for a non-relativistic point particle in an electromagnetic potential

Question

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 $\phi$?

Answer 1

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.