# Gauge invariance

### Electric potential

We normally write the electric potential as $V = V(x,y,z)$, such that the field, $\vec{E}$, is $$\vec{E} = -\nabla V$$

The electric force is then $$\vec{F} = q \vec{E} = -q\nabla V = -\nabla (qV)$$ so $$U = qV$$ is the potential we use in the Lagrangian in the definition of the action.

Notice that, we can add any scalar constant, $\lambda$, to the potential and not affect the force, since $$\nabla (U + \lambda) = \nabla U$$ for any $U = U(x,y,z)$.

### Magnetic potential

The magnetic force is \begin{align*}q\left( \vec{v} \times \vec{B} \right)&= \vec{v} \times \left( \vec{qB} \right)\\&= \vec{v} \times \left( q\left( \nabla \times \vec{A} \right) \right)\\&= \vec{v} \times \left( \nabla \times \left( q\vec{A} \right) \right)\end{align*}

We want to include a contribution to the action from the magnetic potential, which appears in the above (in an analogous form to the electric potential) as $q\vec{A}$. This contribution is calculated by integrating the vector potential along the trajectory, from $P \to Q$ say, $$\int_P^Q q \vec{A} \cdot \mathrm{d}\vec{x}$$

To include it as an integral over the time interval $\left [ t_1, t_2 \right ]$, we note that $$\mathrm{d}\vec{x} = \vec{v}\mathrm{d}t$$ thus, the action due the magnetic potential is $$\int_{t_1}^{t_2} q \vec{A} \cdot \vec{v} \mathrm{d}t$$

### Curl-free vectors

Gauge invariance of the vector magnetic potential is analagous to being able to add a constant to the scalar electric potential and not change the physics in any way.

This would mean we want to find a vector quantity, $\vec{C}$, such that, when we add it to the potential, we don't change the equations of motion. That is, $$\nabla \times \left( \vec{A} + \vec{C} \right) = \nabla \times \vec{A} + \nabla \times \vec{C} = \nabla \times \vec{A}$$

This is the same as saying we want $\vec{C}$ to be curl-free $$\nabla \times \vec{C} = 0$$

But the gradient of any scalar function, $\lambda = \lambda(x,y,z)$ is curl-free, since \begin{align*}\nabla \times \nabla\lambda &= \left( \partial_x, \partial_y, \partial_z \right) \times \left( \partial_x \lambda, \partial_y \lambda, \partial_z \lambda \right)\\&= \left( \partial_y \partial_z \lambda - \partial_z \partial_y \lambda, \partial_z \partial_x \lambda - \partial_x \partial_z \lambda, \partial_x \partial_y \lambda - \partial_y \partial_x \lambda \right)\\&= \left( 0,0,0 \right)\end{align*}

So, for a given magnetic force, $\vec{B}$, there exists a whole class of vector potentials, or gauges, $$\left\{ \vec{A}: \nabla \times \vec{A} = \vec{B} \right\}$$