# Gauge Theory

A Lagrangian is called gauge invariant if the Lagrangian does not change under certain transformations. The simplest example for the Lagrange function $L=T-V$ the equations of motion do not change if we use a new potential $V'=V+c$ that differs only by a constant from the $V$. Another classic example is the gauge invariance of the Lagrangian for transformations of the vector potential $\mathbf{A}$ \begin{align} \mathbf{A}' = \mathbf{A} + \nabla s \end{align} By the definition of the magnetic field \begin{align} \mathbf{B}= \nabla \times \mathbf{A} \end{align} and the vector calculus rule that the curl of a gradient is always zero $\nabla \times (\nabla s)=0$ we can always add the gradient of a scalar $s$ to the vector potential and get the same magnetic field. \begin{align} \mathbf{B}= \nabla \times (\mathbf{A} + \nabla s)= \nabla \times \mathbf{A}+0 \end{align} $\mathbf{A}$ is not unique, after the gauge transformation $\mathbf{A}' = \mathbf{A} + \nabla s$ we get the same magnetic field, this is why $\mathbf{A}$ is also called a gauge field. Because the magnetic field is invariant, the Lorentz force doesn't change either under the gauge transformation. Hence the Lagrangian is also invariant under this transformation, like we have demonstrated in detail for the charged particle in the electromagnetic field.
Both vector potentials \begin{align} \mathbf{A}=\begin{pmatrix} 0\\ b \cdot x \\ 0 \end{pmatrix} \end{align} and \begin{align} \mathbf{A}'=\begin{pmatrix} -b \cdot y \\ 0\\ 0 \end{pmatrix} \end{align} lead to the same magnetic field $\mathbf{B}$. We calculate the curl with the Levi-Civita symbol. For $\mathbf{A}'$ the only nonzero entry we get for $\partial_yA_x$ with indices $k=1$ and $j=2$, so $i$ must be $3$. Epsilon is negative for these indices $\epsilon_{321}=-1$. \begin{align} \mathbf{B}'=\nabla \times \mathbf{A}'=\nabla \times \begin{pmatrix} -b \cdot y \\ 0\\ 0 \end{pmatrix}=\epsilon_{ijk} \partial_j A'_k= \begin{pmatrix} 0 \\ 0\\ b \end{pmatrix} \end{align} For $\mathbf{A}$ the only nonzero entry is for indices $k=2$ and $j=1$, so $i$ must be $3$. Epsilon is positive $\epsilon_{312}=1$. \begin{align} \mathbf{B}=\nabla \times \mathbf{A}=\nabla \times \begin{pmatrix} 0\\ b \cdot x \\ 0 \end{pmatrix}=\epsilon_{ijk} \partial_j A_k=\begin{pmatrix} 0 \\ 0\\ b \end{pmatrix} \end{align} Like we expected we get the same magnetic field for both vector potentials $\mathbf{B}'=\mathbf{B}$.
In the article A Point Charge in the Electromagnetic Field, we arrived at the Lagrangian \begin{align} L(x,\dot{x})=\frac{1}{2}m \dot{x}^2 - e \cdot (\varphi+\frac{1}{c}\mathbf{v} \cdot \mathbf{A}') \end{align} How does a gauge transformation $\mathbf{A}' = \mathbf{A} + \nabla s$ influence the action integral? \begin{align} \int_{t_0}^{t_1} L'(x,\dot{x}) dt &=\int_{t_0}^{t_1} \frac{1}{2}m \dot{x}^2 - e \cdot \varphi+ \frac{e}{c} \cdot(\mathbf{A}+ \nabla s) \cdot \mathbf{v}dt \end{align} We can rewrite the velocity as derivative \begin{align} \int_{t_0}^{t_1} L'(x,\dot{x}) dt &=\int_{t_0}^{t_1} \frac{1}{2}m \dot{x}^2 - e \cdot \varphi+ \frac{e}{c} \cdot(\mathbf{A}+ \nabla s) \cdot \frac{d\mathbf{x}}{dt}dt \end{align} Now remember the $\nabla$ is really a derivative with respect to $\mathbf{x}$, we rewrite and find \begin{align} \int_{t_0}^{t_1} L'(x,\dot{x}) dt &=\int_{t_0}^{t_1} \frac{1}{2}m \dot{x}^2 - e \cdot \varphi+ \frac{e}{c} \cdot \mathbf{v}\cdot \mathbf{A}+ \frac{e}{c} \cdot \nabla_\mathbf{x} s \cdot \frac{d\mathbf{x}}{dt}dt \end{align} $\nabla_\mathbf{x}$ cancels with $\mathbf{x}$, by the fundamental theorem of calculus the solution is simply \begin{align} \int_{t_0}^{t_1} \frac{e}{c} \cdot \nabla_\mathbf{x} s \cdot \frac{d\mathbf{x}}{dt}dt=\frac{e}{c} \left(s(t_0)-s(t_1)\right) \end{align} Thus, the answer is YES the action does change, but only by a constant. \begin{align} L'=L+\frac{e}{c} \left(s(t_0)-s(t_1)\right) \end{align} This means that the variation of the path $\mathbf{x}(t)$ does not alter the action integral, since $s$ only depends on the boundary conditions $t_0,t_1$. Thus it follows \begin{align} \delta \int L' \, dt= \delta \int L \, dt \end{align}