## Show Solution

## Gauge Transformations do not Change the Field Equations

Let us make ourselves first a little familiar with gauge transformation calculations by confirming that **Maxwell's equations** are **invariant** under such transformations. Again, the gauge transformations (as proclaimed) are given by\[\begin{eqnarray*} \phi\left(x^{\mu}\right)\rightarrow\phi^{\prime}\left(x^{\mu}\right)&=&\phi\left(x^{\mu}\right)-\partial_{t}\lambda\left(x^{\mu}\right)\ \text{and}\\\mathbf{A}\left(x^{\mu}\right)\rightarrow\mathbf{A}^{\prime}\left(x^{\mu}\right)&=&\mathbf{A}\left(x^{\mu}\right)+\nabla\lambda\left(x^{\mu}\right)\ . \end{eqnarray*}\]We now have to check, if the fields are invariant under such transformations given their relations to the potentials. First for the **magnetic field**:\[\begin{eqnarray*} \mathbf{B}^{\prime}\left(x^{\mu}\right)&=&\nabla\times\mathbf{A}^{\prime}\left(x^{\mu}\right)\\&=&\nabla\times\left[\mathbf{A}\left(x^{\mu}\right)+\nabla\lambda\left(x^{\mu}\right)\right]\\&=&\nabla\times\mathbf{A}\left(x^{\mu}\right)\\&=&\mathbf{B}\left(x^{\mu}\right)\quad\surd \end{eqnarray*}\]and now for the **electric field**:\[\begin{eqnarray*} \mathbf{E}^{\prime}\left(x^{\mu}\right)&=&-\nabla\phi^{\prime}\left(x^{\mu}\right)-\partial_{t}\mathbf{A}^{\prime}\left(x^{\mu}\right)\\&=&-\nabla\left[\phi\left(x^{\mu}\right)-\partial_{t}\lambda\left(x^{\mu}\right)\right]-\partial_{t}\left[\mathbf{A}\left(x^{\mu}\right)+\nabla\lambda\left(x^{\mu}\right)\right]\\&=&-\nabla\phi\left(x^{\mu}\right)-\partial_{t}\mathbf{A}\left(x^{\mu}\right)\\&=&\mathbf{E}\left(x^{\mu}\right)\quad\surd \end{eqnarray*}\]Note that we did not have to show that gauge transformations do not change Maxwell's equations if they do not have an impact on the fields.

Let us now find out how gauge invariance implies charge conservation.

## From Gauge Transformations to Charge Conservation

We know that that the free-space Lagrangian density in terms of the **potentials** is given by\[\begin{eqnarray*} \mathcal{L}\left(x^{\mu}\right)&=&\frac{\varepsilon_{0}}{2}\left(\nabla\phi\left(x^{\mu}\right)+\partial_{t}\mathbf{A}\left(x^{\mu}\right)\right)^{2}-\frac{1}{2\mu_{0}}\left(\nabla\times\mathbf{A}\left(x^{\mu}\right)\right)^{2}\\&&-\phi\left(x^{\mu}\right)\rho\left(x^{\mu}\right)+\mathbf{A}\left(x^{\mu}\right)\cdot\mathbf{j}\left(x^{\mu}\right) \end{eqnarray*}\]and the action is just its four-dimensional integral,\[S = \int\mathcal{L}\left(x^{\mu}\right)dVdt\ .\]From the Lagrangian formulation, Maxwell's equations can be derived by the requirement that the action is extremal, \(\delta S=0\), which leads to the Euler-Lagrange equations. I particularly like \(\delta S=0\), because it's the summation of all of physics that we know...

Nevertheless, since we know that the **field equations** are **not changed by gauge transformations**, this also implies that the action remains unchanged under such a transformation. To find the relation to charge conservation, we just have to relate the actions in the different gauges and look what is left. First, we express the transformed Lagrangian in the old gauge:\[\begin{eqnarray*} \mathcal{L}^{\prime}\left(x^{\mu}\right)&=&\frac{\varepsilon_{0}}{2}\left(\nabla\phi^{\prime}\left(x^{\mu}\right)+\partial_{t}\mathbf{A}^{\prime}\left(x^{\mu}\right)\right)^{2}-\frac{1}{2\mu_{0}}\left(\nabla\times\mathbf{A}^{\prime}\left(x^{\mu}\right)\right)^{2}\\&=&-\phi^{\prime}\left(x^{\mu}\right)\rho\left(x^{\mu}\right)+\mathbf{A}^{\prime}\left(x^{\mu}\right)\cdot\mathbf{j}\left(x^{\mu}\right)\\&=&\frac{\varepsilon_{0}}{2}\left(\nabla\left[\phi\left(x^{\mu}\right)-\partial_{t}\lambda\left(x^{\mu}\right)\right]+\partial_{t}\left[\mathbf{A}\left(x^{\mu}\right)+\nabla\lambda\left(x^{\mu}\right)\right]\right)^{2}\\&&-\frac{1}{2\mu_{0}}\left(\nabla\times\left[\mathbf{A}\left(x^{\mu}\right)+\nabla\lambda\left(x^{\mu}\right)\right]\right)^{2}\\&&-\left(\phi\left(x^{\mu}\right)-\partial_{t}\lambda\left(x^{\mu}\right)\right)\rho\left(x^{\mu}\right)+\left[\mathbf{A}\left(x^{\mu}\right)+\nabla\lambda\left(x^{\mu}\right)\right]\cdot\mathbf{j}\left(x^{\mu}\right)\ . \end{eqnarray*}\]Ok, this step was more like copy and paste, but we see directly that a lot of terms cancel and we are left with\[\begin{eqnarray*} \mathcal{L}^{\prime}\left(x^{\mu}\right)&=&\frac{\varepsilon_{0}}{2}\left(\nabla\phi\left(x^{\mu}\right)+\partial_{t}\mathbf{A}\left(x^{\mu}\right)\right)^{2}-\frac{1}{2\mu_{0}}\left(\nabla\times\mathbf{A}\left(x^{\mu}\right)\right)^{2}\\&&-\phi\left(x^{\mu}\right)\rho\left(x^{\mu}\right)+\mathbf{A}\left(x^{\mu}\right)\cdot\mathbf{j}\left(x^{\mu}\right)\\&&+\partial_{t}\lambda\left(x^{\mu}\right)\rho\left(x^{\mu}\right)+\nabla\lambda\left(x^{\mu}\right)\cdot\mathbf{j}\left(x^{\mu}\right)\\&=&\mathcal{L}\left(x^{\mu}\right)+\partial_{t}\lambda\left(x^{\mu}\right)\rho\left(x^{\mu}\right)+\nabla\lambda\left(x^{\mu}\right)\cdot\mathbf{j}\left(x^{\mu}\right)\ . \end{eqnarray*}\]The last two terms look already very familiar - they remind us of the **continuity equation**, but how can we make this relation obvious? We have to remember that we are talking from the perspective of the **action**, \(S=\int\mathcal{L}d^{4}x\) and that this is the invariant quantity. With our result, we can calculate the difference between actions under a general gauge transformation:\[\begin{eqnarray*} S^{\prime}-S&=&\int\partial_{t}\lambda\left(x^{\mu}\right)\rho\left(x^{\mu}\right)+\nabla\lambda\left(x^{\mu}\right)\cdot\mathbf{j}\left(x^{\mu}\right)dVdt\ . \end{eqnarray*}\]The point is now, that the action cannot change under gauge transformations! Remember, \(S\) is a **physical quantity** whereas gauge transformations are merely and adjustment of the potentials. Using partial integration, we find\[\begin{eqnarray*} 0\overset{!}{=}S^{\prime}-S&=&-\int\lambda\left(x^{\mu}\right)\left[\partial_{t}\rho\left(x^{\mu}\right)+\nabla\mathbf{j}\left(x^{\mu}\right)\right]dVdt \end{eqnarray*}\]which is really just the **continuity equation** - derived from the gauge invariance of the action!

Of course, the continuity equation now implies **charge conservation**, since\[\begin{eqnarray*} \frac{d}{dt}\int\partial_{t}\rho\left(x^{\mu}\right)dVdt&=&\frac{d}{dt}\int\rho\left(x^{\mu}\right)dV\\&=&-\int\nabla\mathbf{j}\left(x^{\mu}\right)dV\\&=&-\int\mathbf{j}\left(x^{\mu}\right)\cdot d\mathbf{A}=0 \end{eqnarray*}\]if the current density is localized, or has a compact support as we may naturally assume.

Our little calcuation is really just an application of **Noether's theorem** from which we could have directly reasoned the continuity equation \(\partial_{\mu}j^{\mu}=0\).