In a draft answer to another question about gauge transformations, I played around with demonstrating the action of a gauge transformation on the Lagrangian density. Beginning with the classical electromagnetism Lagrangian density \begin{align} \mathcal{L}_{\text{E&M}} &= \frac{\epsilon_0}{2}\mathbf{E}^2 + \frac{1}{2\mu_0}\mathbf{B}^2 - \rho\Phi + \mathbf{J}\cdot\mathbf{A} + \mathcal{L}_{Q}(\rho,\mathbf{J}) \\ & = \frac{\epsilon_0}{2}\left(-\nabla\Phi - \frac{\partial \mathbf{A}}{\partial t}\right)^2 + \frac{1}{2\mu_0}(\nabla\times\mathbf{A})^2 - \rho\Phi + \mathbf{J}\cdot\mathbf{A} + \mathcal{L}_{Q}(\rho,\mathbf{J}), \end{align} where $\mathcal{L}_Q$ is the Lagrangian density for the charge carriers that produce $\rho$ and $\mathbf{J}$. Under a gauge transformation $\Phi\rightarrow\Phi - \partial_t\Lambda$ and $\mathbf{A}\rightarrow \mathbf{A} + \nabla\Lambda$. By construction, the first two terms and $\mathcal{L}_Q$ are gauge invariant [this does not hold in QFT, but let's set that aside for the moment]. Under the gauge transformation, we get \begin{align} \mathcal{L}_{\text{E&M}} & \rightarrow \mathcal{L}_{\text{E&M}} + \rho \partial_t\Lambda + \mathbf{J}\cdot\nabla\Lambda \\ & = \mathcal{L}_{\text{E&M}} + \left[\frac{\partial\rho\Lambda}{\partial t} + \nabla \cdot (\mathbf{J}\Lambda)\right] - \Lambda \left(\frac{\partial \rho}{\partial t} + \nabla\cdot \mathbf{J}\right).\tag1 \end{align} The part of (1) in square brackets is just a collection of surface terms, and the part in large parentheses vanishes by the continuity equation for charge $$\frac{\partial \rho}{\partial t} + \nabla\cdot\mathbf{J} = 0$$ that defines what it means for charge to be locally conserved. So, we can say that as long as charge is locally conserved, the action is invariant under gauge transformations.

Can we not also say, though, that $\Lambda$ plays the role of a Lagrange multiplier that enforces charge conservation? Were I to introduce a Lagrange multiplier to enforce conservation to $\mathcal{L}_{\text{E&M}}$ it would look exactly like (1), up to the surface terms. Admittedly, treating it as such in theories where charge is already conserved is superfluous, but does it produce any interesting results or insights into gauge theory, regardless?

For convenience, the equations of motion that come from the gauge transformed Lagrangian where we treat $\Lambda$ as a Lagrange multiplier are \begin{align} \frac{\delta S}{\delta \Lambda} &=0\Rightarrow & \frac{\partial\rho}{\partial t} + \nabla \cdot \mathbf{J} & = 0 \\ \frac{\delta S}{\delta \Phi} & = 0\Rightarrow & -\epsilon_0\nabla^2\Phi - \nabla \cdot \frac{\partial \mathbf{A}}{\partial t} &= \rho & \Leftrightarrow \nabla\cdot\mathbf{E} & = \frac{\rho}{\epsilon_0} \\ \frac{\delta S}{\delta \mathbf{A}} & = 0\Rightarrow & -\frac{1}{\mu_0} \nabla\times\nabla\times\mathbf{A} - \epsilon_0 \frac{\partial \nabla\Phi}{\partial t} - \epsilon_0 \frac{\partial^2\mathbf{A}}{\partial t^2} & = -\mathbf{J} & \Leftrightarrow \nabla\times\mathbf{B} & = \mu_0\mathbf{J} + \mu_0\epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \\ \frac{\delta S}{\delta \rho} & = 0\Rightarrow & -\Phi + \partial_t\Lambda + \frac{\partial \mathcal{L}_Q}{\partial \rho} &=0 \\ \frac{\delta S}{\delta \mathbf{J}} & = 0 \Rightarrow & \mathbf{A}+ \nabla\Lambda + \frac{\partial \mathcal{L}_Q}{\partial \mathbf{J}} & = 0, \end{align} where derivatives with respect to vector valued quantities are understood to be a form of gradient. Admittedly, those last two equations are not actually equations of motion, since in pretty much every theory of E&M $\rho$ and $\mathbf{J}$ are derived from other variables. Thus, the real equations of motion would either be derived from them using chain rules (with possible space and time derivative shifts), or the relationships have to be inverted to make $\rho$ and $\mathbf{J}$ independent while some other 4-vector field is rendered auxiliary.

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    $\begingroup$ The invariance of the action under a gauge transformation, up to a surface term, leading to a conservation law for electromagnetic charge is a consequence of Noether's theorem. The working scenarios I can see for the introduction of such Lagrange multipliers with conservation laws as constraints are ones in which the multipliers themselves are part of the transformations that leave the action invariant (not necessarily in just gauge theories). $\endgroup$ Jul 12, 2018 at 19:55

2 Answers 2


So, we can say that as long as charge is locally conserved, the action is invariant under gauge transformations.

Yes, gauge-invariance implies that we can only consider$^1$ gauge fields in matter configurations $(\rho,{\bf J})$ that satisfy the electric charge continuity equation.

Can we not also say, though, that $\Lambda$ plays the role of a Lagrange multiplier that enforces charge conservation?

No, charge conservation follows from global gauge symmetry of the action via Noether's first theorem, cf. e.g. this Phys.SE post


$^1$ There is a dual question of how to determine the matter configuration from a give gauge field configuration.


I'm going to float "no" for an answer, but for different reasons than those suggested so far. The reason is that the gauge transformation is a symmetry transformation, and not an independent variable. To the extent that the gauge transformation can be treated as an independent variable, it has to replace one of the existing variables to keep the overall number of degrees of freedom constant.

Introducing a Lagrange multiplier, on the other hand, must introduce new degrees of freedom to the action, as suggested by the need to produce an equation of motion from taking the derivative with respect to the multiplier.

So the correct answer is that the gauge transformation field, $\Lambda$, is not a Lagrange multiplier. Any Lagrange multiplier introduced to enforce the conservation law, however, will not be gauge invariant. If $\lambda$ is the Lagrange multiplier that enforces local charge conservation, then it transforms with the gauge transformation as $$\lambda \rightarrow \lambda + \Lambda$$ in order to make the overall action gauge invariant.

Where this can provide some insight is that Lagrange multipliers are commonly understood to be the force that makes the system obey the constraint (e.g. for a bead constrained to be on a ring the Lagrange multiplier describes the force the ring exerts on the bead). So a gauge transformation introduces a shift in the force that holds charge conserved in a parameterization of the physics that does not naturally conserve charge without the multiplier.

Note that just because all known physical systems permit parameterizations that conserve charge naturally doesn't mean that it is meaningless to consider what happens when that is relaxed and the constraint is enforced with a multiplier; again, I refer to the ring example where you can either work with generalized coordinates that hold the bead on the ring and use no multipliers, or work in a higher dimensional space and use a Lagrange multiplier. Admittedly, in that case the force that a physical ring has to exert to keep the bead on it is of physical interest, if for no other reason that to make sure that the forces aren't greater than the ring and bead can tolerate.

As @ArjitSeth notes in a comment, this pattern may be completely general for symmetries/conservation laws.


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