I think that I don't fully understand concept of gauge invariance. Suppose we have a Lagrangian for classical ED which is: $$\mathcal{L} = -\frac{1}{4} (F_{\mu \nu})^2 - j^{\mu}A_{\mu}.$$ First part with Maxwell tensor is of course gauge invariant. After I transform my A like $A^{\mu} \rightarrow A^{\mu} + \partial^{\mu} f$, $f$ function couples to current etc and our Lagrangian is no longer gauge invariant, equations of motion still work and are independent of f.

My questions are:

  1. Do we require full Lagrangian to be gauge invariant or only equations of motion?

  2. What's the case in other theories like $SU(2)$ or $SU(3)$ gauge theories?

  • 2
    $\begingroup$ $j^\mu$ are conserved, so the Lagrangian is indeed invariant. $\endgroup$
    – Meng Cheng
    Mar 16, 2016 at 19:06

3 Answers 3


Do we require full Lagrangian to be gauge invariant or only equations of motion?

With new potential $A'^\mu$ also a new Lagrangian density $\mathcal{L}'$ is implied which is the same function of its parameters (potential), but is to be used in the new gauge with new arguments (quantities that are put in to evaluate the function). Namely $A'^\mu$ is used instead of $A^\mu$. So it is the same function, but may have different value for a particular spacetime point $\mathbf x,t$ because different potential is used.

With the gauge transformation $$ A^\mu \rightarrow A'^\mu = A^\mu + \partial^\mu f $$

the value of the new Lagrangian density for the point $\mathbf x,t$ is

$$ \mathcal{L}' = -\frac{1}{4}F'^{\mu\nu}F'_{\mu\nu} + j^\mu A'_\mu $$

while the value of the old Lagrangian density for the same point is $$ \mathcal{L} = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} + j^\mu A_\mu. $$

The FF terms have the same value since FF term is a function of gauge-invariant fields $\mathbf E,\mathbf B$, but the jA terms have different value. The new Lagrangian density has higher value by the amount

$$ j^\mu\partial_\mu f. $$

However, this difference in value among the two Lagrangian densities does not lead to any difference in values of actions

$$ S[A^\mu] =\int_V\, d^3 \mathbf x \int_{t_1}^{t_2}\, dt \mathcal{L}(A^\mu) $$

$$ S'[A'^\mu] =\int_V\, d^3 \mathbf x \int_{t_1}^{t_2}\, dt \mathcal{L}'(A'^\mu) $$

if $\mathbf j$ or $\nabla f$ vanish on the boundary of the region $V$ all the time and $\rho$ or $\partial_tf$ vanish at times $t_1,t_2$ everywhere. This is because the difference

$$ \int_V\, d^3\mathbf x \int_{t_1}^{t_2}\,dt~ j^\mu\partial_\mu f. $$

can be transformed into

$$ \int_V\, d^3\mathbf x \int_{t_1}^{t_2}\,dt~ \partial_\mu(j^\mu f) -\int_V\, d^3\mathbf x \int_{t_1}^{t_2}\,dt~ \partial_\mu j^\mu f. $$

The second integral is zero because of the equation $$ \partial_\mu j^\mu = 0 $$ that current density obeys and the first integral can be transformed into surface integral (via Gauss theorem) and is zero if above conditions hold.

Summary: both Lagrangian density and action are gauge-independent as functions of their parameters - potentials. Lagrangian density value, however, is not gauge-independent, because the parameter has different value - $A'^\mu$ instead of $A^\mu$. Action value is gauge-independent, as difference in Lagrangian density integrates to give zero contribution.


If you write the Lagrangian with an external source current $j^\mu$, then you need to use its conservation law $\partial_\mu j^\mu = 0$ to conclude invariance of the Lagrangian. This is not "on-shell" (which would make $\delta S=0$ true by definition under any infinitesimal transformation) because here $j^\mu$ is not a dynamical variable, and its conservation law is given to us as additional off-shell information to ensure the gauge invariance of the model.

  • $\begingroup$ So is $j^{\mu}$ U(1) current which we can get from Noether's theorem applied to a free theory and then we add it to the lagrangian as linear term in $A^{\mu}$ field? $\endgroup$
    – Caims
    Mar 16, 2016 at 19:20
  • $\begingroup$ @Caims: Yeah, that's one way. Another is to not give an external current at all, but to minimally couple your gauge field to a given matter field by replacing $\partial_\mu\mapsto\partial_\mu + A_\mu$ (signs and factors of $\mathrm{i}$ omitted) in the Lagrangian for the matter field, and the just inspect what the resulting $j^\mu$ is (then it also is indeed the conserved current for the global version of the gauge symmetry). $\endgroup$
    – ACuriousMind
    Mar 16, 2016 at 19:27

The equations of motion are what is gauge invariant. But people will say the Lagrangian is gauge invariant when what they mean is that Lagrangian changes by a total derivative (which forces the equations of motion to be the same).

Lots of Lagrangians give the same equations of motion. Add a constant. Multiply by a nonzero scalar. Add a function that is a total derivative. Same equations of motion.

But since we think of all these Lagrangians as essentially the same Lagrangians (because they give the same equations of motion) when we say a Lagrangian is gauge invariant, we don't mean that it's the same Lagrangian when you change the gauge. We mean that when you change the gauge you get a Lagrangian that differs by a total derivative.

Specifically in your case when you change the gauge, the electromagnetic field $F$ doesn't change so the difference in the two Lagrangians is just the 4-current $J$ times a derivative of $f$ and since the 4-current satisfies a continuity equation an integration by parts gives that this term is equal to a total derivative.

So your Lagrangian is literally a different function when you change the gauge. It happens to be one that differs by a total deriavtive (and hence gives you the same equations of motion). And by defining the phrase "gauge invariance of a Lagrangian" to mean that it changes to something that differs by a total derivative, then we can say the Lagrangian is gauge invariant (even though it changes).

It changes to something that is close enough. The gauge transformation isn't a symmetry of the Lagrangian, but you could talk about a quasi-symmetry of the action. Yet another different thing. You can see this post about the action, as suggested by ACuriousMind.

An action and a Lagrangian are different. They even have different units. And in the linked post you'll see that a quasi-symmetry of the action differs by a boundary integral and here we talk about Lagrangians differing by a total derivative. But it's the same issue. When you integrate a Lagrangian then a total derivative term in the Lagrangian turns into a boundary integral of the action.

  • $\begingroup$ Invariance of the equations of motion is a distinct concept from invariance of the Lagrangian up to total derivatives, cf. this answer by Qmechanic. In particular, it is the notion of Lagrangian (quasi-)symmetry (use "quasi" if you want to make the "up to total derivative" clear) that forces conservation laws by Noether's theorem, while a mere symmetry of the equations of motion does not. $\endgroup$
    – ACuriousMind
    Mar 16, 2016 at 23:20

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