Gauge invariance for classical fields

Question

I recently did some exercises in classical field theory and tried to think deeply about the gauge symmetry of the free electromagnetic field described by the Lagrangian $$ \mathcal L = -\frac 1 4 F^{\mu\nu}F_{\mu\nu}-j_\mu A^\mu $$ which implies the equation of motion $$ \partial_\mu F^{\mu\nu} = j^\nu \quad. $$ Now i used the continuity equation $\partial_\nu j^\nu = 0$ to get $$ \partial_\nu j^\nu = 0 = \Box \partial_\nu A^\nu - \Box \partial_\mu A^\mu = 0 $$ so there is no constraint imposed on $A_\mu$. When adding a mass term to the theory things change a little. So the new Lagrangian i was looking at is $$ \mathcal L = -\frac 1 4 F^{\mu\nu}F_{\mu\nu}+ \frac 1 2 m^2 A_\mu A^\mu-j_\mu A^\mu $$ when now calculating the equations of motion and using the continuity equation to get a constraint on $A_\mu$. One can obtain that it imposes the Lorenz Gauge $$ \partial_\mu A^\mu = 0 $$

So in this second theory there is no gauge invariance.

Question:

After reading these questions 1 & 2 i kind of get why a gauge partly fixes the degrees of freedom of the field, but what implies the gauge invariance?

Especially in this example what implies the loss of this gauge invariance? And because of the lost gauge invariance, can one conclude that the parameter $m$ of the mass term must vanish? If yes, why?

Edit:(to be more precise i hope)

What excactly is the physical implication of gauge invariance (for the electromagnetic field or in general)? Or rather what is the "physical truth" implied by gauge invariance and what are the consequences when this invariance is broken by a term in the Lagrangian (here a mass term in the electromagnetic free field lagrangian)?

Answer 1

What excactly is the physical implication of gauge invariance (for the electromagnetic field or in general)? Or rather what is the "physical truth" implied by gauge invariance and what are the consequences when this invariance is broken by a term in the Lagrangian (here a mass term in the electromagnetic free field lagrangian)?

There are kind of two parts to this. First, gauge invariance implies conservation of charge (the continuity equation) via Neother's theorem, even classically. Second, ignoring the philosophical stance that $A_\mu$ isn't real, it means that we can break down $A_\mu$ into parts that are independent and behave in very different ways. Specifically:

1. The solenoidal part of $\mathbf{A}$,
2. The irrotational part of $\mathbf{E}$, and
3. The electric potential $\phi$.

By solenoidal and irrotational, something related to Helmholtz decomposition of the vector field. You can think of the solenoidal part as being the part generated by a Biot-Savart law applied to the field's curl, and the irrotational part is the total field minus the solenoidal part.

Basically, if you examine the free field Lagrangian you'll notice three things. First, $\dot\phi$ doesn't appear anywhere. This is important, because it means that $\phi$ isn't really a physical field, in some sense. It's sort of like a massless spring - it can instantaneously adjust into any configuration because it doesn't have any inertia (inertia $\Leftrightarrow$ kinetic energy). This makes it akin to something known as an "auxiliary field", and the Weyl gauge just sets it identically to zero.

Second, the only place that the irrotational part of $\mathbf{A}$ appears is in the electric field, contributing to $\mathbf{E} = -\nabla\phi - \dot{\mathbf{A}}$. This means that part of the field has a kinetic term and no potential term. It's like a free particle where it has a translates at a constant speed, but without any way to interact with its position, we have no idea what value it has. Thus, the only quantity we can definitively label as "physical" is the irrotational part of $\mathbf{E}$. Notice that gauge transformations generate translations in exactly this part of $\mathbf{A}$, leaving its solenoidal part untouched.

Finally, the solenoidal part of $\mathbf{A}$ has both potential and kinetic terms. Recall that the solenoidal part of $\mathbf{E}$ is just the negative time derivative of the solenoidal part of $\mathbf{A}$. Because both the potential and kinetic terms are quadratic, they support simple harmonic oscillator solutions, so it is this part of the field that carries waves/photons.

Breaking gauge invariance as described in the OP will make it so that the irrotational part of $\mathbf{A}$ is no longer free to just translate (as it is in non-Coulomb gauges), making it possible to measure its value. $\phi$, though, will still play the role of an auxiliary field because its time derivative is still absent. Note, also, that you can give the photon mass without breaking gauge invariance, you'll just have to throw Lorrentz invariance, and possibly locality, out the window (i.e. use a term like $-\frac{m^2 }{2} \mathbf{A}_{\mathrm{solenoidal}} \cdot \mathbf{A}_{\mathrm{solenoidal}}$).

Answer 2

The mass term in the Lagrangian is not gauge-invariant. Under gauge transformation:

$$A_\mu \rightarrow A_\mu - \frac{\partial \eta}{\partial x^\mu}$$

$$m^2 A_\mu A^\mu \neq m^2 A_\mu A^\mu - m^2\left( \frac{\partial \eta}{\partial x^\mu} A^\mu +\frac{\partial \eta}{\partial x_\mu}A_\mu -\frac{\partial \eta}{\partial x^\mu}\frac{\partial \eta}{\partial x_\mu} \right)$$

Thus, you get equations of motion which are not gauge invariant. The equation you have written is closely related to the Proca Lagrangian.

To answer your second question, yes, the photon mass $m$ must be zero for gauge invariance.

Edit: To answer the edited question-

There is no "physical" field $A_\mu$ as it cannot be observed (that's the present consensus). The physical fields $E$ and $B$ are gauge invariant by definition. In general, $E$ and $B$ fields together have 6 degrees of freedom (each is a 3-vector) but the four constraints imposed by Maxwell's equations (for a photon/EM wave), lead to only 2 (=6-4) degrees of freedom. So, the photon should have only two degrees of freedom(dofs) (the two states of polarisation) in any theory used to describe it. But when we use $A_\mu$ (a 4 vector) to describe the photon field, we have 4 dofs in the theory. So, 2 dofs are redundant. One dof is lost when the equation of motion constrains $A_\mu$. In order to remove the one remaining dof, a gauge is fixed. Now, if this gauge invariance were not present, then $A_\mu$ would have another dof that would be part of the physical description. Hence, $E$ and $B$ fields would have net 3 dofs instead of 2, and the four Maxwell's equations would not be satisfied. Since in real life, photons do not have 3 dofs, it would have to be a different particle. The Proca lagrangian describes one such particle. The presence of mass term that forces the Lorentz gauge condition, could imply Lorentz invariance of the particle. An answer that explains the usefulness of Lorentz gauge very well is here.