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Consider a real massive vector field with lagrangian density

$$\begin{align}\mathcal{L}&=-\frac{1}{4}(\partial_\mu A_\nu-\partial_\nu A_\mu)(\partial^\mu A^\nu-\partial^\nu A^\mu)+\frac{1}{2}m^2 A^\mu A_\mu\\&=-\frac{1}{2}(\partial_\mu A_\nu-\partial_\nu A_\mu)\partial^\mu A^\nu+\frac{1}{2}m^2 A^\mu A_\mu.\tag{1}\end{align}$$

Under a local gauge transformation $A_\mu\longrightarrow A_\mu+\partial_\mu f$ the lagrangian density is not invariant

$$\begin{align}\mathcal{L}\longrightarrow&~\mathcal{L}+\frac{1}{2}m^2\left(2A^\mu+\partial^\mu f\right)\partial_\mu f\\=&\,\mathcal{L}-\frac{1}{2}m^2f\left(2\partial_\mu A^\mu+\square f\right)\end{align}\tag{2}$$

unless we take an $f$ that satisfies $$\square f=-2\partial_\mu A^\mu\tag{3}.$$


Equations of motion for the vector field are

$$(\square+m^2) A^\mu-\partial^\mu\partial_\nu A^\nu=0,\tag{4}$$ and taking its divergence $\partial_\mu$ we obtain $$\partial_\mu A^\mu=0\tag{5}.$$

This identity is the Lorenz gauge, but it is satisfied automatically. Then, from $(3),$ $A_\mu\longrightarrow A_\mu+\partial_\mu f$ is a symmetry of the theory as long as $\square f=0$. Can this be considered a kind of "partial" or "reduced" gauge symmetry?

Edit: I realized one thing.

With $(5)$, the eom $(4)$ becomes $$(\square+m^2)A_\mu=0,\tag{6}$$ but this equation is not invariant under the transformation $A_\mu\longrightarrow A_\mu+\partial_\mu f$ even if $\square f=0$. Why is this equation not invariant but the lagrangian density is?

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1 Answer 1

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  1. Symmetries of the action must be considered without use of the equations of motion. An on-shell symmetry is a vacuous notion - if you use the equations of motion, as you do when using eq. (5) to conclude that $\square f = 0$ should be a symmetry, then any infinitesimal transformation of the action will leave the action invariant, precisely because the equations of motion mark a stationary point of the action, and the definition of a stationary point is more or less that all infinitesimal variations vanish. See also this answer and its linked answers by Qmechanic.

  2. Precisely because symmetries must be considered off-shell, your eq. (3) is inconsistent. Since we are off-shell, $A^\mu$ is an arbitrary field not taking any specific values, but $f$ needs to be a fixed function. Since eq. (3) cannot be fulfilled off-shell for arbitrary $A^\mu$, there is no gauge invariance of the action of a massive vector field.

  3. As discussed in this answer of mine, the Hamiltonian theory of the massive vector field has two constraints: \begin{align} \pi^0 & \approx 0 \\ \partial_i \pi^i + m^2 A_0 & \approx 0 \end{align} The Poisson bracket of these constraints is non-vanishing, so they are both second-class constraints, but only first-class constraints can possibly generate gauge transformations, see points 4. and 5. in this answer of mine. Therefore, the theory of the massive vector field is constrained, but it is not a gauge theory.

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