"Deriving" Lorenz Gauge Condition from the Lagrangian

I am studying QFT with Mandl & Shaw, Quantum Field Theory and ran into Problem 2.3 (page 37 in the Second Edition) that says:

Problem 2.3: Show that the Lagrangian density

$$\mathscr{L} = -\frac{1}{2}\big[\partial_\alpha\phi_\beta(x)\big]\big[\partial^\alpha\phi^\beta(x)\big] + \frac{1}{2}\big[\partial_\alpha\phi^\alpha(x)\big]\big[\partial_\beta\phi^\beta(x)\big] + \frac{\mu^2}{2}\phi_\alpha(x)\phi^\alpha(x)$$

for the real vector field $\phi^\alpha(x)$ leads to the field equations

$$\left[g_{\alpha\beta}\left(\Box+\mu^2\right) - \partial_\alpha\partial_\beta\right] \phi^\beta(x) = 0$$

and that the field $\phi^\alpha(x)$ satisfies the Lorenz condition

$$\partial_\alpha \phi^\alpha(x) = 0.$$

I have done the first part, but am not sure how to approach the second part---deriving the Lorenz gauge condition. I now know how to derive both the results.

However, why/how is this possible? The Lagrangian looks very much like a Klein-Gordon Lagrangian for each field component, except for the second-term. However, if the Lorenz condition was indeed derivable from the given Lagrangian, the second term will be identically zero, and the Lagrangian will reduce completely to a Klein-Gordon Lagrangian for each component. Then, what was the point of the second term in the Lagrangian from the beginning? Its contribution to the equation of motion will also be identically zero, if Lorenz condition held.

Is there a meaningful physical interpretation to be learned here?

• What happens if you take the derivative of the field equation with respect to $\alpha$? Aug 7 '17 at 8:46
• @gj255 okay, wow, that does the job. It was embarrassingly simple. I still don't understand what this implies, however. Aug 7 '17 at 9:01
• you cannot plug e.o.m. back into the Lagrangian unless the field is auxiliary (cf. e.g. the on-shell Lagrangian vanishes). Aug 7 '17 at 9:17
• If you had a theory of four independent scalar fields satisfying the Klein-Gordon equation, there would be no Lorenz condition relating them. The fact that you've used the Lorenz condition to reduce the Lagrangian or equations of motion in a certain way doesn't mean you can forget that you've still got the Lorenz condition. Aug 7 '17 at 9:17

As @gj255 already noted, $$\partial^\alpha$$ obtains $$\mu^2\partial_\beta\phi^\beta=0$$, which for $$\mu^2\ne 0$$ cancels to the Lorenz gauge. Substituting this back into the equation of motion gives $$(\square +\mu^2)\phi_\alpha=0$$, a Klein-Gordon equation as you guessed. In other words, we can think of our single Euler-Lagrange equation as two equations in one, a KGE and the Lorenz gauge.
You asked what this means physically. What it means is that, whereas $$\phi_\alpha$$ has $$4$$ degrees of freedom, the gauge reduces that to $$3$$, the number expected of a massive spin-$$1$$ particle. (The second term's function, which you asked about, is therefore to create this DOF reduction. A Lagrangian with it deleted isn't equivalent, because we lose the Lorenz gauge.)
You can see this in the Appendix to Chapter I.5 of A. Zee's Quantum Field Theory in a Nutshell, where without assuming any Lagrangian he motivates the Lorenz gauge to reduce the number of DOFs and shows this + the KGE is equivalent to your equation of motion. He then notes that multiplying the EOM's left-hand side by $$\frac{\phi^\alpha}{2}$$ gives a massive Lagrangian density, which (up to integration by parts) is just the massive Maxwell Lagrangian density.