# Gauge the symmetry $φ \to φ + a(x)$ for a free massless real scalar field

How does one alter the Lagrangian density for a real scalar field

$$\frac{∂_μφ∂^μφ}{2}$$

such that is will be invariant under the gauge transformation $φ → φ + a(x)$? For a complex scalar field with internal transformation $φ → e^{i \lambda (x)}φ$ the lagrangian can be altered by using the covariant derivative instead of the usual $∂_\mu$ to remain locally invariant (correct me if I'm wrong). I've just can't find in any textbook the $φ \to φ + a(x)$ transformation, apart for the global transformation when $a$ is a constant.

Any pointers would be much appreciated!

• Did this transformation come up somewhere or are you just wondering? It seems a little weird to me, considering that you can use it to make any field configuration equal to any other (or all of them equal to zero). – Javier Nov 15 '15 at 14:39
• The symmetry you want to introduce is huge (the largest possible), and will hence make everything trivial, I think. – Danu Nov 15 '15 at 15:06

If $$\tag{1} \delta\varphi~=~\varepsilon$$ is a global shift symmetry, we can gauge the symmetry, i.e. enhance it to a local symmetry by (i) introducing a gauge field $A_{\mu}$ with gauge symmetry $$\tag{2} \delta A_{\mu} ~=~\partial_{\mu}\varepsilon,$$ and (ii) replace partial derivatives $\partial_{\mu}\varphi$ with covariant derivatives $$\tag{3} D_{\mu}\varphi~=~\partial_{\mu}\varphi- A_{\mu}$$ in the action. The latter is known as minimal coupling.