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In the lecture we were given the lagrangian for this case, which is:

$\mathcal{L}=-\frac{1}{4}F^{\mu\nu}(x)F_{\mu\nu}(x) + \frac{m^2}{2}A_\mu (x) A^\mu(x)$

We then considered a transformation of the form $A^\mu(x)\rightarrow A^\mu(x) -\partial^\mu f(x)$, where $f(x)$ is an arbitrary smooth function

We were told that the invariance occurs when we consider Lorentz gauge and also that $\Box f(x)=0$.

Assuming that $F^{\mu\nu}(x)F_{\mu\nu}(x)$ is invariant under this transformation, something which I believe can be shown by inserting the transformation, for the mass term one would have:

$$\frac{m^2}{2}(A_\mu(x) -\partial_\mu f(x))(A^\mu(x) -\partial^\mu f(x))=$$ $$\frac{m^2}{2}(A_\mu A^\mu - A_\mu \partial^\mu f -\partial_\mu f A^\mu + \partial_\mu f \partial^\mu f)=$$

$$\frac{m^2}{2}(A_\mu A^\mu - A_\mu \partial^\mu f -(\partial_\mu f) A^\mu - f(\partial_\mu A^\mu) + (\partial_\mu f)\partial^\mu f + f\Box f )$$

The first term $A_\mu A^\mu$ is the mass term of the original lagrangian.

The third term $f(\partial_\mu A^\mu)$ contains the lorentz gauge, so it vanishes.

The last term $f\Box f =0$ because we are given the condition that $\Box f=0$.

But there are 2 problem here, from what I can tell:

  1. The terms $2A_\mu\partial^\mu f$ is not zero and it should be if we want the lagrangian to be invariant.

  2. The claim was made that $f\Box f =0$ but we were told that, $f(x)$ is an arbitrary smooth function And we know that: A smooth function is a function that has continuous derivatives of all orders. This property ensures that the function behaves nicely and can be differentiated repeatedly without encountering any abrupt changes or discontinuities

So, am I misunderstanding something or there's a mistake somewhere?

imbAF
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1 Answers1

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  1. Gauge invariance occurs at the level of the action, not the Lagrangian. In the action, you can integrate by parts $$ S = \int d^4 x B^\mu \partial_\mu C= - \int d^4 x C\partial_\mu B^\mu + {\rm boundary\ term} $$ where $B^\mu$ and $C$ are a vector and scalar valued function respectively, and where the boundary term is $\int d^4 x \partial_\mu \left(C B^\mu\right)$, which can be converted to a boundary integral over the boundary of the spacetime volume integrated over in $S$ using Stoke's theorem, and which we typically drop assuming the fields die off at asymptotically spatial infinity, and that the initial and final configuration on the field on the initial and final time slices aren't varied. When you account for integration by parts, then you see that $A^\mu \partial_\mu f$ is equivalent to $-f \partial_\mu A^\mu$, in the sense that they will lead to the same action. And, $\partial_\mu A^\mu=0$ by assumption.

  2. You are right that $\square f = 0$ is a stronger condition than the condition that $f$ is a smooth function. However, $\square{f}=0$ when you assumed when you said "We were told that the invariance occurs when we consider Lorentz gauge and also that $\square f = 0$." So you are allowed to use this stronger condition.

Andrew
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