What does it mean for an action to be invariant under $x \to x'$, $\phi \to \phi'$?

Question

I'm suddenly getting very confused about a basic question. Suppose somebody tells you that the action is invariant under the transformation $$x \to x', \quad \phi(x) \to \phi'(x').$$ I realize this notation is ambiguous, but it seems to be common. For example, one might define a Lorentz transformation in this sloppy fashion as $$x \to \Lambda x, \quad \phi \to \phi(\Lambda^{-1}x)$$ or a dilation transformation as $$x \to \lambda x, \quad \phi \to \lambda^\alpha \phi(x/\lambda).$$

Now suppose the action is $$S_{000}^0 = \int_a^b dx \, h(\phi(x)).$$ Then I can think of fifteen things "the action is invariant" could naively mean. Define $$S^1_{111} = \int_{f(a)}^{f(b)} dx' \, h(\phi'(x')), \quad S^0_{101} = \int_a^b dx'\, h(\phi(x')), \quad S^1_{010} = \int_{f(a)}^{f(b)} dx \, h(\phi'(x))$$ along with twelve other quantities in what is hopefully a self-explanatory notation. Then one of these quantities is equal to $S_{000}^0$, but which one is typically meant?

Answer 1

1. Noether's theorem works even for non-geometric theories, so to be as general and simple as possible, we shall not use notions & concepts from differential geometry. For the purpose of Noether's theorem, it is enough to discuss infinitesimal variations: $$ \delta x^{\mu} ~:=~ x^{\prime\mu} - x^{\mu} ~=~ \varepsilon~ X^{\mu}(x),\tag{1}$$ $$ \delta\phi^{\alpha}(x) ~:=~\phi^{\prime\alpha}(x^{\prime})-\phi^{\alpha}(x) ~=~ \varepsilon~ Y^{\alpha}(\phi(x),\partial\phi(x), x),\tag{2}$$ where $\varepsilon$ is an infinitesimal ($x$-independent) parameter, and $X^{\mu}$ and $Y^{\alpha}$ are generators.

2. If $V~\subseteq~\mathbb{R}^4$ is a spacetime region, let $$ V^{\prime}~:=~\{ x^{\prime}\in \mathbb{R}^4 \mid x \in V \} ~\subseteq~\mathbb{R}^4 \tag{3}$$ denote the varied spacetime region.

3. The infinitesimal variation of the action is by definition $$\begin{align}\delta S_V~:=~& S_{V^{\prime}}[\phi^{\prime}] -S_V[\phi]\cr ~:=~& \int_{V^{\prime}}\! d^4x^{\prime}~{\cal L}(\phi^{\prime}(x^{\prime}),\partial^{\prime}\phi^{\prime}(x^{\prime}),x^{\prime}) \cr &-\int_V\! d^4x~{\cal L}(\phi(x),\partial\phi(x),x).\end{align}\tag{4}$$ Formula (4) is $S^1_{111}-S^0_{000}$ in OP's notation. See e.g. Refs. 1 & 2.

4. The infinitesimal variation (1) & (2) are called a quasi-symmetry of the action if the infinitesimal variation (4) is a boundary integral, cf. my Phys.SE answer here. In the affirmative case, Noether's theorem leads to an on-shell conservation law.

References:

1. H. Goldstein, Classical Mechanics, 2nd edition, Section 12.7.

2. H. Goldstein, Classical Mechanics, 3rd edition, Sections 13.7.

Answer 2

I'm making the assumption that $S[\phi(x)]$ is some action functional for a field theory of $\phi$. It is important to note that symmetries act only on fields, not on coordinates. You should think of the coordinates as dummy variables, with a coordinate transformation being equivalent to a relabeling. With this fact in mind, if under a field transformation $\delta$ $$ \delta: \phi(x) \mapsto \phi'(x) $$ the action satisfies $$ S[\phi'(x)] = S[\phi(x)], $$ we say that $\delta$ is a symmetry of the theory.

Now, this being said, it is sometimes useful to think of the transformation as a coordinate transformation that then induces a transformation on the fields. In this picture, the coordinate transformation $\delta'$, with $$ \delta': x \mapsto x' $$ naturally induces a transformation on the field by $$ \phi'(x') = \phi(x). $$ Of course, due to this statement, it is always true that $$ S[\phi'(x')] = S[\phi(x)]. $$ Note the difference between the previous statement about the action. The first follows from a symmetry of the theory, whereas the second is always true. Intuitively, this means one can undo a field transformation through a relabeling (coordinate transformation).