Why is Noether's theorem not guaranteed by calculus?

Question

The action of a system, say a scalar field is

$$ S = \int_{\mathcal{M}} {\rm d}^4 x ~ \mathcal{L}(\phi(x),\partial \phi(x)). $$

Now, if one does a variable transformation $x \to x'$, then

$$ S' = \int_{\mathcal{M}'} {\rm d}^4 x' ~ \mathcal{L}(\phi'(x'),\partial' \phi'(x')). $$

Then why isn't $S'$ guaranteed to be equal to $S$ because of the fact that integrations remain unchanged due to changes in variables by the rules of multivariable calculus? Even though the Lagrangian density is shrouded in mystery by being a composite function of $\phi$ and $\partial\phi$ etc., it is ultimately just a function of $x\in \mathcal{M}$ given the history for the field configuration. That is why I do not understand why isn't the invariance of integral under variable change does not apply here and we have to rely on symmetry principles and satisfaction of the equation of motion along with them.

EDIT: An interesting point has been clarified by @Gold in the comments that the scope of Noether's theorem is much broader than only the symmetry corresponding to the coordinate transformations, for example, the theorem is also applicable for the internal symmetries which are different from the symmetries in the physical spacetime. However, my question, which is now to be only limited in the case of changes in the field due to coordinate transformations, still remains. Consider for example the energy-momentum tensor $T^{\mu}_{~~\nu}$ and angular momentum tensor $\left(\mathcal{J}^\mu\right)^{\rho\sigma}$, which are claimed to arise from symmetry under translations and Lorentz transformations, respectively (see Prof. David Tong's lecture note on QFT for their definitions). My question is why do we need these symmetries if, under a change of coordinates, the action remains invariant just because of the rules of calculus? I hope these examples clarify my question.

EDIT 2: @Prahar raises a very point and I believe that probably can be ultimately what answers this question. However, there's still room for further discussion. For example in the link to Prahar's answer that is provided in the comments below, I find the following assertion: "This is crucial and often a something that a lot of people confuse with. In field theories, all symmetry transformations act only on the fields, not on the coordinates. One often like to talk about spacetime symmetries which are described as acting on coordinates in some way $x \to x'$. However, it is crucial to remember that that is simply a tool to package information about how fields transform. For instance, you might like to talk about translations. This is described by the field redefinition $\phi(x) \to \phi'(x)$ where $\phi'(x+a) = \phi(x)$. Note that the equation $\phi'(x+a) = \phi(x)$ is to be understood as a way to deduce what is $\phi'(x)$ in terms of $\phi(x)$ and not as translations acting on the coordinates in some way."

My objection is to the last line. The field redefinitions $\phi(x) \to \phi'(x)$ is not a statement independent of what happened to the underlying coordinate system. If it were the underlying coordinate transformation $x\to x'=x+a$ could not be used as a "tool" (so-called) to "deduce what is $\phi'(x)$ in terms of $\phi(x)$." The deduction of how the field changes $\delta\phi(x)=\phi'(x)-\phi(x)$ is different if different things were to happen to the underlying coordinates (say if instead of $x\to x'=x+a$ we had $x\to x'=\Lambda x$). Therefore, what happens to the underlying coordinates is related to the field redefinition and vice versa and are not independent of each other. If this is not true then, this point requires further elaboration as opposed to assertions of fact that these are independent. A Lorentz transformation of the field or a translation of the field is associated with some corresponding transformation in the coordinates too.

Answer 1

Since the problem here appear to be coordinates, let's just stop using coordinates, and for simplicity consider the theory of a single scalar field on space(time) $M$:

Our field is a function $\phi : M\to\mathbb{R}$, and the action is a functional $S : [M,\mathbb{R}] \to \mathbb{R}$, where $[M,\mathbb{R}]$ is the notation for some space of sufficiently nice functions. Noether's theorem is a statement about continuous transformations in field space where to each $\epsilon\in[-1,1]$ we have a flow that associates to a pair $(\phi,\epsilon)$ of a field and parameter a deformation $\phi_\epsilon$ and $\phi_0 = \phi$. This is the "field theory version" of the flow of a vector field on phase space.

Such a transformation is an (off-shell) symmetry of the action if $S[\phi_\epsilon] = S[\phi]$ for all $\phi$ and all $\epsilon$. Since this statement used coordinates nowhere, it is clear that this is not some trivial statement about how coordinates behave inside integrals (we didn't even write $S$ in the form of an integral).

But how does it now make sense to speak about things like translations in this context? A continuous transformation of coordinates $x\mapsto x_\epsilon$ is generated as the flow of a vector field $\delta x^\mu = \epsilon X^\mu$. Every function on $M$ is "dragged along" this flow by the Lie derivative $\mathcal{L}_X$, and so this induces a continuous transformation on the function space $[M,\mathbb{R}]$ $$ (\phi,\epsilon) \mapsto \phi + \epsilon \mathcal{L}_X\phi,$$ and it is the invariance of the action under this transformation that we mean when we speak of e.g. translation invariance.

"Continuous transformation of coordinates" in the previous sentence is just intended to make contact with the usual language in physics texts here. We can again phrase this entirely in coordinate-free language:

A continuous family $f_\epsilon$ of diffeomorphisms of $M$ is a map $[-1,1]\to [M,M], \epsilon\mapsto f_\epsilon$ such that all $f_\epsilon$ are diffemorphisms and $f_0 = \mathrm{id}_M$. This induces a corresponding continuous transformation on field space via the pullback $\phi_\epsilon = f^\ast_\epsilon(\phi) = \phi\circ f_\epsilon$, and it is this transformation on field space that we mean when we speak of the invariance of the action under the family of diffeomorphisms $f_\epsilon$.

Since we're already being pedantically rigorous: Note that not all continuous families of diffeomorphisms in the above sense are generated as the flow of vector fields, but merely are finite products of flows of vector fields, cf. this MO question. When $f_\epsilon$ is generated as the flow of a vector field, then the Lie derivative is the infinitesimal version of the pullback - rigorously, in that $\mathcal{L}_X\phi = \lim_{\epsilon\to 0} \frac{\phi_\epsilon(x) - \phi(x)}{\epsilon}$ - and we recover the claim about the Lie derivative from before.

As for the confusion in the question about the "mystery" of what the Lagrangian is actually a function of - which is not actually relevant here because we can state the notion of symmetry on the level of the action without ever mentioning a Lagrangian - see this answer of mine, in particular the last paragraph.

Answer 2

It might be easier to see what's going on by making a few simplifications:

• First, we can work in $0+1$ dimensions -- in other words, we can work with ordinary particle mechanics, where the action is a time integral of the Lagrangian.
• Second, we can have the Lagrangian explicitly break time translation invariance, and see how this results in breaking Noether's theorem.

Combining these two bullet points, let's consider the following action \begin{equation} S = \int_{t_1}^{t_2}{\rm d}t f(t)\left(\frac{m}{2}\dot{x}^2 - V(x)\right) \end{equation} where $f(t)$ is some arbitrary function that we put into the action. If $f={\rm const}$, then the action will be time translation invariant, and we can run Noether's theorem to derive energy conservation. For any other choice of $f$, Noether's theorem will not apply and energy will not be conserved.

At a high level, we can give some plausibility arguments. Clearly, the value of $S$ will depend on our choice of $f$, so it's at least possible that properties of $f$ can affect conservation laws. Second, we can check if energy is conserved explicitly by deriving the equations of motion by varying $x$ \begin{equation} m \ddot{x} + \frac{\dot f}{f} m \dot{x} + V' = 0 \end{equation} The term $\sim \dot{x}$ is a friction term which removes energy from the system (and this friction term goes away if $\dot{f}=0$). This shows you that something must be wrong with your logic, because your logic would have led you to conclude that Noether's theorem must apply in this example, and energy must be conserved. However, we haven't pinpointed exactly what's wrong with your argument.

To really answer the question, we should get into the nitty gritty details and try to run Noether's theorem in this example and see where it goes wrong. As @Prahar stated in the comments, we have to understand what time translation invariance really means in this context. It does not mean that we perform a coordinate transformation $t\rightarrow \tilde{t}(t)$ (which indeed cannot change anything physical about the system). Physically, the idea is that we move the entire physical system from time $t$ to time $t+\delta t$ (eg, by waiting a time $\delta t$ before starting the experiment). Then, the particle's motion will change from $x(t)$ to $x(t+\delta t)$, but we don't also shift our clocks to adjust for this change, so any anywhere $t$ appears explicitly in $L$ we do not transform it.

Therefore, when we transform $L$, we get \begin{eqnarray} \delta L &=& L(x(t+\delta t), \dot{x}(t+\delta t), t) - L(x(t), \dot{x}(t), t) \\ &=& \frac{\partial L}{\partial x} \dot x \delta t + \frac{\partial L}{\partial \dot{x}} \ddot{x} \delta t \\ &\neq& \frac{{\rm d} L}{{\rm d} t} \delta t \end{eqnarray} because we do not transform the explicit $t$ dependence. The failure of $L$ to transform as a total derivative in this situation means the action does change and Noether's theorem does not hold.

Answer 3

The action is a functional of the field, $$ S[\phi] = \int d^4 x {\cal L}(\phi(x),\partial_\mu \phi(x) , \cdots ) . $$ You are of course free to change the integration variable $x \to x'(x)$ as you wish and you'd get $$ S[\phi] = \int d^4 x' {\cal L}(\phi(x'),\partial'_\mu \phi(x') , \cdots ) . $$ Notice that the field has not changed so $\phi$ remains $\phi$ on both sides of the equation. What you do NOT get when you do $x \to x'$ is $\int d^4 x' {\cal L}(\phi'(x'),\partial'_\mu \phi'(x') , \cdots )$ (as you have mentioned in your question).

To discuss symmetries you need to transform the fields of the theory, not the spacetime coordinates. For details on how to deal with spacetime symmetries you can refer to my previous answers I have linked in the question comments.

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Let me consider the specific example of translations to clarify my point. Translations is parameterized by a vector $a^\mu$ and it act on the fields as $$ \phi(x) \to \phi'_a(x) , \qquad \phi'_a(x) = \sum_{n=0}^\infty \frac{1}{n!} a^{\mu_1} \cdots a^{\mu_n} \partial_{\mu_1} \cdots \partial_{\mu_n} \phi(x) \tag{1} $$ Notice how I defined the primed field $\phi'_a$ at the point $x$ in terms of an infinite sum of fields at the same point $x$. Now, due to special property of the infinite sum, I can equivalently write $$ \phi'_a(x-a) = \phi(x) \tag{2} $$ where now it looks like I've acted on the coordinates, but this way of writing the transformation is a simple way to package the infinite sum (1).

So far, we have discussed translation as a transformation on the fields. We now discuss translations as a symmetry of the action. To do this, we consider the action $S[\phi'_a]$. By definition, this is $$ S[\phi'_a] = \int d^4 x {\cal L}(\phi'_a(x),\partial_\mu \phi'_a(x) , \cdots ) $$ Notice that everywhere I only changed the field, NOT the coordinates. We now change the dummy integration variable to $x \to x - a$ which implies $$ S[\phi'_a] = \int d^4 x {\cal L}(\phi'_a(x-a),\partial_\mu \phi'_a(x-a) , \cdots ) $$ We now simplify using property (2), $$ S[\phi'_a] = \int d^4 x {\cal L}(\phi(x),\partial_\mu \phi(x) , \cdots ) = S[\phi]. $$ We have therefore shown that translations is a symmetry of the action because the action evaluated in the field configuration $\phi'_a(x)$ is the same as the action evaluated in the field configuration $\phi(x)$.

Note for instance, that if the action was $$ \int d^4 x \sqrt{g(x)} {\cal L}(\phi(x),\partial_\mu \phi(x) , \cdots ) $$ as it is in general relativity, then translations would no longer be a symmetry!!! even though its action on the fields would continue to be (1).

Answer 4

Noether's theorem is about invariance of the action under active transformations. Let's say you have an action that is a fixed function of the field trajectory $S[\phi (x, t) ]$. Noether's theorem looks for a family of field trajectories indexed by a parameter $\alpha$, $\phi _{\alpha} (x, t)$, such that the action is invariant for the family. Now, note that not just any family of field trajectories will have an invariant action.

Where does Noether's derivation assume this?

It's in the very beginning when it considers a fixed function of a famiy of field trajectories indexed by a parameter, $S[\phi (x, t, \alpha)]$. In the OP's notion of passive co-ordinate transformations, the functional form of the action can change, so Noether's argument does not even begin to apply.