# Why is Noether's theorem not guaranteed by calculus?

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.

• Noether's theorem has more generally to do with field transformations instead of just spacetime coordinate transformations. What we are interested in is not $x\to x'$, but rather in $\phi(x)\to \phi'(x)$. One particular field transformation can be induced by a coordinate transformation, but then it is indeed clear that if ${\cal L}$ just depends on the coordinates through the fields the action will indeed be invariant. More generally we may consider field transformations $\phi(x)\to \phi'(x)$ that have nothing to do with coordinate transformations.
– Gold
Apr 16 at 1:54
• One such example: suppose that you have a complex scalar field $\phi(x)$. Then we can transform $\phi(x)\to e^{i\alpha(x)}\phi(x)$, which is not a coordinate transformation. One may then discuss whether or not the action is invariant or not under this transformation. Some actions will be invariant only for constant $\alpha$, others for local $\alpha$ as well, and some others will simply not be invariant at all.
– Gold
Apr 16 at 1:57
• In a field theory, the symmetry transformation (even the coordinate dependent ones) acts only on the fields, NOT on the coordinates. Coordinates are used as a tool to describe the symmetry transformation in a concise way (see footnote in my answer - physics.stackexchange.com/questions/327999/… and physics.stackexchange.com/questions/328473/…) Apr 16 at 2:24
• My answer in physics.stackexchange.com/questions/612945/… is also relevant. Apr 16 at 2:26
• @Gold sounds like an answer. Comments are only for suggesting post improvements or song clarifying questions Apr 16 at 2:57

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.

• +1 Great answer! Perfect thing to do in this confusing situation. Apr 16 at 11:57
• +1 But we should also note that OP's issue is not fundamentally about co-ordinates, and the issue can easily extend to the infinite dimensional configuration space of fields too. e.g. If we consider passive transformations on the configuration space of field variables, the functional form of the action would change and the action would remain trivially invariant under all such transformations. I think core issue here is about active vs passive transformations, rather than field vs co-ordinate transformations Apr 16 at 12:23
• +1 Thanks for this great answer. I think I now understand why my initial thinking about a change of variables has nothing to do with Noether's theorem. However, it seems to me that what @Prahar insists is also incorrect that we never change coordinates in field theory. Yes, symmetry of the action in general only concerns the change in field $\phi\to\phi_{\epsilon}$, however in some specific cases these changes are generated due to actual change in the coordinates as you write $\delta x^{\mu}=\epsilon X^{\mu}$. I don't see how we can say "coordinates do NOT change in FIELD theory" Apr 16 at 12:40
• @FaberBosch I've added another part showing we can eschew the notion of "coordinate change" also in this instance. Apr 16 at 12:59
• @FaberBosch - There maybe other points of view that suit your understanding better. The point of view that I have suggested is the least confusing one (to me at least) and generalizes trivially to more complicated theories like CFTs, gravity and their quantum mechanical counterparts like string theory as well. As long as you understand the point using whichever method best suits you, that's good! Apr 17 at 7:53

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 $$$$S = \int_{t_1}^{t_2}{\rm d}t f(t)\left(\frac{m}{2}\dot{x}^2 - V(x)\right)$$$$ where $$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 $$x$$ $$$$m \ddot{x} + \frac{\dot f}{f} m \dot{x} + V' = 0$$$$ The term $$\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.

• "It does not mean that we perform a coordinate transformation $t→\tilde{t}(t)$ (which indeed cannot change anything physical about the system)" Not sure I understand this. Changing the time coordinate does change the action too unless you can prove that it doesn't. Time reparametrization invariance is an additional requirement and leads to zero Hamiltonian. A very different scenario. Apr 16 at 3:20
• Not clear why you write "$\delta L = L(x(t+\delta t), \dot{x}(t+\delta t), t) - L(x(t), \dot{x}(t), t)$" and NOT $\delta L = L(x(t+\delta t), \dot{x}(t+\delta t), t+\delta t) - L(x(t), \dot{x}(t), t)$? Because "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" But why is that? Why would you choose not to do it. Needs more elaboration. Apr 16 at 4:18
• @FaberBosch The idea is that we are shifting the particle from $t$ to $t+\delta t$, keeping the coordinates fixed. We aren't relabeling coordinates. Apr 16 at 14:46

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.

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).

• We should not that, while this issue is about active vs passive transformations, it is not about field transformations vs spacetime transformations. It is obviously possible to define passive transformations of the field variables that would leave the action invariant in the sense of OP.. E.g. you can define passive transformations on the infinite dimensional configuration space of fields. The action would just change its functional form under any passive transformation on the confirguation space, and hence remain invariant. Apr 16 at 5:16
• Even on the configuration space, passive transformations of field variables always leave the action invariant. The solution here is to interpret all transformations on the configuration space as active transformations, i.e. we fix the functional form of the action and then we investigate the symmetries of the theory. Apr 16 at 5:29
• @Prahar Let's $x \to x' = \Lambda x$. Then even if we only relabel the coordinates, we can interpret this as a field redefinition too, for we have $\phi(x)\to\phi(x')=\phi(\Lambda x)=\phi'(x)~\text{for some }\phi'$, then this implies a field redefinition $\phi(x)\to\phi'(x)$. To say that we never transform coordinates and only the fields, is only an assertion that is neither self-sufficient nor obvious and requires justification. Apr 16 at 5:50
• @FaberBosch - What you have done is what I am saying one should do. Transform the fields and use the coordinate transformations as a tool to describe how the new field is defined in terms of the old field. I will edit my answer to clarify this point. Apr 16 at 11:39

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.

• Doesn't answer my question. Apr 16 at 3:11
• @FaberBosch In that case, you should look at the derivation of Noether's theorem. It assumes a fixed function $L(\phi, \partial ^{\mu} \phi, \alpha)$. Noether's derivation would fail with your notion of invariance of action. Apr 16 at 3:33
• This is dubious. The Lagrangian need not be invariant! It can indeed differ utmost by a total derivative term, c.f. David Tong's QFT lecture note (page 14). My question is not that though but rather the question is why the rule of calculus regarding the change of variable is not sufficient here. Apr 16 at 3:43
• @FaberBosch Yes, the more precise thing to say is to allow for the functional form of the action to change upto a total derivative. The rest of the argument is the same. Again, you should look at Noether's derivation to see why your passive change of variable transformations do not derive the theorem. Noether's theorem is about the invariance of laws under active transformations. The functional form of the action does not change under an active transformation. So your notion of invariance of action is ruled out from the beginning in Noether's derivation. Apr 16 at 3:47
• @FaberBosch I have edited the answer to make this clear. Again, the english statement of Noether's theorem may be misleading. But the derivation of Noether's theorem makes it clear that we're dealing with the same function $S[\phi, \partial ^{\mu} \phi, \alpha]$ Apr 16 at 3:59