How is energy conservation & Noether's theorem a non-trivial statement?

Question

Noether's theorem says that energy conservation is a result of temporal translation symmetry of the laws of physics. This is implied to be - and I'm not saying it's not - a very non-trivial statement. So then how come it doesn't seem to be so non-trivial when we try to formalize this informal statement in the following seemingly very natural and intuitive way?

In order to make sense of the claimed statement formally, we need to make sense of two things: what a "law of physics" is, and what "temporal translation symmetry" is. And theoretical mathematics seems to provide a rather neat answer for the first - what is called a dynamical system.

A dynamical system is a triple $(M, T, \Phi)$ consisting of a set $M$ called phase space, a monoid (a set where you can add elements together, roughly speaking) $T$ called time space, and finally a family of maps $\{ \Phi^t \}_{t \in T}$, $\Phi^t : M \rightarrow M$, called evolution, flow, or if we like, physics, which satisfy the following semigroup property: for all $P \in M$ and times $s, t \in T$,

$$\Phi^0(P) = P$$ $$\Phi^s(\Phi^t(P)) = \Phi^{t + s}(P)$$

. I say this is very intuitive because it rather directly corresponds to what we imagine laws of physics to do: they take what we know about the present, i.e. the phase space point $P$, and they predict the future, i.e. what will be the case after some interval: that is $\Phi^t(P)$.

(Also, a note: in physics, at least $T$ is typically taken as the real numbers $\mathbb{R}$, together with addition. In specifically classical mechanics, $M$ is a differentiable manifold and more specifically a "symplectic manifold", so it has some additional structure beyond purely topological and analytical. For a simple system consisting of a moving point-like object, the elements of $M$ are pairs $(X, \mathbf{p})$ of a position $X$ and momentum $\mathbf{p}$.)

We can then define a symmetry of the dynamical system as a bijective self-map $S: M \rightarrow M$ of the phase space into itself that respects the dynamics, i.e. for all $P \in M$ and $t \in T$,

$$\Phi^t(S(P)) = S(\Phi^t(P))$$

.

Put another way, if we transform phase space by the symmetry, the transformed point generates a history that is simply the transform of the history of the original point, i.e. both are generated by the same law of physics $\Phi$. Or to put it another way, symmetries are some kind of "automorphism" of the dynamical system.

But here's the thing: each time translator, $\Phi^s$, is a symmetry of $\Phi$ by this definition, and thus that means there should always be a conserved energy by implication, no? And if we were given a time series that could not be given by a dynamical system as described above, e.g. two values in the series were equal but with different values following them so the semigroup property fails or equivalently the trajectory self-intersects, we can always just enlarge the phase space. And then once more we can express some kind of conserved "energy".

In other words, all dynamical systems are both time-symmetric and conserve some quantity as a result thereof which we could potentially call an energy.

So why then is the statement that energy is conserved a "nontrivial" statement about the universe if we can always come up with something to call a "conserved energy"? Or, to put it another way, how is it that temporal translation symmetry is nontrivial, when it is essentially baked in to the definition of dynamical systems and phase spaces (in particular, it follows from the non-crossing of phase-space trajectories)?

Answer 1

Here are a few observations, that I think together form an answer to your question (although I'm not sure I followed all of the notation in the body of the question).

1. Any solution to a (scalar) second order differential equation has two constants associated with the solution, namely the two integration constants. However, in general, we usually don't call these constants conserved quantities, because they can't be computed using only the dynamical degrees of freedom at a given time. In other words: to compute the energy of a harmonic oscillator at time $t$, we only need to know the position and velocity at time $t$ (plus constants like the frequency of oscillation). However, to know the initial conditions at some time $t_0$, we need to integrate the equations of motion back in time. Of course, for a harmonic oscillator, we can and often do this and relate the initial conditions to the energy, in which case the initial conditions can be recovered knowing the state of the system at only the current time (up to a phase). But for a system without a conserved energy, you would only be able to compute the initial conditions by evolving the system backward; you can't directly compute the initial conditions given only the state of the system at some later time.

2. Of course, there are dynamical systems without conserved quantities. Namely, any system without time-translation invariance (such as a harmonic oscillator where the frequency of oscillation changes with time). Your formalism should be able to see this (although it's not 100% clear to me how), but for a system with a time-dependent Hamiltonian, in order to define the evolution operator (which you called $\Phi^s$) you should have to specify both the initial time $t_0$ and the final time $t$ to which you are evolving the system.

To summarize, the non-trivial aspect about energy is that it is a statement about a quantity that can be computed using only the state of the system at a single time. We can also label phase space trajectories using initial conditions, but (unlike energy) the only way to compute the initial conditions (given the state at a later time $t$ is to evolve the system back to the initial time.

Answer 2

Energy conservation means that the equations of motion have a first integral. This is a non-trivial statement, since it selects a subset of equations out of all imaginable types of them.

While many equations of motion obtained empiriclaly were known to have such an integral, the Noether's theorem makes this statement in the most general form, as a result of the homogenuity of time.

Answer 3

1. On one hand, Noether's theorem in its original formulation assumes an action formulation. OP's setup (v1) lacks this. The action formulation leads among other things to the standard formula for energy (which is of course the Noether charge for time translations), cf. e.g. this Phys.SE post.

2. On the other hand, given only OP's triple, there is not a clear definition of energy. One is usually not allowed to pick any old integration constant of the EOMs and claim it's the energy.

Answer 4

This is a part of an answer; not yet a full answer.

The issue with the question as I've framed it is that this definition of dynamics may not be what a lot of physicists have in mind, even though it is a very intuitive one because it basically says that a dynamical or physical law is something you can use to predict the future from the present, given the present state and how far ahead you want to go.

In my comments under @Andrew's post I discussed this part a bit, and the problem is somewhat subtle: the phase space is the present state of the system, however we may conceptualize a history generated by applying one $\Phi$ family for a certain range of times, then apply another after that, and then another, and so on and so forth. This would not be a dynamical system by the definition given, but would represent a "changing law of physics" in exactly the way we think of via the contrapositive of Noether's theorem, given that we take $\Phi$ to be the object that makes precise the notion of the "law of physics".

Hence, to discuss Noether's theorem in both temporal and spatial translation symmetry, as a non-trivial statement, we need to admit that the definition of dynamics we gave was, in a sense, too strict. One potential looser definition to capture the above is that the map family $\Phi$ has both a time increment and a starting time at which said increment is to be applied: we might change the notation to $\Phi_{t, \Delta t}$ instead, to emphasize it's no longer a straightforward iterative process. The semantic meaning of this map is "interpret the passed physical state as holding at the specific time instant $t$. Then evolve it forward $\Delta t$ according to the relevant laws of the time." And we require the following: again,

$$\Phi_{t,0}(P) = P$$

but now we modify the semigroup law to

$$\Phi_{t+\Delta t,\Delta s}(\Phi_{t, \Delta t}(P)) = \Phi_{t, \Delta t + \Delta s}(P)$$

where we note now the further evolution by $\Delta s$ has to be started explicitly at the time point where that the first evolution by $\Delta t$ left off, i.e. $t + \Delta t$ from the initial time $t$ at which state $P$ is valid. That part is needed because the laws may have shifted by that point.

In this formalism, temporal translation symmetry can be defined as stating that $\Phi_{t,\Delta t}(P)$ is independent of the starting time $t$, and thus now no longer is a trivial statement - we can have cases where it fails. Or to put it another way, temporal translation symmetry is in effect the statement that the dynamics on the given phase space forms a dynamical system as we defined it earlier.

That said, it still does seem possible we can enlarge the phase space, and then once more we're at the same problem. However - this is where the answer by @Qmechanic comes in: energy is not just any old random quantity we can make conserved with time. It is a very specific such conserved quantity - and that would require further explication and is where my answer must reveal its partiality.

Answer 5

Here's my take on the title question How is energy conservation & Noether's theorem a non-trivial statement?

Consider this quote from Spivak's Calculus on manifolds

Stokes' theorem shares three important attributes with many fully evolved major theorems:

1. It is trivial
2. It is trivial because the terms appearing in it have been properly defined.
3. It has significant consequences

I feel like Noether's theorem is definitely of the same sort mentioned here, and indeed I have seen it referred to as "trivial" in certain formulations. For example, in Baez's paper Getting to the bottom of Noether's theorem he makes remarks like

"Having reduced Noether’s theorem to a triviality by working only with generators..."

"What, however, is the meaning of the bracket’s antisymmetry? So far we have simply posited it in the definition of Poisson algebra, thus essentially building Noether’s theorem in from the start."

Also see Hudgins Understanding Noether’s theorem with symplectic geometry

This paper will show that Noether’s theorem, like all great theorems, is trivial to formulate and prove once the proper mathematical theory is developed, which, in this case, will be the theory of symplectic manifolds

To me this has all the hallmarks of Spivak's view of Stokes' theorem. Anyone with a good grasp of all the machinery used to prove Noether's theorem may well say that it seems trivial, and this is precisely because they have the benefit of the right machinery.

Lie algebra, in which Noether was an expert, certainly contains all the seeds for the insights needed to prove the theorem, but I think symplectic geometry on manifolds, while extant, was not well-known at the time. But after the 1960's symplectic geometry is quite well-known. Since then, mathematicians have been massaging the statement and definitions for years to maximize its potential for explaining things, including conservation of energy (via the trivial equation $\{H, H\}=0$.) So we are apt to fall prey to the "curse of knowledge" and wonder how earlier readers could have been worked up about something that seems trivial to us.

So considering all this, I think most people will agree Noether's theorem is a nontrivial result in the sense that the underlying ideas tie together in a useful and not entirely obvious theorem, even if its proof can be written very succinctly with modern concepts. I don't think anyone has any illusions about its difficulty.

Answer 6

There are two main ways to formulate classical mechanics: with a Lagrangian or with a Hamiltonian. In your question, you have essentially given a definition of classical mechanics which is highly reminiscent of Hamiltonian mechanics (without defining the symplectic form, but let's not get into that).

The thing is that, in Hamiltonian mechanics, time translational symmetry is indeed sort of trivial, you are correct. More specifically, time translation is generated by the vector field $X_H$ on phase space, where $H$ is your energy function, defined by $$ X_H(f) \equiv \{f, H\}. $$ By the trivial equation $\frac{d}{dt} H = \{H,H\} = 0$, we can see that energy indeed conserved.

So, in Hamiltonian mechanics, time translational symmetry is indeed "baked into the cake," and you are indeed correct.

However, in the Lagrangian formulation, it is not nearly so straight forward. If you have a completely general Lagrangian, $L(q_i, p_i, t)$, that depends explicitly on time $t$, then energy will not be conserved. Only when the Lagrangian is independent of time, $L = L(q_i,p_i)$, will energy be conserved.

Indeed, the way Emmy Noether formulated her original theorem was in the Lagangian framework.

Actually, in the Hamiltonian framework, you can also have time dependent Hamiltonians as well, $H(t)$. In these cases, energy will similarly not be conserved.

For more on this, check out my "Manifesto: What really is a symmetry?" in this answer here:

https://physics.stackexchange.com/a/461762/157704