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This kind of relates to my prior question about the non-triviality of temporal translation symmetry and will use some of the same concepts:

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

Because it seems that, in fact, temporal translation symmetry, when taken under the formalism there which I consider to be a good and rigorous way to make sense of the idea of a "law of physics", there are many conserved quantities associated with that symmetry.

Suppose that we have a dynamical system $(M, \Phi^t)$, with phase manifold $M$ and evolution map ("law of physics") $\Phi^t$, $t \in \mathbb{R}$. Any such system is temporal translation symmetric in that a point $P \in M$ uniquely determines the trajectory $t \mapsto \Phi^t(P)$, whether the dynamical object is imagined to be at that point $P$ at time 0, or at any other time. In particular, this means that no two trajectories cross.

In such a formalism, a "quantity" is just a real-valued function of the phase point $P$. It follows rather simply from the temporal translation symmetry's property above - that no phase trajectories cross - and also the fact that every phase point generates a trajectory under $\Phi^t$, that we can create a conserved quantity by, for each trajectory, picking a real number value freely, and then assigning every point on that trajectory the same value. When we have done enough such choices as to cover the entire phase space, we have a conserved quantity, whose conservation is thus due to temporal translation symmetry.

Formally, let $S \subseteq M$ be a subset of the phase space such that the orbits, i.e. the sets $\mathrm{Orb}(P) := \{ \Phi^t(P)\ |\ t \in \mathbb{R} \}$ are pairwise disjoint, i.e. $\mathrm{Orb}(P_1) \cap \mathrm{Orb}(P_2) = \emptyset$ for all $P_1, P_2 \in S$ with $P_1 \ne P_2$. Call such a subset an initial subset because, if it contains a point that is different from a given one but that appears on the same orbit as the previous, it will generate a furtherance of that orbit and thus the two sets will not be disjoint - at the maximal possible way. Hence it can only contain one point from each orbit involved. Further, call such an $S$ maximal or generator if $\Phi^t[S]$, i.e. applying it to every point in $S$ elementwise, generates the whole manifold $M$.

Then for such a generator, pick an arbitrary function $f_S: S \rightarrow \mathbb{R}$ define a quantity $Q: M \rightarrow \mathbb{R}$ via the following: if $P \in S$, then $Q(P) := f_S(P)$, but if $P \notin S$, then figure out the time delta $\Delta t$ required so that $\Phi^{\Delta t}(P) \in S$ (it could be either positive or negative, and moreover one must exist by the above argumentation). Then define $Q(P) := f_S(\Phi^{\Delta t}(P))$.

This quantity $Q$ is then conserved via temporal translation symmetry.

There are $\beth_2$ such quantities, and $\beth_1$ if we require continuity.

These numbers are uncountably infinite. $\beth_2$ is enormous; much bigger than the real number continuum.

Why then do we speak of the energy as "the" conserved quantity associated with this symmetry? What, among that huge infinity of other potential quantities, singles it out over all the different contenders? I do note that all such quantities will be functions of each other, but then why is energy the "canonical" or "go-to" representative of this class?

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    $\begingroup$ It is true that any function of a constant is also a constant. But at some point you are just making your life difficult if you take the sine of the hamiltonian and want to call that the "energy." $\endgroup$
    – hft
    Apr 29 at 0:24
  • $\begingroup$ @hft : The question though is, if you were only given $\Phi^t$ and $M$, how would you then deduce that, of all the conserved functions related to temporal translation symmetry, to take $H$ as your definition of conserved energy if you didn't know of it before? $\endgroup$ Apr 29 at 0:56
  • $\begingroup$ I don't really know what those things are, so I'm not sure. $\endgroup$
    – hft
    Apr 29 at 2:54
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    $\begingroup$ What makes $Q$ special from all the various $f(Q)$ is that it's additive. If you consider a larger system consisting of two subsystems, their energies will add linearly to yield the total energy of the larger system. Any other function of energy (up to global constant factor) won't add linearly. $\endgroup$
    – Ruslan
    Apr 29 at 12:58
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    $\begingroup$ I had thought about this topic a a little bit before and have come to the conlusion that our selection of energy (1/2 mv^2 or integral F * dx) was probably just more of a historical coincidence than anything. We were bound to discover SOME conserved quantity in that equivalence class and that happens to be hte one we found first. We even continued to go so far as to invent concepts like "potential energy" to make it stay conserved because these are USEFUL concepts. as Ruslan points out, energy is nice because it can be calculated using addition: maybe thats why we discovered it first. $\endgroup$ Apr 29 at 21:38

7 Answers 7

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The OP's question is basically stating that in a system with time-translation invariant dynamics, we can define a conserved quantity by arbitrarily assigning a real number to each orbit; when the system is in a particular state, the value of the quantity is the number assigned to the orbit that the state belongs to. The OP is asking why one particular quantity, the energy, is the conserved quantity associated with the time-translation symmetry and not any of the other possible arbitrary assignments.

In general, we are not interested in most of the possible conserved quantities that could be defined using this approach. Conserved quantities themselves are not useful; conservation laws are useful in the following sense: if $Q$ is a conserved quantity and $P$ is the initial state of the system, then you know that the system cannot evolve into any state $P'$ such that $Q(P) \ne Q(P')$. This statement is of use to the physicist only if there is a method to calculate $Q$ without having to integrate the equations of motion.

If the only way to calculate the conserved quantity is to work out the entire orbit that a particular state belongs to, so that, only once the orbit is identified in this manner, it is possible to see what $Q$ value was assigned to that orbit, then knowing the value of $Q$ no longer adds anything to your understanding. You have already done the hard work of calculating the entire orbit explicitly, and you don't need $Q$.

Noether's system gives an explicit formula for a conserved quantity if the system is described by a known action and there is a known continuous symmetry of that action. Essentially, if you write down the action and the symmetry, you just plug and chug to get the formula for the corresponding conserved quantity. For time-translation symmetry, the corresponding conserved quantity is the energy. This is extremely useful because the energy may generally be computed easily from this formula even if you don't know the entire orbit. In general, Noether's theorem will not help you determine how long it takes for the system to reach a particular state, but it will help you place strong constraints on which states the system might visit at all, assuming that the (known) action actually has a known symmetry.

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  • $\begingroup$ You understood the question right. I am not sure why this didn't seem to come across clear, perhaps, to the others. $\endgroup$ Apr 29 at 19:56
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    $\begingroup$ I suspect your question did not come across clear because not that many physicists - even professional ones, and not everybody on this site is a professional physicist - are familiar with the rather abstract mathematical formalism you used to state it. $\endgroup$ May 1 at 9:57
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What is special about the conserved quantity $Q(x, p) := \frac{1}{2} (x^2 + p^2)$, when also the quantity $Q_2(x, p) = \sin(x^2 + p^2)$ is conserved too, by the same temporal translational symmetry?

If $x^2+p^2$ is conserved, then so is any $f(x^2+p^2)$ such as $sine(x^2 + p^2)$. But manipulation like this is a little baroque.

You may remember where the Hamiltonian ($H$ or "energy") comes from in the first place. It is a useful quantity because we know the dynamics of the system can be determined from the Lagrangian, via the Euler-Lagrange equations: $$ \frac{d}{dt}\frac{\partial L}{\partial \dot q} - \frac{\partial L}{\partial q} = 0 $$

This means that (if there is no explicit time-dependence): $$ \frac{dL}{dt} = \frac{\partial L}{\partial q}\dot q + \frac{\partial L}{\partial \dot q}\frac{d\dot q}{dt} $$ $$ =\dot q\frac{d}{dt}\frac{\partial L}{\partial \dot q}+ \frac{\partial L}{\partial \dot q}\frac{d\dot q}{dt} $$ $$ =\frac{d}{dt}\left(\frac{\partial L}{\partial\dot q}\dot q\right)\;. $$ Or, in other words: $$ \frac{d}{dt}\left(L - \dot q\frac{\partial L}{\partial \dot q}\right) = 0 $$ Or, in other words: $$ H = \dot q\frac{\partial L}{\partial \dot q} - L $$ is conserved (i.e., constant in time).

Or, put another way: If there is "time translation symmetry" then $H$ is conserved.

Yes, any other function of $H$, $f(H)$, is also conserved if $H$ is conserved, but it is $H$ that is useful since we can use it to get Hamiltonian's form of the equations of motion. OTOH, you are certainly free to consider any other quantity you like and free to call it what you like. But if you call something other than $H$ the "energy" it will just not coincide with the common and conventional usage of the terms (common for at least over a hundred years or so).


An update based on comments

Given a Lagrangian $L$, a different Lagrangian $L'$ will generate the same equations of motions (same dynamics) if the two are related by: $$ L'(q,\dot q, t) = L(q,\dot q, t) + \frac{dF}{dt} $$ $$ =L(q,\dot q, t) + \frac{\partial F}{dq}\dot q + \frac{\partial F}{\partial t} \;, $$ where $F$ is any function of $q$ and $t$, but not $\dot q$.

For the time-translation invariant systems of interest here $F=F(q)$.

The relationship between the momentum $p=\frac{\partial L}{\partial \dot q}$ and the momentum $p' = \frac{\partial L'}{\partial \dot q}$ is $$ p' = p + \frac{\partial F}{\partial q} $$

The relationship between the Hamiltonian $H$ and $H'$ is found from $$ H' = \dot q p' - L' $$ $$ =\dot q(p + \frac{\partial F}{\partial q}) - L - \frac{dF}{dt} $$ $$ =\dot q (p + \frac{\partial F}{\partial q}) - L - \frac{\partial F}{\partial q}\dot q $$ $$ =\dot q p-L $$ $$ = H $$

That is, H and H' are the same.

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  • $\begingroup$ So the better way to say it is that Noether's theorem only implies the conservation of energy, but it does not motivate the specific choice of the definition of energy in its whole, then? $\endgroup$ Apr 29 at 1:00
  • $\begingroup$ However, aren't there many Lagrangians that can generate the same dynamics? So I'm not sure how this solves the problem. $\endgroup$ Apr 29 at 1:30
  • $\begingroup$ To any Lagrangian you can add any function of the form $dF/dt$, where F=F(q,t), and you get the same equations of motion. If you have time translation invariance you can only have F=F(q), and in that case the Hamiltonians are the same. $\endgroup$
    – hft
    Apr 29 at 3:02
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    $\begingroup$ The usual definition is $H=\dot{q}p-L$. $\endgroup$
    – J.G.
    Apr 30 at 17:00
  • $\begingroup$ Ha! Thanks for the correction, I updated it to use the correct definition of H. $\endgroup$
    – hft
    Apr 30 at 18:09
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  1. For what it's worth, OP seems to be mischaracterizing Noether's theorem by seemingly ignoring one of its main assumptions: That the physical system is equipped with an action formulation $$S~=~\int\! dt~ L(q, \dot{q},t).\tag{1}$$

  2. The action formulation in turn leads to a notion of momentum $$ p_i ~:=~\frac{\partial L}{\partial \dot{q}^i },\tag{2} $$ and energy $$ h~:=~p_i\dot{q}^i-L.\tag{3}$$

  3. And a time translation symmetry of the action leads to energy conservation, cf. e.g. this Phys.SE post.

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  • $\begingroup$ Yeah, this may be an important point, but that's because most of the introductions were given talking of the idea of a "physical law", which suggests a sense of generality to the result, and the dynamical concept mentioned there is what to my ears "general" means. I generally like to try and see if there's some way to put rigor to all those "physicist flourishes" of how that hard maths is talked about and maybe that missed something. $\endgroup$ Apr 29 at 19:54
  • $\begingroup$ I'm not sure about the first assumption. If I recall correctly, one only needs that the system obeys the Euler-Lagrange equations, which afaik is a weaker condition than that of having an action formulation (I may be rusty on this though). $\endgroup$ Apr 30 at 19:39
  • $\begingroup$ Hi @Massimo Ortolano. Thanks for the feedback. Noether's original article uses an action formulation. If you think it is in general unnecessary consider to provide a reference that supports your claim. $\endgroup$
    – Qmechanic
    Apr 30 at 20:06
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If I understand the question correctly, then a flow $\Phi^t$ uniquely determines a Hamiltonian vector field $\mathbf X := \mathrm d\Phi^t\big|_{t=0}$. Such a vector field can be written as $$\mathbf X = \{\cdot, F\} \iff X^\mu = \omega^{\mu\nu} (\mathrm dF)_\nu \iff (\mathrm dF)_\mu = \omega_{\mu\nu} X^\nu$$ where $\{\cdot,\cdot\}$ is the Poisson bracket, $\omega$ is the symplectic form, $F$ is a smooth function on the phase space, and $\cdot$ denotes an empty slot. We say that $F$ generates the vector field $\mathbf X$, and by extension the flow $\Phi^t$.


Example of the non-uniqueness: Let the $M$ be given by the 2-dimensional real phase space with points $(x, p)$. Define the time-evolve map via $$\Phi^t(x, p) = \left(x \cos(t) + p \sin(t), -x \sin(t) + p \cos(t)\right)$$

This flow induces the Hamltonian vector field $$\mathbf X = \mathrm d\Phi^t\big|_{t=0} = p \frac{\partial}{\partial x} - x \frac{\partial}{\partial p}$$ Comparing this with $\{\cdot, H\}= \frac{\partial H}{\partial p} \frac{\partial}{\partial x} - \frac{\partial H}{\partial x} \frac{\partial}{\partial p}$ for some smooth function $H$, we observe that $\frac{\partial H}{\partial p} = p$ and $\frac{\partial H}{\partial x} = x$, which implies that $H(x,p) = \frac{1}{2}(p^2+x^2) + C$ for some arbitrary constant $C$ which we may set to zero with no loss of generality.


From the expression $\mathrm dF = \omega( \cdot, \mathbf X) = \omega(\cdot,\mathrm d\Phi^t\big|_{t=0}\big)$ and the fact that $\omega$ is non-degenerate, we see that given a smooth flow $\Phi^t$ we obtain a function $F$ which generates it (up to a constant) and vice-versa.

QUESTION 1: What is special about the conserved quantity $Q(x,p)=\frac{1}{2}(x^2+p^2)$, when also the quantity $Q_2(x,p)=\sin(x^2+p^2)$ is conserved too, by the same temporal translational symmetry?

$Q$ and $Q_2$ are both conserved by the flow you specify, $Q$ actually generates that flow while $Q_2$ does not.

QUESTION 2: Is a better statement of Noether's theorem perhaps that every symmetry law of a dynamical system implies the existence of a naturally-associated set of conserved quantities?

In the context of Hamiltonian dynamics, Noether's theorem may be expressed as follows. If the flow $\Phi_F$ generated by $F$ preserves the Hamiltonian, then $H\circ\Phi_F^t = H$ and so $\frac{d}{dt}(H\circ \Phi_F^t)\big|_{t=0}=0$. Such a flow is called a symmetry of the system. This implies immediately that time evolution (i.e. the flow generated by $H$) preserves $F$, i.e. $F\circ \Phi_H^t = F \implies \frac{d}{dt}(F\circ \Phi_H^t)\big|_{t=0} = 0$. The proof consists of a single line:

$$\frac{d}{dt} \big(H \circ \Phi_F^t\big)\big|_{t=0} = \mathbf X_F (H) = \{H,F\} = -\{F,H\} = -\mathbf X_H(F) = -\frac{d}{dt}(F\circ\Phi_H^t)\big|_{t=0}$$

If $F$ is such a conserved quantity, then obviously $f(F)$ will be conserved. However, note that $F$ and $f(F)$ will induce different flows.

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There are nice answers already. I just want to add one small observation.

If $H$ is the energy, ordinary time is defined by the Hamiltonian flow with $H$, so observables evolve according to

$$\frac{dO}{dt} = \{H,O\}.$$

For any $f: \mathbb{R} \to \mathbb{R}$, $f(H)$ will be a constant of motion. If we consider the Hamiltonian flow by $f(H)$, which has a new parameter $\tau$, we have

$$\frac{dO}{d\tau} = \{f(H),O\} = f'(H)\{H,O\} = f'(H) \frac{dO}{dt}.$$

So if you like, $f(H)$ generates time translation but with a strange definition of the time,

$$f'(H) d\tau = dt.$$

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I think this is an interesting question. One should keep in mind that the symmetries are in the action S, which is in many cases composed of dual Fourier pairs (position and momentum, or energy and time, for instance). We focus a lot on invariance of time translation leading to conservation of energy in 4-momentum space, but I think you are pointing to a dual relationship, invariance of a quantity under translation in 4-momentum space (such as the zero of energy can be arbitrarily moved anywhere in 4-momentum space) leading to a conserved quantity in spacetime, as you say "the trajectory". It's called the "center-of-mass theorem".

The following snippet is from my original research on this, which I think is relevant here:

"Noether's theorem provides a link between physical symmetries and conservation laws. A conservation law will exist for a property if the Lagrangian $\mathcal{L}$ is not dependent on that property, i.e., if $\mathcal{L}$ is not a function of time, then energy is conserved over time, and similarly for the space/momentum pair. If both are true, then Noether's theorem results in the relation (Neuenschwander 2011) $$p_\mu \xi^\mu - H\tau = \text{constant}$$ where $\xi^\mu$ and $\tau$ are generators of translation. Symmetries in spacetime (time translation) result in conserved coordinates in momentum-space (energy conservation). What does this look like in momentum-space? Just as invariance of the action under infinitesimal translations in spacetime leads to conserved quantities in momentum-space, invariance of the phase distribution in momentum-space under infinitesimal boosts leads to a conservation law in spacetime. The result thus obtained is known as the center-of-mass theorem (\url{http://users.physik.fu-berlin.de/~kleinert/b6/psfiles/Chapter-7-conslaw.pdf}). Space and time are not individually conserved in this manner, but a certain relationship between them is indeed conserved, a ``conservation of trajectory''." (ref: https://www.mdpi.com/2624-960X/3/1/2)

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The answer is that the statement that there is such a thing as "the" conserved quantity associated to temporal translation - or to any symmetry - of a dynamical system, is simply wrong: instead, the existence of a symmetry implies, by the given argumentation, an enormous class of conserved quantities.

What the correct statement is is that given the quantity we know of as "energy", we can determine it belongs to the class generated by temporal translation symmetry, and as a result its conservation is a consequence of said symmetry.

What distinguishes energy from the other conserved quantities in its class really is, I believe, what @J. Murray mentions: it, or more accurately the quantity "Hamiltonian", $H$, generates the flow $\Phi^t$ in that an infinitesimal temporal advance $\Phi^{dt}$ can be given as (informally)

$$\Phi^{dt}(P) = P + [\nabla H](P)\ dt.$$

The reason we like to use energy is given by @Brian Bi's answer, though it is crucial to note that while this makes energy useful, it is not the sole quantity that has the property he describes as, say, $\sin(H)$, would also have the same property. And if we were not aware we were working with $\sin(H)$, we would still be able to do the manipulations he describes with it and it would not betray that it was not the "real" energy in that way. However, taken together, these two reasons provide a compelling argument for distinguishing it from the rest of the pack.

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  • $\begingroup$ I think this is a reasonable summary. Note that while every continuous symmetry gives rise to an infinity of conserved quantities, Noether's theorem (in the Lagrangian formulation) provides an explicit formula for one in particular, which is what we refer to when we casually talk about the link between continuous symmetries and conserved quantities. In other words, "energy is the conserved quantity associated to time translation symmetry [via Noether's theorem]" is a correct statement. Brian mentions this in the last paragraph of his answer. $\endgroup$
    – J. Murray
    May 4 at 14:28

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