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I have recently been trying to see what consequences Noether's theorem would have if our world was setup with different symmetries.

It is a quite elegant result that the invariance of the Lagrangian due to translation (translational symmetry) is the same as the conservation of momentum. Which leads me to ask under what symmetries would a different quantity be conserved; like $m \frac{d^2q}{dt^2}$, or more generally $m \frac{d^nq}{dt^n}$. Which would lead Newton's Second Law to read $F=m \frac{d^{n+1}q}{dt^{n+1}}$? Would it be something like 'the Lagrangian is invariant under a change in velocity', I am trying to run through the Math but I keep going around in circles with the definition of the Lagrangian itself.

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  • $\begingroup$ Can acceleration (or its components along axes, rather) even be considered as an observable (in the sense of a smooth function on the cotangent bundle)? It seems like one can't determine the acceleration given position and momentum only, just as one can't determine momentum from position only. The reason I ask is that the first thing that occurred to me was to use the technique mentioned in Baez's paper that takes an observable and recovers a flow on the cotangent bundle for it. $\endgroup$ – rschwieb Jan 27 at 15:07
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Try and think of a Noether charge besides a momenta (linear, angular, Hamiltonian). It's rather hard to do in point particle mechanics because there really aren't any we talk about (it's easier in field theory to come up with things that aren't just momentum).

So, let me pose the following question, which I think will make the point I want to get at: without looking at your Lagrangian, can you tell me what the correct expression for the momentum is? If you say $mv$, that's not always correct. For example, if your point particle charged and in the presence of an electromagnetic field the standard momentum ends up being something like $m\boldsymbol v-q\boldsymbol v\cdot\boldsymbol A/c$ where $\boldsymbol A$ is the vector potential and $c$ is the speed of light.

So the point here is that the actual expression for momentum depends entirely on the Lagrangian you're looking at. In this view, the word "momentum" is nothing more than indicating the Noether charge associated to translation invariance. I could very easily come up with a different transform and then give its associated Noether charge a special name, but at the end of the day, it's nothing more than "the Noether charge associated to the given symmetry."

As a result, the statement "momentum is the Noether charge associated to translations" is of a different nature than "what symmetry would give rise to acceleration conservation." The latter question is more like "what symmetry gives rise to the conservation of $mv$." The answer is going to depend on the details of the Lagrangian you have on hand. For many standard Lagrangians, the answer to the question "what symmetry gives rise to the conservation of $mv$" will be "translations," but as I've pointed out above, if our Lagrangian is the one which gives us the Lorentz force equations, $mv$ is not conserved and even if it was, it wouldn't be the momentum because it differs from the actual Noether charge by an addition of the vector potential.

So when you ask "what symmetry would give rise to acceleration conservation," really you're asking a question that requires a specific Lagrangian in order to answer.

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There is of course the cheap answer: Sure, when you have conservation of velocity $\frac{\textrm{d}q}{\textrm{d}t} = V$, then you get conservation of all higher derivatives for free, $\frac{\textrm{d}^n q}{\textrm{d}t^n} = 0$ for $n\ge2$. Of course, conservation of velocity isn't really a statement in Noether's theorem, as Richard Myers explains in his elegant answer.

But I suppose you won't be very pleased with that, so let me go a bit deeper. Assume you have a Lagrangian depending on some generalized coordinate and its derivative with respect to time (or any continuous parameter for a more general case) and possibly time itself, $L = L(q,\dot{q},t)$. You will find in your texts that the conserved quantity (sometimes dubbed the Noether charge) is a function related to other functions that depend only on $q$, $\dot{q}$ and $t$, cf. the wiki-page on Noether's theorem. There is no symmetry that gives you an a priori relation of these to the acceleration. So these are the quantities that can be used to define conserved quantities, and acceleration isn't one of them.

There is an important observation hidden here. The information you are looking for exists on different levels of the description. The Noether theorem is beautifully general: given a symmetry of the action (often expressed as a symmetry of the Lagrangian), you get information about conserved quantities. The symmetry sure is a restriction, but there are still many distinct forms the Lagrangian could take. At this level of description, position, velocity and time are the only available physical variables. Only after you supplied a definition of the Lagrangian $L$, you can gain information about how a certain setup of position, velocity and time leads to an acceleration. Clearly, this is a much greater specification/restriction of the Lagrangian.

Just as a side-note: The fact that velocities and positions are more qualified to be the determining factors of conserved quantities also makes sense when considering the Euler-Lagrange equation of this prolem. It will be a second order differential equation $$F(\ddot{q},\dot{q},q,t) = 0.$$ The solutions to this ordinary differential equation are completely determined by $q(t_0)$ and $\dot{q}(t_0)$ for all $t$ with $t_0 \le t < T$ (in physics, we often take $T = \infty$). So those are the quantities that a conserved quantity should be related to.

Everything I said can be circumvented by considering Lagrangians with a higher order of derivatives, see for example Derivation of Euler-Lagrange equations for Lagrangian with dependence on second order derivatives. In these cases, the Noether charges will depend on functions that also have an explicit dependence on higher derivatives, so you can use symmetries to directly fix your acceleration.

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OP's question would be easy to answer if there exists an inverse Noether's theorem. Such inverse theorem works best in the Hamiltonian formulation. But to formulate acceleration $\ddot{q}$ in the Hamiltonian formulation, we would have to get rid of time derivatives via Hamilton's equations.

  1. In other words, if we have successfully performed a Legendre transformation $L(q,\dot{q},t)\longrightarrow H(q,p.t)$, we should apparently rewrite acceleration $$\ddot{q}\longrightarrow \{\{q,H\},H\}\tag{A}$$ as a double Poisson bracket. Then it superficially seems that we can answer OP's question: The corresponding infinitesimal quasi-symmetry is then $$ \delta ~=~\epsilon\{\cdot, \{\{q,H\},H\}\},\tag{B}$$ where $\epsilon$ is an infinitesimal parameter. Well, not quite. There are at most $2n$ independent constants of motion (COM), and a higher-order quantity like acceleration is not counted as an independent COM. In practice the infinitesimal symmetry (B) vanishes identically, cf. below example.

    However, there is a remedy. Instead of asking if $\ddot{q}$ is a COM, we could equivalently ask if $\dot{q}-\ddot{q}t$ is a COM. In other words, playing the same game as before, we should consider the quantity $$ Q~:=~\{q,H\}-\{\{q,H\},H\}t. \tag{C}$$ The corresponding infinitesimal quasisymmetry is then $$ \delta ~=~\epsilon\{ \cdot, Q\}.\tag{D}$$ Eq. (D) is the main answer to OP's title question.

    Example: $H=\frac{p^2}{2m} + mgq$. The acceleration $-g$ is constant. Then $Q=\frac{p}{m} + gt$ and the quasisymmetry $\delta$ is $q$-translations.

  2. If the Lagrangian $L(q,\dot{q},\ddot{q},t)$ depends on acceleration $\ddot{q}$, we should use the Ostrogradsky formalism, cf. e.g. my Phys.SE answer here. Since velocity $v$ is now a phase space variable, it is enough to rewrite acceleration $$\dot{v}\longrightarrow \{v,H\}\tag{E}$$ as a single Poisson bracket. The corresponding infinitesimal quasi-symmetry is then $$ \delta ~=~\epsilon\{\cdot, \{v,H\}\}\tag{F}$$ by the same token. In a higher-order theory, this is non-trivial.

  3. The further strategy is now clear. If the Lagrangian $L(q,\dot{q},\ddot{q},\dddot{q},t)$ depends on jerk $\dddot{q}$, then acceleration $a$ becomes a phase space variable in the next-order Ostrogradsky formalism. The corresponding infinitesimal quasi-symmetry is then $$ \delta ~=~\epsilon\{\cdot, a\},\tag{G}$$ and so forth.

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I am not sure, whether it would be an exhaustive answer to the question. But the thought that came to me is to consider the Lagrangian with the derivatives of higher order. Let us for simplicity look at the Lagrangian with the derivatives up to 2nd order $L(\phi, \dot{\phi}, \ddot{\phi})$. The variation of the action gives with respect to some change of the generalized coordinate $\phi \rightarrow \phi + \delta \phi$ gives: $$ \delta L(\phi, \dot{\phi}, \ddot{\phi}) = \frac{\delta L}{\delta \phi} \delta \phi + \frac{\delta L}{\delta \dot{\phi}} \delta \dot{\phi} + \frac{\delta L}{\delta \ddot{\phi}} \delta \ddot{\phi} = $$ $$ = \left(\frac{\delta L}{\delta \phi} - \frac{d}{dt}\frac{\delta L}{\delta \dot{\phi}} + \frac{d^2}{dt^2}\frac{\delta L}{\delta \ddot{\phi}} \right) \delta \phi + \frac{d}{dt} \left(\frac{\delta L}{\delta \dot{\phi}} \delta \phi -2 \frac{d}{dt} \frac{\delta L}{\delta \ddot{\phi}} \delta \phi\right) + \frac{d^2}{dt^2} \left(\frac{\delta L}{\delta \ddot{\phi}} \delta \phi\right) $$ Assuming that the action is invariant $\delta S = \int L \ dt$, would follow from following conditions:

  1. Euler-Lagrange action (the first bracket) for any variation vanishing on the boundary: $$ \delta q(t_1) = \delta q(t_2) = 0 $$
  2. Conservation of some current (Noether's current) along the trajectory: $$ \frac{\delta L}{\delta \dot{\phi}} \delta \phi -2 \frac{d}{dt} \frac{\delta L}{\delta \ddot{\phi}} \delta \phi = \text{const} $$ For the Lagrangian, depending only on the first order derivatives the first term gives the familiar expression
  1. Linear dependence on time of the following quantity: $$ \frac{\delta L}{\delta \ddot{\phi}} = a t + b $$ The dumbest thing which comes to mind and has these properties is : $$ L = m \ddot{q}^2 $$ Then conditions $2,3$ would imply that: $$ \ddot{q} = at + b \rightarrow \frac{d^3}{dt^3} q = \text{const} $$ However, this theory hardly corresponds to anything physical.

In general, Lagrangians having higher order derivatives are thought to be unsuitable to account for any physical system due to the Ostorgrsky instability https://en.wikipedia.org/wiki/Ostrogradsky_instability.

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interesting question. I don't know if there is a good answer. I mean to conserve acceleration you should have a lagrangian like $L=...\ddot{q}+Fq$ with F constant force. So the request is like $L \to L+Fa$ if $q\to q+a$ so it's not a symmetry but more like you want that the symmetry is broken in a specific way. I don't know if it's possible to make it to a symmetry of $L$ to use Noether. I could suggest to look how it goes in the Hamiltonian formalism there's a Noether version for it and maybe you don't bite the tail since $q$ and $P$ are independent.

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