Energy conservation without action principle?

Question

The normal tagline for energy conservation is that it's a conserved quantity associated to time-translation invariance. I understand how this works for theories coming from a Lagrangian, and that this is the context that the above statement is intended to refer to, but I'm curious as to whether or not it's true in greater generality (i.e. is true in a wider context than can be shown through Noether's theorem). I'll stick to single ODEs, since this case is already unclear to me. If we have a differential equation

$$\ddot{x}=f(x,\dot{x})$$

for general $f$ this clearly possesses time translation symmetry. This does not conserve energy as the term is normally used, since this includes examples such as a damped harmonic oscillator. However is there actually no conserved quantity of any kind associated to the symmetry? If there's no dependence on $\dot{x}$ we can easily find an integral of motion, but I'm not sure why any dependence on $\dot{x}$ would ruin this.

Answer 1

(After possibly introducing more variables) then OP is essentially considering an autonomous system of $n$ coupled 1st order ODEs

$$\tag{1} \frac{d\vec{z}(t)}{dt}~=\vec{f}(\vec{z}(t)), \qquad f: U \to \mathbb{R}^n , \qquad U\subseteq \mathbb{R}^n, $$

i.e. without explicit time dependence, so that the system (1) possesses time translation symmetry.

OP is now pondering whether there exists a non-trivial function $h:U\to\mathbb{R}$ such that $t\mapsto h(\vec{z}(t))$ is constant along each solution to the system (1)?

The caveat is that $h$ is only allowed to depend on the structure function $f$ of the system (1), but not e.g. on the initial conditions for a solution.

For $n\geq 3$, such $h$ does not exist in general. However, there are two important exceptions:

1. If the system has an autonomous action functional formulation, then Noether's theorem shows that the corresponding energy function is conserved on-shell. (This case is already mentioned by OP.)

2. In the case $n=2$, which is in fact the case that OP wanted to focus on, then there always exists (in sufficiently small open neighborhoods) an autonomous Hamiltonian formulation of the system (1), cf. e.g. my Phys.SE answer here. The Hamiltonian $H$ itself is a conserved quantity. (The case $n=1$ also has a conserved quantity.)

From a physics perspective, it is natural to associate the function $h$ with the energy of the system. If this identification was correct, then all dissipative systems would be counterexamples. However $h$ does not need to be the actual physical energy of the system. See e.g. this non-standard action functional formulation of a dissipative system. Nevertheless, such small-$n$ accidents become rarer as $n$ grows.

Answer 2

I think I can remember the derivation for a conservative force field in classical mechanics, wich is a somewhat stronger assumption than pure time-translation invariance.

Let $\vec{F}$ be a conservative force field, that is $$ \nabla \times \vec{F} = 0 $$ or alternatively $$ \phi := -\int_\gamma \vec{F} \cdot d\vec{a} $$ does not depend on the path $\gamma$ and its parametrization $\vec{a}(s)$, but only on the "endpoints".

Take Newton's equation of motion: $$ m \ddot{\vec{r}} = \vec{F} $$ multiply by $\dot{\vec{r}}$ and integrate over time $$ m \int_{t_0}^t \frac{d \dot{\vec{r}}}{dt} \cdot \dot{\vec{r}} ~dt = \int_{t_0}^t \vec{F} \cdot \dot{\vec{r}} ~dt$$ $$ m \int_{\vec{v}_0}^{\vec{v}} \dot{\vec{r}} \cdot d \dot{\vec{r}} = \int_{\vec{r}_0}^{\vec{r}} \vec{F} \cdot d \vec{r} $$ $$ m \int_{v_0}^{v} v ~dv + m \int_{\vec{n}_0}^{\vec{n}} v^2 \vec{n} \cdot d \vec{n} = -\phi(\vec{r}) + \phi(\vec{r}_0) $$ $$ \frac{1}{2} m v^2 - \frac{1}{2} m v_0^2 + m \int_{\vec{n}_0}^{\vec{n}} v^2 \vec{n} \cdot d \vec{n} = -\phi(\vec{r}) + \phi(\vec{r}_0) $$ with $\vec{n}$ the unit vector pointing in the direction of the velocitiy $\vec{v} = \dot{\vec{r}} = v\cdot\vec{n}$. The remaining integral is a path integral over a path on the unit circle / sphere. That is: $\vec{n} \perp \frac{d\vec{n}}{ds}$ for any parametrization $\vec{n}(s)$. Thus, we have: $$ \int_{\vec{n}_0}^{\vec{n}} v^2 \vec{n} \cdot d \vec{n} = 0 $$ and we are left with:

$$ \frac{1}{2} m v^2 + \phi(\vec{r}) = \frac{1}{2} m v_0^2 + \phi(\vec{r_0}) = \text{const.} $$

Answer 3

I believe this is discussed in Nonlinear Dynamics and Chaos by Strogatz. I don't remember the details, but it's worth looking at his discussion of an energy function.

Answer 4

Here it explains what I think you are asking:

We work with a formulation of Noether-symmetry analysis which uses the properties of infinitesimal point transformations in the space-time variables to establish the association between symmetries and conservation laws of a dynamical system. Here symmetries are expressed in the form of generators. We have studied the variational or Noether symmetries of the damped harmonic oscillator representing it by an explicitly time-dependent Lagrangian and found that a five-parameter group of transformations leaves the action integral invariant. Amongst the associated conserved quantities only two are found to be functionally independent. These two conserved quantities determine the solution of the problem and correspond to a two-parameter Abelian subgroup.

so, in summary, what is found is that the differential operator equation

$$ \ddot{x} + \beta \dot{x} + \gamma x = 0$$

is not a self-adjoint operator, so by Helmoltz criterion it cannot have a Lagrangian, but by adding a factor of $e^{\lambda t}$ it can be made self-adjoint by the Lagrangian

$$ L = e^{\lambda t} \Big ( \frac{ \dot{x}^2}{2} - \frac{ \omega^2 x^2}{2} \Big )$$

that leaves explicit what is the time dependence of the Lagrangian with time