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Per Noether's theorem, time-translational symmetry leads to energy conservation. This identity works for any system not just free particles. However one could consider a dynamical system (breaking time-translation) but where kinetic energy remains the same. What would be the symmetry here? Any guide on how to reverse Noether theorem would help or any indication on why there is no such a symmetry.

From Is the converse of Noether's first theorem true: Every conservation law has a symmetry? I can work out that action has to be invariant to the transformation $q\to q+p \epsilon$ where $q$ is the position and $p$ the momentum and $\epsilon$ is small. Does this transformation have any meaning?

Mauricio
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I will consider autonomous systems. Noether’s theorem is easier in the Hamiltonian setting. Let $H$ be the Hamiltonian on a $2n$ dimensional phase space with canonical coordinates $x_i,p_i$ and $K$ the kinetic energy: $$ K=\sum \frac12p_i^2 $$ Conservation of energy means that in terms of Poisson brackets: $$ \{H,K\}=0 $$ This means that $H$ is invariant by the symmetry generated by $K$: $$ \dot x_i=\{x_i,K\}=p_i \quad \dot p_i=\{p_i,K\}=0 $$ i.e. it is invariant by your transvections: $$ x_i\to x_i+p_is\quad p_i\to p_i $$ Physically, this symmetry is just ballistic motion. Thus your Hamiltonian is invariant by ballistic motion: $$ H(x_i,p_i) = H(x_i+sp_i,p_i) $$

It is not hard to find all of such systems. A simple way is choose a transverse hypersurface to the flow lines. A suitable candidate is the hyperplane: $$ \sum x_i\frac{p_i}{\sqrt{2K}}=0 $$ Thus such Hamiltonians are of the form for any function $h$ defined on the hyperplane: $$ H=h\left(x_i-p_i\sum_j x_j\frac{p_j}{\sqrt{2K}},p_i\right) $$ Geometrically, this means that $H$ is a function of $p$ and the normal components of $x$ to $p$.

For physical applications, this is rarely the case. In non relativistic mechanics, you typically have $H = K+V$ with $V$ a position dependent potential. The only $V$ compatible with this symmetry is a constant, so conservation of kinetic energy in this case is trivial since it is just the Hamiltonian. Even adding magnetism will not help.

LPZ
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