6

Noether's Inverse Theorem says that there must be a symmetry $\delta q^{i}$ given a conserved quantity $C$.

$$\delta q^{i}=\epsilon W^{ij}\frac{\partial C}{\partial q^{j}},$$

where $\epsilon$ is an arbitrary constant and $W^{ij}$ is the Hessian of the Lagrangian.

Basically, one can guess some $C$ to obtain $\delta q^{i}$ from this equation, but usually one knows from the Lagrangian which $C$ is suitable for doing this.

My question is: Is there a way of knowing what is not a conserved quantity for the system? Or a way to make this equation fail given a $C$?


In particular, I am working with the Lagrangian:

$$L=\frac{m}{2}(\dot{x}^2+\dot{y}^2+\dot{z}^2)-V(x^{2}+y^{2}+z^{2})-\frac{m}{2}[(\dot{x}y-\dot{y}x)\omega^{2}+\omega^{2}(x^{2}+y^{2})]$$

where $V$ is a derivable function.

I found that the energy $E$ gives a specific symmetry. But, I am afraid of inserting the angular momentum and find that there is another symmetry, and actually afraid that anything that I would insert in the equation would give more symmetries... I think there should be something that tells me that some of those are actually not symmetries.

Qmechanic
  • 220,844
rsvdr
  • 872

1 Answers1

4

It is best/most systematic to discuss the inverse Noether's theorem in the context of Hamiltonian formalism (as opposed to Lagrangian formalism), cf. e.g. this and this Phys.SE posts, which also provide Lagrangian counterexamples. Therefore OP's Lagrangian of the form $$ L~=~\frac{m}{2}(\dot{q}_x^2+\dot{q}_y^2+\dot{q}_z^2)+B(\dot{q}_yq_x-\dot{q}_xq_y)-V(q) $$ should preferably first be Legendre transformed into the corresponding Hamiltonian

$$ H~=~\frac{1}{2m}\left( \left(p_x +Bq_y \right)^2+\left(p_y -Bq_x \right)^2 +p_z^2 \right) + V(q). $$

Then as shown in Statement 3 in my Phys.SE answer here, we have an inverse Noether's theorem: If $Q=Q(q,p,t)$ is a constant of motion, then the corresponding Hamiltonian vector field $$-X_Q~:=~ -\{Q,\cdot \} ~=~\{\cdot ,Q\}$$ will generate a quasisymmetry of the phase space action.

Qmechanic
  • 220,844