6
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ That doesn't look like any Noether's theorem formula that I've ever seen... $\endgroup$
    – tparker
    Sep 13, 2016 at 0:08
  • $\begingroup$ Looks like maybe axial symmetry around z. You can try. Not obvious to me if the xy dot- yxdot is symmetric, and I probably need to do some more thinking. Try the rotation around z. I'll look if any procedure other than guessing - and easier by trying the symmetry, eg rotation around one axis, than the conserved quantity, $L_z$ $\endgroup$
    – Bob Bee
    Sep 13, 2016 at 0:21
  • $\begingroup$ @tparker my teacher called it "Noether's Inverse Theorem" in the sense that you can obtain a symmetry from a conserved quantity... (I haven't seen it anywhere in the internet too) $\endgroup$
    – rsaavedra
    Sep 13, 2016 at 0:47
  • $\begingroup$ Ok, if you can guess C. I just think that's harder, but you could try $L_z$. I'm not sure it'll work, maybe the X dot y etc term needs to be square for that, can't think on my feet. I think the other terms are axially symmetric, but I could be wrong. Need to try it. $\endgroup$
    – Bob Bee
    Sep 13, 2016 at 0:51

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.