# Conservation law for any potential field?

1. Dec 29, 2015

### maline

Consider a free particle moving in a general time-dependent scalar potential. Energy & momentum are not conserved. However, there is a symmetry in the lagrangian: the velocity appears only as its square, so we can rotate it without affecting the value of L. What conservation law results from this symmetry?

2. Dec 29, 2015

### Staff: Mentor

Conservation of angular momentum.

3. Dec 29, 2015

### maline

No, that's in the case of a central potential, where the symmetry is a rotation of both the position & velocity. I am asking about a completely arbitrary potential, and noting that rotation of the velocity alone should still be a symmetry.

4. Dec 29, 2015

### Staff: Mentor

Hmm, that is a good point that I missed. You would still calculate the conserved quantity using Noether's theorem, but I don't know what it would be offhand.

Last edited: Dec 29, 2015
5. Dec 31, 2015

### maline

One bump...

6. Dec 31, 2015

### Jilang

Isn't this just time reversal symmetry?

7. Dec 31, 2015

### maline

No, I'm talking about rotating the velocity by a general angle in any direction. This should be a continuous symmetry.

8. Dec 31, 2015

### Orodruin

Staff Emeritus
You cannot just rotate the velocity without rotating the coordinate system itself. The transformations covered by Noether's theorem are of the form $t \to t +ks$ and $\vec x \to \vec X(t,s,\vec x)$, not transformations of the velocities.

You can do more general canonical transformations in Hamiltonian mechanics, but based on the symmetries of the Lagrangian this is not the case. In order to have a symmetry of the Lagrangian you therefore need to have rotational symmetry of the potential as well, resulting in conservation of angular momentum.