- #1

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**static**magnetic field is:

##\frac{d}{dt}[m{\dot{\vec{r}}}]=q\ \dot{\vec{r}}\times \vec{B}##

From this, we can see that ##\frac{d}{dt}[m\dot{\vec{r}}-q \vec{r}\times \vec{B}]=0##

and so the following quantity is conserved:

##m\dot{\vec{r}}-q \vec{r}\times \vec{B}##

My question is how can I derive the above conserved quantity through Noether's theorem?

I attempted to do this by a simultaneous spatial transformation and a gauge transformation for a

*static*and

*homogeneous*magnetic field, so that the Lagrangian remains invariant. Through doing this, and finding this more "generalized" translation symmetry (generalized due to the presence of the magnetic field, or just a gauge field), I found that the quantity ##m\dot{\vec{r}}+q\vec{A}## , where ##\vec{A}## is the vector potential, is conserved. But that is the canonical momentum of the particle, which we know that it is not conserved and this is evident by the fact that its (total) time derivative is not zero, as can be seen from the equation of motion of the particle (As given above).

So, how does one arrive at the desirable conserved quantity through Noether's theorem?

Thanks in advance.