The Lagrangian of a charged particle in a magnetic field reads: $$ L=\frac{m}{2}\dot{\bf{r}}\cdot \dot{\bf{r}} + q\bf{A}\cdot \dot{\bf{r}} $$ This is the Lagrangian in the reference frame $Oxyz$.

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I would like to know how this Lagrangian modifies if one moves to a rotating frame $Ox^\prime y^\prime z^\prime$ such that, at $t=0$, the two frames coincide and the primed frame rotates with respect to the first one with angular velocity $$ {\bf\Omega}=(0,0,\Omega). $$
For simplicity, you can assume that vector potential ${\bf A}$ generates a uniform magnetic field ${\bf B}$ along the $\hat{z}$ direction.

  • $\begingroup$ $q(\dot{\vec r}\cdot \vec A)$....you need a dot on the "r". $\endgroup$ – JEB Feb 28 at 19:19
  • $\begingroup$ yes of course, sorry for the typo! $\endgroup$ – AndreaPaco Feb 28 at 19:20
  • $\begingroup$ I don't think you can do it properly in 3d. State the transformation rules for the full spacetime in two frames. The reason is that vector potential is part of a four-potential, and you are trying to involve time in your transformation, so all the components of four-velocity and four-potential will be involved $\endgroup$ – Cryo Feb 28 at 19:38
  • $\begingroup$ Better still, write your lagrangian density in terms of four-vector contractions, and the only transformation you will be able to pick up is a jacobian $\endgroup$ – Cryo Feb 28 at 19:41
  • $\begingroup$ Thanks a lot for your comment. I apologize but I have a basic knowledge of relativity...! So I don’t fully understand your explanation. So, basically, it is not possible to properly define the system lagrangian in the rotating frame? $\endgroup$ – AndreaPaco Feb 28 at 19:53

The trouble with talking about moving reference frames and 'charged particle in a magnetic field' is that the latter point will not be shared by all the reference frames.

Imagine you are in the region with some static magnetic field $\mathbf{B}=\mathbf{B}\left(\mathbf{r}\right)$, and now you 'move to moving reference frame', that moves with velocity $\mathbf{v}$. Naively one may say that the magnetic field in this new reference frame will be $\mathbf{B}=\mathbf{B}\left(\mathbf{r}-t\mathbf{v}\right)$. But we also know, from Maxwell's Equations, that:

$\partial\mathbf{B}/\partial t=-\boldsymbol{\nabla}\times\mathbf{E}$

So you should get some electric field! Now, whilst my explanation is a bit hand-wavey the last point is true. If one observer sees pure static magnetic field in his/hers reference frame, another observer, in motion relative to the first one, will see both electric and magnetic field.

So to make progress in your task, finding the Lagrangian in 'rotating frame', you need to start with a more general Lagrangian, that includes both electric and magnetic fields, or , equivalently, both scalar and vector potentials.

Next we come to the action that you are trying to minimize. Currently you are considering problem that looks like this:

$S\left[\bar{x}\right]=\int_{t_a}^{t_b} dt \,L\left(\bar{x},\,\dot{\bar{x}},\,t\right)$

Where $\bar{x}$ is the trajectory of the particle. Geometrically speaking, you are integrating on a 3-dimensional sub-space in 4-d spacetime , in the region bounded by constraints $t=t_b$ and $t=t_a$. This is frame-specific, in a different (moving) frame (call it $S'$), the region your integrals is running over will still be bounded by two planes, but they will no longer be given by $t'=???$ constraint, instead the constraint will be something like $\dots t'-\dots x'=const$, i.e. it will involve both space and time.

So instead of singling out the temporal dimension you can write something like:

$ S\left[\bar{x}\right]=\int_\Xi cd^4x \,\mathcal{L}\left(\bar{x},\,\dot{\bar{x}},\,x\right)$

Where the integral now runs over some space-time hyper-volume $\Xi$ and $\mathcal{L}$ is the Lagrangian density:

$L=\int_{V_{\Xi}} d^3 r \mathcal{L} $

Now these are just pre-lims (see wiki). By the way, a sensible way to work with your particle, in this case, is to express it through four-current density:

$J^\mu\left(x;\,\bar{x}\right) = \int cd\tau \,qu^\mu\,\delta^{(4)}\left(x-\bar{x}(\tau)\right)$

Where $c$ is the speed of light, $q$ is the charge of the particle, $\tau$ is the proper time, and $\delta^{(4)}$ is the 4d delta function.

Now you need to state the rules for transforming between the coordinates of the moving frame and the static frame.

Only then can you look at your question

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