# A charged particle moves in a plane subject to the oscillatory potential

A charged particle moves in a plane subject to the oscillatory potential:

$U(r)=\frac{m\omega^2 r^2}{2}$

There is also a constant EM-field described by:

$\vec{A}=\frac{1}{2}[\vec{B}\times\vec{r}]$

where B is normal to the plane.

This produces the Lagrangian:

$L=\frac{m}{2}\dot{\vec{r}}^2+\frac{e}{2}\dot{\vec{r}}\vec{A}-U(r)$

Now my friend says we need to transform this into polar coordinates and that produces:

$L=\frac{m}{2}(\dot{r}^2+r^2\dot{\phi}^2)-mr^2\omega_L\dot{\phi}-U(r)$

where $\omega_L$ is the Larmor precession frequency:

$\omega_L=-\frac{eB}{2mc}$

My question is, How does he get this transformation? I don't really understand where the second term is coming from in the mechanical kinetic energy.

In polar coordinates $d\vec{r}=\hat{e}_r dr+\hat{e}_{\phi}rd\phi$. Devide it by $dt$ and you will have the particle velocity $\dot{\vec{r}}$. Square the latter and you will get the kinetic energy.

• Okay this makes a lot of sense. Thanks. This notation is much easier to read. Apr 21 '12 at 21:50
• @mnky9800n: Note that you can "accept" and answer by clicking the green tick if you feel that it helped you (and you don't want to wait for another answer). Apr 22 '12 at 2:46

$\newcommand{\er}{\hat e_r} \newcommand{\et}{\hat e_\tau} \newcommand{\d}{\dot} \newcommand{\m}{\frac{1}{2}m}$

This one gave me a feeling of déjà vu, since I's already answered a similar one. Here's the relevant part of the derivation:

My $\theta$ is your $\phi$\ (usually $\phi$ is used for the azimuthal angle in spherical coordinates--which are a 3D extension of polar coordinates)

In radial coordinates, $\d\er=\d\theta \et$, and (useless here) $\d\et= -\d r \er$. $\er,\et$ are unit vectors in radial and tangential directions respectively. Due to this mixing of unit vectors (they move along with the particle), things get a little more complicated than plain 'ol cartesian system, where the unit vectors are constant. $$\vec p= r\er$$ $$\therefore \vec v=\d{\vec p}= \d r\er + r\d\er=\d r \er + r\d\theta\et$$ $$\therefore v^2= \vec v\cdot\vec v= \d r^2+r^2\d\theta^2$$

$$\therefore KE=\frac12m\vec v\cdot\vec v=\frac12m|\vec v|^2=\frac12m (\d r^2+r^2\d\theta^2)$$

So basically it's just a few steps of math that he neglected (IIRC this is usually considered an identity).