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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.

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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.

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  • $\begingroup$ Okay this makes a lot of sense. Thanks. This notation is much easier to read. $\endgroup$ – mnky9800n Apr 21 '12 at 21:50
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    $\begingroup$ @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). $\endgroup$ – Manishearth Apr 22 '12 at 2:46
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$\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).

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