What are the normal modes of a velocity-dependent equation of motion? I'm trying to find the normal models of a particle with charge $q$ and mass $m$ in a
$3$-dimensional harmonic oscillator potential with an applied uniform magnetic field $B=B_0 \hat{z}$. The potential has a natural frequency $\omega_0$. The Lagrangian is therefore
$$L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) - \frac{1}{2}m\omega_0^2(x^2 + y^2 + z^2) - \frac{q B_0}{2c}(\dot{x}y - \dot{y}x)$$
The resulting equations of motion are
\begin{align}
&\ddot{x} =  - \omega_0^2 x + \frac{q B_0}{mc} \dot{y} \\
&\ddot{y} =  - \omega_0^2 y -\frac{q B_0}{mc} \dot{x} \\
&\ddot{z} = -\omega_0^2 z
\end{align}
My understanding of finding the normal modes is that I should now fit these equations into the general form $m \ddot{\textbf{x}} = -k \textbf{x}$ for some matrix-valued $m$ and $k$; from there it's straightforward to find the normal modes as the eigenvectors of the matrix $-m^{-1}k$.
Unfortunately, the $x$ and $y$ equations of motion are not in a basic Hooke's law form -- they have a velocity term. I'm not sure how to proceed from the equations of motion, though I suspect I'm just missing something simple.
 A: The solution by David Z is an excellent case of pedagogy for my particular situation, but I want to give a more general answer to the question I asked.
The issue here is, the $x$ and $y$ equations of motion are not in the Hooke's law form $\ddot{x} = -\alpha x$ because they have a velocity-dependent term $b \dot{x}$. We have to find a way to rewrite them that gets rid of the velocity dependence. David Z suggested taking advantage of the symmetry of the problem to write it in cylindrical coordinates. Naively it isn't clear how this helps, but going ahead with the suggestion we turn the Lagrangian in $(x,y,z)$ coordinates
$$L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) - \frac{1}{2}m\omega_0^2(x^2 + y^2 + z^2) - \frac{q B_0}{2c}(\dot{x}y - \dot{y}x)$$
into a Lagrangian in $(r,\theta,z)$ coordinates
$$L = \frac{1}{2}m(\dot{r}^2 + \dot{z}^2) - \frac{1}{2}m\omega_0^2(r^2 + z^2) - \frac{q B_0}{2c}r^2 \dot{\theta}$$
Because this Lagrangian lacks $\theta$ (it only has $\dot{\theta}$), the $\theta$ equation of motion will become a conservation law. The three equations of motion turn out to be
\begin{align}
&\ddot{r} = -\omega_0^2r - \omega_c \dot{\theta} r\\
&\frac{d}{dt} \left( \frac{-q B_0}{2c} r^2 \right) = 0\\
&\ddot{z} = -\omega_0^2 z
\end{align}
for $\omega_c = \frac{q B_0}{m c}$, where we can clearly see only two equations of motion, the $r$ and $z$ equations, can conceivably be cast in the form of Hooke's law. Indeed -- crucially -- there's no evolution of $\dot{\theta}$ at all! Therefore, it can be taken to be a parameter of the system; I renamed it $\omega = \dot{\theta}$. At this point the $r$ equation is clearly of the correct form, $\ddot{r} = -(\omega_0^2 + \omega_c \omega)r$. We can now proceed with our normal mode analysis on $r$ and $z$.
In a sense we got lucky in this problem. It wasn't enough to find a symmetry $\theta$ of the system: we had to hope that that coordinate would have no evolution whatsoever. (It's possible it is enough just to have a conservation law, but I'll leave that as an exercise for the reader.) But because $\dot{\theta}$ had no way of changing, we could take it to be a parameter of the system.
A: A change of variables can sometimes be used to eliminate a term of a particular order in an expression. For an algebraic example:
$$x^2 + 2x + 4 = (x + 1)^2 + 3 = y^2 + 3$$
By using the substitution $y = x + 1$, you've made the linear term "disappear". Of course it hasn't really disappeared; it's more like, whether it's there or not is a matter of how you choose to write the expression.
Similarly, you may be able to use a suitable change of variables to make the linear terms in your equations "disappear" - that is, write them in terms of $u = f(x, y, z)$ and so on in such a way that there are no terms containing $\dot{u}$.
Variable transformations like this are a common technique when manipulating low-order algebraic or differential equations. In fact, the transformation you wind up using can tell you something useful about the physics. In my algebraic example above, the fact that $x + 1$ was the quantity that had to be replaced with $y$ meant that $x + 1$ might be special in some way in the physical system that expression came from.
