Using Lagrangian to show a particle has a circular orbit

[gromit]

Homework Statement

The motion of an electron of mass m and charge (-e) moving in a magnetic
field: $B = \nabla \times A(r)$ is described by the Lagrangian
$L = \frac{1}{2}mv^2 - ev \cdot A(r)$

Working in cylindrical polar coordinates, consider the vector potential:
$A = (0, rg(z), 0)$
where $g(z) > 0$:
1.) Obtain two constants of motion
2.) Show that if the electron is projected from a point $(r_0, θ_0, z_0)$ with velocity $\dot{r} = \dot{z} = 0$ and $\dot{θ} = \frac{2eg(z_0)}{m}$, then it will describe a circular orbit provided that $g'(z_0) = 0$
3.) Show that these orbits are stable to shifts along the z axis if $tg′′ > 0$

Relevant Equations

$L = \frac{1}{2}mv^2 - ev \cdot A(r)$
$A = (0, rg(z), 0)$

Hi :) This is a problem from David Tong's Classical Dynamics course, found here: http://www.damtp.cam.ac.uk/user/tong/dynamics.html. In particular it is problem 6ii in the first problem sheet.

Firstly we can see the lagrangian is $L = \frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2+\dot{z}^2) - er^2g(z)\dot{\theta}$. From this it has symmetry in time and $\theta$ so the hamiltonian and generalised momentum for $\theta$ must be conserved. I found these to be:
$$H = \frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}+\dot{z}^2)$$
$$p_{\theta} = mr^2\dot{\theta} - er^2g(z)$$
So these will be the two constants of motion. The next part is where I get a little shaky. I assume that you could consider the circle:
$$r(t) = r_0$$
$$\theta(t) = \frac{2eg(z_0)}{m}t + \theta_0$$
$$z(t) = z_0$$
And show that this satisfies the equations of motion derived from the Lagrangian (as well as the given boundary conditions). Namely:
$$m\ddot{r}+2erg(z)\dot{\theta} - mr^2\dot{\theta} = 0$$
$$(mr^2\dot{\theta} - er^2g(z))' = 0$$
$$m\ddot{z} + er^2g'(z)\dot{\theta} = 0$$
Then this part is solved (please correct me if I'm wrong). Substitution does show that the described solutions works for the first two, and it must be assumed that $g'(z) = 0$ for the last equation of motion to be satisfied. On a side note, if this is a valid method, could I just use the constants of motion instead and that would still be sufficient?

The final part is where I am completely stuck. It seems to me that the trajectory remaining a circular orbit for a shift in the z-axis relies on $g'(z) = 0$, but the inequality provided does not guarantee this for an arbitrary shift?

Thank you so much for your help :)

 

[ergospherical]

I wonder what is meant here by "stable under shifts along the z axis"? I'm sure it cannot be, say, a change in the initial conditions $z_0 \mapsto z_0 + \alpha$, because $z_0$ is already arbitrary making the statement obvious.

So is he talking about making a small pertubation $z = z_0 + \xi$, and then showing that $\xi$ oscillates simple harmonically provided the condition is satisfied?

 

[gromit]

I wonder what is meant here by "stable under shifts along the z axis"? I'm sure it cannot be, say, a change in the initial conditions $z_0 \mapsto z_0 + \alpha$, because $z_0$ is already arbitrary making the statement obvious.

So is he talking about making a small pertubation $z = z_0 + \xi$, and then showing that $\xi$ oscillates simple harmonically provided the condition is satisfied?

That makes much more sense, thank you!

 

[ergospherical]

As a hint, you could substitute $z = z_0 + \xi$ into the equation of motion $m\ddot{z} + er^2g'(z)\dot{\theta} = 0$ for $z$ and then Taylor expand $g(z_0 + \xi)$ to first order in $\xi$. If $g''(z_0) > 0$ then the resulting equation of motion for $\xi$ is one of SHM with an equilibrium position displaced from the origin, and a substitution allows you to calculate $\xi$. This is assuming that we can take $\dot{\theta}$ to be approximately constant in the perturbed case.

Again I'm not sure if that's correct, but it seems reasonable...?

 

[gromit]

As a hint, you could substitute $z = z_0 + \xi$ into the equation of motion $m\ddot{z} + er^2g'(z)\dot{\theta} = 0$ for $z$ and then Taylor expand $g(z_0 + \xi)$ to first order in $\xi$. If $g''(z_0) > 0$ then the resulting equation of motion is one of SHM with an equilibrium position displaced from the origin, and a substitution allows you to calculate $\xi$. This is assuming that we can take $\dot{\theta}$ to be approximately constant in the perturbed case.

Again I'm not sure if that's correct, but it seems reasonable...?

I just went through this roughly by hand, and it seems to answer it perfectly. Thanks again!