- #1
gromit
- 3
- 1
- 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 :)
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 :)