# Using Lagrangian to show a particle has a circular orbit

• gromit
In summary, the conversation discusses a problem from David Tong's Classical Dynamics course, specifically problem 6ii in the first problem sheet. The lagrangian for the problem has symmetry in time and theta, leading to the conservation of the hamiltonian and generalised momentum for theta. The next part involves considering a circle as a solution to the equations of motion derived from the lagrangian. The final part involves showing that the trajectory remains a circular orbit for a shift in the z-axis, and the hint provided suggests using a Taylor expansion to show that the resulting equation of motion for the perturbed case is one of simple harmonic motion with an equilibrium position displaced from the origin.
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 :)

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
ergospherical said:
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!

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...?

Last edited by a moderator:
vanhees71
ergospherical said:
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!

Last edited by a moderator:
vanhees71

## 1. How does the Lagrangian method show that a particle has a circular orbit?

The Lagrangian method is a mathematical approach used in classical mechanics to analyze the motion of a particle. In this method, the Lagrangian function, which is a combination of the kinetic and potential energies of the particle, is minimized to find the path of the particle that satisfies the equations of motion. If the Lagrangian function is minimized to a constant value, then it can be concluded that the particle is in a circular orbit.

## 2. What is the difference between using Lagrangian and Newtonian mechanics to analyze circular orbits?

The main difference between the Lagrangian and Newtonian methods is that the Lagrangian method takes into account the total energy of the particle, while the Newtonian method only considers the gravitational force acting on the particle. This makes the Lagrangian method more accurate and versatile in analyzing complex systems, such as multiple particles in orbit.

## 3. Can the Lagrangian method be used to show circular orbits for any type of force?

Yes, the Lagrangian method can be used to show circular orbits for any type of force, as long as the force is conservative. This means that the force depends only on the position of the particle and not on its velocity. Examples of conservative forces include gravity, electric and magnetic forces.

## 4. Are there any limitations to using Lagrangian to show circular orbits?

One limitation of using the Lagrangian method is that it assumes the system is in equilibrium, meaning that the forces acting on the particle are balanced. This may not always be the case in real-world situations, such as when the particle is experiencing external forces. Additionally, the Lagrangian method may become more complex and difficult to solve for systems with more than two particles.

## 5. How does using Lagrangian to show circular orbits contribute to our understanding of celestial bodies?

Using the Lagrangian method to show circular orbits allows us to accurately predict and analyze the motion of celestial bodies, such as planets and satellites, in our solar system. It also helps us understand the underlying principles of orbital mechanics, which are crucial in space exploration and satellite technology. By studying circular orbits using the Lagrangian method, we can gain a deeper understanding of the laws of motion and the forces that govern the motion of objects in space.

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