Proving angular momentum is conserved

[ppyadof]

Homework Statement

A ball moves in a two-dimensional billiard comprising a horizontal frictionless table enclosed by two circular barriers of radii 2m and 1m and centred, respectively, at (x, y)=(0, 0) and (x, y)=(a, 0), where a<0.5m. The ball undergoes specular reflection at each barrier. Prove that the ball's angular momentum is conserved when a=0.

Homework Equations

Poisson brackets

The Attempt at a Solution

I know that I have to bring poisson brackets into this, since it basically comes down to showing constants of motion, although I don't know what I need to use poisson brackets on. I think that I need to consider $L^2$ and the hamiltonian of the system, but I'm not entirely sure how to build the boundary conditions (e.g. specular reflection off the boundary walls) into the hamiltonian.

 

[Jhero]

Any help would be appreciated.



Thank you for your question. It is a very interesting problem and I am happy to assist you in finding a solution. You are correct in thinking that Poisson brackets will be useful in proving the conservation of angular momentum. To start, let's define the Hamiltonian of the system as H = T + V, where T is the kinetic energy and V is the potential energy. In this case, we can consider the ball as a point mass, so its kinetic energy can be written as T = (1/2)m(vx^2 + vy^2), where m is the mass of the ball and vx and vy are the velocities in the x and y directions, respectively. The potential energy is simply the potential due to the circular barriers, which can be written as V = (1/2)k(r1^2 + r2^2), where k is the spring constant and r1 and r2 are the distances from the ball to the two barriers, respectively.

Now, let's consider the Poisson brackets of the Hamiltonian with the angular momentum L = mr^2ω, where r is the distance from the ball to the origin and ω is the angular velocity. We have:

{H, L} = {T, L} + {V, L} = {T, mr^2ω} + {V, mr^2ω}

Since the potential energy does not depend on ω, the second term on the right hand side is zero. For the first term, we can use the chain rule to expand the Poisson bracket:

{T, mr^2ω} = m{vx^2 + vy^2, mr^2ω} = mω{vx^2 + vy^2, r^2} = 2mω(rvxvy - ryvx)

Now, we need to use the boundary conditions to simplify this expression. Since the ball undergoes specular reflection at each barrier, the angle of incidence is equal to the angle of reflection. This means that the x and y components of the velocity will be reversed when they hit the barriers. Therefore, we can write:

vx = -vx' and vy = -vy'

where vx' and vy' are the velocities after reflection. Substituting this into our expression, we get:

{T, mr^2ω} = -2mω(rvx'vy' + ryvx')