Hamiltonian Mechanics: Central Potential V(r) w/ Spherical Coordinates

Logarythmic
Messages
277
Reaction score
0

Homework Statement



Using spherical coordinates (r, \theta, \phi), obtain the Hamiltonian and the Hamilton equations of motion for a particle in a central potential V(r).
Study how the Hamilton equations of motion simplify when one imposes the initial conditions p_{\phi}(0) = 0 and \phi (0) = 0


The Attempt at a Solution



I have obtained a Hamiltonian

H = \frac{1}{2m} \left( p_r^2 + \frac{p_{\theta}^2}{r^2} + \frac{p_{\phi}^2}{r^2 \sin^2{(\theta)}} \right) + V(r)

and from this also the equations of motion

\dot{r} = \frac{p_r}{m}
\dot{\theta} = \frac{p_{\theta}}{mr^2}
\dot{\phi} = \frac{p_{\phi}^2}{r^2 \sin^2{(\theta)}}

m \ddot{r} = \frac{1}{m} \left( \frac{p_{\theta}^2}{r^3} + \frac{p_{\phi}^2}{r^3 \sin^2{(\theta)}} \right) - \frac{\partial V}{\partial r}
m^2 \left( 2r \dot{r} \dot{\theta} + r^2 \ddot{\theta} \right) = \frac{p_{\phi}^2 \cos{(\theta)}}{r^3 \sin^3{(\theta)}}
2 \dot{r} \sin{(\theta)} \dot{\phi} + r \left( 2 \cos{(\theta)}\dot{\theta} \dot{\phi} + \sin{(\theta)} \ddot{\phi} \right) = 0

But how should I proceed with the last part of the problem?
 
Physics news on Phys.org
I'm not too sure how you've got the last 3 eqns. It would be easier to just use the fact that \dot{p_i}=-\frac{\partial H}{\partial q_i} Try doing it this way, then you will have the momentum terms on the LHS. It might be clear from the phi eqn obtained in this way how to use the initial conditions.
 
That's EXACTLY what my equations are. ;)
 
Ok, well consider the last equation. It can be written \dot{p_\phi} = -\frac{\partial H}{\partial \phi} = 0 Integrating this, and using the initial conditions will allow you to simplify the other equations
 
Integrate with respect to what? I don't see this at all..
 
This equation can also be written as \frac{d p_\phi}{dt}=0. Can you integrate this?
 
That would take me back to

p_\phi = mr^2 \sin^2{\theta} \dot{\phi} = 0?
 
Well... we get the equation p_\phi= C where C is a constant of integration. Then, the intial conditions will imply that p_\phi is zero for all t. Do a similar thing for \phi. Remember that you are trying to show how the equations *simplify* on imposing the initial conditions
 
I don't get it. How does this simplify the equations?
 
  • #10
Try putting p_\phi = 0 into the equations you've posted in your question. You'll see it simplifies the equations! Then do a similar thing for the third equation, using the inital condition for phi
 
  • #11
But what about the \phi-part?
 
  • #12
I'm not sure what else I can say without doing it for you! What equation do you get when you sub p_{phi}=0 into the third equation? How can you solve this, and what is the answer given the initial conditions? [hint.. posts #6 and #8]
 
  • #13
Then \phi = const. = 0? Aha! ;)
 
  • #14
Correct. Now the equations will look a lot simpler!
 
  • #15
Yeah thanks for your help. I'm blaming on the fact that it's sunday today. ;)
 
Back
Top