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

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Homework Help Overview

The discussion revolves around deriving the Hamiltonian and Hamilton equations of motion for a particle in a central potential using spherical coordinates. The original poster has formulated the Hamiltonian and equations of motion and is exploring how these equations simplify under specific initial conditions.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the derivation of the Hamiltonian and equations of motion, with some questioning the methods used to arrive at certain equations. There are suggestions to utilize the relationship between momentum and the Hamiltonian to clarify the equations.

Discussion Status

The discussion is active, with participants providing guidance on how to approach the simplification of equations under the given initial conditions. There is an exploration of integrating equations and the implications of setting certain momentum terms to zero.

Contextual Notes

Participants are navigating the complexities of Hamiltonian mechanics and the specific conditions imposed on the problem, such as the initial conditions for momentum and angular position.

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Homework Statement



Using spherical coordinates [tex](r, \theta, \phi)[/tex], 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 [tex]p_{\phi}(0) = 0[/tex] and [tex]\phi (0) = 0[/tex]


The Attempt at a Solution



I have obtained a Hamiltonian

[tex]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)[/tex]

and from this also the equations of motion

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

[tex]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}[/tex]
[tex]m^2 \left( 2r \dot{r} \dot{\theta} + r^2 \ddot{\theta} \right) = \frac{p_{\phi}^2 \cos{(\theta)}}{r^3 \sin^3{(\theta)}}[/tex]
[tex]2 \dot{r} \sin{(\theta)} \dot{\phi} + r \left( 2 \cos{(\theta)}\dot{\theta} \dot{\phi} + \sin{(\theta)} \ddot{\phi} \right) = 0[/tex]

But how should I proceed with the last part of the problem?
 
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I'm not too sure how you've got the last 3 eqns. It would be easier to just use the fact that [tex]\dot{p_i}=-\frac{\partial H}{\partial q_i}[/tex] 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 [tex]\dot{p_\phi} = -\frac{\partial H}{\partial \phi} = 0[/tex] 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 [tex]\frac{d p_\phi}{dt}=0[/tex]. Can you integrate this?
 
That would take me back to

[tex]p_\phi = mr^2 \sin^2{\theta} \dot{\phi} = 0[/tex]?
 
Well... we get the equation [tex]p_\phi= C[/tex] where C is a constant of integration. Then, the intial conditions will imply that [tex]p_\phi[/tex] is zero for all t. Do a similar thing for [tex]\phi[/tex]. 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 [tex]p_\phi = 0[/tex] 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 [tex]\phi[/tex]-part?
 
  • #12
I'm not sure what else I can say without doing it for you! What equation do you get when you sub [tex]p_{phi}=0[/tex] 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 [tex]\phi = const. = 0[/tex]? 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. ;)
 

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