Hamiltonian in cylindrical coordinates

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SUMMARY

The discussion focuses on deriving the Hamiltonian for a mechanical system in cylindrical coordinates, starting from the Lagrangian \( L = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2 + \dot{z}^2) - U(r, \theta, z) \). The user initially calculates the Hamiltonian as \( H = \frac{1}{2m}[p_r^2 + (2-r^2)p_\theta^2 + p_z^2] + U(r, \theta, z) \), but realizes the standard form is \( H = \frac{1}{2m}(p_r^2 + \frac{p_\theta^2}{r^2} + p_z^2) + U(r, \theta, z) \). The discrepancy arises from the incorrect treatment of the angular momentum term \( p_\theta \).

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  • Understanding of Lagrangian mechanics
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  • Knowledge of cylindrical coordinate systems
  • Proficiency in calculus and differential equations
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  • Study the derivation of Hamiltonian mechanics from Lagrangian mechanics
  • Learn about angular momentum in cylindrical coordinates
  • Explore potential energy functions \( U(r, \theta, z) \) in mechanical systems
  • Investigate common mistakes in transitioning from Lagrangian to Hamiltonian formulations
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Physicists, mechanical engineers, and students studying classical mechanics, particularly those interested in the application of Hamiltonian mechanics in cylindrical coordinates.

Piano man
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Hi, I'm trying to find the Hamiltonian for a system using cylindrical coordinates.

I start of with the Lagrangian [tex]L=\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2+\dot{z}^2)-U(r,\theta,z)[/tex]

From that, using [tex]H=\sum p\dot{q}-L[/tex]
[tex]=p_r\dot{r}+p_\theta\dot{\theta}+p_z\dot{z}-\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2+\dot{z}^2)+U(r,\theta,z)[/tex]
[tex]=\frac{1}{m}(p_r^2+p_\theta^2+p_z^2)-\frac{1}{2m}(p_r^2+p_\theta^2r^2+p_z^2)+U(r,\theta,z)[/tex]
[tex]=\frac{1}{2m}[p_r^2+(2-r^2)p_\theta^2+p_z^2]+U(r,\theta,z)[/tex]

But the standard answer is
[tex]H=\frac{1}{2m}(p_r^2+\frac{p_\theta^2}{r^2}+p_z^2)+U(r,\theta,z)[/tex]

So where did I go wrong?

Thanks for any help :)
 
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What did you use for pθ?

Don't forget that

[tex]p_j= \frac{\partial L}{\partial \dot{q}_j}.[/tex]
 
Ah, of course!
Kind of obvious really...
:redface:

Thanks for the help :)
 

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