Lagrange - Mass under potential in spherical

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Homework Help Overview

The problem involves a particle of mass m moving in a force field with a specified potential in spherical coordinates. The task is to identify the two constants of motion for the system based on the provided Lagrangian.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the constants of motion from the Lagrangian but expresses confusion regarding the absence of the variable φ. Some participants suggest that the lack of explicit time dependence in the Lagrangian implies a conserved quantity, while others mention the energy function or Hamiltonian as a potential constant of motion.

Discussion Status

The discussion is exploring different interpretations of the constants of motion related to the Lagrangian. Some guidance has been offered regarding the implications of the Lagrangian's structure, but there is no explicit consensus on the identification of the two constants of motion.

Contextual Notes

Participants are considering the implications of the Lagrangian's dependence on the coordinates and whether the absence of φ leads to a specific conclusion about constants of motion.

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Homework Statement



A particle of mass m moves in a force field whose potential in spherical coordinates is,

U = \frac{-K \cos \theta}{r^3}

where K is constant.

Identify the two constants of motion of the system.

The Attempt at a Solution



L = T - V = \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \sin^2 \theta ~\dot{\phi}^2) + \frac{K \cos \theta}{r^3}

I don't see how there are two constants of motion if the Lagrangian is missing only \phi, i.e.,

\frac{ \partial L}{\partial \phi} = 0 \Rightarrow \frac{\partial L}{\partial \dot{\phi}} = constant
 
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I'm not 100% sure that this is what the questioner has in mind, but I can think of one quantity that is always a constant of motion whenever the Lagrangian has no explicit time dependence...:wink:
 
Energy function/Hamiltonian?

\frac{\partial L}{\partial t} = 0 = - \frac{dH}{dt}

So H = constant.
 
Yup.:smile:
 

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