How to Find ∂L/∂q: A Simple Guide for Calculating Lagrangian Equations

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Discussion Overview

The discussion revolves around the calculation of the partial derivative of the Lagrangian with respect to a generalized coordinate, specifically exploring the relationship between \(\frac{∂L}{∂q}\) and momentum (\(p\)). The context includes theoretical derivations related to Lagrangian and Hamiltonian mechanics.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant requests clarification on how \(\frac{∂L}{∂q} = \dot{p}\), indicating that \(L\) is the Lagrangian, \(p\) is momentum, and \(q\) is a generalized coordinate.
  • Another participant expresses that their inquiry is not related to homework but is part of deriving the Hamiltonian equation.
  • A participant asserts that \(\frac{∂L}{∂q} = \dot{p}\) because the Lagrangian \(L = T - U\) depends on \(q\) only through the potential energy \(U\), linking this to Newton's second law expressed in terms of potential.
  • One participant asks about identifying translational symmetry and what quantity must be zero to determine it.
  • Another participant responds that for translational symmetry, the condition \(p = \text{constant}\) (or \(\dot{p} = 0\)) must hold for the momentum conjugate to the coordinate.
  • A participant seeks clarification on the term "momentum that is conjugate to that coordinate," confirming it refers to the momentum corresponding to the specific coordinate in question.
  • One participant defines the canonical momentum of the generalized coordinate \(q\) as \(p = \frac{\partial L}{\partial \dot{q}}\) and notes the distinction between canonical momentum and mechanical momentum, providing an example involving a particle in a magnetic field.

Areas of Agreement / Disagreement

Participants express varying degrees of understanding and clarification on the relationship between Lagrangian mechanics and momentum, but no consensus is reached on the implications of these relationships or the conditions for translational symmetry.

Contextual Notes

Some participants highlight the distinction between canonical and mechanical momentum, suggesting that this may depend on the context of the problem, such as the presence of external fields.

iScience
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hi, silly question but would someone please show me how [itex]\frac{∂L}{∂q}[/itex]=[itex]\dot{p}[/itex]?

L being the lagrangian, p being the momentum, and q being the general coordinate.
 
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(i hope this doesn't qualify as a homework question this actually has nothing to do with my homework.. I'm just trying to derive the hamiltonian equation and this was just part of the steps)
 
iScience said:
please show me how [itex]\frac{∂L}{∂q}[/itex]=[itex]\dot{p}[/itex]?

[itex]\frac{∂L}{∂q}[/itex]=[itex]\dot{p}[/itex] because the Lagrangian, L=T-U, depends on q only in the potential energy, U.

[itex]-\frac{∂U}{∂q}[/itex]=[itex]\dot{p}[/itex] is Newton's 2nd law of motion expressed in terms of the potential.
 
thanks!

also, i had another question i hope i can just ask it in the same thread; if not let me know (moderators/admins) and i'll just make a new thread.



how do i know when there is translational symmetry? in other words what quantity has to be zero?
 
You want p=constant (p-dot = 0) for the momentum conjugate to that coordinate.
 
"momentum that is conjugate to that coordinate" meaning just the momentum corresponding to the particular coordinate at hand right? (just checking)
 
The canonical momentum of the generalized coordinate [itex]q[/itex] is by definition given by
[tex]p=\frac{\partial L}{\partial \dot{q}}.[/tex]
It's important to keep in mind that also for a Cartesian coordinate it is the canonical and not necessarily the mechanical momentum. An interesting example for that both need not be the same is the motion of a particle in a magnetic field.
 

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