Simple lagrangian question

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  • #1
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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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  • #2
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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)
 
  • #3
UltrafastPED
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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.
 
  • #4
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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?
 
  • #5
UltrafastPED
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You want p=constant (p-dot = 0) for the momentum conjugate to that coordinate.
 
  • #6
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"momentum that is conjugate to that coordinate" meaning just the momentum corresponding to the particular coordinate at hand right? (just checking)
 
  • #7
vanhees71
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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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