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Simple lagrangian question

  1. Oct 11, 2013 #1
    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.
  2. jcsd
  3. Oct 11, 2013 #2
    (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)
  4. Oct 11, 2013 #3


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    [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.
  5. Oct 11, 2013 #4

    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?
  6. Oct 11, 2013 #5


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    You want p=constant (p-dot = 0) for the momentum conjugate to that coordinate.
  7. Oct 11, 2013 #6
    "momentum that is conjugate to that coordinate" meaning just the momentum corresponding to the particular coordinate at hand right? (just checking)
  8. Oct 12, 2013 #7


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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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