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Angular momentum from the Lagrangian

  1. Feb 15, 2012 #1


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    In Landau and Lifgarbagez, Vol. 1, it says "the component of angular momentum along any axis (say the z-axis) can be found by differentiation of the Lagrangian:

    M[itex]_{z}[/itex] = [itex]\Sigma_{a}[/itex] [itex]\partial L[/itex]/[itex]\partial \dot{\varphi_{a}}[/itex]

    where [itex]\varphi[/itex] is the angle of rotation about the z axis. This is evident from the proof of the law of conservation of momentum..."

    They are referring to the proof on pg. 19, where a finite rotation is applied to an arbitrary isolated system, the resulting change in the Lagrangian is required to be zero (from isotropy of space), and the time derivative of the angular momentum is thereby shown to be zero.

    My question: How is the former evident from the latter? I've spent quite a while thinking about it and it seems anything but evident right now!
  2. jcsd
  3. Feb 15, 2012 #2
    Applying a finite rotation is the same as adding up many many many small rotations. (Infinitesimally) Small rotation is related to the partial derivation:
    [tex]\frac{\partial L}{\partial \phi_j}[/tex]
    If the Lagrangian is invariant under rotation, then the above partial derivative is zero. Using Euler-Lagrange equation, this means the total time derivative:
    [tex]\frac{d}{dt}\frac{\partial L}{\partial \dot{\phi}_j} = \frac{\partial L}{\partial \phi_j} = 0[/tex]
    So angular momentum in the j-th coordinate is a conserved quantity.
  4. Feb 15, 2012 #3


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    I think I see it now, but I made a mistake earlier - L&L apply an infinitesimal rotation in their derivation of the constancy of angular momentum.
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