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Angular momentum in Lagrangian Mechanics

  1. Oct 15, 2012 #1
    In Newton's problem,and other central force problems in Classical Mechanics, you can get with decreasing the center of mass movement to the lagrangian:
    [itex]L=1/2m(r' ^2+r^2 \varphi'^2)-V(r) [/itex]
    because [itex]\varphi [/itex] is cyclic, you can write:
    [itex]\frac{d}{dt}(mr^2 \varphi')=0 [/itex]
    or, defining the angular momentum:
    [itex]mr^2 \varphi'=l [/itex]

    In order for you to understand my question we will go through two ways:

    Way no.1) Now if you go ahead and use euler-lagrange equation to r you will get:
    [itex]\frac{d}{dt}(mr')=mr\varphi'^2-\frac{∂V(r)}{∂r} [/itex]
    and exchanging for l you can get:
    [itex]\frac{d}{dt}(mr')=\frac{l^2}{mr^3}-\frac{∂V(r)}{∂r} [/itex]

    Way no.2) But if you first excange phi and then use euler-lagrange to r you get:
    [itex]L=1/2m(r' ^2+\frac{l^2}{m^2r^2})-V(r)=1/2mr' ^2+\frac{l^2}{2mr^2}-V(r) [/itex]
    and then euler lagrange for r:
    [itex]\frac{d}{dt}(mr')=\frac{∂}{∂r}(\frac{l^2}{2mr^2})-\frac{∂V(r)}{∂r}=-\frac{l^2}{mr^3}-\frac{∂V(r)}{∂r} [/itex]

    The plus in way no.1 became a minus in way no.2. How can you explain it?
    If you can, don't show me a solution using energy because I already know that one. I want to use only lagrangian formalism.
     
  2. jcsd
  3. Oct 20, 2012 #2
    The answer to the question is that the angular momentum is a constant in relation to time, but the partial derivative in r is not zero. In other words:
    [itex]\frac{dl}{dt}=0[/itex],

    [itex]\frac{∂l}{∂r}\neq0[/itex].

    I do think that:
    [itex]\frac{dl}{dr}=0[/itex].
     
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