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Some lagrangian question

  1. May 10, 2012 #1
    If we considered some coordinate as being a generalized one, like when we are considering spherical coordinates-let us suppose that I chose theta and phi as generalized coordinates. After deriving the Lagrangian equation it turned out that the equation doesnt depend on phi. Which means that derivative of the Lagrangian by phi is zero. Does this mean it is not a generalized coordinate? If not, what does it mean? And lastly, whats the difference if the case was that the equation doesnt depend on phi dot(the time derivative of phi)
    thanks in advance
     
  2. jcsd
  3. May 17, 2012 #2

    jfizzix

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    Let's say we have some Lagrangian

    [itex]L(\theta, \phi, \dot{\theta}, \dot{\phi})[/itex]

    Let's look at the equation of motion for [itex]\phi[/itex]. Recall that the generalized conjugate momentum to the coordinate [itex]\phi[/itex] is

    [itex]p_{\phi}=(\frac{\partial L}{\partial \dot{\phi}})[/itex]

    and so our equation of motion becomes

    [itex]\frac{d}{d t}(\frac{\partial L}{\partial \dot{\phi}}) -(\frac{\partial L}{\partial \phi}) = \frac{d}{d t} p_{\phi} - (\frac{\partial L}{\partial \phi})= 0[/itex]

    If L does not depend on [itex]\phi[/itex], then [itex] (\frac{\partial L}{\partial \phi})= 0[/itex], and so [itex]p_{\phi}[/itex] is constant in time; it is a conserved quantity.


    As for what happens when L does not depend on [itex]\dot{\phi}[/itex] we can look to the same equation of motion. This would be saying the same thing as [itex]p_{\phi} = 0[/itex]. As far as dynamics go, I think what it means is that there would be no relevant dynamics in the [itex]\phi[/itex] direction, and you would instead look at how things are changing in the [itex]\theta[/itex] direction.


    hope this helps,

    -James
     
  4. Jun 6, 2012 #3
    Phi is still a generalized coordinate, and when you're considering your final equations of motion you will need to consider the time-dependence of the phi coordinate in your answer.

    The fact that the derivative of the Lagrangian is zero means, as was mentioned above, that you have a conserved quantity. For example, consider the Lagrangian for a small planet orbiting a very large star: assuming that Newtonian gravity is valid, you will get a potential energy that will have no angular dependence at all. If you were free to choose whatever reference frame you wanted, then you would be foolish not to choose one in which one of the generalized coordinates is zero. However, you can imagine a potential that is slightly more complicated in which there is no "phi" dependence, but some "theta" dependence (like the gravitational potential energy of a galactic disk). In this case, the fact that dL/d(phi) = 0 means that angular momentum is conserved about the axis of symmetry. This does not reduce the effective dimensionality of the solution set though.
     
  5. Jun 30, 2012 #4
    I apologize for the REAL delay in replying, but I thank you both.
     
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