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How to find the metric given the interval

  1. Jan 10, 2010 #1
    If we have a Lagrangian which looks like this:




    If we are told that:

    [tex]ds^2=d\phi^2 +(sin^2 \phi) d\theta^2[/tex]

    How can we show that the Lagrangian is:

    [tex]L=\frac{m}{2}[\dot{\phi}^2 +(sin^2 \phi) \dot{\theta}^2][/tex]

    Is there a general way of determing the metric from the interval?

  2. jcsd
  3. Jan 10, 2010 #2


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    It is easy. The term in the interval proportional to [itex]d\phi^2[/itex] comes from the term [itex]g_{11}[/itex]. Therefore you can read off g_{11} = 1. Similarly you can read off [itex]g_{22} = sin^2(\phi)[/itex]. There are no terms in the interval proportional to [itex]d\phi d\theta[/itex], so you can conclude the off-diagonal terms are 0 in the metric.

    If none of this is obvious, write out the summation [itex]g_{ij}dx^{i}dx^{j}[/itex] and convince yourself of it.
  4. Jan 10, 2010 #3
    thanks nicksauce:)
  5. Jan 10, 2010 #4
    In what subject area is this topic addressed? It sounds very interesting.
  6. Jan 10, 2010 #5
    Classical Mechanics...
  7. Jan 11, 2010 #6
    Bummer, in my junior mechanics class, we didn't do this.

    (i'm not talking about lagrangian stuff but, rather, the metric stuff.)
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