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When does a Lagrangian have (1/2)?

  1. Apr 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Not really a homework question: just a general query. About half the time when working examples, I see a (1/2) thrown in to a Lagrangian for use with Euler-Lagrange, but I can't seem to find out why. Is the (1/2) present (or not?) only for the case of a non-symmetric metric?

    Like if I had:

    [tex]ds^2 = dx^2 + 2dxdy + dy^2[/tex],

    would the Lagrangian then be:

    [tex]L = \frac{1}{2}(\dot{x}+...)[/tex]?

    Whereas, the Lagrangian for the Schwartzschild metric doesn't have the (1/2):

    [tex]L = -(1 - \frac{2m}{r})\dot{t} + ... [/tex]

    because the metric that describes Schwartzschild space time is symmetric?
     
  2. jcsd
  3. Apr 9, 2012 #2
    Usually when I've seen a 1/2 in a Lagrangian, it's because it's quadratic in the variable of interest, so when you take the derivative of it as part of the E-L equation, you'll pull in a factor of 2. Adding the 1/2 therefore gives you equations of motion with no numeric coefficient.
     
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