When Is (1/2) Used in a Lagrangian?

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SUMMARY

The presence of the factor (1/2) in a Lagrangian is primarily associated with the quadratic nature of the variable of interest, which simplifies the resulting equations of motion by eliminating numeric coefficients upon differentiation. In cases where the metric is non-symmetric, such as with the metric ds² = dx² + 2dxdy + dy², the Lagrangian can include this factor. Conversely, for symmetric metrics like the Schwarzschild metric, the Lagrangian does not require the (1/2) factor, as seen in L = -(1 - 2m/r)dot{t} + ... . This distinction is crucial for correctly applying the Euler-Lagrange equations.

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Homework Statement



Not really a homework question: just a general query. About half the time when working examples, I see a (1/2) thrown into 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?
 
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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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