Is it always the case that the square root of a lagrangian gives the same equations of motion as the lagrangian itself?

Question

Inspired by the Phys.SE post Geodesic Equation from variation: Is the squared lagrangian equivalent? I was wondering if it is always the case that the square root of a lagrangian gives the same equations of motion as the lagrangian itself? Are there specific counterexamples, or is there any way to derive a set of conditions the lagrangian has to satisfy, for it to have this particular property?

Are you talking about the Lagrangian formalism for particles or for fields? — J. Murray, Jun 24 '20 at 06:34

Qmechanic · Accepted Answer · 2020-06-24T08:32:56.547

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That is rarely$^1$ the case. Sufficient conditions and examples are given in this duplicate Math.SE post.

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$^1$ An instructive example is perhaps this Phys.SE post: The coefficients of each squared term in the Lagrangian have been carefully fine-tuned to make it work. It is not just $(T-V)^2$.

edited Jun 24 '20 at 08:32

answered Jun 24 '20 at 08:13

Qmechanic

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Is it always the case that the square root of a lagrangian gives the same equations of motion as the lagrangian itself?

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