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Re: [Phys-l] Landau on Lagrangian

>Those arguments only attempt to show that L = L(v^2).

No, they do not. Please pay attention.

One of my implicit intents in starting this thread was to talk about the Lagrangian alone. As such, it is the function L that appears in the kernel of the integral assigned to S (the action). The question was a fleshing out of L&L's (and other's) treatment of the form L would take for a free particle, based on considerations of symmetry and the key point about Galilean boosts.

By "form" was meant its dependence on r, t, r-dot, r-doubledot (or not), not the question of whether it was ever to be written as KE (Bob's sticking point I think) or (I hope I'm right here) JD's sticking point thinking I meant that L&L was using these arguments to derive explicit versions of L.

To me, the question has been answered really well by DB. Once asserting (yes, an assertion, but a really interesting one) a stationary action, the math alone suffices to yield the Euler-Lagrange equation, which itself suggests that constants of motion exist, given certain explicit versions of L.

This is technically the end of the question. To suggest that L is written in the form of T - U, and/or that T=KE and U=PE, is a further step. But the mere fact that L=L(v^2) has been "deduced" is highly suggestive of what at least one of them (T or U) must look like for a particle. To go further and find other explicit additions to L, such as jdotA, is a separate question, not addressed. However, I'm willing to bet that symmetry arguments can also be used to deduce analogous generalities for L wrt fields?

The question of how must one write L so that the Euler-Lagrange equation yields N2, or anything else desired, is unresolved. But to me, the seemingly trivial step of "deducing" that L=L(v^2), without invoking N2, is at least incrementally better than asserting, without any justification other than it gives N2, that L = 1/2mv^2 (free particle).

If I've still missed some subtleties, I'm happy to be told so!

Stefan Jeglinski