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# Re: [Phys-l] Landau on Lagrangian

David wrote:
If we have any other nonlinear function of v^2 (not proportional to the above linear one or with an affine offset) then substituting v' + v_0 = v into the nonlinear function of v^2 will not result in the >same function of v'^2 plus a total time derivative of some function of the dynamical variables, and thus the resulting EOM will not be frame invariant under GTs.

This does not seem to exclude a linear function of v (ex., L=mv). Or do I misread you're argument?

P.S.
Does EOM mean "energy of motion"?

Bob Sciamanda
Physics, Edinboro Univ of PA (Em)
treborsci@verizon.net
http://mysite.verizon.net/res12merh/

--------------------------------------------------
From: "David Bowman" <David_Bowman@georgetowncollege.edu>
Sent: Friday, January 22, 2010 5:01 PM
To: "'Forum for Physics Educators'" <phys-l@carnot.physics.buffalo.edu>
Subject: Re: [Phys-l] Landau on Lagrangian

Regarding Bob S's question:

Another (different) objection:
Quoting Landau, Stefan wrote:

. . . L must also be independent of the direction of v, and is
therefore a function only of its magnitude, i.e. v^2."
Ergo: L = L(v^2) .

Why not :
Ergo: L=L( | v | ) ?

Bob Sciamanda

Indeed.

No generality is lost here. Note that |v| is a function of v^2.
In particular |v| = sqrt(v^2). So any real function of |v| is
also a real function of v^2 as well (because a function of a
function is a function).

to which Stefan responded:

OK, but the spirit of the question survives, unless I
misunderstand you. In the "first principles" approach that's
been discussed in this thread, the argument does not appear to
preclude one from supposing the free classical particle Lagrangian
to be (1/2)m|v|. This would seem justified (Occam's Razor?), other
than the rather glaring inability to get N2, which takes us back
ultimately to the only argument left: "in our universe, this
apparently just doesn't seem to be the case."

Stefan Jeglinski

and Bob S also responded with:

To be sure!
I merely wanted to point out that Landau's treatment (even if
accepted as valid) does not demonstrate that the translational
kinetic energy must be proportional to the SQUARE of the speed.
The opposite is easily (but incorrectly) inferred from the
presentation.

Bob Sciamanda

I can't seem to locate my copy of L & L book at the moment. But I thought I remembered that their argument also included (besides Hamilton's Principle, spatial homogeneity, temporal homogeneity, spatial isotropy) the additional requirement that the EOM be invariant under Galilean Transformations so that Galilean relativity holds. The Lagrangian EOM will be invariant if the difference between the Lagrangian functions L(v^2) and L(v'^2) is a total derivative w.r.t. time of some function of the dynamical variables. This causes the difference in the corresponding actions to depend only on the fixed endpoints of the dynamical path and that guarantees that both Lagrangians are made stationary by the same physical path regardless of which frame of reference it is denominated in. Now it just so happens that only the function of v being v^2 (and constant multiples of it) has the property that when v = v' + v_0 (with constant velocity v_0 of frame K' w.r.t. frame K) then:

v^2 = v'^2 + d/dt(t*v_0^2 + 2*v0.r')

where r' is the particle's dynamical position vector in the K' frame and . is the dot product.

If we have any other nonlinear function of v^2 (not proportional to the above linear one or with an affine offset) then substituting v' + v_0 = v into the nonlinear function of v^2 will not result in the same function of v'^2 plus a total time derivative of some function of the dynamical variables, and thus the resulting EOM will not be frame invariant under GTs.

(But maybe I'm misremembering this latter requirement from some other author(s) than L & L.)

Dave Bowman
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