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Re: [Phys-l] Landau on Lagrangian

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

Not for me. I
took it to mean *equation(s)* of motion.
"""""""""
Ah, yes. I'm likely the one to have written EOM when he meant Equations of Motion. Sorry about that.



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________________________________
From: David Bowman <David_Bowman@georgetowncollege.edu>
To: Forum for Physics Educators <phys-l@carnot.physics.buffalo.edu>
Sent: Sat, January 23, 2010 8:39:22 AM
Subject: Re: [Phys-l] Landau on Lagrangian

Regarding Bob S's questions:

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 your argument?

You are correct that invariance under Galilean boosts *by itself* does not exclude a linear function of v. But that was *already* excluded by the isotropy of space with the argument that L had to be a function of *square speed* v^2. Taking the sqrt of this, namely |v| is *not* a linear function of the velocity vector. It is a nonlinear function of v^2. Such a nonlinear function does not preserve the EOM for Galilean boosts. Note that there is no function f of the dynamical state (r',v',t) such that:

|v| = sqrt(v^2) = sqrt(v'^2) + d/dt(f(r',v',t)) = |v'| + df/dt

when v = v' + v_0 for constant boost velocity v_0.

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

Not for me. I took it to mean *equation(s)* of motion.

Bob Sciamanda

David Bowman
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