Re: [Phys-l] Landau on Lagrangian

["Bob Sciamanda" <treborsci@verizon.net>]

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

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From: "David Bowman" <David_Bowman@georgetowncollege.edu>
Sent: Thursday, January 21, 2010 5:34 PM
To: "'Forum for Physics Educators'" <phys-l@carnot.physics.buffalo.edu>
Subject: Re: [Phys-l] Landau on Lagrangian

> Regarding Bob S's objection:
>
> > > . . .  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 | )   ?
>
> 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).
>
> David Bowman
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