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*From*: "Bob Sciamanda" <treborsci@verizon.net>*Date*: Thu, 21 Jan 2010 17:43:39 -0500

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/

--------------------------------------------------

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

_______________________________________________

Forum for Physics Educators

Phys-l@carnot.physics.buffalo.edu

https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l

**References**:**[Phys-l] Landau on Lagrangian***From:*Stefan Jeglinski <jeglin@4pi.com>

**Re: [Phys-l] Landau on Lagrangian***From:*"Bob Sciamanda" <treborsci@verizon.net>

**Re: [Phys-l] Landau on Lagrangian***From:*David Bowman <David_Bowman@georgetowncollege.edu>

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