Re: Langrange strange

From: Dirk Van de moortel (dirkvandemoortel_at_ThankS-NO-SperM.hotmail.com)
Date: 01/30/05 

Date: Sun, 30 Jan 2005 16:04:02 GMT

"Tom Capizzi" <etianshrldu@verizon.net> wrote in message news:kk7Ld.2372$6i4.1180@trndny02...
>
> "Bilge" <dubious@radioactivex.lebesque-al.net> wrote in message
> news:slrncvpba0.al9.dubious@radioactivex.lebesque-al.net...
> > DavidBowman:
> > >Mathworld.com has about 5 definitions for "Langrangian", all of which
> > >are like trying to find out what division is and being presented with 5
> > >algorithms for doing it, when the student really was looking for:
> > >"division is the ratio of two numbers".
> > >
> > >I can tell that the Langragian is basic and general, which makes it
> > >interesting. It seems to be some kind of abstract discrete structure
> > >like a group or a lattice, and it's used kind of to mean "system".
> > >
> > >Is it possible to describe the Langrangian such that even a
> > >simpleminded fellow like myself can understand?
> >
> > In the most simpleminded (i.e., non-relativistic, classical)
> > terms, the lagrangian is the kinetic energy minus the potential
> > energy, L = T - U. For example, a mass on a spring has a kinetic
> > energy, T = (1/2)mv^2 and a potential energy, U = (1/2)kx^2
> > (assuming that equilibrium is at x = 0). Then,
> > . .
> > L = (1/2) [mx^2 - kx^2], where v == x = dx/dt
>
> How do you justify this assumption? It appears to be true only if
> all terms equal zero. In general, as the displacement x increases
> the velocity v decreases. When x is zero, |velocity| is maximum,
> and when velocity is zero, |x| is maximum. Haven't you already
> assumed a solution for the equation of motion by setting x = dx/dt?

There is a dot missing. That should be
       L = (1/2) [m{xdot}^2 - kx^2], where v == {xdot} = dx/dt

Dirk Vdm

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