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Message from discussion Hamiltonian Dynamics = Adiabatic Processes Only?
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squark  
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 More options Jun 24 2000, 3:00 am
Newsgroups: sci.physics.research
From: squ...@my-deja.com
Date: 2000/06/24
Subject: Re: Hamiltonian Dynamics = Adiabatic Processes Only?
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In article <395242FD.E9448...@xs4all.nl>,
  Gerard Westendorp <west...@xs4all.nl> wrote:

> Squark wrote:

> > Hello all readers.

> > I have recently thought what is the Lagrangian for a harmonic
> > oscillator  with damping. For instance, consider the equation of
> > motion

> > (A) x'' = (-k/m)x - ax'

> > Actually, the problem arises for a simple Newtonian body moving with
> > friction. If I'm right, many of the readers probabely know this,
> > but the conclusion I arrived at is that there is no Lagrangian.

> How about:

>   L = 1/2 (m(x')^2 - k x^2) + max't

Cool!

> > And that's why:
> > assume otherwise, i.e., that the above models may be formulated
> > using a Lagrangian, and therefore, eventually put into a
> > Hamiltonian form.

> This one fails. The canonical conjugate of x is (mx' + mat)

> So the Hamiltonian:

>   H = 1/2 (  m (x')^2 - k x^2 )

> The time dependent term has canceled.

That's because you are using the wrong variables! Lets express the
Hamiltonian through the canonically conjugate x and p = mx' + mat:

(B) H = 1/2 ( (p - mat) / m^2 - k x^2 )

As you see, there is time dependance. When we use x and x', the
sympletic structure carries the time dependance. The funny thing is,
that we started with the time-translation invariant equation (A), and
got the time-translation-asymmetric Hamiltonian (B). That's where the
non-adiabatic nature reveals itself. This also explains where the
Noether theorem fails.

> > This is interesting, how may we describe non-adiabatic models
> > after all, and so, address the issue of quantum entropy.

Hmm, my plan failed. The entropy -k Tr(rho ln rho) doesn't rise.

Best regards, squark.

Sent via Deja.com http://www.deja.com/
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