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
> > (A) x'' = (-k/m)x - ax'
> > Actually, the problem arises for a simple Newtonian body moving with
> How about:
> L = 1/2 (m(x')^2 - k x^2) + max't
> > And that's why:That's because you are using the wrong variables! Lets express the
> > 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.
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
> > This is interesting, how may we describe non-adiabatic modelsHmm, my plan failed. The entropy -k Tr(rho ln rho) doesn't rise.
> > after all, and so, address the issue of quantum entropy.
Best regards, squark.
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