Natural Science Forum / Physics / Research / December 2007

In a different thread (http://groups.google.com.au/group/
sci.physics.research/browse_frm/thread/836f60ae8caec5a8/
cd0d72b214648c59?hl=en#cd0d72b214648c59), as somewhat as an aside,
Igor Khavkine posed the following problem:

While I am not myself married to the Hamiltonian formalism, such a
construction is not difficult.  In particular, the Hamiltonian
    H(x,p) = p^2 / (2M) + (1/2) M w^2 x^2   (*)
yields the equation of motion
     d^2 x / t^2 + b dx/dt + w^2 x = 0           (**)
as required, providing that one defines the "mass" M to be the time-
dependent quantity
     M(t) := exp[ \int_0^t  ds b(s) ],
so that b(t) = d (log M)/dt.    The construction also goes through
when the "frequency" w is time-dependent.  Note that M is strictly
positive.

One can further generalise to obtain a Hamiltonian for the case where
the first term in (**) is multiplied by an arbitrary function m(t).
(as one need only then divide througout by m(t) to get an equation of
the same original form).

The corresponding Lagrangian is interesting, as it shows that the
damping may be interpreted as a redefinition of the time coordinate
and frequency of an _undamped_ oscillator.  In particular, one has the
action
      A = (1/2) \int dt M [ (dx/dt)^2 - w^2 ]
         = (1/2) \int dT [ (dx/dT)^2 - W^2 ] ,
where one defines T and W via
     dT/dt := 1 / sqrt{M}
and
     W(T) = sqrt{M} w .

I believe there have been papers suggesting that one can't
consistently quantise the damped oscillator, but the above shows that
this would be the case if and only if one couldn't consistently
quantise a time-dependent oscillator.