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I recently found out that it is possible to formulate a Hamiltonian for a system with time-dependent coordinates such that the Hamiltonian is not the same as the energy When is the Hamiltonian of a system not equal to its total energy? and that has me wondering if it is possible to formulate a Hamiltonian for a damped system under these conditions. I know that Hamilton's equations require that energy be conserved, but if the coordinates are time-dependent, would it still be possible to formulate and solve the problem?

I started trying to do it for a damped simple harmonic oscillator by starting with the Lagrangian for the system

$$L=e^{\gamma * t}*(\frac{mv^2}{2}-\frac{kx^2}{2}),$$

but I keep on coming up with a Hamiltonian that is just equal to the energy

$$H=e^{\gamma * t}*(\frac{mv^2}{2}+\frac{kx^2}{2}).$$

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    $\begingroup$ Uh...why do you want a Hamiltonian that is not the energy? $\endgroup$
    – ACuriousMind ♦
    Commented Apr 5, 2015 at 17:28
  • $\begingroup$ That's what I'm trying to figure out. I believe the Hamiltonian can be different from the energy if the coordinates are time-dependant. (see the link) $\endgroup$
    – Alexandra Ellis
    Commented Apr 5, 2015 at 18:47
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    $\begingroup$ No, the hamiltonian is still equal to the energy, only the energy changes with time. A damped system loses energy to the surroundings. $\endgroup$
    – Ihle
    Commented Apr 5, 2015 at 19:24
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/147341/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Apr 5, 2015 at 19:29
  • $\begingroup$ Qmechanic, would you mind explaining what you mean by "The caveat is that the Hamiltonian (7) does not represent the traditional notion of total energy." in your "Unconventional approach"? I'm still a bit unclear how that is not the expression for total energy as a function of time. Also, should the e(t) be in the numerator for both terms in your expression for the Hamiltonian? $\endgroup$
    – Alexandra Ellis
    Commented Apr 6, 2015 at 0:25

1 Answer 1

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The following might help:

$H = \frac{1}{2}(mv^2 + kx^2) + \gamma mkvx$

decays exponentially with time along the solution of the damped system. Check by differentiating $H$ with respect to $t$ and using the equations of the system. So the "energy" $H$ decays exponentially instead of remaining constant.

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  • $\begingroup$ Can you write down the equations of motion? I think you did your math wrong. Time independent Hamiltonians are always constant in time $ \dot{H} = \{ H, H \} = 0 $. $\endgroup$
    – user92177
    Commented Jan 25, 2018 at 1:08
  • $\begingroup$ The system is damped, so their is no preserved quantity. The "Hamiltonian" i mentioned is not preserved as expected but it decays exponentially, probably the next best thing you would expect. $\overdot{H}= e^{-\gamma t}H$. $\endgroup$
    – user30850
    Commented Jan 25, 2018 at 7:38
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    $\begingroup$ You aren't using Hamilton's equations correctly. You are forgetting that the third term gets differentiated for both equations. Hamiltonians are by construction constant in time. $\endgroup$
    – user92177
    Commented Jan 25, 2018 at 17:19

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