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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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Alexandra Ellis
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    Uh...why do you want a Hamiltonian that is not the energy? – ACuriousMind Apr 05 '15 at 17:28
  • 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) – Alexandra Ellis Apr 05 '15 at 18:47
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    No, the hamiltonian is still equal to the energy, only the energy changes with time. A damped system loses energy to the surroundings. – Ihle Apr 05 '15 at 19:24
  • Possible duplicates: http://physics.stackexchange.com/q/147341/2451 and links therein. – Qmechanic Apr 05 '15 at 19:29
  • 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?

     – Alexandra Ellis Apr 06 '15 at 00:25

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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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user30850
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  • 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 $. –  Jan 25 '18 at 01:08
  • 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. $\overdot{H}= e^{-\gamma t}H$. – user30850 Jan 25 '18 at 07:38
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    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. –  Jan 25 '18 at 17:19