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The equation of motion of a damped oscillator $$\frac{d^2x}{dt^2}+\gamma\frac{dx}{dt}+\omega_0^2x=0$$ which is invariant under time-translation $t\rightarrow t+a$, but not under time reversal $t\rightarrow -t$. We also find that this system does not conserve energy. However, we know from classical mechanics that the time-translation invariace is related to the conservation of the energy not the time reversal. Why does the system dissipate energy?

Time-reversal invariance is related to reversibility, and as I understand not to dissipation directly.

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SRS
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2 Answers2

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To apply Noether's theorem, which is what you are alluding to here, one needs to look at continuous symmetries of a Lagrangian description of a system's dynamics.

The damped oscillation equation you have written, although it is invariant with respect to a time translation as you rightly say, is not a Lagrangian description. If you write the Lagrangian for this system, you'll find that it is not time shift invariant, so this lack of symmetry is where your implied argument ("my description does not depend on time $\rightarrow$ energy is conserved") breaks down.

Some dissipative systems have Lagrangian descriptions but they always wind up with time varying Lagrangians, so there is no contradiction of Noether's theorem: See Joseph Johnson's answer and link to read about some interesting historical attempts to broaden the Lagrangian notion to all systems here.

Incidentally, if you want to model this kind of thing in quantum mechanics, a way to do this is to embed your oscillator in a huge set of coupled quantum mechanical oscillators where your initially excited oscillator is coupled weakly to all the others. The system as a whole varies unitarily with time, so there is no overall dissipation, but if one oscillator begins in a raised state with all others in their ground state, the amplitude to find the oscillator in its excited state dwindles exponentially with time and the energy becomes inexorably spread throughout the whole system of oscillators - a lovely illustration of a simple macroscopically "irreversible" (but microscopically reversible) system. You can do the same in theory with a classical harmonic oscillator too - embed it in a huge system of oscillators and couple it weakly to a huge reservoir of others: the same thing happens. The oscillator on its own follows a damped equation, even though the whole system is not dissipative.

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Selene Routley
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  • +1. Maybe also mention of the Fluctuation-dissipation theorem for the classical case? – CuriousOne May 28 '16 at 08:47
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    Indeed, the wanted Lagrangian is $L(t, x, \dot{x})= \frac{e^{\gamma t/m}}{2}(m \dot{x}^2 - \omega_0^2 x^2)$, which depends on time as suggested by Rod. +1 – Valter Moretti May 28 '16 at 09:06
  • What is the significance of time-reversal invariance in this equation? – SRS May 28 '16 at 09:20
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    @SRS There's not really a "physics"significance that I can put my finger on: there's the mathematical one that if you replace $t$ by $-t$ in any solution, you have another solution. But one needs then to step in and explicitly impose one's physical isight to choose which solutions are the "physical" ones : in a spring on a mass we'd say that the $exp(+\gamma,t)\cos(\sqrt{\omega^2 - \gamma^2},t)$ solution, with $\gamma > 0$ is unphysical because we're not putting energy in; if we got this equation from a system of laser rate equations, ...... – Selene Routley May 28 '16 at 10:43
  • ....we'd keep the solution because it is describing the buildup of a wave in a pumped medium. – Selene Routley May 28 '16 at 10:43
  • Dear @WetSavannaAnimalakaRodVance There is again a confusion. You said, "If you write the Lagrangian for this system, you'll find that it is not time shift invariant". I agree. But note that the Lagrangian given by Valter Moretti changes under time-translation only by an overall constant! If the Lagrangian changes just by a multiplicative constant under time-translation will that lead to energy nonconservation? Isn't this change a trivial one? – SRS Nov 06 '18 at 23:03
  • @ValterMoretti You might want to look at my comment above for WetSavannaAnimalakaRodVance. If a Lagrangian changes by a multiplicative constant only will that lead to still lead to energy nonconservation by Noether's theorem? – SRS Nov 06 '18 at 23:12
  • What matters here is not Noether theorem but Jacobi theorem: if $\frac{\partial L}{\partial t}=0$, then ${H} = \frac{\partial L}{\partial \dot{q}} \dot{q}-L$ is conserved alnog the solutions of Euler-Lagrange equations. – Valter Moretti Nov 07 '18 at 10:07
  • Well, a direct computation proves that $\frac{d H}{dt} = -\frac{\partial L}{\partial t}$, in the general case. So, in the considered case $\frac{d H}{dt} = c L$. It does not vanish so that $H$ is not conserved. Finally, it is worth observing that $H$ has not the standard form $T+U$ in view of the non-standard form of the Lagrangian. – Valter Moretti Nov 07 '18 at 10:08
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Conservation of energy is related to time translation invariance for systems that can be described by a Lagrangian. Dissipative systems in general are not describable by Lagrangian mechanics (without altering the formalism that is) and so Noether's theorem cannot be applied to check whether or not energy is conserved.

EDIT: the dissipative system the OP refers to admits lagrangians that enable to recover the right equation of motion as pointed out by Valter Moretti, but then the hypotheses under which Noether's theorem holds are not met by such lagrangians.

gatsu
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    A Lagrangian producing the equation $\frac{d^2x}{dt^2}+\gamma\frac{dx}{dt}+\omega_0^2x=0$ is just $L(t, x, \dot{x})= \frac{e^{\gamma t/m}}{2}(m \dot{x}^2 - \omega_0^2 x^2)$. The point is that it depends on time. Jacobi (Noether) theorem holds in principle but the hypotheses are not true here. – Valter Moretti May 28 '16 at 09:06
  • Do you mean that to prove Noether's theorem we need to assume that the system to be non-dissipative? @gatsu – SRS Nov 06 '18 at 23:15
  • @SRS For dissipative systems which admit some Lagrangian description, arguably a bit artificial in their form, Noether's theorem can still apply but the conserved quantities have no straightforward interpretation and since the Lagrangian is time-dependent, it is clearly not the Hamiltonian of the system that is conserved – gatsu Nov 09 '18 at 20:49