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Ostrogradsky's instability theorem says that under some conditions, a system governed by a Lagrangian which depends on time derivatives beyond the first is "unstable". In the proof, one computes the Hamiltonian and shows that the dependence on one of the canonical momenta is linear.

I don't understand either the statement or the proof. I think this is because I don't remember my Hamiltonian mechanics very well.

Questions:

  1. What is meant by saying that the system is unstable? There are several notions of stability out there. Is the theorem saying the for any choice of initial conditions, the system is, say, Lyapunov unstable?

  2. What does the linearity of the Hamiltonian have to do with stability?

  3. My sense is that Ostrogradsky's theorem is sometimes taken to justify the idea that Lagrangians in physics shouldn't depend on higher time derivatives. Why is this? What's wrong with studying systems which are unstable? If I understand what "unstable" means, isn't it just another word for "chaotic"? And certainly there are chaotic systems in the real world...

stafusa
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tcamps
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    We usually require the Hamiltonian to be bounded from below in quantum mechanics, otherwise the system will evolve as to spontaneously create an infinite number of particles. I look forward to someone coming to answer about this question in a purely classical regime though. – Diffycue Jul 06 '19 at 14:32

1 Answers1

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Not a specialist here either, but I think the Scholarpedia page linked in the OP already provides the answers.

  1. What is meant by saying that the system is unstable? There are several notions of stability out there. Is the theorem saying the for _any_ choice of initial conditions, the system is, say, Lyapunov unstable?

It's unstable in the sense of presenting explosive vacuum decay:

In fact the system instantly evaporates into a maelstrom of positive and negative energy particles.

  1. What does the linearity of the Hamiltonian have to do with stability?

    The root of the problem seems to be that:

Because the Hamiltonian is linear in all but one of the conjugate momenta it is possible to arbitrarily increase or decrease the energy by moving different directions in phase space.

A similar explanation is found in this answer:

$H$ has only a linear dependence on $P_1$, and so can be arbitrarily negative. In an interacting system this means that we can excite positive energy modes by transferring energy from the negative energy modes, and in doing so we would increase the entropy — there would simply be more particles, and so a need to put them somewhere. Thus such a system could never reach equilibrium, exploding instantly in an orgy of particle creation.

  1. My sense is that Ostrogradsky's theorem is sometimes taken to justify the idea that Lagrangians in physics _shouldn't_ depend on higher time derivatives. Why is this? What's wrong with studying systems which are unstable? If I understand what "unstable" means, isn't it just another word for "chaotic"? And certainly there are chaotic systems in the real world...

It's certainly not because one wants to avoid chaotic systems, but rather that

[explosive vacuum decay] certainly does not describe the universe of human experience in which all particles have positive energy and empty space remains empty.

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stafusa
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  • Thanks! I'd still be interested in a classical explanation. But even in the quantum picture, I don't understand the claim that the system would "instantly" explode into particle creation. Surely there's some timescale this occurs at, having to do with the relative magnitudes of the other terms in the Hamiltonian? – tcamps Jul 06 '19 at 14:57
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    @tcamps I hope that someone who actually knows the field also answers, but, once again citing the Scholarpedia entry, Woodard claims that: "Atomic decays have just the fixed energy difference between the two states to apportion, so they are chiefly driven by the arbitrary directions which can be taken by the decay products. In contrast, the decay of an interacting, nondegenerate higher derivative field theory can involve particles of {\it any} energy, as long as the total sums to zero." ... – stafusa Jul 06 '19 at 16:41
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    ... "So one should think of the decay rate as having the same sort of angular factors as an atomic decay at some fixed energy, followed by one or more integrals -- all the way to infinity --- over the magnitudes of the various energies. The volume of phase space is so large that these integrations cause the decay to be instantaneous." – stafusa Jul 06 '19 at 16:42
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    And it continues, arguing that "Indeed, the only way people derive finite decay rates for particles with a kinetic instability is by cutting off the phase space at some point, in which case the rate is dominated by the cutoff, for example see Cline et al., 2004. Such a cutoff might make sense if the kinetic instability appeared in some nonlocal effective field theory, but it has no place in fundamental physics." This is all from the section 3.3 Instability/Vacuum decay. – stafusa Jul 06 '19 at 16:45
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    As for the classical explanation, all I've found or a cursive search is that a linear term would not be in agreement with Newton's laws. Didn't check it myself, tough, so I don't want to include that in the answer - besides, apparently it's continuum field aspect that seems to be considered relevant here. – stafusa Jul 06 '19 at 16:47