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When one consider a lagrangian and construct hamiltonian, we expect to be bounded below. While looking to the Hamiltonian formulation of general relativity, I have difficulties to see how it can be bounded.

  1. How this can be shown?

  2. Is it related to the positive energy theorem?

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  • $\begingroup$ What Lagranian and Hamiltonian are you considering? $\endgroup$ – md2perpe Jul 19 '18 at 10:19
  • $\begingroup$ The Lagrangian of general relativity $L=\sqrt{-g}R$ and the ADM hamiltonian $\endgroup$ – anubis Jul 19 '18 at 11:14
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This is exactly the content of the positive energy theorem:

Let $(\Sigma, h_{ab}, K_{ab})$ be an initial data set that is geodesically complete and asymptotically flat. Assume that the energy-momentum tensor satisfies the dominant energy condition. Then $E_{ADM} \geq \sqrt{P_i P_i}$, with equality if and only if the initial data set arises from a surface in Minkowski spacetime.

The proof is not easy. You can find one in Witten's 1981 paper, "A new proof of the positive energy theorem".

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  • $\begingroup$ Thx that will help me to be more specific. The positive energy theorem works only under some conditions, asymptotically flat and strong energy condition. But that means that I can't show generically that Hamiltonian is bounded below? Let us e.g. suppose a positive cosmological constant, the theorem doesn't apply, what should we conclude? I guess just that this theorem doesn't apply and not that Hamiltonian is not bounded below in this case. Am I right? If yes, is there any generic proof $\endgroup$ – anubis Jul 20 '18 at 15:39
  • $\begingroup$ If the spacetime is asymptotically flat, you can define the ADM energy. If any of the other two conditions is violated (geodesic completeness of initial data, dominant energy condition) then you can find spacetimes with arbitrarily negative energy (for instance negative M Schwarzschild) $\endgroup$ – John Donne Jul 20 '18 at 21:03

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