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This is a thought I had a while ago, and I was wondering if it was satisfactory as a physicist's proof of the positive mass theorem.

The positive mass theorem was proven by Schoen and Yau using complicated methods that don't work in 8 dimensions or more, and by Witten using other complicated methods that don't work for non-spin manifolds. Recently Choquet-Bruhat published a proof for all dimensions, which I did not read in detail.

To see that you can't get zero mass or negative mass, view the space-time in the ADM rest frame, and consider viewing the spacetime from a slowly accelerated frame going to the right. This introduces a Rindler horizon somewhere far to the left. As you continue accelerating, the whole thing falls into your horizon. If you like, you can imagine that the horizon is an enormous black hole far, far away from everything else.

The horizon starts out flat and far away before the thing falls in, and ends up flat and far away after. If the total mass is negative, it is easy to see that the total geodesic flow on the outer boundary brings area in, meaning that the horizon scrunched up a little bit. This is even easier to see if you have a black hole far away, it just gets smaller because it absorbed the negative mass. But this contradicts the area theorem.

There is an argument for the positive mass theorem in a recent paper by Penrose which is similar.

Questions:

  1. Does this argument prove positive mass?
  2. Does this mean that the positive mass theorem holds assuming only the weak energy condition?
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  • $\begingroup$ Choquet-Bruhat's paper is a streamlined, cleaned-up version of the spin-manifold proof of Witten. It does not work for non-spin manifolds. $\endgroup$
    – Willie Wong
    Aug 24, 2011 at 13:19
  • $\begingroup$ Joachim Lohkamp, however, has initiated a project to prove the positive mass theorem without the spin assumption in all dimensions. The path he plans to take looks like a modification of the Schoen and Yau proof, since he is now concerned with regularisation of minimal surfaces. The jury is still out on whether this program will succeed or not. $\endgroup$
    – Willie Wong
    Aug 24, 2011 at 13:23
  • $\begingroup$ The Penrose et al proof alluded to is this one, I think. A quick look at that paper, there is one thing that stand out (also for your argument): for this type of proof seems to require strong conditions on $\mathscr{I}$, which I don't think is guaranteed by the statements of the PMT with regards to initial data sets. So it is possible that the weakening of the energy condition is logically compensated by assuming better control on the geometry near infinity. $\endgroup$
    – Willie Wong
    Aug 24, 2011 at 13:49
  • $\begingroup$ I only need standard 1/r decay at infinity to get flat horizons, as far as I can see. I will think about it some more. Thanks for digging up all the references. $\endgroup$
    – Ron Maimon
    Aug 27, 2011 at 8:06
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    $\begingroup$ I don't see the link to ADM/Bondi energy in this argument. I've always thought of the mass in a spacetime as being defined by boundary terms in the action, not information about how particles move. @Willie Wong: You're going to at least need asymptotic flatness at infinity to make any sense of mass/energy, right? $\endgroup$
    – Zo the Relativist
    Sep 7, 2011 at 15:46

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Given that your argument requires moving in an accelerating frame and considering its Rindler horizon, I wonder if what you stated is more similar to statements about asymptotically hyperbolic hypersurfaces in a space-time. In which case, that the analogue of the positive mass theorem can be derived using only the weak energy condition has been shown before (for example, see this paper of Anderson, Cai, and Galloway).

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  • $\begingroup$ The paper you linked is using Schoen Yau methods. The point of this argument is that it uses only geodesic deviation, and no action minimizing surface, and the proof you linked does not do that. If you introduce minimizing surfaces, you're back in Schoen-land. The argument I gave is not far from rigorous as is. $\endgroup$
    – Ron Maimon
    Aug 27, 2011 at 8:05

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