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.


  1. Does this argument prove positive mass?
  2. Does this mean that the positive mass theorem holds assuming only the weak energy condition?
  • $\begingroup$ 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. $\endgroup$ Aug 24 '11 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. $\endgroup$ Aug 24 '11 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. $\endgroup$ Aug 24 '11 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 '11 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? $\endgroup$ Sep 7 '11 at 15:46

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).

  • $\begingroup$ 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 '11 at 8:05

In order to achieve mass polarity, parameters of interactive relativity must be met. Stasis attenuation of partical binding force have to be amplified through conductivity of said force. This method passes most limits mathematically and physically. Just as with quantum parameter expression, there are few limits outside the box. The threshold is the limitation of the parameter limit. As with the assumption that all things are finite, this area, as far as is known, has no limitation. Welcome to 'dark energy'. Mechanics involved are simple, requires little power, and is more than readily available. The 'feed' uses the stasis relationship between control and chaos, where probability of order from chaos becomes its own basis. Plasma is a prime example of exactly this. Directing the forces involved with plasma elude classical physics because of the impossibility to formulate control/regulation of the resultants. Just as with the self regulation of a lazer, plasma regulation depends directly on its manner of propagation. Within the 'parameter' of perception, answers to these questions all reside outside the prameters of the accepted norm. Only when all things are understood by passing or exceptions to known 'laws' will there be any practical application. These conditions do not negate known laws, rather they compliment them, and are the resultants being searched for. A direct tap of source, forced propagation, and creation are The options presently. Indirect, stasis conductivity, and creation are the others. As experiments have proven, anything outside these exact limits takes everything outside any parameters known. 1/ 0.000137 = 7299.270073 A relationship few have any idea of as to application or any practical use, etc. Yet the cubic 9, perfect cube, is contained within it, and on both sides of the base formulation. When classic and fringe join together does the resultant reflect the whole. As with Pythagoras triangulation formula is incomplete as it is one sided and very limited. When going into reality, the whole must be applied. A^2 + B^2 = C^2 is exactly half an equilateral triangle, not a whole of of equal dimension. In the same light, a partial truth has been propounded for ages in the 'equality' math method using 1 - 1 = 0...again, a partial rendition. This time for good reason. Just as all electronic and atomic sciences are falsified intentionally, there is however an answer to the problem. It is nothing more than the fact that our mathematical capability is unequal to the task. No calculation used by us can work out a positive result from a negative perspective, and is stated so in all elecronics texts and some physics texts... When dealing with solid objects (spherical dynamics) our math falls short, One very obvious limit is the fact of our ability to 'read' results in electonics, where the spherical limit of 'hemisphere' comes into play. Square root of 2 = 1.414213562, this has to be divided by 2. 1.414 / 2 = 0.707106781 The seven db down point of all signals used in an analog manner. WHY? All calculus format is based on a 180 Deg limit, and can not pass the said limit without going negative. The fractal expression is The problem. A sphere is not calculated as a whole. Therefore, a new perception is required. Within that framework is found the real 'equivalence' formula of 1.0.1 Whole numerics on both sides is the only way it will work. I call this spherical mathematics. The second part deals with vibratory physics in relation to the above.

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    $\begingroup$ A question like this requires someone with a PhD standard of education to answer it with some authority. Your answer looks like self indulgent jibberish from someone who has never read a graduate level physics text book. $\endgroup$ Sep 7 '11 at 14:01
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    $\begingroup$ I don't have a PhD, but thankfully authority is of no value on the internet. This answer isn't bad because it lacks authority. It is bad because it is using terms without understanding. $\endgroup$
    – Ron Maimon
    Sep 7 '11 at 14:53
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    $\begingroup$ Amen @Ron. Just a tiny quibble, user2146 said "PhD standard of education", not "holding a PhD"; one can of course have the former and not the latter, as you seem to indicate. $\endgroup$ Sep 7 '11 at 19:45

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