3

Since neither the object nor its field could exist without the other, it would seem strange not to include the field energy as part of the object. But how exactly does the accounting go? How is the mass of the system divided between the rest mass and the field mass?

For a Schwarzschild black hole, the mass appears to be shifted completely to the field.

According to Lynden-Bell and Katz, http://adsabs.harvard.edu/full/1985MNRAS.213P..21L, the total energy distributed in the gravitational field of a Schwarzschild black hole is mc^2. In other words, all the mass of the black hole resides outside the event horizon.

dcgeorge
  • 513
  • 1
    First you need to define the energy of the gravitational field which itself is a thorny subject. There is no nontrivial local energy-momentum tensor for the gravitational field in GR, so the best you can get is a pseudo-tensor (i.e. something which depends on your coordinate choice) or a global property like the ADM mass which is only applicable to asymptotically flat spacetimes (this is the case for the Schwarzschild black hole). This is an extra layer of difficulty over and above the self-energy problem in, say, electromagnetism where the term "energy" is unambiguous. – Michael Sep 13 '13 at 15:52

2 Answers2

4

GR does not have a local, tensorial measure of energy density in the gravitational field. This is because of the equivalence principle, which tells us that the Newtonian gravitational field $\textbf{g}$ can't be a tensor. (It vanishes in an inertial frame, and any four-vector that vanishes in one frame vanishes in all frames.)

GR does have various measures of the total mass in an asymptotically flat spacetime, such as the ADM mass. These are not local things, so they don't give us a way to define where the energy resides.

According to Lynden-Bell and Katz, http://adsabs.harvard.edu/full/1985MNRAS.213P..21L, the total energy distributed in the gravitational field of a Schwarzschild black hole is mc^2.

The Schwarzschild spacetime is a vacuum solution, so if you want to attribute its mass-energy to something, you're going to have to attribute it to some property of the empty space..

In other words, all the mass of the black hole resides outside the event horizon.

No, this is wrong. There is no logical connection to the event horizon.

  • "There is no logical connection to the event horizon." I should have quoted more of the paper's summary. The authors claim: "We show by physical arguments that static spherical systems have a coordinate-independent field energy density. ... the field energy outside a Schwarzschild black hole totals Mc^2. In this sense all the energy remains outside the hole." I was surprised by this conclusion so I was hoping someone would weigh in on their arguments. In any case, I can see more clearly now how confusing the whole issue of field mass is. Thank you for your insights. – dcgeorge Sep 17 '13 at 16:53
  • @dcgeorge: In "Relationship to other expressions," they list many different ways of writing this energy density. They all disagree on the location of the energy. Lynden-Bell believes his is the best physically motivated, but that is not a widely accepted view. In general, MNRAS is a crap journal that publishes lots of papers that are nonsense (e.g., arxiv.org/abs/1109.5189 ). If their measure of energy density implies that all the energy of a black hole is outside the horizon, then that's in fact an argument against taking their result as having any deep significance. –  Sep 17 '13 at 22:38
  • "... MNRAS is a crap journal that publishes lots of papers that are nonsense ..." That may be true but you might want to take a look at Lynden-Bell's Wikipedia page before you conclude the paper is nonsense. http://en.wikipedia.org/wiki/Donald_Lynden-Bell – dcgeorge Sep 20 '13 at 17:50
  • "... If their measure of energy density implies that all the energy of a black hole is outside the horizon, then that's in fact an argument against taking their result as having any deep significance." It does indeed imply that. But, if it disagrees with the general consensus and is correct, it's likely of great significance. Lynden-Bell and Katz are not exactly lightweights in the field. – dcgeorge Sep 20 '13 at 17:53
  • @dcgeorge: The paper I linked to is the one that I claimed was nonsense. This one is IMO just a paper that is unimportant and whose authors have greatly oversold its importance. –  Sep 21 '13 at 17:05
0

First of all, dcgeorge’s contrast of the rest mass vs the field mass is erroneous. The concept of rest mass requires the center-of-momentum frame of reference. Rest mass is indifferent about whether the massive object is a “particle”, a “field”, or both things combined. The problem is that curved spacetime doesn’t admit inertial frames of reference. The only known workaround is to define a reference frame that is inertial asymptotically (on the infinite distance from the object), and this is the way to define masses of strongly gravitating bodies (such as black holes). We can measure lengths, time intervals, and velocities far away of the body, and that’s how we can determine how is it massive.

Also, field energy of gravitation is, in general, very ill-defined concept in General Relativity. One reason is the same as above: it disrupts translational symmetry of the spacetime, and hence hinders the “part of the mass/momentum/energy lies here and another part lies there” discourse. This not only makes a general definition of gravitational energy impossible, but hinders consistent definitions of momentum/energy distribution in a curved spacetime (time is a conjugated to energy, and spatial length – to momentum).

Lynden-Bell and Katz introduce their definition of mass/energy density for some cases. Where the outer space is asymptotically flat, we can unambiguously define the total mass/energy (relatively to this asymptotically flat universe), as well as momentum. But the question “how is this mass/energy distributed inside” does not have a universal solution for several reasons. Should I explicate it? Lynden-Bell and Katz said themselves that “their” mass distribution is different from Penrose’s one.