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In article <1992Aug8.010716.16818@nuscc.nus.sg> matmcinn@nuscc.nus.sg
(Mcinnes B T (Dr)) writes:
> [ referring to the question of the "mass" of Schwarzschild spacetime ]
>OK, now that I have your attention: let me be more precise. Certainly
>there is a parameter in the S solution which tells you what the mass of
>this thing would be, calculated a la Kepler, if it were a Newtonian
>source...which it isn't. For certain purposes this is no doubt useful.

Let's be clear here.  In general relativity, the "energy-momentum"
of a spacetime is defined *only* for asymptotically flat spacetimes.
For these spacetimes (which include both the Schwarzschild and
Kerr spacetimes, but not most cosmological models), the energy-
momentum of the spacetime is *defined* by any of several different
surface integrals over a 2-sphere at infinity.  (Under suitable
technical assumptions, the several different definitions can be
shown to all be equivalent.)  So defined, the energy-momentum of
a spacetime is a Lorentz 4-vector in the far-field Minkowski
spacetime.  In a coordinate system in which it's spatial (momentum)
components are zero (this defines the "asymptotic rest frame" of
the spacetime), it's time (mass-energy) component has a value which
(modulo minus signs) we call the "mass" of the spacetime.

(After ~60 years of attempts by various people, it was finally
proved by Schoen and Yau in 1987 that the mass of a spacetime, so
defined, is in fact positive semidefinite, i.e. it's always greater
than or equal to zero, and is zero only for a slice of Minkowski
spacetime.  This is a Very Good Thing, since we probably only want
to call this quantity "mass" if this holds.  Not to mention the
fact that one could extract an *infinite* amount from a "negative
mass" system.)

Given these definitions, one can prove that the mass of the spacetime
is in fact equal to the value which would be obtained from the usual
far-field Kepler formulas, applied in the asymptotic rest frame of
the spacetime.  We can thus, if we prefer, take the far-field Kepler
formulas as an alternative operational definition of the mass of the
spacetime.

But it's important to realise that these are all *definitions*.  We
are free to define the "mass" to be whatever we want.  A definition
can't be "right" or "wrong" (though it can be inconsistent with
itself or some other useful formula/definition/etc) -- the test of
whether a definition is "good" or not is whether or not it's *useful*.

In this regard, it's important to note that the definition given
here of "mass" is only for the spacetime as a whole -- there's *no*
definition of the "mass" of anything smaller, of any compact "piece"
of a spacetime.  Various people (notably both Penrose and Hawking)
have proposed candidate definitions for "quasi-local mass", but
these often don't agree with each other, or have other counterintuitive
properties.  This illustrates the subtelty of "mass" in general
relativity.

>But I am thinking of mass as a measure of "quantity of matter".

Again, this is not what the word "mass" means in general relativity.
Hereinafter I'll use the phrase "quantity-of-matter" for the concept
you're talking about.  Note that quantity-of-matter is *not* a
conserved quantity, since (eg) it increases in pair production
and decreases in radioactive decay and matter-antimatter reactions.
The quasi-local mass definitions mentioned above are all attempts
to construct local conserved quantities which agree with
quantity-of-matter for simple situations.

>So let me ask again:
>How much matter is there in the whatever-it-is that the Schwarzschild
>solution describes? [People have objected, quite rightly, to my use of
>the word "object"...another piece of bait, I must confess.] After all,
>there must be some matter there, right? I mean, "matter tells spacetime
>how to curve", right? And so there must be some matter there, "causing"
>spacetime to curve, otherwise it would not know how to curve...right?

Well, not really.  Gravitation is *nonlinear*.  Fundamentally,
*all* mass-energy gravitates [is that word in the OED???],
including the mass-energy density of various fields.  Thus, if
you have (say) an electromagnetic field, the mass-energy in the
EM field is a source for gravitation.  More to the point here
(since black holes in nature don't have enough electric charge
or magnetic field to significantly affect their gravitational
fields), if you have a gravitational field, the mass-energy of
the gravitational field *is* *itself* a source for gravitation.
This is the basic cause of gravitation's nonlinearity.

It's thus possible to have a "pull-yourself-up-by-your-bootstraps"
self-consistent solution of the *vacuum* Einstein equations wherein
spacetime is entirely empty (= no matter, EM fields, or anything at
all except for the gravitational field) but contains a gravitational
field sufficient to "sustain itself".  Wheeler calls such a system
a "geon".  Back in the 1950s he and others did a lot of research to
see whether geons could be models of elementary particles, but this
didn't pan out.  (I don't know why.)  The term "geon" has also become
archaic.

>All right, to quit fooling, what I am driving at is this: The amount of
>matter in this spacetime is zero. Exactly zero. 

In modern terminology, Schwarzschild spacetime is a (the unique!)
solution of the Einstein equations which is vacuum, asymptotically
flat, and spherically symmetric.  The "point" $r = 0$ (its actual
topology is rather more complicated, but probably irrelevant here)
is *not* part of this spacetime.  It can't be, since the Einstein
equations themselves aren't well-defined there, indeed the definition
of "spacetime" is that a *smooth* differential manifold.

That said, you won't go too far wrong if you say that $r = 0$ *is*
a part of the spacetime, and that the mass density in the spacetime
is given by $\rho(t,r) = M \delta(r)$, where $\delta()$ is the Dirac
delta-function, and $M$ is the legally-defined "mass" of the spacetime.
Except for the minor fact that the Einstein equations don't work very
well for delta-functions, this works fine.  And you can probably fudge
the Einstein equations via the usual limit-of-narrower-and-narrower-
-Gaussians trick.  (I don't know how the Einstein equations fare
in distribution theory.)

One useful thing to remember is that the spacetime *exterior* to
any compact spherically symmetric region *must* be Schwarzschild.
That is, the gravitational field of any spherically symmetric
mass-energy distribution is identical to that of a Schwarzschild
spacetime, once you go outside into the vacuum region.  (This is
"Birkhoff's theorem.)  Thus, for example, the spacetime "left
behind" by a star that collapses into a black hole is (assuming
the collapse to be spherically symmetric) Schwarzschild spacetime.
The fact that the interior of the star doesn't the least resemble
the neighborhood of $r = 0$ in Schwarzschild spacetime is irrelevant
-- the exterior spacetime is indentical.

One can of course ask "what happens to the *matter* in the former
star in this process, especially once it reaches $r = 0$?"  This
question is beyond the scope of general relativity, because quantum
gravity effects become important in the immediate neighborhood of
$r = 0$.  Unfortunately, despite ~60 years of effort by a lot of
very bright people, we don't yet know how quantum gravity works.
So for now about all we can do is carefully look the other way,
and not inquire too closely about physics near singularities...

The various definitions of quasi-local mass I mentioned above are
all actually definitions of the mass interior to any specified
"Gaussian surface", i.e. "balloon".  They are all designed so that,
for Schwarzschild spacetime, if the Gaussian surface contains $r = 0$
the quasi-local mass is equal to the mass of the spacetime, while
if the Gaussian surface doesn't contain $r = 0$ the quasi-local
mass is zero.  So in this sense, one can say that all the mass of
Schwarzschild spacetime "lives" at $r = 0$.

But again, this is an imprecise statement.  The classical (non-quantum)
Einstein equations are valid only away from singularities, so we
can't *rigorously* say much of anything about $r = 0$.

One final point:  Real astrophysical black holes, formed from
the collapse of massive stars in supernova explosions, closely
resemble Schwarzschild (Kerr if rotating) spacetime in their
$r > 0$ geometry, for the reasons outlined above.  We refer to
them as "Schwarzschild (Kerr) black holes".  But for most
astrophysical purposes the fact that we don't know *what* lurks
at $r = 0$ is irrelevant.

In contrast, one could concieve of "primordial" black holes,
present in the fabric of spacetime right from the big bang.  If
they exist, these may have very different $r = 0$ structures.
(This is where "wormholes" come in, but that's a topic I really
don't want to get into just now, or my thesis will *never* get
done...)  But it's difficult (impossible, so far as we know) to
observationally tell the difference from $r > 0$, since all one
sees in either case is the simple Schwarzschild gravitational
field.  So long as one is only dealing with $r > 0$ processes,
one can thus pretend that a black hole is of either type
interchangebly, as a matter of convenience.

- Jonathan Thornburg
  <jonathan@einstein.ph.utexas.edu> or <jonathan@hermes.chpc.utexas.edu>
  University of Texas at Austin / Physics Dept / Center for Relativity
  and (for a few more months) U of British Columbia / {Astronomy,Physics}