🚀 go-pugleaf

In article <1992Aug9.055643.1254@chpc.utexas.edu>
jonathan@chpc.utexas.edu (Jonathan Thornburg) writes [wrote]:
>
> [ discussing where the mass "is" in Schwarzschild spacetime ]
>
>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.
>
>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.)
>
>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 $r = 0$.

A useful analogy here is that of contour integration around a
pole in complex analysis.  The function $f(z) = 1 / z$ is analytic
for all $z \neq 0$, but has a pole at $z = 0$.  Hence by the
Residue theorem, if we integrate $f$ around any closed contour
$C$ in the complex plane, the result will be $0$ if $C$ does
*not* contain the origin, but $2 \pi i$ if it *does* contain
the origin.  But $f$ isn't even *defined* at $z = 0$, so really
the pole there isn't "part of $f$", i.e. the domain of $f$ is only
the set of all *nonzero* complex numbers.

- 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}