Okay, I think I've pretty much got my head around
the concept of local conservation of energy, which if I'm
not very much mistaken could be stated thusly:
"The integral of the stress-energy tensor over any
finite 3-manifold without boundary* is zero."
This is in fact a stronger statement than what I
originally thought would be the case, since I am coming
from a background in mathematics and my understanding of
local properties derives therefrom. A weaker local
conservation law would be:
"For every point in 4-space (the time-space
continuum), there exists a neighborhood of that point such
that the boundary of that neighborhood is a finite 3-manifold
without boundary over which the integral of the
stress-energy tensor is zero."
Actually, these two statements are equivalent in
most circumstances. The only case in which there is a
meaningful difference is when there is a discontinuity in
the stress-energy tensor. One could define the point or
points where this discontinuity exists as being outside the
space-time continuum. In such a case, a 3-manifold
'containing' this discontinuity would be included in the
condition of the first statement, but would not be included in
the second, so using the weakened conservation law, the
integral of the stress-energy tensor over such a manifold
would not necessarily be zero.
So this raises the question of what, exactly, a
discontinuity in the stress-energy tensor means, exactly.
The first and most obvious candidate would be a naked
singularity. To resolve a bet, Kip Thorne and another
physicist whose name escapes me showed that a naked
singularity can be created by a spherical gravity-wave
implosion (I won't even pretend to understand how), so we
know that they're at least theoretically possible. Would a
regular non-naked singularity also qualify as such a
discontinuity? Is this entire flight of fancy completely
ridiculous?
--
e^(i*pi)+1=0
George W. Harris For actual email address, replace each 'u' with an 'i'.
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