Re: Misinterpretation of the radial parameter in the Schwarzshild solution?
- From: carlip-nospam@xxxxxxxxxxxxxxxxxxx
- Date: Tue, 4 Jul 2006 03:57:29 +0000 (UTC)
LEJ Brouwer <intuitionist1@xxxxxxxxx> wrote:
[...]
The reason I am interested is because the following papers claim that
there is an error in the interpretation of the radial coordinate 'r' in
the standard Schwarzschild metric:
L. S. Abrams, "Black holes: The legacy of Hilbert's error", Can. J.
Phys. 23 (1923) 43, http://arxiv.org/abs/gr-qc/0102055
S. Antoci, "David Hilbert and the origin of the 'Schwarzschild
solution'", http://arxiv.org/abs/physics/0310104
S. J. Crothers, "On the general solution to Einstein's vacuum field and
its implications for relativistic degeneracy", Prog. Phys. 1 (2006)
68-73.
In particular they show, in a rather simple fashion, that the event
horizon is at radius zero, coinciding with the position the point mass
itself, and actually appears pointlike to an external observer.
These papers are complete nonsense. In particular, the authors seem
not to understand the basic fact that physics does not depend on what
coordinates one chooses to use.
It is trivially true that if one changes coordinates in the standard
Schwarzschild solution from r to r-2m, then the horizon is at r=0.
It is also trivially true that this does not change the spacetime
geometry -- the horizon is still a lightlike surface, with an area at
fixed time of 4m^2. Choosing a coordinate that makes the horizon
look like a point doesn't make it a point -- it just means that you've
made a dumb coordinate choice.
They claim that the reason that the original misinterpretation occurred
is because Hilbert incorrectly assumed a priori that the 'r' which
appears in the metric must be the radial coordinate (in fact, it need
only parametrise the radii to ensure a spherically symmetric solution).
It is radial in the sense that the set of points at constant r and t
is a two-sphere of area 4pi r^2. It is not "radial distance," but
no one has claimed that it is.
The careful analysis of Abrams et al shows that the point mass actually
resides at r=2m, which therefore corresponds to the true origin, so
that there is in fact no 'interior' solution.
This analysis is not "careful" -- it's mathematically awful. How can
a "point mass" reside at a two-sphere of finite area? What sense does
it make to say that a mass resides at a position at which the Ricci
tensor is zero?
Abrams makes an elementary mistake. He concludes that r=2m (in standard
Schwarzschild coordinates) is singular because the "radius" of a circle
around this "point" goes to zero as r->2m while its "circumference"
does not. But this is not a singularity -- it's just a reflection
of the fact that r=2m is a two-sphere, not a point.
If the event horizon is at the origin, and there is no interior
solution, then this tends to raise the question, "well, where does a
radially infalling particle actually go?". Does it just bounce off the
'brick wall' (or rather, 'brick point')?
To answer this, you just compute the motion. You find that it falls
right past the "origin," with nothing peculiar happening there. (Of
course, you can insist on using bad coordinates, but that's your own
fault...).
Have we really all been making this silly mathematical error, and is
our present understanding of the simplest classical black hole way off
the mark?
No.
Steve Carlip
.
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