Re: More on the controversy about the Schwarzschild radius and black holes.

LEJ Brouwer wrote:
Suppose I am the planet Earth
and all the stars start converging upon me.

In a spherically symmetric manner.

Whatever the metric it is
in my vicinity, it has a spherical symmetry, and will be approximately
Schwarzschild.

Neglecting minor deviations from spherical symmetry (sun, moon, earth rotation, earth non-sphericity, etc.), and neglecting the atmosphere, it will be _exactly_ Schwarzschild -- that's Birkhoff-s theorem.

I also assume you are not "the planet earth", but are an observer on its surface (in a space suit; see above).

I also assume that by the "Schw. interior metric" you mean Schwarzschild's solution for the interior of a spherically symmetric uniform density star or planet. I cannot make sense of your statements any other way. This is (bizarrely) consistent with your claim to "be" the planet earth (though its density is by no means uniform, but ignore that).


There will be no net gravitational field acting on me
due to the stars due to the symmetry of the problem. When the event
horizon forms, nothing happens locally to change this sphercal
symmetry.

Right. You can have no local knowledge of when the horizon expands around you and the earth.


We have already determined that the Schwarzschild interior
solution has an SO(3) symmetry, but it does not have the central
spherical symmetry.

There is no "Schw. interior solution" anywhere in this problem, except inside the volume of the earth itself. You are on its surface, and the metric there is the Schw. _exterior_ solution, corresponding to the mass of the earth -- again, that is Birkhoff's theorem (given my caveats above). The metric in this region _HAS_ spherical symmetry.


Therefore the interior solution cannot coincide
with the metric around me when the horizon forms.

You remain confused -- the interior solution is irrelevant. See above.


Rather, I will
remain quite oblivious of the formation of the horizon, and whatever
the metric was before horizon formation will not change greatly after
it.

True -- the metric where you are standing on the earth's surface will not change at all until those stars reach you, at which time you will be obliterated. Ditto for any point of the earth's interior (since you seem hung up on the interior solution, and that's the only place it applies).


But once the stars cross the event horizon, these are not spherical
shells in the usual sense, but 'temporal' spheres, which hit the
singularity at a given time relative to passing through the event
horizon.

You are confused. This is NOT the Schw. manifold, but because it is spherically symmetric (by my caveats above), Birkhoff's theorem applies to its vacuum regions.

Consider the period of time between:
a) the stars cross the spatial locus where their horizon will
ultimately be.
and
b) the stars engulf you on the surface of the earth.
During this time the metric at the surface of the earth is unchanged by the impending doom of the incoming stars.

After the incoming stars engulf you and the earth, the spatial region where you used to be will be inside the horizon and will be vacuum, so Birkhoff's theorem again applies to this region. This applies all the way down to the center of the earth.

[Hmmm. I'm glossing over the difficulty of identifying
spatial regions before and after.... But then, afterwards
the entire manifold is vacuum, so it's OK.]


There is no way that this can be identified with the
spherically symmetric region in which I sit.

I repeat: you are confused. See above.


Although the assumption
that the stars continue to fall 'radially inwards' upon crossing the
horizon seems natural,

It is well known, and is not an assumption, it is the conclusion of a _calculation_. Oppenheimer and Snyder computed this for a quite similar case (no earth at the center) in the 1930's.


it seems to me to be invalid as the
Schwarzschild interior metric does not describe a space with spatial
spherically symmetry.

You REALLY should start doubting your ability to GUESS about such things -- you have been wrong EVERY time you try. Moreover, the interior Schw. region _IS_ spherically symmetric. Manifestly so -- the metric has an SO(3) symmetry (or in the usual Schw. coordinates of this region, the metric components are all independent of \theta and \phi).


> [... further elaboration of the above errors]


Tom Roberts
.

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