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From: JanPB on 7 Apr 2006 20:52
I keep following my posts (first sign of crankhood):
JanPB wrote:
>
> Now look inside r<2M. Not only isn't this portion spacelike anymore, it
> is also funnel-shaped which means it doesn't match the "Einstein-Rosen"
> portion outside the horizon without a crease (the funnel surface
> reaches r=2M from the inside sloping "off-vertical" while the upper
> half of the Einstein-Rosen-shaped surface reaches r=2M from the outside
> "vertically").

I took a closer look at this funnel and it is an oddly-shaped one
(insert your own joke here). In books like d'Inverno (p. 208) it's
pictured like an (apparently) infinitely deep and sort of exponential
in shape. I don't know where they get this shape from, perhaps I made a
mistake in my calculation.

First of all, I don't know how d'Inverno can say "the curved geometry
of this two-dimensional surface is best understood if it is embedded in
the flat geometry of a three-dimensional Euclidean manifold". But this
surface (r<2M) has an indefinite metric so it can't be embedded in
*any* Euclidean space. Then he says "this is depicted in [the usual
funnel picture]".

The only reasonable way to show the curvature of the region r<2M I
could think of was to embed it in a 2+1 Minkowski space and then find
out which surface of revolution in there inherits the metric equal to
Schwarzschild's.

This is easily done if one imagines the Minkowski space as an xyz-space
with the metric -dz^2+dx^2+dy^2. Any surface of revolution obtained by
revolving the graph of (yet unknown) z(r) around the z-axis can be
parametrised in the usual fashion:

f(r,phi) = (r cos(phi), r sin(phi), z(r))

The standard Minkowski metric then induces on this surface the pullback
metric:

-f*(dz)^2 + f*(dx)^2 + f*(dy)^2 = (simplify) =

= (1 - (dz/dr)^2) dr^2 + r^2 dphi^2

Comparing it to Schwarzschild (t = 0, theta = pi/2, r<2M):

ds^2 = 1/(1 - 2M/r) dr^2 + r^2 dphi^2

we get:

1 - (dz/dr)^2 = 1/(1 - 2M/r)

....which is easily solved:

z(r) = -sqrt(8M(2M - r))

This is the curve that's revolved around the z-axis and it does look
like a funnel but it's of finite "depth" (-4M, corresponding to r=0).
It "looks" like a positively curved surface but due to the ambient
Minkowski metric it's negatively curved, as required (-M/r^3).

So where does d'Inverno & Co. get their infinite Euclidean funnels
from?
(Yes, I know. But I need a different joke now.)

--
Jan Bielawski

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