Re: A misconception conquered!

[David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>]

Regarding the quote from a post of Skip's (that provoked Leigh to comment
on a misconception related to a confusion of a gravitational potential
well with a curvature of space that is sometimes pictured as curving into
an extra dimension that embeds it):

> > Since light can't slow down, as it climbs out of the gravity well, it loses
> > energy by decreasing frequency. Another way of saying this is that space is
> > stretched near the event horizon.

Even though the first statement here seems to be a non sequitur, I don't
want to pursue that matter here.  Instead I want to comment on the 2nd
statement above.  I can't think of any way in which one could
legitimately describe the space around a black hole's event horizon as
"stretched" relative to the space farther away.  If anything, one could
say that the space near the horizon is radially *compressed*
gravitationally, with the local degree of compression decreasing with
distance from it.  The reason we can think of space as compressed around
an isolated gravitating mass is because if we imagine enclosing the a
region of space including the mass with a closed surface (say, a sphere
for, instance) and carefully measure the surface area of this bounding
surface and then (based on the ordinary geometry of flat space) predict
some interior properties of the space so enclosed one finds that there is
just more space crammed inside this sphere than we would have expected
based on our flat-space geometric experience.

For instance, suppose we take the properly measured area A of the
sphere and calculate a predicted radius R of this sphere according to
R = sqrt(A/(4*[pi])) and then actually descend into the sphere with a
meter stick and measure the actual radius R_actual.  It will be found
that R_actual > R.  The actual direct distance across its interior
(i.e. 2*R_actual) is greater than we would have predicted from flat
space geometry.  Similarly, suppose we predict the interior volume V of
the sphere according to the flat space formula:

V = (4*[pi]/3)*R^3 = (4*[pi]/3)*(A/(4*[pi]))^(3/2) .

[Ben suppose we carefully investigate the interior of the sphere and
carefully measure its actual proper volume (again using our meter stick
or some other means that carefully integrates the interior volume using
a proper tiny volume element).  It would be found that this measured
volume V_actual is *larger* than our predicted volume V.

BTW, in order to make the measurements described above we need to make
sure that our gravitating mass does not possess an event horizon because
if it did we would not be able to go in and measure the distance across
nor measure its volume, nor be able to return with our measurements.

So, we see that somehow the curvature of space around the gravitating
mass successfully crams more spatial volume inside our sphere than would
be the case if the space was uncurved.  Since more space (i.e. volume,
radius, diameter, etc.) fits inside the sphere than would be the case if
there were no curvature, it is quite legitimate to say that the space
has been "compressed".  If we allowed a picture that had this extra
space "hump up" into a 4th dimension (where our 3-d space is embedded as
a 3-d "surface" in it) the "compression" could be relieved.  But if
we insist on using coordinates (e.g. spherical coordinates) suited to a
flat 3-space the curvature of the space manifests itself as a
"compression" of the space inside.  Such distortions are similar to
what happens by projecting the globe onto a flat plane to make a map of
the world.

A diagram such as Leigh referred to in MTW tries to illustrate this
inclusion and compression of more interior space by making a
2-dimensional analogy drawn as showing the curvature making the surface
"dip below" or "hump up" into a 3rd dimension that embeds the original
2-d spatial "surface".

Such a curvature is similar to how the 2-d surface of a (Northern)
hemisphere is curved relative to a flat (equatorial) plane.  If you draw
a circle about the equatorial base of the hemisphere and divide by 2*[pi]
you would predict the radius of the hemispherical surface.  But if you
actually measure the actual radius along the curved surface starting at
the North Pole and follow a meridian of longitude to the equator you
would discover that the actual measured radius from the north pole to the
equator is 1/4 of the equatorial circumference not 1/(2*[pi]) of it.  It
is this extra interior spatial surface inside the hemisphere's equatorial
circle that makes the hemisphere "hump up" into a third dimension above
the flat equatorial plane.

If you imagined smashing the hemisphere flat into the equatorial plane
the extra area would have to be compressed in order to be accomodated
inside the equatorial circle.  When one observed the hemisphere from the
from the outer edge of its equator and didn't account its curvature one
would be led to think that the amount of spatial surface area comprising
it would be just the area of the flat equatorial circle.  But in fact
the real proper area of the hemisphere is precisely *twice* as large.
Somehow (i.e. by humping into the 3rd dimension) the curvature of the
hemispherical surface allows twice as much interior area inside the
equatorial circle as would be the case for a flat circle case.  If we
insisted on suppressing the 3rd dimension and required that the the
hemisphere be described using coordinates suited for a smashed-flat
version of it, we would have to include the effect of this extra interior
region inside the equator as a radial "compression" of the spatial
surface inside.

David Bowman
David_Bowman@georgetowncollege.edu