Re: black hole is an oximoron
- From: "Randy M. Dumse" <rmd@xxxxxxxxxxxxx>
- Date: Sat, 7 May 2005 21:29:51 -0500
"solar plexus" <r384923hr823h28eh@xxxxxx> wrote in message
news:1115511107.957730.327120@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> i understand that the space is extremely
> bendend by black holes, then the gravity
> has no effect over for light, mostly because
> photons have no mass
Yes, space near black holes is extremely curved. However, according to
GR, all forms of mass-energy gravitate. So both mass and light is
effected by gravity.
In fact, it's worse than you might imagine, because light is effected
twice as much by gravity than is the equivalent amount of mass.
> so photons follow their geodesic path (bended)
> but gravity has no power over for them
Light follows geodesic paths, but so does mass. A geodesic path simply
means one free from accelerations, a natural falling motion. Light
follows null geodesics. As I said, these curve twice as much as the
geodesics massive objects can follow. Isn't that interesting/puzzling?
Things without mass move at the speed of light, and fall twice as much
as things with mass do. Things with mass move less than that speed of
light and fall half as much as light.
So you see, it is very wrong to think that gravity has no power over
light and other massless particles.
> why couldnt photons be reflected from
> the surface of such a high density / condenset
> object?
Light will indeed be at least partially reflected from a neutron star
surface, the most dense form of matter we know, short of gravitational
collapse. To the best of our knowledge a neutron star would be still 2
to 10 times bigger (less dense) than matter that would collapse.. So you
could see a neutron star.
But once the matter gets below its Schwarzschild radius, it is so
compact nothing can overcome the force of gravity pulling it in and full
collapse is thought to follow..
Let's imagine for a moment, the mass actually could resist infall, any
further collapse, just when the star was exactly at the black hole
horizon (by current theory this can't happen, but just to make the
point, we'll imagine it for a moment). Now let's say a photon falls on
the star, and is reflected. (That only makes the impossible situation
for matter trying to stay on the horizon worse, because when the light
reflects, it imparts more moment to the mass, kicking it below the
horizon!) Where is the reflected light going to go? Out? No. It would be
trapped there on the horizon. Unless something tipped it over, it would
just hang there. But it wouldn't come out. And if it was tipped over, it
could only fall in. In fact, that's what current theory holds happens
with all the mass and energy that gets to the horizon. The most it could
possibly do is be turned to light, reflected, and hang on the horizon
_or_ fall on in.
> is it because the space cant escape ?
No, not the space, space is considered empty, but the mass-energy, it
can't escape.
In fact, they call this solution which describes the field outside a
black hole, "the Schwarzschild vacuum equation" cause empty space is
considered a vacuum, and this equation describes the exterior field of
any symmetrical, spherical (non charged non spinning) gravitating body.
Once you get inside the star, they use other equations, because you're
not in a vacuum anymore. But if you're talking outside, in "space", the
vacuum equation is it.
--
Randy M. Dumse
Caution: Objects in mirror are more confused than they appear.
.
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