Re: Two Schwarzschild radii

carlip-nospam@xxxxxxxxxxxxxxxxxxx <carlip-nospam@xxxxxxxxxxxxxxxxxxx> wrote in message
h0ooga$meo$2@xxxxxxxxxxxxxxxxxxx
mluttgens@xxxxxxxxx wrote:

You are claiming that an object (or a light ray) at the
Schwarzschild radius of a black hole, on the other hand, can
never move outward at all.

I am stating that this is a completely unambiguous prediction
of general relativity. Whether it is physically the correct
description of a black hole is a matter for experiment and
observation, but so far the observations match the predictions
well; see http://arxiv.org/abs/0903.1105, for example.

Your claim comes from the following GR formula, where the
Schwarzschild radius is given by Rs = 2GM/c^2, a formula that
can be written 2GM/c^2Rs = 1.

No, it certainly doesn't!

For Marcel Luttgens is does - even if you stand on your
head. Even if you stand on *his* head for that matter.

It comes from an analysis of the behavior
of null geodesics in the Schwarzschild geometry. To do this analysis
you have to write the Schwarzschild solution in coordinates that
include the horizon (Schwarzschild coordinates won't do, but you
can use Kruskal-Szekeres or Painleve-Gullstrand coordinates, for
instance). Then you have to write down the equations for the null
geodesics and solve them.

You seem to have missed the point of my last post.

Luttgens barely understands high school algebra. He hasn't
even reached basic calculus.

Let me try one
more time.

You are a patient man, so you are in good company, as you
are talking to a patient troll :-)

You *cannot* understand general relativity by merely
pulling an equation or two out of context or by reading simple
popularizations.

Luttgens can do just that.

Horrible as it is, you have to actually learn to do
the math.

:-)


Yhen, a light of frequency fe emitted at Rs will have a frequency
fr = 0 / sqrt(1-RS/(Rs+H)) = 0 at a distance Rs+H from the center
of the object considered as a black hole. This *could* be interpreted
as the impossibility for a light ray to leave Rs.

Bur another hallmarks of crank physics is to accept any conclusion
from formulae derived from a pet theory, and this is the case of GR.

Indeed, when a light of frequency fr emitted at a distance Rs+H
from the center of a blackhole, the received frequency at the
distance Rs is, according to GR, infinite, according to GR equation
fr = sqrt(1-Rs(Rs+H)) / 0.

This is the frequency *as measured by an observer at rest.*

Or, to an infinite frequency corresponds an infinite energy.

When you drive toward a streetlight, the light that reaches you is
blue shifted. Do you think that your driving is somehow affecting
the intrinsic properties of the photons? I hope not. It's true that
the photons strike you with more energy than they would if you
were at rest, but that's because of your motion, not the properties
of the light.

Marcel has a well known quibble about this. Not sure if he'll
bring it up here.


Frequency is a *relative* concept; it depends both on the light and
on the observer's motion. An observer who is trying to remain at
rest outside a black hole must expend energy to do so -- she has to
fire her rockets to keep from falling toward the black hole. Relative
to her, an incoming photon is blue shifted and has increased energy,
just as a photon from a streetlight is blue shifted and has increased
energy relative to a car driving toward the source. A freely falling
observer, on the other hand, would see no blue shift. (You can check
this yourself, but -- sorry, again -- you'll have to learn the math.)

A thousand times he has been pointed to the freely available
first chapters of Taylor and Wheeler's little book. He hasn't
even looked at them. There's limits, McClaurin series and
integrals in there. Marcel doesn't do those.


An observer remaining at rest at a black hole horizon would have to
expend an infinite amount of energy and would be moving at the speed
of light relative to a freely falling observer. This observer would,
indeed, see an infinite blue shift. But that's a property of the observer,
not of the light. Again, a freely falling observer would see no blue
shift or energy increase.

Indeed, if there *were* a "surface" just outside the Schwarzschild
radius, rather than an event horizon, then relative to that surface,
infalling matter and light would, certainly have a very large energy.
(There's nothing very strange about this -- if you drop a ball from
a tall building, it hits the ground with a lot of energy.) Stuff hitting
such a surface would emit a huge amount of energy, which we would
see. The preprint I cited above puts very strong observational limits
on the existence of such a surface for the massive object at the center
of our galaxy. Matter falling in *doesn't* give off the energy it would
if it were hitting a surface; rather, the energy simply vanishes, as it
would if there were an event horizon.

Steve Carlip

Nice post - alas, a pearl for a swine :-)

Dirk Vdm
.

Relevant Pages

  • Black holes and red shift
    ... Obviously an outside observer cannot see inside the event horizon, ... Our universe was formed inside a black hole by an energy bundle ...
    (sci.physics)
  • Re: Two Schwarzschild radii
    ... Schwarzschild radius of a black hole, on the other hand, can ... This is the frequency *as measured by an observer at rest.* ... rest outside a black hole must expend energy to do so -- she has to ...
    (sci.physics.relativity)
  • Re: Two Schwarzschild radii
    ... Schwarzschild radius of a black hole, on the other hand, can ... This is the frequency *as measured by an observer at rest.* ... rest outside a black hole must expend energy to do so -- she has to ...
    (sci.physics.relativity)
  • Re: a relative question
    ... When calculating the total energy of a moving object, ... relative velocity to the observer. ... slower processes in accordance with the special theory of relativity. ... object at a steady rate, the acceleration ...
    (sci.physics.relativity)
  • Re: Two Schwarzschild radii
    ... Schwarzschild radius of a black hole, on the other hand, can ... of general relativity. ... This is the frequency *as measured by an observer at rest.* ... rest outside a black hole must expend energy to do so -- she has to ...
    (sci.physics.relativity)