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Warp wrote:
> OTOH GR has resisted the test of time rather well.
BTW, the article I read recently that talked about this was discussing a
gravity-wave detector where the amount of random noise they were getting was
pretty close to the amount of noise predicted by this whole
holographic-universe kind of thing.
It was a popular-science article, but google turns up
http://arxiv.org/abs/gr-qc/9906003
that pretty much sums it up, given what we're talking about. :-)
Basically, since the "hologram" of the universe is at 37 billion lightyears,
or whatever it is, but space is much bigger than that surface area, you get
some "blurring" at the quantum level that makes the "pixels" big enough to
actually measure, closer to 10^-15 than 10^-35 or some such. Way over my
head, but interesting.
--
Darren New, San Diego CA, USA (PST)
Why is there a chainsaw in DOOM?
There aren't any trees on Mars.
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Warp <war### [at] tagpovrayorg> wrote:
> I have heard this, I have a very hard time understanding how it would be
> possible.
>
> And after you cross the event horizon... Who knows. But you certainly
> would *not* see the universe in any normal way, if at all. You probably
> wouldn't be able to measure anything of the universe at all (because,
> if the GR equations are right, *all* geodesics inside the event horizon
> point directly towards the center).
I saw a simulation of this a while back. Your field of view increases as you
pass through the horizon, until 360+ degrees shrink to a circle directly above
you. As you go further and further into the black hole, the whole universe
above you shrinks to a point. Strictly speaking, though, you can observe
anything in the universe that you could before since you are still traveling
slower than the speed of light. The event horizon is only the point at which
light cones tip past vertical so that even light moving directly outward can no
longer make it. There really isn't anything special to the person passing
through it; it's just the point of no return.
> Some people seem to think that there could be objects "floating" around
> inside the event horizon, and that someone could be there and see nothing
> unusual. However, if all geodesics are pointing directly towards the
> center.
They don't really have to point directly toward the center, do they?. It's only
that NONE of them point outward. If you're inside the event horizon, you still
have an infinite number of choices, but 'out' is not one of them.
- Ricky
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Darren New <dne### [at] sanrrcom> wrote:
> > From a timescale point of view, an observer falling to a black hole
> > would not see any change in timescales for himself. How he sees the
> > rest of the universe, however, is another story.
> I think that's what I was saying. I don't understand why anyone inside *or*
> outside would think it takes infinite time to fall in.
Things get weird on the extremes with relativity.
For example a photon might travel billions of years from a star and
collide on Earth. However, from the point of view of the photon itself,
what is the distance it has travelled and how long it took for it to
travel it?
> > For example if the falling object had a clock (and let's
> > forget those tidal forces ripping it apart) and the external observer
> > would look at this clock with a telescope, the clock would slow down
> > indefinitely. Well, until no photons arrive anymore from the object and
> > it could not be observed anymore.
> But inside, he'd still be falling. And outside, he'd still be falling, yes?
> He'd just look stopped yet still falling to those outside.
Not stopped. Asymptotically slowing down.
(But as I said in my original post, I don't think it would be possible to
observe the object indefinitely from the outside. It just stops sending
photons at some point.)
> > Does QM say that space is not continuous?
> I believe that's correct. Or that gravity at least is not continuous
> (because nothing is continuous).
Unless I'm mistaken, the "graviton" is one of those particles invented
by quantum mechanics for the simple reason that "because everything seems
to consist of particles, then there must exist a particle for every
phenomenon we can see, including gravity".
In other words, it feels more a question of principle than a result of
measurements or predictions of some theory. Gravitons must exist. Why?
Because.
I really can't understand why quantum mechanists have such a problem
in thinking about gravity as timespace geometry rather than as the exchange
of some exotic particles which cannot be detected. I don't understand
what would be the problem. Why do we *need* a gravity particle? What
would it explain that gravity-as-timespace-geometry doesn't?
Timespace curvature *is* something which can be (and has been) measured.
A graviton isn't.
> I don't really understand it, but my
> layman understanding is that GR's math only works if "forces" are
> continuous, and QM says that "forces" are not.
How is, for example, momentum quantized? What is the particle which is
the smallest unit of momentum? Or kinetic energy? Or inertia?
Must everything be quantized?
> > Assuming Hawking radiation indeed exists...
> If it doesn't, you've still lost the information. :-)
If you add two pieces of information together, has some information been
lost?
> >> With a big enough black hole, you'll never know when you cross the event
> >> horizon.
> >
> > I have heard this, I have a very hard time understanding how it would be
> > possible.
> >
> > Space is *really* warped near the event horizon.
> Not necessarily. It's almost really warped just outside, and it's just a
> little bit more warped inside. So while the slope is steep, the second
> derivative isn't. At least, that's how I understand it. :-)
Well, given that the event horizon is, by definition, the area of space
were the curvature of space is so large that even photons have no way out,
and that this curvature of space increases as you approach the event
horizon, I have hard time believing you wouldn't notice anything when you
approach and cross the event horizon.
When you start approaching the event horizon light starts visibly bending
because of the curvature of space. Any light which traverses closer to the
event horizon gets "sqeezed", and thus everything seems to flatten.
Wikipedia has a good image about this: Assuming GR is right, what a
"bare" black hole (ie. one with no accretion disc or other stuff orbiting
around) would look like against a starry background, from a relatively
close distance:
http://en.wikipedia.org/wiki/File:Black_Hole_Milkyway.jpg
I suppose the size of the black hole only affects the rate at which
the change happens as you get closer to the event horizon, rather than
the strength of the visual effect.
(Btw, that image might not be totally correct, as it seems to lack
red-shifting of the light closer to the event horizon...)
--
- Warp
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triple_r <nomail@nomail> wrote:
> I saw a simulation of this a while back. Your field of view increases as you
> pass through the horizon, until 360+ degrees shrink to a circle directly above
> you. As you go further and further into the black hole, the whole universe
> above you shrinks to a point.
I believe you are talking about this:
http://www.spacetimetravel.org/expeditionsl/expeditionsl.html
However, if you read the explanation carefully, none of the images
simulates what happens *inside* the event horizon. All of them are from
the outside, closer and closer to the event horizon. The last image is
"taken" from a distance of 1.005rs, where 'rs' is the Schwarzschild radius,
ie. the radius at which the event horizon is located.
(OTOH, I still think those images don't take red/blue-shifting into
account, so they might not be completely accurate in their coloration,
even if they are in their geometry.)
> > Some people seem to think that there could be objects "floating" around
> > inside the event horizon, and that someone could be there and see nothing
> > unusual. However, if all geodesics are pointing directly towards the
> > center.
> They don't really have to point directly toward the center, do they?.
I'm a complete layman on the subject and the GR equations go *well* over
my head. The Schwarzschild solution to those equations goes even more
beyond my comprehension, so everything I know about the subject is from
reading things put into layman terms.
I have understood that all geodesics point directly towards the center
(at least in the Schwarzschild solution, ie. a non-rotating non-charged
black hole), but I may well have understood wrongly.
I'm sure, though, that the equations predict that it's physically
impossible to avoid advancing towards the singularity. Even trying to
stay still, without moving, is impossible. (I suppose the only possibility
to avoid this would be some weird quantum mechanical effect.)
--
- Warp
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Warp wrote:
> (Btw, that image might not be totally correct, as it seems to lack
> red-shifting of the light closer to the event horizon...)
There's clearly redshift a ways out from the horizon. I believe what you see
between the red ring and the horizon is light that has been "slingshotted"
around the hole - that is, you're seeing stars from behind your back. That
light would blue-shift going in, and red-shift coming out again, to balance
out.
--
Darren New, San Diego CA, USA (PST)
Why is there a chainsaw in DOOM?
There aren't any trees on Mars.
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Warp wrote:
> Things get weird on the extremes with relativity.
Sure.
>> But inside, he'd still be falling. And outside, he'd still be falling, yes?
>> He'd just look stopped yet still falling to those outside.
>
> Not stopped. Asymptotically slowing down.
So, the steeper the gravity gradient, the slower he falls? That doesn't make
sense? He actually *slows down* as he falls? Nothing ever goes thru the
event horizon?
His *clock* slows down. His actual progress thru space doesn't. Yes?
> Unless I'm mistaken, the "graviton" is one of those particles invented
> by quantum mechanics for the simple reason that "because everything seems
> to consist of particles, then there must exist a particle for every
> phenomenon we can see, including gravity".
I wouldn't be surprised if that's the case. I never heard of anyone even
proposing properties of a graviton, let alone finding evidence.
> I really can't understand why quantum mechanists have such a problem
> in thinking about gravity as timespace geometry rather than as the exchange
> of some exotic particles which cannot be detected.
I don't know they have a problem with it. That was my comment.
>> I don't really understand it, but my
>> layman understanding is that GR's math only works if "forces" are
>> continuous, and QM says that "forces" are not.
>
> How is, for example, momentum quantized?
The uncertainty principle comes into play. I'm not sure I understand all of
it. Possibly something to do with renormalization in the QM equations: if
you don't assume there's a lower limit on scale, you get infinities in all
the QM equations (as divergent summation series). It doesn't seem to matter
much *what* the lower limit is, which is what Feynman got the Nobel for
proving. But Plank length has always been bandied around in the stuff I've
read. (Plank length is something like ten orders of magnitude smaller than a
proton or some such, so ...)
> Must everything be quantized?
I don't understand it well enough. I'm just saying what I've heard said - GR
assumes space that's continuously differentiable everywhere outside an event
horizon, and QM requires it not to be that way.
>>> Assuming Hawking radiation indeed exists...
>
>> If it doesn't, you've still lost the information. :-)
>
> If you add two pieces of information together, has some information been
> lost?
If hawking radiation doesn't exist, then you have no way of seeing what's
inside the event horizon, and hence the information on the spin of the
particles in there is lost?
> I suppose the size of the black hole only affects the rate at which
> the change happens as you get closer to the event horizon, rather than
> the strength of the visual effect.
I think that's what I was trying to express. With a sufficiently large black
hole, you might not know you crossed the event horizon because the rate of
change of curvature is so slow. Of course you know you're in trouble.
There's just no "bump" as you cross over. And if you're in free-fall
anyway, what might you notice?
--
Darren New, San Diego CA, USA (PST)
Why is there a chainsaw in DOOM?
There aren't any trees on Mars.
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Darren New wrote:
> Possibly something to do with renormalization in the QM
To be a little more clear:
The way you calculate if particle P goes from place X to place Y is to
calculate first the probability of oging there directly, then the
probability of going there by going X->A->Y, then X->A->B->Y, then
X->A->B->C->Y, for all possible A, B, C, etc.
In QED (quantum mechanics of photons and electrons), the -> bit multiplies
by a rather small number, so the infinite summation comes to a total pretty
fast, so you don't have to do too many before you get numbers that match
experiements.
In QCD (quantum mechanics of nuclear reactions), the -> bit multiplies by a
much larger number, so you have to trace out the possibilities eight or ten
deep to get decent numbers.
In either case, you really have to add up an infinite number of
possibilities (including where B is a spacelike distance from both X and Y)
to get the 100% correct answer, but the formula diverges. If you stop at any
given distance, then divide the result by that length (for some meaning of
the word "divide"), you get the answer that agrees with experiment. (That
step is "renormalization".)
So it *seems* like there's a lower limit to the number of times you can
divide a space up and come up with the right numbers in QM. And the
mathematics of GR are (I'm led to believe) don't work without continuous
differentiability. So ... there ya go.
--
Darren New, San Diego CA, USA (PST)
Why is there a chainsaw in DOOM?
There aren't any trees on Mars.
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Warp <war### [at] tagpovrayorg> wrote:
> I'm a complete layman on the subject and the GR equations go *well* over
> my head. The Schwarzschild solution to those equations goes even more
> beyond my comprehension, so everything I know about the subject is from
> reading things put into layman terms.
I think we all have to agree on this point. I pretty much stopped reading these
layman's explanations on topics like relativity and quantum mechanics, figuring
that all the effort would be better spent learning the real thing. I picked up
a textbook on the subject a couple weeks ago, but haven't gotten far enough to
say anything more intelligent. Here's an interesting set of notes, though:
http://preposterousuniverse.com/grnotes/ It has a pretty neat explanation of
the Lorentz transformations, but it takes a bit more effort to dig into GR.
- Ricky
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Darren New <dne### [at] sanrrcom> wrote:
> > Not stopped. Asymptotically slowing down.
> So, the steeper the gravity gradient, the slower he falls? That doesn't make
> sense? He actually *slows down* as he falls? Nothing ever goes thru the
> event horizon?
No, it looks *from the outside*, from the point of view of an external
observer, like his time is slowing down.
Remember that speed and time is always relative to who is observing.
> > Unless I'm mistaken, the "graviton" is one of those particles invented
> > by quantum mechanics for the simple reason that "because everything seems
> > to consist of particles, then there must exist a particle for every
> > phenomenon we can see, including gravity".
> I wouldn't be surprised if that's the case. I never heard of anyone even
> proposing properties of a graviton, let alone finding evidence.
Actually there are a few things which can be said of gravitons. In other
words, "if gravitons existed, they must have these properties". For example,
they would have spin 2.
(And no, I haven't the faintest idea what "spin" means. I just read
wikipedia. :P )
> >>> Assuming Hawking radiation indeed exists...
> >
> >> If it doesn't, you've still lost the information. :-)
> >
> > If you add two pieces of information together, has some information been
> > lost?
> If hawking radiation doesn't exist, then you have no way of seeing what's
> inside the event horizon, and hence the information on the spin of the
> particles in there is lost?
Why is it lost? Just because you can't see it doesn't mean it doesn't
exist.
If something moves to the other side of the cosmological horizon (from
our perspective), is the information "lost"?
> > I suppose the size of the black hole only affects the rate at which
> > the change happens as you get closer to the event horizon, rather than
> > the strength of the visual effect.
> I think that's what I was trying to express. With a sufficiently large black
> hole, you might not know you crossed the event horizon because the rate of
> change of curvature is so slow. Of course you know you're in trouble.
> There's just no "bump" as you cross over. And if you're in free-fall
> anyway, what might you notice?
I think that you would notice everything being warped.
--
- Warp
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Warp <war### [at] tagpovrayorg> wrote:
> However, if you read the explanation carefully, none of the images
> simulates what happens *inside* the event horizon. All of them are from
> the outside, closer and closer to the event horizon. The last image is
> "taken" from a distance of 1.005rs, where 'rs' is the Schwarzschild radius,
> ie. the radius at which the event horizon is located.
Btw, if it's not clear why the universe seems to shrink into a small
circle above as you get closer to the event horizon, I drew a picture.
It demonstrates the reverse light paths which arrive at the observer
(ie. like reverse raytracing), which shows why most of the field of
view is covered by the event horizon when you are close enough to it:
http://warp.povusers.org/snaps/EventHorizon.png
--
- Warp
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