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clipka wrote:
> So from this, and hearing that it would actually take an eternity (as seen from
> an outside observer) to *reach* the event horizon,
This still doesn't make sense to me, I fear.
You give one particle that is constantly emitting photons in all directions
a push due south towards the singularity. You have an observer due north of
the singularity looking south. The observer due north will continue to see
photons coming off the particle indefinitely?
> it sounds logical to me that
> in fact the event horizon *is* the singularity...
But it's not.
> If you'd take that trip, you'd experience it as being torn to
> pieces by tidal forces,
Tidal forces are a difference in gravity gradient between head and feet.
It's not an absolute number. A sufficiently small pebble would experience
very little tidal effect regardless of how close to the primary it orbits. :-)
--
Darren New, San Diego CA, USA (PST)
"Ouch ouch ouch!"
"What's wrong? Noodles too hot?"
"No, I have Chopstick Tunnel Syndrome."
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Warp <war### [at] tag povrayorg> wrote:
> (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.)
Strictly speaking I guess it doesn't - the probability of it sending photons
just decreases asymptotically to zero...
> (Btw, that image might not be totally correct, as it seems to lack
> red-shifting of the light closer to the event horizon...)
No red-shifting necessary, because before it gets out of the gravitational well
- experiencing redshift - it gets into it first, experiencing a corresponding
ammount of blueshift.
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Darren New wrote:
>> it sounds logical to me that
>> in fact the event horizon *is* the singularity...
>
> But it's not.
By which I mean:
"singularity" has a specific meaning in mathematics, and the event horizon
doesn't meet that definition, while the "center" of the black hole does.
Take the graph of tan(x). At pi/2, there's a singularity. At 0, it just goes
thru a 45-degree slope (and a point of inflection). That's kind of the
difference.
--
Darren New, San Diego CA, USA (PST)
"Ouch ouch ouch!"
"What's wrong? Noodles too hot?"
"No, I have Chopstick Tunnel Syndrome."
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Darren New escreveu:
>> If you'd take that trip, you'd experience it as being torn to
>> pieces by tidal forces,
>
> Tidal forces are a difference in gravity gradient between head and feet.
> It's not an absolute number. A sufficiently small pebble would
> experience very little tidal effect regardless of how close to the
> primary it orbits. :-)
hah! Thanks for the tip! I'll roll up into fetal position when nearing
a black hole... :)
Anyway, trying to understand something that until now has not been
observed and is only but a prediction from a math model that may be
incomplete -- which is how many scientists see the singularities of GR
equations -- seems a bit off. Some scientists are suggesting that
blackholes are in fact stars made out of dark matter and energy.
http://www.newscientist.com/article/mg18925423.600-three-cosmic-enigmas-one-audacious-answer.html
whoa, head spins...
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Darren New <dne### [at] sanrrcom> wrote:
> "Spin" is a fascinating concept related to symmetry. A "spin 1/2" particle
> means if you turn it 180 degrees, it looks the same. So a "two of diamonds"
> playing card would be "spin 1/2" - it looks the same right side up as upside
> down. (Discounting, of course, that it's 3D, so has a back that's different
> from the front.)
IIRC it's just the other way round - so you have to turn quite a lot of
subatomic particles around twice to get them look the same again...?
> It's lost because it cannot influence other particles in the local area. Of
> course, that's because the particles outside the horizon are a spacelike
> distance from the particles inside the horizon, so maybe you're right. Maybe
> it's the equivalent of being outside the cosmological horizon.
Does there exist such a thing? It doesn't have this "one way ticket" thing to it
like black holes do.
AIUI the cosmological horizon is just the distance we can see *now*. Everything
beyond will become visible to us over time. Theoretically speaking.
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clipka wrote:
> IIRC it's just the other way round - so you have to turn quite a lot of
> subatomic particles around twice to get them look the same again...?
It's possible I got that wrong. Indeed, it's possible I don't know anything
at all about the subject that I think I do. :-) Were this an actual
scientific forum, I'd have shut up a long time ago.
>> It's lost because it cannot influence other particles in the local area. Of
>> course, that's because the particles outside the horizon are a spacelike
>> distance from the particles inside the horizon, so maybe you're right. Maybe
>> it's the equivalent of being outside the cosmological horizon.
>
> Does there exist such a thing? It doesn't have this "one way ticket" thing to it
> like black holes do.
I'm not sure what you speak of.
> AIUI the cosmological horizon is just the distance we can see *now*. Everything
> beyond will become visible to us over time. Theoretically speaking.
Oh, the event horizon. No, I think it's caused by space itself expanding,
due to the big bang sort of thing. If it all collapses again, that's a
different question.
--
Darren New, San Diego CA, USA (PST)
"Ouch ouch ouch!"
"What's wrong? Noodles too hot?"
"No, I have Chopstick Tunnel Syndrome."
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nemesis wrote:
> hah! Thanks for the tip! I'll roll up into fetal position when nearing
> a black hole... :)
You're not kidding. Larry Niven actually did a sci-fi story about that.
> equations -- seems a bit off. Some scientists are suggesting that
> blackholes are in fact stars made out of dark matter and energy.
Dude. Scroll up to the first post in this conversation. :-)
--
Darren New, San Diego CA, USA (PST)
"Ouch ouch ouch!"
"What's wrong? Noodles too hot?"
"No, I have Chopstick Tunnel Syndrome."
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Darren New escreveu:
> Dude. Scroll up to the first post in this conversation. :-)
oh, crap! So that's where this tab got open... :P
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Darren New <dne### [at] sanrrcom> wrote:
> If you take the Earth and condense it down to a black hole, then stop a
> particle in orbit 100 miles above where the surface used to be, the particle
> will accelerate towards the black hole, right? But to avoid crossing the
> event horizon, it must logically decelerate, as seen from the POV of someone
> in orbit (say). You wouldn't stop at the original altitude of the surface,
> but you would stop before you get to the event horizon, so in this case,
> gravity must be causing you to decelerate. Yes?
Depends on definition if speed, but as it is usually specified based on the
observer's timeframe: Yes. Gravitation near a black hole is so strong that -
from an observer's POV - through time delation and the warping of space it
causes "that poor old sod over there" to slow to a halt.
From that poor old sod's perspective, gravitation near a black hole is so strong
that it keeps speeding him up - his feet much stronger than his head (if he
plunges in feet first) - until he is ripped to isolated subatomic particles.
Paradoxically, while this happens, I think that he will at the same time see his
feet slow down to a standstill, just as an outside observer would. So maybe he's
not torn apart after all..!?!
(In a similarly paradoxical fashion, the particles at his head can still easily
influence those at his feet - but the influence of the feet particles on the
head particles will drastically diminish (remember them sending out less
photons?)
This actually shows that gravitation is a quite strange beast in the zoo of
forces: The influence of gravitation does *not* diminish while closing in on a
black hole.
So either gravitation cannot be a "normal" QM-style force (else the object would
send out fewer and fewer gravitons as its own timeframe slows down with respect
to an outside observer) - or the object's e=mc^2 increases towards infinity as
it comes close to the event horizon, but why should it? After all, we have seen
that from an outsider's point of view the object actually slows to a halt.
Hum... sounds to me like bad news for the people at the LHC - and speaks for the
"gravitation = spacetime geometry" view of GR.
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nemesis <nam### [at] gmailcom> wrote:
> Darren New escreveu:
> > nemesis wrote:
> >> kind of after image that seems to ever approach the event horizon
> >> without ever touching it, but in reality the crossing of the EH
> >> already took place and nothing of what happens in the inside is seen...
> >
> > Hmmmm.... I'll have to think on it.
>
> I think it has something to do with the last photons coming out of the
> object entering the EH are severely slowed down by the massive gravity
> and only reach you after much more time than normal has passed.
Yeah - interestingly, the distance between a point near the EH and a point far
away is shorter when moving towards the EH than when moving away from it on the
very same route...
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