Can light orbit a black hole?

Mar 25, 2014 by Fraser Cain, Universe Today
The curvature of space due to gravity.

Since black holes are the most powerful gravitational spots in the entire Universe, can they distort light so much that it actually goes into orbit? And what would it look like if you could survive and follow light in this trip around a black hole?

I had this great question in from a viewer. Is it possible for light to orbit a black hole?

Consider this , first explained by Newton. Imagine you had cannon that could shoot a cannonball far away. The ball would fly downrange and then crash into the dirt. If you shot the cannonball harder it would fly further before slamming into the ground. And if you could shoot the cannonball hard enough and ignore air resistance – it would travel all the way around the Earth. The cannonball would be in orbit. It's falling towards the Earth, but the curvature of the Earth means that it's constantly falling just over the horizon.

This works not only with cannonballs, astronauts and satellites, but with light too. This was one of the big discoveries that Einstein made about the nature of gravity. Gravity isn't an attractive force between masses, it's actually a distortion of spacetime. When light falls into the gravity well of a massive object, it bends to follow the curvature of spacetime.

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Distant galaxies, the Sun, and even our own Earth will cause light to be deflected from its path by their distortion of spacetime. But it's the incredible gravity of a black hole that can tie spacetime in knots. And yes, there is a region around a black hole where even photons are forced to travel in an orbit. In fact, this region is known as the "photon sphere".

From far enough away, act like any massive object. If you replaced the Sun with a black hole of the same mass, our Earth would continue to orbit in exactly the same way. But as you get closer and closer to the black hole, the orbiting object needs to go faster and faster as it whips around the massive object. The photon sphere is the final stable you can have around a black hole. And only light, moving at, well, light speed, can actually exist at this altitude.

Imagine you could exist right at the photon sphere of a black hole. Which you can't, so don't try. You could point your flashlight in one direction, and see the light behind you, after it had fully orbited the black hole. You would also be bathed in the radiation of all the photons captured in this region. The visible light might be pretty, but the x-ray and gamma radiation would cook you like an oven.

Artist impression of a black hole. Credit: ESO/L. Calçada

Below the photon sphere you would see only darkness. Down there is the , 's point of no return. And up above you'd see the Universe distorted by the massive gravity of the black hole. You'd see the entire sky in your view, even stars that would be normally obscured by the black hole, as they wrap around its . It would be an awesome and deadly place to be, but it'd sure beat falling down below the event horizon.

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big_hairy_jimbo
5 / 5 (4) Mar 25, 2014
Wouldn't the photon sphere be regularly disrupted by in falling matter as the light would excite matter? I guess the photon sphere evaporates when the black hole is gorging, but then rebuilds when the black hole is quiet. Perhaps this is another energy input for black holes with jets. Perhaps the photon sphere is absorbed as energy/momentum and helps to eject matter at high velocity as a jet. On another note, I would have thought the point where photons go into a stable orbit, would be the event horizon. Also, wouldn't a black hole be constantly bathed in photons from the surrounding galaxy and universe? If so, then wouldn't a blackhole be constantly growing as it ingests photons? Black holes would be constant photon parasites. I'm confused this morning!!

Anyway, just thinking aloud.
Jizby
Mar 25, 2014
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what_the_hell
1 / 5 (1) Mar 25, 2014
Could the photons form an exotic state of matter, like a soup? Also, is there a theoretical limit to how much energy a black hole can absorb before it overcomes the gravity field? Or, could black holes be pumping energy from within our universe to just outside the perimeter of our universe, and could that explain why the expansion of our universe isn't uniform? One could anticipate that expansion would be the slowest nearer clusters of black holes if that were the case.
Jizby
Mar 25, 2014
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what_the_hell
2.5 / 5 (2) Mar 25, 2014
how much energy a black hole can absorb before it overcomes the gravity field
?
Would there be stratified, exotic elements within a black hole? Ones that would be unstable otherwise? Are neutrinos orbiting black holes? Would these orbits looks like the rings of Saturn? If so, would be be able to detect this as matter entered those rings, disturbing them to break free of their orbits? I'm loony; I know.
RobertKarlStonjek
5 / 5 (4) Mar 25, 2014
They have failed to consider time dilation in their description of the view from the event horizon. The distant stars would be blue shifted so much as to appear super-hot and events would appear to occur instantly. Indeed, the interval from the big bang to the present would appear to take less than a millisecond.

This is just the reverse of the view of the event horizon from space ie that events occur ever slower and more red shifted as you view objects ever closer to the event horizon until they are frozen.
Returners
3.4 / 5 (5) Mar 25, 2014
RobertKarlStonjek:

Prevailing theories aside, Light actually should be red shifted by gravity regardless of the direction it is moving relative to the massive object.

You can prove this using a piece of paper and a few simple diagrams within about 5 minutes or so.

Red shift inwards:
If a beam of light is directed to a black hole, and you are near the EH looking up. You will see red shift. Imagine several points along a line leading out from the BH and pick any group of points. Points closer to the BH are accelerated faster than points farther away. Thus as the beam of light moves past each point, the "leading" photons are attracted slightly faster than the "trailing" photons, stretching them out in space. this means the time between the first and last photon striking your eye will be greater than the time between the first and last photon leaving their original source. This is a form of Red Shift.

Red Shift Outwards:
It works equal and opposite.
vlaaing peerd
5 / 5 (2) Mar 26, 2014
I thought light - unlike matter - doesn't fall in by the gravity pull of the black hole but purely follows the path of bent space-time (also into the black hole), which essentially should be a straight line.

As a photon doesn't have mass, how could it be affected by gravity?

Did I completely misunderstand this or is there something missing in the article?
antialias_physorg
5 / 5 (7) Mar 26, 2014
As a photon doesn't have mass, how could it be affected by gravity?

It's the space that is curved. Light follows a geodesic
http://en.wikiped...Geodesic
where a gedoesic is defined as the least time path between any two points given a constant speed. If the geodesic reconnects with itself you have an orbit.

On an unrelated note: Does the photonsphere exist for rotating black holes? If the space is not completely uniformely curved (inversely proportional to r^2) than any orbit that is not exactly equatorial should not be stable for a photon. So no photonsphere but a photonring (if that)?
there a theoretical limit to how much energy a black hole can absorb before it overcomes the gravity field?

No. There's no theory yet that puts an upper limit on that.
Are neutrinos orbiting black holes?

Yes they should - just like photons. Possibly a tiny bit inside the photonsphere/ring.
Jizby
Mar 26, 2014
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Jizby
Mar 26, 2014
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GSwift7
5 / 5 (5) Mar 26, 2014
than any orbit that is not exactly equatorial should not be stable for a photon. So no photonsphere but a photonring (if that)?


Unless the black hole is isolated enough to make outside gravity trivial, it's center of gravity will shift around in space. It should wobble back and forth if the black hole is orbiting with another massive object, and the event horizon and photosphere should bulge out towards any nearby or infalling mass as well.

I'd say that the photosphere is not likely to hold any one photon for long, but due to the abundance of photons constantly streaming in from all angles simultaneously, it'll constantly be replennished with a stream of photons entering and eventually falling into the black hole.

As far as anything escaping the photosphere once it reaches the point where a stable orbit is possible, I'd say only quantum particles or photons set free by shifting of the center of gravity or changing shape of the EH. Theory is likely different than reality tho.
GSwift7
5 / 5 (5) Mar 26, 2014
Wouldn't the photon sphere be regularly disrupted by in falling matter as the light would excite matter?


There's a thing called the roche limit, where anything held together by gravity will be torn apart by tidal forces if it gets too close to a larger massive object. With a black hole, this effect is magnified many times over, especially when you get near the event horizon or photosphere. The effect of tidal force is so strong near the photosphere that not only do objects get torn apart, the matter they are made of gets torn apart. As an atom nears the event horizon, the pull of gravity on the near side is stronger than the pull of gravity on the far side of the atom. The difference is so great that it will overcome subatomic bonds and rip the atom into subatomic fundamental particles. That's why the article above says "And only light, moving at, well, light speed, can actually exist at this altitude". No matter will be possible at the photosphere. According to theory.
Gawad
5 / 5 (4) Mar 26, 2014
As an atom nears the event horizon, the pull of gravity on the near side is stronger than the pull of gravity on the far side of the atom. The difference is so great that it will overcome subatomic bonds and rip the atom into subatomic fundamental particles.


Ah, I just want to throw in the caveat that as far as happening at or near the event horizon this "spaghettification" applies to stellar mass class black holes, but not to the much larger ones in the centers of galaxies. For the latter it happens much closer to the singularity.
Jizby
Mar 26, 2014
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antialias_physorg
5 / 5 (5) Mar 26, 2014
As an atom nears the event horizon, the pull of gravity on the near side is stronger than the pull of gravity on the far side of the atom.

I was under the impression that the place where the gradient gets that severe was somewhere inside the event horizon.
IIRC for supermassive black holes you could survive falling through the event horizon (not much longer, but at that point tidal forces aren't going to kill you yet)
GSwift7
4.3 / 5 (6) Mar 26, 2014
this "spaghettification" applies to stellar mass class black holes, but not to the much larger ones in the centers of galaxies. For the latter it happens much closer to the singularity


IIRC for supermassive black holes you could survive falling through the event horizon


good points!!!!

I've heard that too. And I've never really thought about this, but those are conflicting!

That makes me wonder.

speculation alert!!! for those who read this, beware

As for surviving the PoNR of a SMBH, all bets are off once you cross the EH, since GR breaks after that point. For example, if you're a photon moving at your usual speed and you cross, the gravity will accelerate you, right? What does that mean? You can't really go faster, so what gives? There couldn't be a photosphere inside the EH because the escape velocity inside the EH is beyond the speed of light. So, when we're talking about the photosphere we MUST be talking about a smaller black hole. pass the smell test?
GSwift7
4 / 5 (4) Mar 26, 2014
continued:

as for surviving the crossing of any event horizon, I've heard that proposed, but I don't think it makes sense.

The instant you cross, it takes velocity above the speed of light to move anything outwards, so how do your subatomic bonds hold together when there's no signal coming from immidiately next to you?

So yes, the speghettification doeson't happen till farther in, but that's when the tidal force actually wraps you around the entire circumferenece a billion times in an instant. I think the matter you're made of would dissociate long before that. It wouldn't be tidal force, but the difference in time dilation/distance distortion/mass distortin would also compete for your bonds, wouldn't they? No matter where the spighettification line would be, there's nothing but black when you look down from inside the EH. You wouldn't get any subatomic bonds coming from below, so poof go your atoms.

There shouldn't be any electron orbits inside the EH, or any waves.
GSwift7
4 / 5 (4) Mar 26, 2014
I know, TLDR, but here's one more thought (REALLY speculative now, btw)

So, if you accept that there's no way anything can move outward from the center once you cross the EH, then that leaves only one logical alternative. Conservation of momentum means that you can't just stop the oscilation of a wave-like particle like a photon, and it can't oscillate outward, so mustn't it just make a bee-line to the center? In that case, it will reach the center and then INSTANTLY stop, right? It can't go even a little bit past the center, so it must reach the center and stop, dead cold. But that breaks quantum certainty, right? You would know the exact location and speed of anything at the center. That's messed up. I need a beer and a Goody's powder. lol.

speaking of oscillating photons, if it travels at the speed of light, but it also moves back and forth, then how fast is it really moving? Hmmm.
big_hairy_jimbo
not rated yet Mar 27, 2014
Hmmm, I think I need to clarify a few things.
First of all, Gravity influences all objects that possess momentum. Momentum for normal matter is dependant on mass and velocity, hence the idea that gravity only influences objects with mass. However light also has momentum in terms of DeBroglie Wavelength. Hence light is also affected by gravity. The alternative view is that light follows the curvature of spacetime, and that momentum curves spacetime.

Secondly, light does NOT slow down, even in the presence of a blackhole. What does happen is that light is stretched or contracted. So imagine a light beam coming from the singularity towards the EH at light speed. The light is stretched, ie redshifted and loses energy, but then curves back around and is then blue shifted back towards the singularity. Energy conserved!!
big_hairy_jimbo
not rated yet Mar 27, 2014
I do see a few problems with my red shifting of light though travelling from singularity to EH.
Sure it makes sense that the light will be red shifted. But, red shifting means the light wave grows longer and longer. If it gets near the Event Horizon (which it should by definition), then wouldn't the quantum tunneling probability increase as the wavelength increases. So, perhaps there is a mechanism by which light COULD escape the EH, BUT the photon is SO red shifted that the wave is essentially flattened. Hence the photon can ONLY exist within the EH. Now ignoring what happens AT the singularity, the photon should race around inside the EH travelling back and fourth through the singularity, experiencing red shifting, then blue, then red then blue etc. But all this is meaningless, as physics breaksdown inside the EH towards the singularity (according to maths anyway).
big_hairy_jimbo
not rated yet Mar 27, 2014
finally consider this;
A matter particle is falling toward the blackhole and seems doomed. A photon is also racing toward the BH but at a glancing angle and will thus escape capture, but will have its course deviated, much like a comet around the sun. Now the photon is being BLUE shifted as it approaches, and thus accumulating energy. The photon "strikes" the particle and imparts energy/momentum to the particle, and the photon loses energy continues and is red shifted. The particle can now escape due to extra momentum. The photon however continues to be red shifted on its outward trajectory. This mechanism is robbing the gravitational potential energy of the BH and giving it to the escaping particle. Remember all this occurs OUTISDE the EH for argument sake.
Seems to replicate an idea of black hole jets, and also very red shifted photons.

No doubt I've stuffed up my physics in all three posts above, so please feel free to interrogate away!! Remember they are pure speculations!!
antialias_physorg
5 / 5 (4) Mar 27, 2014
if you're a photon moving at your usual speed and you cross, the gravity will accelerate you, right?

Not really. It will just blueshift the photon as it falls in.
There couldn't be a photosphere inside the EH

For a photonsphere you need closed geodesics.But no geodesic inside the BH can be closed (as they all point towards the center and none can go any further from it)

So, when we're talking about the photosphere we MUST be talking about a smaller black hole

The conditons (given a fully static black hole) should be satisfiable for small and black holes. Cosider that the EH is the point where a photon pointing AWAY cannot escape. The photonsphere is where the photons are in orbit but still tangential to the orbiting surface (so that is always further out than the EH)
antialias_physorg
5 / 5 (2) Mar 27, 2014
In that case, it will reach the center and then INSTANTLY stop, right?

That really depends on what happens there. Current assumption is that space holds at incredibly high energy densities - and that isn't nevcessariyl so. There may be a new unification happening there (call it a "space-energy unification"). Or you may get to a point where space inside a black hole stretches as fast or faster than light. If the latter happens a photon could fall in essentially forever without reaching the center.

But as you say: speculating what happens inside a black hole (particularly close to the center) is sort of problematic...as we can't go take a look.
Jizby
Mar 27, 2014
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Jizby
Mar 27, 2014
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GSwift7
3.7 / 5 (3) Mar 27, 2014
The conditons (given a fully static black hole) should be satisfiable for small and (large) black holes


Doh, sorry, that's right. I don't know what I was thinking.

Jizby:

What in the world (not pun intended) was all that supposed to mean? I can follow some alternative rants, at least to the point of understanding what they're trying to say, but that one you just posted is like a an impressionist painting translated into science-ish words. It's like there's a ghost of an image of an intelligent thought, which leaves you with the impression that there's something there, but you can't really pick any details out of it, and when you look at it closely, it's just a bunch of abstract parts thrown together in a way that suggests something non-abstract.
Jizby
Mar 27, 2014
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Jizby
Mar 27, 2014
This comment has been removed by a moderator.
Jizby
Mar 27, 2014
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GSwift7
1 / 5 (1) Mar 27, 2014
Not really. It will just blueshift the photon as it falls in


I'm not arguing this, just asking if you'd like to comment on a thought that keeps popping into my mind.

In order to have frequency, don't you need a wave? Can there be a wave inside the EH?
Gawad
5 / 5 (3) Mar 27, 2014
The instant you cross, it takes velocity above the speed of light to move anything outwards, so how do your subatomic bonds hold together when there's no signal coming from immidiately next to you?
You're still thinking of this as SR; it's GR. You're in free fall, not using reaction mass to accelerate through space. Notwithstanding tidal forces (which isn't a problem if the BH is big enough compared to you), all the particles that make you up are in freefall together including the ones that mediate forces. So even if there's no more going "back out" in absolute terms, in relative terms they can still move "less down" or "more down" with respect to adjacent particles and mediate between them. A better analogy is to think of space-time as flowing toward the singularity, and you as a swimmer splashing around in the water. The water can flow FTL as you go over the falls (the EH) while you still swim around in–to you–agitated but otherwise still water. Well, until the spag thing...
antialias_physorg
5 / 5 (3) Mar 27, 2014
In order to have frequency, don't you need a wave? Can there be a wave inside the EH?

Sure. Why not? It's just the oscilaltion of the electric and magnetic field (which are at right angles to the line of propagation of the photon).
So this happens
http://en.wikiped...redshift
(well, the blueshift part. Note that an observer falling inside the EH could only see blueshifted photons from behind. Any kind of photons from further in cannot get as far out as the observer, so he sees no redshifted ones. There actually shouldn't be any redshifted ones at all - no matter which direction they are emitted from a body falling in, as all geodesics point to areas of higher gravity)
Also note: Falling through the EH does not mean that one is moving at the speed of light at that point. You can cross it as slowly as you like given good drives.
antialias_physorg
5 / 5 (2) Mar 27, 2014
Quoting myself here because here's a quirky thought:
all geodesics [originating from any point within the EH] point to areas of higher gravity

Which basically means that from your point of view you, once you cross the EH, you always appear to be sitting in a gravity well. BUT there is no force pushing you back to the that point when you move from it (as would be the case if you were really at the center of a gravity well)...weird.
Gawad
5 / 5 (2) Mar 27, 2014
Falling through the EH does not mean that one is moving at the speed of light at that point. You can cross it as slowly as you like given good drives.
An excellent point! In my earlier comment to G I was imagining only free fall from standstill with no acceleration applied though space, but it's definitely worth mentioning that you can have scenarios where the "swimmer" is fighting the ST flow at up to c and they experience all the effects of SR.
Jizby
Mar 27, 2014
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GSwift7
3.7 / 5 (3) Mar 27, 2014
BUT there is no force pushing you back to the that point when you move from it (as would be the case if you were really at the center of a gravity well)...weird


And no matter which way you try to move, even though it may appear that you are moving in some other direction, as all your accelerometers and such would tell you, you'd actually be moving towards the center, right?
Gawad
5 / 5 (2) Mar 27, 2014
The instant you cross, it takes velocity above the speed of light to move anything outwards, so how do your subatomic bonds hold together when there's no signal coming from immidiately next to you?
You're still thinking of this as SR; it's GR. You're in free fall, not using reaction mass to accelerate through space.[Blah, blah, blah myself]
On rereading that, I don't think I was being very clear. I guess what I was trying to point out, G, is that what you're saying makes sense in a context where acceleration is the result of particle interactions in space(time)–SR. IOW, ye 'ol rocketship. But what you're saying doesn't apply when the acceleration is the result of free fall in curved ST, i.e.,GR, and which isn't limited by c. (Sigh, I'm not really sure that's any more clear....)
GSwift7
3.7 / 5 (3) Mar 27, 2014
Falling through the EH does not mean that one is moving at the speed of light at that point. You can cross it as slowly as you like given good drives


You could stop at exactly the point before you cross, but the escape velocity at the EH is C, so the escape velocity immediately beyond the EH has got to be greater than C. Since we believe all forces, even the fundamental subatomic forces, propagate at C, as soon as you cross the EH, your warp drive engine will disintegrate. Though I believe that you could show mathematically that the bonds which hold molecules together would fail before you actually reach the EH, even with a SMBH. Weakly held molecules would fail first, as you start to red-shift apart.

G, is that what you're saying makes sense in a context where acceleration is the result of particle interactions in space


Immediately beyond the EH, each and every atom you're made of cannot detect any of your atoms farther towards the center. Inertia is meaningless.
Gawad
5 / 5 (2) Mar 27, 2014
BUT there is no force pushing you back to the that point when you move from it (as would be the case if you were really at the center of a gravity well)...weird


And no matter which way you try to move, even though it may appear that you are moving in some other direction, as all your accelerometers and such would tell you, you'd actually be moving towards the center, right?
Yes, inevitably. To go back to the waterfall analogy: if you think of space-time as flowing faster than c towards the singularity beyond the EH, it no longer matters in what direction you try to move in space or how fast (your own limit being c in the flow as it were), "DOWN" is in your future; there are no more paths "up" or "out".
Gawad
5 / 5 (2) Mar 27, 2014
Since we believe all forces, even the fundamental subatomic forces, propagate at C, as soon as you cross the EH, your warp drive engine will disintegrate
Nope. If the gravitational gradient is SMOOTH enough at that point (i.e., no significant TIDAL forces) your SR limited drives won't even feel it and neither will you.

An note that I say SR limited drives, because "warp drives" is just *undefined*.
antialias_physorg
5 / 5 (2) Mar 27, 2014
you'd actually be moving towards the center, right?

Yes, the only thing you could mitigate would be the acceleration at which you move towards it (though no matter which path you choose: you' always accelerate up to (close to) the speed of light...until relativistic mass increase sets in, you get ripped apart into radiation, or you hit singularity - whichever comes first)

I don't know if atoms would be stable once you go in. Electrons would probably be. Atomic nuclei might not. The force that holds them together would not be able to mediate (nuclear force). Then again: neither would the force that pushes them apart (electrostatic force)
big_hairy_jimbo
not rated yet Mar 27, 2014
keep in mind that a proton may disintegrate via the following. IF the BH attempts to rip apart a proton, then the energy this requires to overcome the strong force, soon mounts up to the energy of creating another quark. Hence no quark is found alone, the minimum state is a doublet (meson) instead of the protons triplet. Therefore ripping a proton apart should create two mesons at least. But here's the weird bit. This process would rob gravitational energy from the BH, however this energy is then transformed into mass as it creates a new quark. Would this new mass then create enough gravitational potential to counter what was lost when the proton was ripped apart? If so a BH is indeed a self sustaining matter blender. At best the inside would be a quark-gluon-plasma, OR as was recently postulated on PhysOrg, a Plank Star <-- my current favourite hypothesis. Makes sense as extreme objects in space demonstrate transitions through exclusion boundaries as is the case with a Neutron Star.
RealityCheck
1 / 5 (1) Mar 27, 2014
Hi GSwift7. :)
@Jizby: What in the world was all that supposed to mean?
I think Jizby is pointing out that even at the centre of Earth the gravitational FIELD ENERGY is 'balanced' but still 'present and active', except the balanced directional effects produce no 'gravitational acceleration' as such, trying to 'tear you apart from all directions' expand your matter body.

Your E-M forces keep you 'intact', but they are under 'static strain' radially directed outwards at Earth center. Much like at 'L' points in space between earth and moon where bodies experience 'micro g accelerations' (because of minor 'perturbations' in the otherwise 'balanced' tugs from Earth and Moon at that "l" location). You actually have to use a little rocket thrust to leave that "L" point!

See what Jizby is indicating now? While your accelerating motion may be zero, gravitational ENERGY flux at such positions still 'there', only 'balanced' in effect. The position/gravity 'fluxes' at such points. :)
Noumenon
3.7 / 5 (3) Mar 27, 2014
a gedoesic is defined as the least time path between any two points given a constant speed


Not important, but I thought I would point out that a geodesic maximizes proper time.
bicubic
5 / 5 (3) Mar 27, 2014
If a photon can orbit a black hole, is it possible for a photon to slingshot around a black hole? If so, can we see earth as it was thousands of years ago ( round trip time to nearest black hole ) with a ("very very" ) good telescope? Are there objects we see in the sky that appear to us in a "slingshot" version that maybe are really in the opposite direction than we think they are?
antialias_physorg
5 / 5 (3) Mar 28, 2014
If a photon can orbit a black hole, is it possible for a photon to slingshot around a black hole? If so, can we see earth as it was thousands of years ago

Yes. If you find a (or a number of) gravitational source(s) so that light emitted from Earth would get bent enough to fall back on Earth eventually -you could. However the image would be very dim (as the chances for any one photon to make that exact trip would be infinitesimal. So you might get such a photon every few (thousand/million) years...but not enough to make a coherent 'live' picture.

If we had a tiny black hole orbiting the Earth (or the other way around) we could prpbably get a good image from "a few minutes back" (depending on how far out we are from it).
GSwift7
3.7 / 5 (3) Mar 28, 2014
As for seeing an image from photons that lensed all the way around a black hole, it would be like looking though a tiny pin-hole fish-eye lense, and the fisheye would be an entire 360 degree view. So, everything would apear very small and distorted, and it would be so overwhelmed by noise that you'd never be able to pick it out, unless there's some way to sort out the photons you want from all the others.
GSwift7
3.7 / 5 (3) Mar 28, 2014
The force that holds them together would not be able to mediate (nuclear force). Then again: neither would the force that pushes them apart (electrostatic force)


Yeah, so as things start to come apart, they should just continue wherever their momentum was carying them at the time? You might not need any repulsive forces. It might all just randomize, obeying the laws of entropy? Maybe?

I can't see how anything other than fundamental particles would be permitted beyond the EH. The gradualness of the transition would be irrelevant. No information can travel 'upstream' inside the EH, so each fundamental particle is completely isolated from anything that's even one planck length farther in than itself. And due to quantum randomness, no two particles could remain exactly side-by-side for more than an instant. Every bit of everything that 'can' come apart will come apart at the instant you cross any event horizon. Hypothetically.
Gawad
4.8 / 5 (5) Mar 28, 2014
I can't see how anything other than fundamental particles would be permitted beyond the EH.

Sorry guys, but you're just wrong about this. It's widely accepted that at least in GR for a sufficiently massive/large black hole NOTHING SPECIAL happens to the observer when they cross the EH. In fact, they never notice they even cross it (to them they never do) and they can even signal someone ahead or behind them in local space:

http://math.ucr.e..._in.html
http://en.wikiped..._horizon
http://sciencelin...key=1842
Benni
1 / 5 (1) Mar 28, 2014
But here's the weird bit. This process would rob gravitational energy from the BH

How? Are you suggesting that a mass/energy transformation within the BH causes a gain or loss of gravity? Gravity is conserved in any mass/energy transformation

however this energy is then transformed into mass as it creates a new quark. Would this new mass then create enough gravitational potential to counter what was lost when the proton was ripped apart?


No new gravity magically appears when energy is transformed to mass, it was already present in the energy field before transformation. When atoms are split they become lighter because they've given up energy, they also give up gravity in a linear relationship corresponding to the quantity of energy that was transformed. Energy is pure "relativistic mass" with its own gravity field which it gives back upon transformation back to mass.
TheGhostofOtto1923
1 / 5 (2) Mar 29, 2014
@gswift
speculation alert!!! for those who read this, beware
-You should add this caveat to all your posts. Also

'Caution - this poster rarely researches his notions before posting!'
Jizby
Mar 29, 2014
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Benni
1 / 5 (1) Mar 30, 2014
No new gravity magically appears when energy is transformed to mass, it was already present in the energy field before transformation
It is correct, for example photons mediate the (portion of) matter of star during supernova explosions. When the mass travels across space in form of photons, it just means that the gravitational waves are present inside of each photon. The mass m, of a photon is related to its energy E or wavelength λ of the photon through the Einstein relation E = ��Žc/λ = mc^2. The fact that the photons can mediate mass has been proven experimentally already: for example the gamma ray photons http://www.ncbi.n...20438075 the mass (of atoms) of some isotopes by their excitation, which can be subsequently detected with mass spectrometer, when these isotopes are deexcited and radiate their energy into gamma ray photons, then the mass of atoms is returning back.


Dead on the money accurate- now when will you dispense with foolish AWT ?
Jizby
Mar 30, 2014
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Osiris1
2.3 / 5 (3) Mar 30, 2014
The 'photon sphere' of orbital lightspeed object will orbit at the distance for which lightspeed is the escape velocity === the event horizon. This could be a sharply defined place such that the top half of the photons would be visible and the bottom halves hidden, as if that makes any sense. Given the 'particulate nature of photonic quasi-matter, we are forced by Einsteinites to accept this horse manure. Any 'photon sphere' would disrupt any matter inbound and disintegrate it like being hit by a laser of immense power and power density on extremely narrow focus....the shere of orbit, one photon thick.. The photonic density would be amazing, and any collisions would result in velocity loss or trajectory changes such that the loss from this 'sphere' would be constant, an almost uniformly down to the 'universal one-ness of the BH. Of course this would be a test of photonic mass. Mass under forces like gravity lose energy.
adam_russell_9615
not rated yet Mar 31, 2014
I would think the photon sphere would be exactly the same radius as the event horizon. If light is in orbit at the photon sphere then anywhere above it light can escape. The way orbits work the greater the radius the lower the required velocity to maintain orbit. So if C is required at the PS, then less than C is required outside of it. So light escapes.
gale_langseth
not rated yet Mar 31, 2014
Yet another recent -- in humantime, not spacetime :) -- story states that "Hawking posted a blog on Jan. 22, 2014, stating that event horizons – the invisible boundaries of black holes – do not exist." (http://phys.org/n...tml#jCp) So where does this leave us, or where do we leave a black hole?
Zachia
not rated yet Mar 31, 2014
Photon sphere has a double radius than the event horizon of black hole, so that it shouldn't interfere with our understanding of black hole as such very much.
antialias_physorg
4.5 / 5 (2) Mar 31, 2014
I would think the photon sphere would be exactly the same radius as the event horizon.
Not really: Photons at the photonsphere are moving tangentially to the BH. Any that wish to escape have to move at slightly off tangentially (for whatever reason. This can be something as minute as the gravitational attraction of infalling matter still outside the photonsphere that weakens the gravity well at the photonsphere).
Photons at (just outide) the EH still capable to get out have to be moving radially away.

From the difference in direction you can see that it must be somewhat easier for photons to escape the photonsphere than those emitted just outside the EH, because the paths possible for the former are less constrained than for the latter. Therefore the photonsphere must be somewhat outside the EH.
bluehigh
1 / 5 (1) Mar 31, 2014
Not that Mr Swift needs or even wants my backup and at the risk of once again playing in the gutter ...

Re:
@gswift
-You should add this caveat to all your posts. Also
'Caution - this poster rarely researches his notions before posting!'

@OttoTard
- YOU should add to all your posts ...
'Caution this poster rarely puts his brain in gear before opening his mouth.'

OttoTard , maybe you can wear that pointy cap and sit in the corner for not contributing anything worthwhile.

For my part ... Space time is not modified geometrically by gravity. Only by mass. No, not by momentum or velocity .. Only Mass interacts gravitationally.

Tell me why .. I don't like Mondays.
bluehigh
1 / 5 (1) Mar 31, 2014
Photons have Mass therefore they exhibit gravitational interaction.

A photon at rest is not physically real.
11791
Mar 31, 2014
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11791
Mar 31, 2014
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GSwift7
1 / 5 (1) Mar 31, 2014
gawad:

Sorry guys, but you're just wrong about this. It's widely accepted, for a sufficiently massive black hole NOTHING SPECIAL happens to the observer when they cross the EH


Yes, I've seen that many times in print, and you've got to understand that is a layperson's simplification. That only applies to the tidal force, and its ability to physically distort you. It does not consider the whole picture of what's going on.

Forgive the math, but that's only the slope of the graph representing the magnitude of the gravitational gradient at a point in space, X. You must also consider the area under the graph from zero to X, which represents the absolute magnitude of the acceleration due to gravity at point X, regardless of the gradient (slope).

Once u cross any EH, the absolute acceleration due to gravity is greater than C/second squared. That is the definition of an EH. No forces can be communicated upstream once you cross an EH. No forces = no atoms. You would not survive this
TheGhostofOtto1923
1 / 5 (1) Mar 31, 2014
Yes, I've seen that many times in print, and you've got to understand that is a layperson's simplification
Well that can't be true because you're a layperson and you don't understand it.

"Observers crossing a black hole event horizon can calculate the moment they have crossed it, but will not actually see or feel anything special happen at that moment."
Once u cross any EH, the absolute acceleration due to gravity is greater than C/second squared. That is the definition of an EH. No forces can be communicated upstream once you cross an EH. No forces = no atoms. You would not survive this
Perhaps before you embarrass yourself any further you ought to read the explanation that real scientists at the univ of Colorado and the NSF have so graciously provided.
http://jila.color...chw.html

-But I suspect you may not understand this either. Oh well.
Gawad
5 / 5 (3) Mar 31, 2014
gawad:

It's widely accepted, for a sufficiently massive black hole NOTHING SPECIAL happens to the observer when they cross the EH
Yes, I've seen that many times in print, and you've got to understand that is a layperson's simplification. That only applies to the tidal force, and its ability to physically distort you. It does not consider the whole picture of what's going on.
I understand it's a simplified picture, and it doesn't account for quantum effects. But this *isn't* just a layperson's explanation, and it doesn't involve anything actually going "upstream" in an absolute sense, only in a *relative* sense (no pun intended) in "local space" even beyond the EH.

There's a reason I included Baez's page in my links. The guy is no slouch with math and he's not presenting a misleading picture just for laypeople. If you were going to be atomized (or worse) at the EH, why not just say so? McIrvin even cites Weinberg and Wheeler's hard-core tomes as references.
Gawad
5 / 5 (3) Mar 31, 2014
gawad:
It's widely accepted, for a sufficiently massive black hole NOTHING SPECIAL happens to the observer when they cross the EH
Yes, I've seen that many times in print, and you've got to understand that is a layperson's simplification. That only applies to the tidal force, and its ability to physically distort you. It does not consider the whole picture of what's going on.
I understand it's simplified (and it doesn't account for quantum effects). But this isn't *just* a layperson's explanation, and it doesn't involve anything actually going "upstream" in an absolute sense, only in a *relative* sense (no pun intended) in "local space" even beyond the EH. I included Baez's page in my links, as the guy is no slouch with math and he's not presenting a misleading picture just for laypeople. If you were going to be atomized (or worse) at the EH, why not just say so? McIrvin even cites Weinberg and Wheeler's hard-core tomes as references. If you can point me to something...
Gawad
5 / 5 (3) Mar 31, 2014
Sorry about the double post; the page normally reloads on clicking "Submit".
osnova
Mar 31, 2014
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11791
Mar 31, 2014
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baudrunner
1 / 5 (1) Mar 31, 2014
Nobody who has posted here seems to understand light because they have all gone in a direction that deviates from reality. Light isn't really anything, it's the excited outer orbital shells of atoms in the propagating medium oscillating at visible optical wavelengths until they resonate with the cones and rods in the retinas of the eyes, which convert those oscillations to bio-electrical impulses that travel via the optic nerve to the occipital lobe in the brain. What we call a photon is the energy lost or gained during the emission and absorption within one atom. Therefore, if visible wavelengths are propagated by the oscillations of the electron orbitals affecting adjacent particles in the transmitting medium, then no light is visible when the electrons have been stripped from the nuclei, but we do detect X-rays. The effect of bending light around large bodies in space is actually the result of refraction. The density of particles are greater the closer you get to a large body.
Gawad
5 / 5 (3) Mar 31, 2014
It's widely accepted, for a sufficiently massive black hole NOTHING SPECIAL happens to the observer when they cross the EH
Until (s)he doesn't hit the firewall or worm hole. The information paradox clearly says, that the information (about falling object) cannot be preserved bellow event horizon. The only question is, where exactly it will be destroyed.

Well, no, that's NOT what the info paradox says. Look it up. There wasn't even a paradox until Hawking radiation. Until then the info WAS considered safely locked up in the BH. THEN comes Hawking and his QM process evaporates the BH in a manner completely unrelated to the BH CONTENTS, making the information DISAPEAR. Which it CAN'T (just like it can't be truly duplicated, a.k.a. "no cloning"). That's the paradox. So then Susskind & Co. come up with a "complimentarity" solution where the info is only in one place at a time: inside if you are or outside, if you are. And the whole deal with the firewall is probably (cont.)
Gawad
5 / 5 (3) Mar 31, 2014
(cont.) just an indication that the "complimentarity" solution doesn't really work as it's based on wrong, err, incompatible assumptions. When Polchinski took another look at the "solution" of complimentarity for BH evaporation through Hawking radiation, assuming holography works (i.e. a description of what happens in a volume with gravity corresponds to an equivalent description on a surface without gravity) they discover that halfway through the evaporation process the BH becomes...FULL! I.e., there's no longer enough surface area to correspond to the description of the interior, so nothing else can get in (hence the "atomizing firewall") and it no longer evaporates.

And NOTE: this so called firewall is actually a *quantum effect* (o.k. in a GR context). So, not the what was originally at issue here.

And most important: the most likely thing this "firewall" is saying is just that the original "solution" was wrong, because no one really expects things to ever go "splat" at the EH.
11791
Mar 31, 2014
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Gawad
5 / 5 (2) Mar 31, 2014
And when I write "splat" I mean "SPLAT", because if Polchinski's right, nothing goes in anymore, not whole, not atomized, not even sub-atomized. SPLAT. And a fair bunch of physicists are up in arms because while they can't find fault (yet anyway) with his paper, they consider the implications are ridiculous. Not to mention that it would mean QM *completely* breaks GR. It's even apparently lead to impressive screaming matches and "howls of flummoxation" at conferences, though apparently no clothes rending yet.
Gawad
5 / 5 (3) Mar 31, 2014
Hawking's old theory had no firewall attached to a big black hole as opposed to quantum sized black holes that emit radiation emanating from virtual particle pair production. What causes the firewall around a big black hole in his new theory? His old theory had very limited virtual pair production and low amounts of hawking radiation released from bigger black holes.

It's not "his new theory". He didn't come up with the firewall problem. If anything Hawking doesn't believe there really is a "firewall". Hell, everybody get this: NOBODY DOES, not even really Polchinski who came up with it! Polchinski's more like, "if what you've proposed so far is correct, THAT'S where it leads, and that's crazy, so find something better." Which is what Hawking and a whole lots more folks are trying to do. There might even be more baseball encyclopaedias at stake.
11791
Mar 31, 2014
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Gawad
5 / 5 (2) Mar 31, 2014
what would make all matter splatter at a black hole event horizon or whatever is limiting entry into its interior? I don't understand how the firewall is supposed to work. Can you explain it?
Well, exactly! But no, sorry, not really. For one, that's WAY above my pay grade :) But mostly, because most physicists are saying the firewall is crazy. It makes it sounds like a brick wall comes out of nowhere all of a sudden. If you want to read the original paper it's at http://arxiv.org/...23v4.pdf and it's actually not too bad a read. And it's only a dozen pages. But basically, the problem is that in addition to the INFO (momentum, mass, spin, etc., etc.) already in a black hole from any normal mass/energy in it, infalling Hawking radiation ADDS to the INFO because of entanglement with the escaping particle, WHILE the BH shrinks until the max info the volume can hold and surface can hold, well, meet. At that point the BH's spacetime simply can't take any more. And SPLAT. Crazy
Gawad
5 / 5 (2) Mar 31, 2014
@GSwift
Yes, I've seen that many times in print, and you've got to understand that is a layperson's simplification. That only applies to the tidal force, and its ability to physically distort you. It does not consider the whole picture of what's going on.
I should probably have included this as well, from my 2004 ed of Penrose's The Road to Reality, p. 711:

"A hapless observer who falls through the event horizon, from the outside to the inside, would not notice anything locally peculiar just as the horizon is crossed. Moreover, the black hole itself is not a ponderable body; we think of it merely as a gravitating region of spacetime from within which no signal can escape."

p. 712: "Although the horizon H has strange properties, the local geometry there is not signifcantly diverent from elsewhere...an observer in a space ship would notice nothing particular happening as the horizon is crossed from the outside to the inside."

TRtR is not aimed at laypeople.
Gawad
3 / 5 (2) Mar 31, 2014
P. 713:

"For a black hole of a few solar masses, the tidal forces would be easily enough to kill a person long before the horizon is even reached, let alone crossed, but for the large black holes of 106M, or more, that are believed to inhabit galactic centres, there would be no particular problem from tidal eVects as the horizon is crossed (the horizon being some millions of kilometres across). In fact, for our own galaxy, the curvature at the horizon of its central black hole is only about twenty times the spacetime curvature here at the surface of the Earth—which we don't even notice! Yet, the relentless dragging of the observer inwards to the singularity at the centre would subsequently cause tidal eVects to mount very rapidly to
inWnity, totally destroying the observer in less than a minute!"
osnova
Mar 31, 2014
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osnova
Mar 31, 2014
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Gawad
5 / 5 (2) Mar 31, 2014
IMO most realistic scenario is


Lovely. Another Zuppet.
osnova
Mar 31, 2014
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Gawad
5 / 5 (2) Apr 01, 2014
But I predicted this outcome before seven years already, when I said, that the black holes are dense neutrino stars.

Yeah, yeah, Zeph. Oh, you forget: even if Polchinski were to turn out to be right, the scenario he describes doesn't actually occur until halfway through the evaporation process by Hawking radiation. Which means *trillions* of years for stellar mass black holes. So once again you've only managed to predict nothing while thinking you have. Go crack some sexist jokes over a beer or ten with Motl why don't you.
osnova
Apr 01, 2014
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GSwift7
3 / 5 (2) Apr 01, 2014
the curvature at the horizon of its central black hole is only about twenty times the spacetime curvature here at the surface of the Earth—which we don't even notice!


As I said above, they're only talking about the instantaneous curvature, which is analogous to the tidal force. They are neglecting to account for what happens due to the observer's own atoms and electrons no longer being able to communicate with eachother. To say that you wouldn't notice this is naive.

Are you proposing that anything can travel up-stream inside the EH? If so, I'd sure like to see a source that says that's possible.

I have actually read some theorists who claim that observers who fall past the EH near each other might continue to communicate, but that's just stupid. Only inwards would be possible, but that limitation would end your elecron orbits and such, so you would already be dead.
11791
Apr 01, 2014
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Gawad
5 / 5 (4) Apr 01, 2014
They are neglecting to account for what happens due to the observer's own atoms and electrons no longer being able to communicate with eachother. To say that you wouldn't notice this is naive.


Are you proposing that anything can travel up-stream inside the EH? If so, I'd sure like to see a source that says that's possible.
Strawman G, I've repeated 3 times that nothing goes upstream in a absolute sense (this is the 4th). In a relative sense, however, yes.

Even if, e.g., 2 electrons and their photons are falling at c+xkm/s inside an EH relative to sm on the outside, they all just in freefall and can exchange relative to each other.

All *textbooks* and my refs say this. The *Equivalence Principle* says this. Nothing simply gets "subatomized" simply because it crosses the EH. Where did you get this??? There are galaxies receding away from ours faster then c because of ST expansion, did they get atomized when they went over our horizon? Did we when we went over theirs?
GSwift7
1 / 5 (1) Apr 01, 2014
In a relative sense, however, yes
they all just in freefall and can exchange relative to each other


No, that's not how it works.

From any point inside the EH, everything farther down appears to be another EH. It doesn't matter if you're in free-fall or not. You cannot see farther down towards the singularity once you cross the EH. Every point farther down will appear to be another EH, like an infinite russian nesting doll of EH's, all the way to the 'bottom'. And if a friend of yours happened to follow you in, you'd see him falling above you, but he wouldn't see you, since you'd be inside another EH from his point of view. You could not wave at him. and if you fell in feet-first, you wouldn't see your own feet, as everything below your eyes would be below another EH. In fact, every planck distance length of your body should be cut off from the rest of your body above it by an apparent EH between them.
11791
Apr 01, 2014
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GSwift7
3 / 5 (2) Apr 01, 2014
There are galaxies receding away from ours faster then c because of ST expansion, did they get atomized when they went over our horizon? Did we when we went over theirs?


At the EH of a black hole, that communication block happens at the planck scale. Electrons and other subatomic particles will be fine, but their ability to stick together depends on them being able to interact with each other. You do understand that they cannot communicate with eachother, right? The gravitational field at the EH of a black hole is a whole different kind of EH than other types of EH. Look up the wiki page on event horizons. There's a short description of various types there. They are not all the same. Beyond the EH of any BH, the gravity field is of such high magnitude that every planck length down is cut off from the outside above it. It is like an infinite, nested series of the cosmological EH's you're talking about.

If you can still communicate, you're not in the EH.
TheGhostofOtto1923
3.7 / 5 (3) Apr 01, 2014
Unbelievable.
They are neglecting to account for what happens due to the observer's own atoms and electrons... have actually read some theorists who claim that observers who fall past the EH near each other might continue to communicate, but that's just stupid... No, that's not how it works.
What makes you think that the things you are arguing about wouldn't have occurred to the hundreds of scientists who have already examined this issue in great depth? What makes YOU think you can come up with something that they haven't already included in their analysis? ESPECIALLY something so obvious and so rudimentary??

Are you really that arrogant? Are you REALLY that self-deluded?? Shouldn't you be asking yourself what it is that you so obviously missed, and doing the research to find out?
Gawad
5 / 5 (2) Apr 01, 2014
O.k., G, you've put out a lot of stuff that I'm, well, unfamiliar with.

There are galaxies receding away from ours faster then c because of ST expansion, did they get atomized when they went over our horizon? Did we when we went over theirs?


At the EH of a black hole, that communication block happens at the planck scale.


Excuse me? Please provide a non-naïve reference for this please. You *must* have gotten this from somewhere. (Cont.)
Gawad
5 / 5 (2) Apr 01, 2014
(Cont.)
Electrons and other subatomic particles will be fine, but their ability to stick together depends on them being able to interact with each other.


If they're accelerating down a gravity well along with their mediating particles, they can. Heck, if Alice and Bob have both crossed the EH of the galactic BH, Alice 100m ahead of Bob and she tries to warn him, her radio signal doesn't have to move "upstream" in an absolute sense to reach Bob. Those photons are sill going "down" but so is Bob and he's going to catch up to them and get the message because those photons still move at c reletive to Bob (though it's to late for him). This will even work if Bob is initially outside the EH but doesn't have the ability to escape. (Cont.)
Gawad
5 / 5 (2) Apr 01, 2014
You do understand that they cannot communicate with each other, right?


No I don't, sorry. And I'd REALLY like a "non-naïve" reference that backs your (and A_P's apparently) claim. Because I can't find or think of any. (lacking that, I'd like to respectfully suggest that you may be misinterpreting the significance of "the area under the curve".)

The gravitational field at the EH of a black hole is a whole different kind of EH than other types of EH.


Is it? How so? Are you aware that even cosmological horizons emit Hawking Radiation? http://math.ucr.e...end.html Paragraph 24. That actually sounds like they are a lot more similar than different. Again, it sounds to me like you're claiming a violation of the Equivalence Principle; please provide a reference.

Look up the wiki page on event horizons.


I did. Come on, this is obvious. I have done so several times over the years.
Gawad
5 / 5 (2) Apr 01, 2014
(Cont.)
There's a short description of various types [of horizons] there. They are not all the same.


If you're thinking of particle horizons, those are purely circumstantial and don't involve Relativity.

If you're thinking of a cosmological horizon, then aside from a BH EH cloaking a singularity, then *how are the two types of EHs themselves* fundamentally different?

Beyond the EH of any BH, the gravity field is of such high magnitude that every Planck length down is cut off from the outside above it. It is like an infinite, nested series of the cosmological EH's you're talking about.


Even when all particles (fermions and bosons) are going "down" anyway? O.k. how about when some are going a little "less down" than the others? Look, just because there are no paths "UP" doesn't prevent paths that *CROSS* when everything is going "down", meaning there's no problem with mediation.
Gawad
5 / 5 (3) Apr 01, 2014
If you can still communicate, you're not in the EH.


Sorry, but I respectfully disagree, and so does the wiki and so does Baez, Penrose and every other physicist I can think of. You never even perceive that you've crossed the horizon; you can still communicate with someone freefalling in local space with you.
Gawad
5 / 5 (3) Apr 01, 2014
Oh and speaking of Wikipedia, you asked for a reference claiming that two astronauts could signal each other inside the EH, well it's right there in the last paragraph of the section "Interacting with an event horizon" of the Event Horizon page you referenced:

"Other objects that had entered the horizon along the same radial path but at an earlier time would appear below the observer but still above the visual position of the horizon [remember, infalling observers never see themselves as crossing the horizon], and if they had fallen in recently enough the observer could exchange messages with them before either one was destroyed by the gravitational singularity."

Sorry, G, but that's gold, man ;^)
Gawad
5 / 5 (3) Apr 01, 2014
@11791:

i'm not sure about you. did you have an ivy league education?


Ouch!

But to the point...11719, are you serious? How shalI put this?

Humm...are you familiar with the show "Cheers"? (If not, you've missed something, IMHO.)

Well, Physorg is the internet "science" version of "Cheers", especially since late 2011. Or that in a hybrid with a quadriplegic version of MMA.

Enjoy...
11791
Apr 01, 2014
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GSwift7
1 / 5 (1) Apr 01, 2014
Sorry, G, but that's gold, man ;^)


Eh, I think that's incorrect. I guess we disagree.

Each observer never sees themselves cross the EH, but anything that crosses the EH before you is essentially gone from your universe, and anything that crosses the EH after you has already seen you disapear forever.

The quote you have from the wiki is a commonly repeated mistake. As I said earlier, I've seen that in a lot of places, and I cringe every time I read it.

It doesn't matter whether you perceive youself as having crossed the EH or not. An outside observer would see you cross and disapear, and as such, anything that crossed the EH before you would appear to be gone, from your point of view.
Gawad
3 / 5 (2) Apr 01, 2014
cheers. whats IMHO mean?

"In my humble opinion"
baudrunner
1 / 5 (1) Apr 01, 2014
You are all in denial. But, as they say, ignorance is bliss. Untruths look better because they are so pure.

Remember Ocham's razor. The truth is simple.
TheGhostofOtto1923
3 / 5 (2) Apr 01, 2014
cheers. whats IMHO mean?
http://www.google.com/
Eh, I think that's incorrect
Well youre obviously WRONG arent you? Why dont you spend some time wrestling with the problem you have in accepting this?
It doesn't matter whether you perceive youself as having crossed the EH or not. An outside observer would see you cross and disapear, and as such, anything that crossed the EH before you would appear to be gone, from your point of view


What makes you think that the things you are arguing about wouldn't have occurred to the hundreds of scientists who have already examined this issue in great depth? What makes YOU think you can come up with something that they haven't already included in their analysis? ESPECIALLY something so obvious and so rudimentary??

Are you really that arrogant? Are you REALLY that self-deluded?? Shouldn't you be asking yourself what it is that you so obviously missed, and doing the research to find out?
Gawad
5 / 5 (2) Apr 01, 2014
Sorry, G, but that's gold, man ;^)
Eh, I think that's incorrect. I guess we disagree.

Well I guess you could say that!
Each observer never sees themselves cross the EH, but anything that crosses the EH before you is essentially gone from your universe, and anything that crosses the EH after you has already seen you disapear forever.
Are you *sure* you don't want to have another look at that?
The quote you have from the wiki is a commonly repeated mistake. As I said earlier, I've seen that in a lot of places.
And you're putting Penrose, Wheeler and Baez in the same basket...

It doesn't matter whether you perceive youself as having crossed the EH or not. An outside observer would see you cross and disapear, and as such, anything that crossed the EH before you would appear to be gone, from your point of view.
No G, an outside observer NEVER sees you cross the EH. That's cannon.

Sooo, until you can provide references, I'll stick with Penrose, Baez & Co.
Gawad
5 / 5 (2) Apr 01, 2014
And BTW, G, why did you ask for any reference claiming that two astronauts could signal each other inside the EH if you were going to *automatically dismiss it* (and from a page you yourself used as a reference)???
Gawad
5 / 5 (3) Apr 01, 2014
You are all in denial. But, as they say, ignorance is bliss. Untruths look better because they are so pure.

Remember Ocham's razor. The truth is simple.


Speaking of Cheers, look, that's Cliffy right there ^^^
GSwift7
3 / 5 (2) Apr 01, 2014
i'm not sure about you. did you have an ivy league education


Auburn University, Auburn Alabama, Aerospace Engineering (though that's not my current field), as if that matters here. Edwin Hubble's big discovery was made with the help of the mule handler from the observatory support staff.

No G, an outside observer NEVER sees you cross the EH. That's cannon


That's easily argued against. If that's the case, then you'd still see everything that's ever been pulled into a black hole, and we would be able to detect them. They wouldn't be black, rather they would be glowing brightly in the deep red. So I'm not buying the idea that you would never perceive something crossing the EH. I think that's an incorrect interpretation of the math.

"In my humble opinion"


I didn't know that. I always thought IMHO was 'in my honest opinion', but I see that urban dictionary has it both ways.
Gawad
5 / 5 (2) Apr 01, 2014
No G, an outside observer NEVER sees you cross the EH. That's cannon


That's easily argued against. If that's the case, then you'd still see everything that's ever been pulled into a black hole, and we would be able to detect them. They wouldn't be black, rather they would be glowing brightly in the deep red.


Euh, well, actually...setting aside the slight incongruity of "glowing brightly in the deep red", that's pretty much exactly what Relativity proposes: the outside observer sees everything that goes in just "fade to black" (going through the deep infrared) at the EH. Like I said, cannon.

So I'm not buying the idea that you would never perceive something crossing the EH.


Well, alright then, I'm not selling! (Hell, I'm giving it away for free!)
TheGhostofOtto1923
3 / 5 (2) Apr 01, 2014
Auburn University, Auburn Alabama, Aerospace Engineering (though that's not my current field), as if that matters here
And yet you have the temerity to say
That's easily argued against
-as if it werent the result of the work of many scientists who together have spent more time on the effort than youve been alive.

Scientists whose work youve been directed to for enlightenment but youve chosen to ignore in favor of your own silly ad hoc notions.

What a freeking idiot you are.
GSwift7
not rated yet Apr 02, 2014
pretty much exactly what Relativity proposes: the outside observer sees everything that goes in just "fade to black" (going through the deep infrared) at the EH. Like I said, cannon


Yeah, that's the most widely accepted interpretation, but not the only one. It can go several different ways, and we don't know for sure if any of our interpretations are correct. When terms in the equation go to infinity (or to zero if arranged in inverse fashion), it's anybody's guess what that means in real life.

There's a similar interpretation gap in regard to universal expansion and what that means with regard to time and distance between distant points. I went round and round with QStar and Antialias in a thread about that once.

I think it's ironic that the very same people will choose to factor out the time-like aspect in regard to expansion (which is mathematically valid, but maybe not realistic) and then they don't factor it out when dealing with an event horizon.
TheGhostofOtto1923
3 / 5 (2) Apr 02, 2014
Yeah, that's the most widely accepted interpretation, but not the only one
Well so far we have that one which is widely accepted by the scientific community, and we have yours. Oh also we have zephyr and cantdrive and lurker et al.

Which one are we to accept as the most reasonable? And which ones are we to shovel into the 'no clue' crank bin?

You won't even take the time to visit the links you were given to find out why you are wrong. Pretty sad g. Pretty ignorant.
osnova
Apr 02, 2014
This comment has been removed by a moderator.
baudrunner
1 / 5 (3) Apr 02, 2014
There is still huge difference in opinions of physicists.
No kidding. There is that school who have a clear understanding of the fundamental nature of reality, and there are the majority, who simply flaunt their degrees and pay lip service to the status quo. That majority is more likely to engage lay people in the reiteration of the ideas they studied in school, but they have still to develop a knowledge of those ideas. I find that about 98% of people in general have no real idea of what a photon is, and are more likely to think that a photon is a magical particle which has no mass but which can be directly influenced by gravity as it races through space. Any actual physicists who believe that ought to have their accreditation revoked.
osnova
Apr 02, 2014
This comment has been removed by a moderator.

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