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 thought experiment, 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.
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, black holes 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 orbit 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.
Below the photon sphere you would see only darkness. Down there is the event horizon, light'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 gravity. It would be an awesome and deadly place to be, but it'd sure beat falling down below the event horizon.
Explore further:
How fast do black holes spin?
big_hairy_jimbo
Anyway, just thinking aloud.
Jizby
Mar 25, 2014what_the_hell
Jizby
Mar 25, 2014what_the_hell
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
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
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
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
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)?
No. There's no theory yet that puts an upper limit on that.
Yes they should - just like photons. Possibly a tiny bit inside the photonsphere/ring.
Jizby
Mar 26, 2014Jizby
Mar 26, 2014GSwift7
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
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
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, 2014antialias_physorg
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
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
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
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
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
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
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
Not really. It will just blueshift the photon as it falls in.
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)
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
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, 2014Jizby
Mar 27, 2014GSwift7
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, 2014Jizby
Mar 27, 2014Jizby
Mar 27, 2014GSwift7
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
antialias_physorg
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
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
Jizby
Mar 27, 2014GSwift7
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
GSwift7
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.
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
Gawad
An note that I say SR limited drives, because "warp drives" is just *undefined*.
antialias_physorg
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
RealityCheck
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
Not important, but I thought I would point out that a geodesic maximizes proper time.
bicubic
antialias_physorg
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
GSwift7
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
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
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
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
'Caution - this poster rarely researches his notions before posting!'
Jizby
Mar 29, 2014Benni
Dead on the money accurate- now when will you dispense with foolish AWT ?
Jizby
Mar 30, 2014Osiris1
adam_russell_9615
gale_langseth
Zachia
antialias_physorg
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
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
A photon at rest is not physically real.
11791
Mar 31, 201411791
Mar 31, 2014GSwift7
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
"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." 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
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
Gawad
osnova
Mar 31, 201411791
Mar 31, 2014baudrunner
Gawad
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
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, 2014Gawad
Gawad
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, 2014Gawad
Gawad
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
"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, 2014osnova
Mar 31, 2014Gawad
Lovely. Another Zuppet.
osnova
Mar 31, 2014Gawad
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, 2014GSwift7
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, 2014Gawad
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
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, 2014GSwift7
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
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
Excuse me? Please provide a non-naïve reference for this please. You *must* have gotten this from somewhere. (Cont.)
Gawad
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
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".)
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.
I did. Come on, this is obvious. I have done so several times over the years.
Gawad
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?
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
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
"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
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, 2014GSwift7
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
"In my humble opinion"
baudrunner
Remember Ocham's razor. The truth is simple.
TheGhostofOtto1923
Well youre obviously WRONG arent you? Why dont you spend some time wrestling with the problem you have in accepting this?
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
Well I guess you could say that!
Are you *sure* you don't want to have another look at that? And you're putting Penrose, Wheeler and Baez in the same basket...
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
Gawad
Speaking of Cheers, look, that's Cliffy right there ^^^
GSwift7
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.
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.
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
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.
Well, alright then, I'm not selling! (Hell, I'm giving it away for free!)
TheGhostofOtto1923
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
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
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, 2014baudrunner
osnova
Apr 02, 2014