An entangled photon enters a black hole. . .?

Question

So what happens to the other photon? Since a black hole is a singularity from which "no information can escape", will we see a change in the state of the photon that is external to the black hole? Will that photon freeze in space? Does changing the state of the entangled external photon change of the state of the photon inside the black hole? How could it? If entangled photons get caught at the event horizon of a black hole, shouldn't there be big clusters of their entangled pairs floating about helplessly some distance from the black hole? Is this the "information" hawking says you can get out of a black hole?

Let's assume the entangled photon travelled to the blackhole from a nearby star, say an orbiting red giant. This just happens to be a natural entangled photon.

The Pauli Exclusion isn't relevant to this question. I'm talking about free entangled photons, not electrons.

???

Can I suggest that you examine the difference between a fermion and a boson? I'm not trying to be rude here, but it seems like you are mocking my question and you really have no idea what I'm talking about. Photons have wave symmetry and full integer spin, that's what makes them a boson and it's also why they are not subject to the Pauli Exclusion principle.

Jared - Entangled pairs can be millions of lightyears apart, given sufficient time. The whole concept is very interesting and I suggest that you look it up as you'll likely find the concept facinating. Please don't answer questions for which you have no knowledge though, since I was hoping to elicit something of an educate response to my question. Thank you!

Answer 1

This is an excellent question, and you have done a good job of exposing a number of morons who clearly haven't the faintest clue about the principles of physics which they profess to be so interested in.

In effect, this is a question about entropy.

Nothing special happens when a particle crosses the event horizon of a black hole - in fact, the black hole believed to be at the centre of our galaxy has an event horizon with spacetime curvature only 20 times higher than here on Earth. So the answer to your first questions are that there will be no change in the state of the photon, and that it will not freeze. In fact, nothing will change.

The question comes, what happens when you outside the black hole make a measurement on your photon. Now the wavefunction collapses, and so you know the state of the photon inside the black hole. But this is not supposed to be possible.

Now the photon in the black hole before your measurement could have been in one of two states. After your measurement it can only be in one. This reduces the degrees of freedom of the black hole, and so reduces its entropy.

And this is unexpected as a black hole is supposed to have maximum entropy per unit volume.

Answer 2

You seem to be an expert on this subject and perhaps you have to answer this one. Yahoo answers very rarely get people with your caliber. So asking such an advanced question in this forum is your mistake. So criticizing the people who is trying help is another mistake. So present this in research forum and try to defend whatever you are defending.

But anyway let me clarify certain concepts. The black hole is big junk of mass having a large gravity. It does pull back photons but still it communicates with others through the gravity. So if you know how the entangled photos communicate then you know the answer. The problem is you don't know how they communicate. This is not a black hole issue. It is the lack of knowledge of entangles photons. Read more or do some research further. Yahoo answer is not the place to do research

Answer 3

'Hawking radiation" is the emission of half of a vacuum fluctuation--like an electron/positron pair, one of which goes in, the other of which comes out. Conservation of energy requires that the black hole provide the energy of the escaping one, and this will gradually cause the hole to shrink. That's why the tiny black holes made at Brookhaven are no cause for alarm. They just evaporate.

The entangled photon question is interesting, because it implies that you can know something about the state of a particle inside a black hole--or even something it interacts with in there.

Answer 4

Suppose I have an entangled pair of photons and I give one to you. I can now measure mine, and when you measure yours, you learn what state both photons are in.

I still can't use this to transmit any information to you, since I have no control over the result of the measurement. So even if one photon was inside the black hole and the other wasn't, no information would be sent out of the black hole.

This is the same reason you can't use entangled pairs of particles to transmit information faster than light.

Answer 5

Hi. The escaping photon (or any virtual particle) is the cause of Hawking Radiation. Actually it IS Hawking Radiation. The problem of entanglement is, I think, time dependent (X time for particle A is the same as Y time for particle B. X=Y). A more fundamental question may involve the apparent violation of symmetry since symmetry collapses as one particle falls closer to the event horizon, accelerates due to the higher orbital speed, and is subject to time dilation. "Back to the Future" anyone?

Answer 6

I need some clarifications:

Where is the entangled photon coming from (a quantum dot?)

Why presume the "other photon" is left on the horizon?

Are you invoking Pauli exclusion to suggest the state change?

Sounds like what you are asking should go to some give-and-take discussion forum rather than Q&A.

Answer 7

for sure one can't "see" a black hollow, yet you will see what's happening on its experience horizon. This marks the radius interior of which easy can now no longer escape, notwithstanding it relatively is no longer the black hollow itself. we would be waiting to observe the exterior for clues yet interior we are able to only theorize. we are able to be sure that remember and capability do no longer behave generally there. Gravity would be so extreme that different forces, electromagnetism, the sturdy and susceptible atomic forces, would be overwhelmed. Atoms would possibly be broken down into person quarks, flowing and colliding, held jointly with the aid of sheer proximity as a results of rigidity of gravity. Photons would preserve their "velocity" yet would possibly no longer be vacationing in a straight away line. it would be extra like a vibration than a vector. it relatively is meaningless to describe what an observer interior the black hollow observes as a results of fact an observer is impossible. remember purely does not behave in a fashion shall we "word". as a results of fact the mass of a black hollow will advance, the progression horizon would strengthen yet we don't have a sparkling concept of the constraints of issues. some issues DO escape from black holes, "geysering" from the poles. the way all of it balances out will take so lots extra learn.

Answer 8

First, has anyone ever visited a black hole or been close enough to it to witness the effects and thereby determine they do exist? I realize there are dead spots in space with no light and whatnot, but it's all just theory.

Second, how is it the external photon is able to avoid the gravity of the black hole? Wouldn't they both go in? If I was standing by one and I stuck my hand in it, would I be able to putt back my severed nub? Wouldn't it be more likely that the rest of me would be drawn in as well?

Answer 9

Whoa! That's the best (& most disturbing!) physics question I have heard asked in a long, long time! My gut feeling is that something essential is missing here (that both you and I are apparently overlooking) that would prevent the scenario you describe!

Answer 10

What's the question here?
Obviously the scientific community, including
S.Hawking, are still confused about black holes.