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The God Effect and the Black Hole

  1. Sep 5, 2012 #1
    Friends, Acquaintances, and Juvenile Delinquents alike lend me your ears...

    As a thought experiment, let us say we have two photons, photon A and photon B. Now our two lovely quanta of electromagnetic radiation are special, because they have been entangled after an atomic cascade. After the atomic collision we immediately trap our two photons and we place each one of them in separate jars. These jars a made perfectly for the transportation of single photons, so that they cannot escape by any means besides opening the lid. Now we place the jar containing photon A on a rocket ship and we fly it to the nearest black hole, while the other stays in a lab. The astronauts on the rocket ship float out of the ship, with the jar in hand, up to the black hole. The astronauts then release the A photon so that it falls directly into the black hole. Back at the lab is photon B still entangled with photon A, who is now taking an exciting trip through the even horizon and onto the singularity.

    So my question to you, dear reader, is what do the scientist at the lab observe about photon B while photon A is taking its trip? Since information from quantum entanglement is non-local and, as so far observed, is instantaneous then the information about photon A should be transmitted to photon B. Right??

    -Do they not observe anything? Just as watching photon A fall through the event horizon would take an infinitely long time
    -Does the observed photon vanish out of existence as its particle twin is annihilated at the singularity?
    -Does the quantum bond between the two photons shatter and the photon B no longer becomes a vessel of information for photon A?

    So what happens????

    Your comments and consideration are much appreciated. Thank you!
     
  2. jcsd
  3. Sep 5, 2012 #2
    Nothing special. For example, you won't be able to tell that one photon has fallen into a black hole by looking at the other photon.

    Only in a very weak sense. For example, you can't use entanglement to actually send a message. In general, you can't change the other photon by changing one photon. It's just that if you measure one photon, you know the result someone else will get when they measure the other photon (and this only works once, for a single measurement).
     
  4. Sep 5, 2012 #3
    The thing with entanglement is that these photons must have opposite polarizations. If we assume these special jars do not interfere with this entanglement or determine a polarization, then when we release the photon into the black hole, it still is in a superposition state of two different polarizations. If we have not made a measurement on the other photon's polarization, they both remain in such undetermined state, all the way until the photon's annihilation via black hole. Nothing special really happens, because information cannot travel faster than light, but it seems that measures to preserve energy conservation CAN.

    A good question to ask would be, if a suicidal (or just extremely curious) physicist were to carry that photon jar into the black hole, with enough equipment to make a measurement of the photon's polarization after passing the event horizon, would it still determine the polarization of the lab photon immediately? I would be inclined to say that it still would. From all we can tell, entanglement seems to be a pretty powerful phenomena.
     
  5. Sep 5, 2012 #4

    PeterDonis

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    Staff: Mentor

    I would too, but that question suggests an even better one: suppose we drop the jar containing photon A into the hole and no measurement is made on it at any time, so that it hits the singularity and is destroyed while still in a superposition of polarizations. Then what happens to photon B?

    This question is really a special case of the question that prompted what Leonard Susskind calls the "Black Hole War" in his recent book of the same name. Photon A falling into the singularity and being destroyed violates unitarity, or at least it seems to. Susskind's solution is that photon A's state gets transferred to the Hawking radiation that gets emitted by the hole as it evaporates.
     
  6. Sep 5, 2012 #5
    Or at least exists on the surface of the black hole's event horizon as a sort of hologram
     
  7. Sep 5, 2012 #6
    You can retrieve information from a black hole. Susskind had some interesting ideas. I kinda liked the idea of tiny perturbations of the event horizon due to in-falling matter being "encoded" on the outgoing Hawking radiation, but I think Hawking solved the paradox via path integrals out to infinity from the black hole, the idea being that you can only probe a black hole via scattering, and no one can really determine if one has formed. Or, to let Hawking say it,

    Personally, I like this explanation better than Susskind's theories

    However, entanglement is not really information transfer. If it were, it would not occur instantaneously, so I wonder how much of this black hole physics would actually factor into the OP's question. Does the act of destroying a particle constitute a measurement of its state? Is the original photon un-entangled if measured after photon A has been destroyed? All good questions.
     
  8. Sep 5, 2012 #7
    My feeling on the matter is that as long as there is at least *one* reference frame in which the measurement on A and B are simultaneous then there will be consistency in all other frames (but the idea that either A or B was measured "first" is obviously invalid by SR.)

    When one of the pair falls into the hole, the entanglement should be broken at the moment that there is no possible frame in which the measurements could coincide. (I'm postulating this; its by no means agreed upon by everyone.) This likely ocurrs at or near the crossing of the EH. Essentially I am suggesting that the entanglement is broken as if a measurement were being made on the infalling particle as it meets the EH.

    Consider a thought experiment where you could measure the QM spin of a large BH. Then the measurement of the BH becomes a proxy for the particle that fell in such that a measurement of A will be consistent with measurement of the BH state. Conservation laws demand this.

    I propose that there is no effective difference between B impacting an atom and having thus been "measured" vs crossing an EH. In both cases as far as A is concerned B no longer exists from the point of view of any and all reference frames. A is no longer in a state of superposition.
     
  9. Sep 5, 2012 #8
    Once an object passes through the event horizon of a black hole, or, for example, the apparent horizon of an accelerating particle forms, the object is causally disconnected from the rest of spacetime.....that is, the outside universe. Whatever happens at the singularity within the black hole has no effect on outside observers.

    Seems like Susskind would probably argue that the characteristics of the lost photon are encoded on the surface of the black hole horizon, but would be unavailable...scrambled...to us on the outside, for practical purposes.
     
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