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Entanglement across different frames of references

  1. Oct 20, 2012 #1
    What is the effect of differing frames of references between entangled particles?

    For example, let's look at the below scenario:

    Photons are entangled in Alice's frame of reference on say earth.

    Photon B is put on Bob's spaceship that travels at say 95% the speed of light.
    The other photon A stays with Alice on earth in a frame of reference moving at zero velocity.

    At 10 am Alice measures her photon and finds it to be spin UP.

    Now in Bob's space-ship (at 10 am, per the Alice's frame of reference) what would Bob find?

    Since photons move at speed of light are the even effected by differing frames of references?

    So instead let's say we take something slower than photons and entangle them and give to Bob and Alice.

    Please make/state the additional assumption/scenarios, if any, that need to be made, before such a question can be answered.
     
  2. jcsd
  3. Oct 20, 2012 #2

    Bill_K

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    Good! Because you can't carry a photon around in a spaceship. And photons can't be "spin up"!

    The answer is, it doesn't matter who measures first, Alice or Bob. And in fact in some reference frames Alice will go first, and in others Bob will. And there will even be a frame in which they measure simultaneously. The measurements will be consistent in any case.
     
  4. Oct 20, 2012 #3
    Clever question.

    It might be that the entanglement exists in some frames of reference and does not exist in others.

    The very existence of the particles might depend on the frame reference. This is the case of accelerating frames relative to each other. If we entangled two particles then moved to a frame where one particle does not exist, then what?

    Finally, the existence of the measurement or the observer himself might be frame-dependent. Suppose we entangle two particles and give them to Alice and Bob. Then we throw Bob into a black hole. From Alice's point of view, the measurement happens after Bob vanished under the horizon, so as well we can say that it never happened. What then?

    I read somewhere that performing observations under the horizon does not have to obey the uncertainity principle in its usual sense. Suppose that Alice measures position of an electron, then throws it along with Bob to a black hole. Then, Bob could be able to measure momentum of the electron with 100% certainity, ostensibly violating uncertainity. However, no laws of physics are violated, since he can not actually tell Alice the result.

    As you see, very much depends on the reference frame.
     
  5. Oct 21, 2012 #4
    First, remove Bill K's primary objection by converting the photons to electrons :D

    Then - He's still right. The spins are intrinsically opposed. Whatever you measure for one, you will get the opposite result for the other (Or, if you measure a different axis - you'll get a result based on the QM-predicted probability for that orientation)

    There are no frames of reference in which the particles 'don't exist'. A frame of reference is only a viewpoint. It has no significance to the existence or any other condition of a particle.

    An event horizon merely represents a highly compressed viewpoint - the particles are still observable from 'outside'. In fact, they never enter it.
    Bob and his electron remain apparently stuck on the horizon in extreme slow motion and deep red-shift.
     
  6. Oct 21, 2012 #5
    Great posts Bill k, haael, AJ Bentley.
    Even though Bob and his electron remain stuck on the horizon in extreme slow motion, the entanglement 'effects', if you will, are instantaneous ?
    In other words:
    If Bob checks the spin of electron B a little after 10 am (per Alice's time) he shall find the spin to be opposite of what Alice's electron A showed at 10 am ?
     
    Last edited: Oct 21, 2012
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