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In the expanding universe the velocity of separation between galaxies depends upon how far they are. If they are much far away will they have relative velocity of separation greater than speed of light and if so how can we even detect such galaxies. There are things like Quantum information that can travel faster than light using entanglement, is there any any possibility to detect such unseen horizons using such effects?

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If they are much far away will they have relative velocity of separation greater than speed of light and if so how can we even detect such galaxies.

We can't detect such galaxies. Redshift goes to infinity at the cosmic horizon, and we cannot see beyond. Note that the cosmic horizon is different from the Hubble sphere: At the former, relative velocities according to parallel transport along the light ray reach $c$, whereas recession velocities reach $c$ at the latter. As far as observable effects go, the Hubble sphere is largely irrelevant.

is there any any possibility to detect such unseen horizons using such effects?

No, entanglement cannot be used that way - it is completely useless without a classical channel of information to 'compare notes'. If one of your particles vanishes behind an event horizon, such a channel is unavailable (Note that from your point of view, the particle will actually never cross the horizon, but gets frozen in time and become unobservable due to redshift).

The fact that the particles are quantum correlated instead of classically doesn't really matter. Instead of using entangled particles and the cosmological horizon, you could take a red ball and a blue ball and put them into two boxes without looking. Then, throw one of the boxes into a black hole and open the remaining one. You'll instantaneously know which ball ended up in the black hole - but what will such an experiment tell you about its interior?

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  • $\begingroup$ can we have a situation like this in which we have three galaxies A,B and C.A farthermost B in middle and C our position.A is beyond Cosmological Event Horizon for us i.e. C but not for B. Can B observe its state can transmit it to us. $\endgroup$ – ReXdean Dec 21 '14 at 19:22
  • $\begingroup$ @ReXdean: no; the cosmic horizon is an event horizon, so you have to take into account the time of arrival: the signal emitted at A will reach B at a point in time that lies beyond C's horizon $\endgroup$ – Christoph Dec 21 '14 at 19:30
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Even in special relativity, the claim that light always travels at a constant speed of c is only true in inertial reference frames, in non-inertial coordinate systems like Rindler coordinates, the coordinate velocity (change in position coordinate with respect to coordinate time) may vary. In general relativity, all coordinate systems covering large regions of curved spacetime are non-inertial, although in infinitesimal regions of spacetime you can have "local" inertial frames, and according to the equivalence principle light must be measured to have a local speed of c in all local inertial frames.

If you use the particular choice of time coordinate most common in cosmology (chosen so that the average density of matter/energy is approximately uniform across large regions of space at each moment of coordinate time), and use proper distance to measure distance between galaxies or light rays (where proper distance is the distance that would be measured by a series of short rulers laid end-to-end at a particular moment of coordinate time), then you can define a "velocity" in terms of the rate proper distance is changing with coordinate time. The total velocity of an object at any proper distance D can be broken down into a sum of "recession velocity" due to the expansion of space (the velocity that an average galaxy at that distance would be expected to be moving away from us) and "peculiar velocity" due to the object's motion relative to what we would expect for an "average" bit of matter at that distance. The recession velocity at proper distance D, i.e. the expected velocity for an average bit of matter at distance D, is given by Hubble's law as v = HD, where H is the Hubble "constant" (the name is a bit misleading since the value is thought to change over time, though it is assumed constant over space). And for light rays, it works out that their peculiar velocity is always equal to c, so if a light ray is emitted directly towards us from a distance D, its initial velocity relative to us will be HD - c.

The recession velocity HD becomes equal to c at the radius of the Hubble sphere, and yet it is possible for light emitted from that distance or greater to eventually reach us. How can this be? As explained in section 3.3 of the paper that PhotonicBoom linked to, Expanding Confusion: Common Misconceptions of Cosmological Horizons and the Superluminal Expansion of the Universe, it's a consequence of the fact that H is thought to be decreasing over time (it may seem confusing that this is possible even in models where physicists say the rate of expansion is accelerating, but what they mean by that is that for a given galaxy expanding away from us its recession velocity will increase over time, which need not conflict with the claim that if we pick a fixed distance D, the recession velocity of an average galaxy at that distance will decrease with time). Quoting from the pdf version of the paper in the link (note that $v_{rec}$ is the recession velocity, $D_H$ is the radius of the Hubble sphere, $a$ is the scale factor, and dots indicate time-derivatives):

We have seen that the speed of photons propagating towards us ... is not constant, but is rather $v_{rec} - c$. Therefore light that is beyond the Hubble sphere has a total velocity away from us. How is it then that we can ever see this light? Although the photons are in the superluminal region and therefore recede from us (in proper distance), the Hubble sphere also recedes. In decelerating universes $H$ decreases as $\dot{a}$ decreases (causing the Hubble sphere to recede). In accelerating universes $H$ also tends to decrease since $\dot{a}$ increases more slowly than $a$. As long as the Hubble sphere recedes faster than the photons immediately outside it, $\dot{D_H} > v_{rec} - c$, the photons end up in a subluminal region and approach us. Thus photons near the Hubble sphere that are receding slowly are overtaken by the more rapidly receding Hubble sphere.

However, the cosmological models currently seen as best supported by evidence do include an event horizon such that light emitted from outside this horizon will never reach us no matter how long we wait, see Fig. 1 of the Davis/Lineweaver paper (particular the third graph, which is a "conformal" one where the paths of light rays are always represented as straight lines at 45 degrees from vertical), along with my answer here.

So the final part of your question could be rephrased as, "would it be possible to use quantum entanglement to communicate (or observe) regions beyond our event horizon?" According to quantum mechanics, the answer is no, because there is a general theorem called the No-communication theorem which shows that measurements of one part of an entangled system can never be used to communicate information to an experimenter measuring a different part of the same entangled system. And there is also this general result in quantum field theory (the relativistic version of quantum mechanics), showing that no form of faster-than-light communication is possible in QFT.

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  • $\begingroup$ I'm glad we agreed in the end. Nice answer. +1 from me! $\endgroup$ – Constandinos Damalas Dec 21 '14 at 19:37
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Such galaxies cannot be detected. and Quantum entanglement is just a correlation between two bits of information and it no way does, and no way can be used to transfer information faster than the speed of light. Any light sent from such galaxies towards ours would show a wavelength of infinity after considering Doppler shift. In short these galaxies can never be detected. This boundary beyond which galaxies move away by the expansion of space time at supersonic speeds is called a Cosmological event horizon. No information from outside this boundary would be able to reach inside if the expansion continued so.

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  • $\begingroup$ So if a entangled particle is implanted on such a galaxies which is going to disappear in the Cosmological event horizon will we still be able to get to know its state after it had disappeared in Cosmological event horizon?? $\endgroup$ – ReXdean Dec 21 '14 at 18:19
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    $\begingroup$ @ReXdean: Sure, but what's the point? Take a red ball and a blue ball and put them into two boxes without looking. Then, throw one of the boxes into a black hole and open the remaining one and you'll know which ball ended up in the black hole. What will such an experiment prove? $\endgroup$ – Christoph Dec 21 '14 at 18:36
  • $\begingroup$ @Christoph Wiki says: Quantum teleportation is a process by which quantum information (e.g. the exact state of an atom or photon) can be transmitted (exactly, in principle) from one location to another, with the help of classical communication and previously shared quantum entanglement between the sending and receiving location.en.wikipedia.org/wiki/Quantum_teleportation $\endgroup$ – ReXdean Dec 21 '14 at 18:44
  • $\begingroup$ @ReXdean: with the help of classical communication $\endgroup$ – Christoph Dec 21 '14 at 18:46
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These galaxies are not out of reach. The Hubble Sphere is the volume of space surrounding an observer where everything inside the sphere moves away from the observer with a speed less than c.

Following this logic it immediately follows that the Hubble Sphere is equivalent to the cosmological event horizon that @Hritik is discussing in his answer. Interestingly enough, this is not the case, as evidence shows that our universe is actually expanding away from us in an accelerating rate.

This implies that the Hubble sphere is expanding as fast as the universe, and thus astronomical bodies which once were receding away from us at a speed faster than $c$ will emit light which will eventually enter out Hubble Sphere. This is discussed in a very interesting Veritasium episode about Universe misconceptions.

It seems that wikipedia's facts might be wrong in the page I have linked above. A paper by Tamara M. Davisa and Charles H. Lineweaver called Expanding Confusion: Common Misconceptions of Cosmological Horizons and the Superluminal Expansion of the Universe linked by Veritasium in the youtube comments section supposedly clears this up but I don't have enough time to read this now.

Also as I stated in the comments, quantum entanglement does not imply faster than light information travel since quantum entanglement is just a correlation between 2 qubits. Not information is travelling at all.

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  • $\begingroup$ The paper by Lineweaver and Davis you link to actually contradicts your claims--for example, p. 2 says "As we will see, the Hubble sphere is not a horizon. Redshift does not go to infinity for objects on our Hubble sphere (in general), and for many cosmological models we can see beyond it." It is true that objects on the Hubble sphere have a recession velocity equal to c, but see section 3.3, titled "Misconception #3: Galaxies with Recession Velocities Exceeding the Speed of Light Exist but We Cannot See Them." $\endgroup$ – Hypnosifl Dec 21 '14 at 18:50
  • $\begingroup$ It seems to me its supporting it. Specifically it says: "In accelerating universes, H also tends to decrease since $\dot{\alpha}$ increases more slowly than $\alpha$... Thus photons near the Hubble sphere that are receding slowly are overtaken by the more rapidly receding Hubble sphere." But, I will read the paper more carefully when I have more time! $\endgroup$ – Constandinos Damalas Dec 21 '14 at 19:02
  • $\begingroup$ note that recession velocities reach $c$ at the Hubble sphere, whereas relative velocities according to parallel transport along the light ray will reach $c$ at the cosmic horizon, so in a way, the horizon is where 'relative veocities' reach speed-of-light; it's just that if spacetime isn't flat, you have to be careful about what you mean by 'relative velocity' $\endgroup$ – Christoph Dec 21 '14 at 19:05
  • $\begingroup$ The part I was disagreeing with was in the initial version of your comment where you said "From this logic it immediately follows that the Hubble Sphere is equivalent to the cosmological event horizon that @Hritik is discussing in his answer"--as seen in Fig. 1 the cosmic event horizon is distinct from the Hubble sphere. But I see you edited your answer to add "Interestingly enough, this is not the case" after that statement, so with that change I don't disagree with what you say. $\endgroup$ – Hypnosifl Dec 21 '14 at 19:34
  • $\begingroup$ @Hypnosifl I thought you misunderstood me so I edited to clarify. $\endgroup$ – Constandinos Damalas Dec 21 '14 at 19:37

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