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Entanglement/event horizon red/blue shift 
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#1
Jan2413, 05:48 PM

P: 1,857

Heres the scenario 2 entangled particles one sent to an event Horizon due to the increase in gravity it will blue shift. The other particle at the same time is sent in the opposite direction. For purpose of this lets say that the force of gravity is simultaneously reducing. In a manner that you have an equal redshift.
What happens on the information of the entangled particles ? For now I'd like to avoid any beyond the event horizon considerations. edit: Will they stay entangled ? 


#2
Jan2413, 06:50 PM

Sci Advisor
PF Gold
P: 5,345

That depends on an assumption about gravity. If it is NOT a quantum force, then the shifting will not break entanglement. If gravity is a quantum force, then the answer is not as clear.



#3
Jan2413, 10:34 PM

P: 1,857

Well that statement definetely side swiped me lol. I would appreciate an expansion on how the quantum gravity will differ vs entanglement compared to the relativity view point of gravity.
In light of that response perhaps asking what occurs when you have two simultaneous, equal but opposite changes in information occurs between two entangled particles. Assuming the entanglement is maintained. Or is that scenario paradoxial in basis? 


#4
Jan2513, 09:40 AM

Sci Advisor
PF Gold
P: 5,345

Entanglement/event horizon red/blue shift
The quantum view of gravity is more complicated. If their were spin interaction and it was possible, in principle, to detect that, then there would be decoherence. 


#5
Jan2513, 11:41 AM

P: 1,857

KK that makes sense, I've been looking over various quantum entanglement articles trying to find the mathematics for entanglement measuring properties other than spin. Turns out thats not very easy to find. One example I found involved transmission lines.
http://arxiv.org/abs/1210.4413 this is the article that had me thinking about the scenario I described above. Judging from searches only and the difficulty in finding examples I have to conclude that the other portion of my question is not really answerable accurately at this time. 


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