Question About Moving Observers and Observable Universes

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[Mentor's Note: Post moved into its own thread from this one]

Hello, I have another newbie relativity question which is somewhat related to this topic. Here it is:

There are 3 observers on 3 different planets each in a different galaxy. Observer 1 stands on the north pole of their planet looking through a telescope looking straight up from the pole and see Observer 2 moving directly away at 51% the speed of light. Observer 1 then travels to the south pole and looks through a similar telescope looking straight up from that pole (the opposite direction of the first observation) and sees Observer 3 moving directly away at 51% the speed of light.

Ignoring there's a planet directly between Observer 2 & 3, could Observer 2 & 3 see each other? Or are they outside each other's observable universe? It would seem from Observer 1's point of view that they are moving away from each other faster than the speed of light.
 
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MassNerd said:
Ignoring there's a planet directly between Observer 2 & 3, could Observer 2 & 3 see each other? Or are they outside each other's observable universe? It would seem from Observer 1's point of view that they are moving away from each other faster than the speed of light.

The rate of separation of Observer 2 and Observer 3 when viewed from Observer 1's frame can indeed be greater than c. However, from Observer 2 or 3's frame, the other is moving away at about 0.81 c. See this page: http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel2.html

So no, they would not be out of each others observable universe assuming that they are closer than several billion light-years to each other. Beyond that it gets...complicated.
 
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Let's break this down into manageable pieces. Observer 1 is stuck in the middle. Let's just call this one earth. Observer 2 is due north of observer 1 and has a proper recession velocity of .51c which corresponds to redshift of .595. and a proper distance now of 7347.5 Mly [assuming the 2013 Plank cosmological parameters]. A galaxy receeding due south in the opposite direction at a redshift of .595 also has a proper distance now of 7347.5 Mly. A galaxy with a proper distance now of 14695 Mly [obtained by adding together the proper distances now of galaxy 2 and 3 from galaxy 1]. would have a redshift of 1.516 and a proper recession velocity of 1.02c. So you see the proper distances now simply add up, but the recession velocities and reshifts do not. The proper recession velocity is not quite double, but, the redshift nearly triples. Of course, we already knew that recession velocity and redshift are not additive, You use SR to add up recession velocities .but .51v is so slow relative to light it very nearly does add up. You use GR for redshifts, and the rules are different. BTW the particle horizon of galaxy 2 and 3 had a radius then of 12554 Mly, but, keep in mind the proper distance back then betrween 2 and 3 was only 5841 Mly, not their current proper distance of 14695 Mly, so they were within each others particle horizon and could both see each other back then with a large enough telescope. This whole then and now business gets a little confusing. Jorrie's cosmological calculator will confirm the values I gave: http://www.einsteins-theory-of-relativity-4engineers.com/cosmocalc_2013.htm
 
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