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Let's say the Hubble constant is constant in the time interval we consider. So subjective "horizon" for any object has the same radius R. Let's say Bob and Charlie are just within the horizon of Alice, but on opposite sides: vectors AB and AC are exactly opposite.

And then Bob accelerates sufficiently close to the speed of light, so as to keep the distance to Alice constant with time. Charlie does the same. Now Bob and Charlie stay within Alice's horizon for as long as they want. Now, Bob can send radio messages to Alice (He's withing her horizon, right?). But then Alice can receive from Bob and forward them to Charlie, effectively acting as a relay.

So, this way Bob can communicate with Charlie, and this channel is a two-way.

So, this way Bob beyond Charlie's horizon, and Charlie is beyond Bob's horizon. But there is a two-way communication possible between them.

Does this actually work? And, if it does, how does "horizon radius" depend on the relative velocities of objects?

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  • $\begingroup$ When we say that B (or C) is within the horizon of A, it doesn't mean that the communication between them is permanently guaranteed and they may exchange information infinitely many times. It means that A may observe at least one event from B's past life (and vice versa). If B is (unnaturally) kept within the horizon volume of A, it may mean that the whole past life of one ultimately becomes accessible to the other, but much later, so that it's too late to get the signal back many times. Your strategy requires a repeated bouncing of the signal which isn't possible. $\endgroup$ – Luboš Motl Jun 24 '14 at 14:45
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They still can't communicate.

The horizon you are talking about is the event horizon. Assuming a spacetime event happens here and now, and let light signal propagate forward in time. The wavefront of the light signal for $t = \infty$ or collapse time is the event horizon. For standard Friedman cosmology, \begin{equation} ds^2 = -dt^2 + R^2(t) \frac{dr^2}{1-kr^2} + r^2 d\Omega^2 \end{equation} only $k = -1$ has event horizon.

Yet without algebra, spacetime diagram is enough to solve this problem.

  • Bob should not have traveled towards Alice. enter image description here

Suppose Bob sent a light signal 1 when he started. Then Bob has to travel slower than light, so he must fall behind signal 1. At any intermediate stage, he sent a light signal 2. Whatever the shape of light cones, signal 2 must arrive latter than signal 1. We can see this from the geodesic equation of the light ray $ds^2 =0 $, the slope \begin{equation} \frac{dt}{dr} = \frac{R(t)}{\sqrt{1-kr^2}} \end{equation} is increasing.

Alternatively, before signal 1 bypass signal 2, they must meet at the spacetime diagram. However the geodesic equation is general covariant, once they merge, they will be a single geodesic forever. Hence signal 1 will arrive no latter then signal 2. Actually, they can't merge at all, because the proper distance of these two light signals will keep fixed.

  • Alice can't help Bob to speed up the signal. enter image description here

Let's analyze signal 1. By assumption, Alice was within the event horizon of Bob. Hence (without any technical problem of course ) she would receive the signal. But the new signal she sent out had the same trajectory on spacetime diagram as if signal 1 were never been blocked. The reason is the trajectory of the light signal is determined by geodesic equation \begin{equation} ds^2 = 0 = g_{\mu\nu} dx^{\mu}dx^{\nu} = g_{\hat{\mu}\hat{\nu}} dx^{\hat{\mu}}dx^{\hat{\nu} } \end{equation} The proper distance is invariant under Lorentz transformation. And a bonus for light ray, the proper distance is zero, it is actually general covariant, which means the slope of the trajectory at any point is independent of the coordinate system we are using.

The light signal forwarded by Alice would propagate as if it was the original signal 1. By assumption, Charlie was outside the event horizon of Bob, the light signal will never reach him.

  • Event horizon is shrinking due to expansion enter image description here

We can see that the signal forwarded by Alice still couldn't reach Charlie, is it a contradiction?

No.

Charlie was in the event horizon of Alice at $t =0$, but at the time when Alice forwarded the massage, Charlie went outside of Alice's event horizon due to the expansion of universe. In other words, event horizon is shrinking. As I pointed above, the slope \begin{equation} \frac{dt}{dr} = \frac{R(t)}{\sqrt{1-kr^2}} \end{equation} is increasing.

I implicitly assumed the universe is expanding. If the universe is shrinking(as is possible for $k =-1$ ), i.e. $R(t) < 0$, then it's possible for Bob to communicate with Charlie. Because by in that case, the event horizon is expanding, and by traveling for a period effectively delay the emission of the signal. Charlie will eventually come into Bob's event horizon, then they can communicate with each other. But the method of ``forwarding messages'' is still of no help.

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  • $\begingroup$ Thank you for this wonderful explanation, and for taking the time to make it so thorough. $\endgroup$ – Klayman Jun 26 '14 at 20:21

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