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The observable universe is limited by a cosmic horizon. Galaxies beyond the horizon move away from us faster than light, so we cannot see them. If we could see a planet close to our horizon, we would see time there strongly dilated and moving slower than ours conceptually coming to a halt at the horizon, as we see it.

Imagine a planet beyond our horizon moving faster than light away from us. We cannot see this planet, as it is shielded from us by the horizon. However, can we theoretically describe how exactly time moves there? If from our viewpoint time stops at the horizon, then what happens to time beyond the horizon?

None of the obvious logical possibilities feel right: time is frozen, starts moving, moves in reverse, becomes a spacelike coordinate. What is the proper answer?

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The answer is surprisingly boring as in the FLRW metric all comoving observers share the same time coordinate. That is, there is a universal time (comoving time) that records the time since the Big Bang, and all comoving observers share this time coordinate. So time is not dilated for distant observers and it doesn't do anything odd as we consider observers the far side of the cosmic horizon.

When you talk ask what happens if we see a planet close to our horizon you need to be clear what you are asking. If you're asking about the light falling on our CCDs and forming an image there then the simple answer is that we would see nothing as the light hasn't had a chance to reach us yet. If you are using the verb to see in the sense of to observe, i.e. the assignment of spacetime events, then my first paragraph applies and observers close to or beyond the horizon share the same time coordinate as us.

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  • $\begingroup$ Consider a clock on a remote planet. The inhabitants there are beaming a ray of light toward us modulated by each tic-tac of the clock. We see this light deeply redshifted, but the number of oscillations between the tic-tac marks cannot change, so we see the time periods between the marks increased as $z+1$, which means we see their time dilated as $z+1$. No? $\endgroup$
    – safesphere
    Commented Oct 21, 2017 at 19:40
  • $\begingroup$ @safesphere: You're using see in the sense of what our eyes/telescopes record, in which this case this is a combination of whatever time dilation may be going on and the travel time of the light. Even in non-relativistic situations where there is no time dilation the clocks run slow if you define seeing the clock in this way, because each successive signal from the clock takes longer to reach us than the previous one. This isn't what we'd normally call time dilation. $\endgroup$
    – John Rennie
    Commented Oct 22, 2017 at 4:51
  • $\begingroup$ Sorry, your comment is not helping. On "the travel time of the light", light is never red shifted in the frame of the emitter, only in the frame of the receiver, so this point is irrelevant. What "non-relativistic situations"? In non-expanding space my example follows Einstein's synchronization. It is not clear why it should be different in expanding space. You may be right, but it is not obvious. Perhaps someone can clarify this with math. $\endgroup$
    – safesphere
    Commented Oct 22, 2017 at 5:39
  • $\begingroup$ Hi John, back to this question since a couple months. The relativistic Doppler shift consists of two effects, the Doppler component 1+β plus time dilation γ or combined z+1=(1+β)γ. According to your answer, there is no time dilation γ=1, only the Doppler component. If this were true, the redshift at the cosmic horizon of β=1 would be z=1 that clearly contradicts observations oz z>11 for distant galaxies. Could you please clarify? Thanks! $\endgroup$
    – safesphere
    Commented Dec 25, 2017 at 22:19
  • $\begingroup$ This is not quite right. It would be more accurate to say that we simply can't define time dilation in a cosmological spacetime in the same way that we could for a static spacetime. $\endgroup$
    – user4552
    Commented Dec 26, 2017 at 3:41

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