3
$\begingroup$

My layman understanding is that the universe is much bigger than that we can observe however due to the cosmological constant the observable universe is the matter etc. that is not moving away from us at a speed faster than light.

It is said that we can never reach matter outside the observable universe which leads me to the following paradox; imagine two people are placed at galaxy A & B in the below diagram: Venn Diagram of two observable universes.

The two circle's indicated are the observable universe from their own perspective, A can never reach B as he is moving away from him faster than light and vice versa. A and B can however reach C, so they both decide to go there to meet up.

My question is would they ever actually reach each other?

From my research I know (without understanding the maths) that even moving towards each other at 0.99c they will still only move towards each other at say 0.9999C (basically never above light speed) but it seems that matter from outside their observable universe could reach them. I have tried to find an answer and found that perhaps when A and B get to C their home universes have now moved away so that neither could ever get home (which means A will never see B's home galaxy and vice versa) but this still doesn't explain how A could reach B (if it even can). I may be misunderstanding the expansion / observable universe but from what I have read the above logic (A never reaches B) should be correct?

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ It seems to me that the answer to the question has to depend on whether the universe is expanding or collapsing and that the problem can not be visualized correctly with Venn diagrams. The correct visualization would require a Penrose diagram. I think that the paper "Penrose Diagram for an uniformly accelerated observer" by Claude Semay may contain part of the answer to your question. I am still looking for Penrose diagrams that show the possible global geometries of the universe. $\endgroup$
    – CuriousOne
    Sep 4, 2015 at 12:43
  • 1
    $\begingroup$ I don't have the time for a comprehensive answer, but note that you don't need to move faster than light to reach an observer which is so far away that expansion pulls her away faster than light. Consider a point emitting light in your direction, but receding faster that light. Initially, the light will move in your direction, but will increase its distance to you. However, it will recede slower and slower until eventually expansion will actually make it recede faster that light from the source toward you. See Ant on a rubber band. $\endgroup$
    – pela
    Sep 4, 2015 at 12:56
  • $\begingroup$ @CuriousOne Thanks for looking, I had a quick look at Penrose diagrams but will need to sit down when I get home from work as I had never heard of them before, it makes sense that it was a conceptual problem (as a lot of physics seems to be) as that is what I thought it would be! $\endgroup$
    – JackFrost
    Sep 4, 2015 at 13:15
  • $\begingroup$ @pela Thanks for the heads up, that example made a lot of sense basically the % of the journey completed will always get smaller so eventually you will reach the end point. One thing that didn't make sense 'it will recede slower and slower until eventually expansion will actually make it recede faster that light' seems to be wrong, do you mean it will recede faster and faster? Also does that mean it would be possible to leave the observable universe given enough time? $\endgroup$
    – JackFrost
    Sep 4, 2015 at 13:17
  • $\begingroup$ @JackFrost: Sorry, I meant as it travels from the emitter, it will of course increase its distance from the emitter, but in the beginning, the expansion will cause it to also increase its distance from us. But in time, this increase of distance from us will be slower, so that it recedes slower and slower, until eventually it will start approaching us. $\endgroup$
    – pela
    Sep 5, 2015 at 19:34

1 Answer 1

Reset to default
0
$\begingroup$

Yes they can meet.

Assuming your diagram has the first circle as all the events that A can reach and the second circle is all the events that B can reach and assuming event C is in the intersection then A and B can meet there.

They can't necessarily "decide" to meet there of waiting for bilateral communicate before you head out makes it so they can't both reach C.

Keep in mind the essential difference between things that can affect you and things you can affect.

If C is in the interaction of things A can affect and things B can affect, then both can affect C.

This is normal. When you look out far away you will see thibgs that you can affect but some of the thibgs that affect it are not things you can affect. This already happens in Special Relativity. Two spacelike separated events can sometimes affect the same event but not each other.

Now if you are a moving observer, then observer A originally at event A could meet observer B originally at event B. And then they can shake hands at event C even though the event A and event B were separated because it is the observers A and B affecting each other, neither is going back in time to change event A or event B.

Which is back to first point, they can't decide to meet up. When they meet up that decision (those two independent decisions) to go to C is/are in the causal past and so already happened.

There isn't a big mystery. When you travel north you can meet someone that was traveling south. If your cars could only go an hour's drive you could thus meet someone that lives 2 hours drive away. Nothing spooky is going on in the slightest and this is totally normal.

No. They can not meet.

Assuming your diagram has the first circle as all the events that can reach observer A and the second circle is all the events that can reach observer B.

Then if A is not in the right circle and B is not in the left circle then there is no place and no way they can meet. By definition.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.