# I Universe is a sphere that is centered on any observer? How?

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1. Jun 2, 2016

### Lohan

I read here, http://www.space.com/24781-big-bang-theory-alternatives-infographic.html , that,

"What we call the "observable universe" (or the "Hubble Volume") is the spherical region, about 90 billion light-years in diameter, that is centered on any given observer. This is the only part of the universe in which light has had time to reach the observer in the 13.8 billion years since the universe began. "

How on earth is it possible to be centered on any given observer??

How can the observable universe for an observer near the boundary of the Hubble Volume also be a 90 billion light-year diameter sphere where that observer is the center??

2. Jun 2, 2016

### ShayanJ

There is your misunderstanding. You're assuming that Hubble volume is an objectively selected part of spacetime, its not!
The definition of Hubble volume is "spherical region of the Universe surrounding an observer beyond which objects recede from that observer at a rate greater than the speed of light due to the expansion of the Universe." So this is a subjective concept and each observer, has a different Hubble volume. So when you say Hubble volume, you should specify whose Hubble volume you're talking about!
So an observer in the edge of another observer's Hubble volume, has his own Hubble volume centered around himself. No contradiction!

3. Jun 2, 2016

### Staff: Mentor

Because it's what the observer can see, due to the finite speed of light.

As an analogy, if you go on top of the Klein Matterhorn (on a clear day), you only see part of the Earth; you could call that the "visible Earth." Another person, on top of the Mont Blanc, will see another "visible Earth." The two parts of the Earth will overlap, but each will see a different bit of the Earth. Same with the Universe.

4. Jun 2, 2016

### Lohan

Can you guys actually visualize this concept of different Hubble volumes for different observers in your minds, or is it something purely mathematical (as a result of mathematics of general relativity) or abstract that we cannot actually visualize?

Is there like a drawing on this you can point to where I can visualize it with my mind, because I sure can't do it now.

For example, if I suddenly, lets say through a wormhole, in a split second travel 44 billion light years. Now, the distance between earth and me is 44 billion light years. So, how come I am not 1 billion light years near Earth's Hubble volume's boundary?

How come I am not only just 1 billion light light years from the boundary of the observable universe?

Like for example, nobody can visualize space-time in their heads, right? We can visualize space and time, but not space-time, right?
It's a purely abstract mathematical physics concept, right?? i.e. There is actually no "thing" called space-time right?? Is this Hubble Volume also a concept like this?

5. Jun 2, 2016

### bapowell

DrClaude's analogy with the mountains is an excellent visualization. Did you read his post?

6. Jun 2, 2016

### Lohan

I did read it but don't get it.

7. Jun 2, 2016

### Bandersnatch

Here's a picture of the same thing, showing two observers at two different places in the universe, with the same time elapsed since big bang:

By the way, these observable regions are not what Hubble volume (aka sphere) is. The definition for Hubble volume was given in post #2 by Shyan. You seem to be talking simply about the radius of observable universe, not the Hubble volume.

8. Jun 2, 2016

### Delta²

You are right that we cant visualize it because our eyes and most importantly our brain, see things in 3 dimensions and perceive the 4th dimension, that is the time dimension, in a very special way. But there IS actually space-time it is a real thing (at least according to relativity theory and string theory) it is not just a mathematical object.

9. Jun 2, 2016

### Chiclayo guy

In the diagram in post #7, assume there is a galaxy in the overlapping observable regions. Will Adam and Barbara both view the galaxy as receding from them?

10. Jun 2, 2016

### Bandersnatch

Yes. And both A and B are of course receding from each other (as are any two points you choose that are not gravitationally bound).

11. Jun 2, 2016

### cristo

Staff Emeritus
Note that the observers themselves are not stationary (this is not a fixed coordinate system), but are receding from one another.

12. Jun 2, 2016

### Chronos

Keep in mind each observer is looking back across time as well as space. A galaxy observed at the same time by observers A and B does not appear to be the same age to both observers [again, due to to the finites speed of light]. Only a galaxy that is equidistant from both observers will appear to be the same age to both of them. It will invariably appear older to whichever observer is less distant. and coversely younger to the more distant observer. By same age, we are talking relative to the apparent age of the universe for each observer.

13. Jun 5, 2016

### sunrah

Cosmology has two underlying principles: no point is special (every point is same) and everything looks the same in all directions. If you think the universe is an infinite flat sheet, then pick any point and draw a circle with radius 45 Gly around it. If you think the universe is like the surface of a sphere, then do something similar. You'll find that in all the possible models, you can choose any point a random and draw a visible universe around it without going over any boundary.

14. Jun 12, 2016

### Andrew Wright

Can Adam take photos of things Barbara can never see and beam them to Barbara?

Edit: ok, I can see from the diagram that Adam and Barbara don't know about each other. What if they were in each other's view?

Last edited: Jun 12, 2016
15. Jun 12, 2016

### Bandersnatch

That's an very nice question.

After all, if we assume Adam being within the event horizon* of Barbara, so that light that is sent from A NOW can reach B some time in the future, does it not mean that A could take a snapshot of the universe HE sees NOW, containing more (or rather, a different region) than what Barbara can see, and send it together with the 'regular' light that's leaving A?

The key thing here is the difference in time - at the moment of taking the snapshot, the extent of A's observable universe is determined by how much time has elapsed since the Big Bang. And so is B's.
While at the time of emission/taking snapshot, B can't see (some of) the stuff A's sending her, she will be able to see all of it (as it was at the time of sending) by the time the snapshot reaches her. It will be equally outdated, and equally redshifted.
One of the ways of seeing it is that there's really no need for making a photo and sending it - the light A sees, that he could make a photo of, is the same light that is passing A on its way to B all by itself.
Another way of seeing this, is that everything tha A sees NOW as his observable universe, will be eventually encompassed by B's radius of the observable universe (as it grows) - as long as they're NOW inside each other's EH.

For the snapshot to contain information that B can't ever see, A would have to wait until he's carried away by the expansion out of B's event horizon before snapping the photo, but then the signal would never reach B by definition of being outside the event horizon.

*the event horizon is neither the radius of the observable universe (aka particle horizon), which tells you from how far (in terms of proper distance) the oldest signals you can see come, regardless of whether a signal sent NOW from the observed object would be able to reach you, nor is it the Hubble radius (aka sphere, volume), which tells you where the recession velocity due to the expansion is c. Rather, the cosmological EH tells you the present distance from which a signal sent NOW will be able to reach you at some time in the future. In our universe, EH grows, asymptotically approaching a set value.
In terms of current size, particle horizon>EH>Hubble radius. Particle horizon grows indefinitely, EH asymptotically approaches a set radius, HR asymptotically approaches EH.

Last edited: Jun 12, 2016
16. Jun 12, 2016

### Stephanus

You don't have to on earth, but anywhere in this universe is the center of the universe

17. Jul 8, 2016

### Lohan

I am trying to get my mind on this but still can't . I just read the definition of the Hubble sphere in Wikipedia: https://en.wikipedia.org/wiki/Hubble_volume
It says this: "More generally, the term "Hubble volume" can be applied to any region of space with a volume of order {\displaystyle (c/H_{0})^{3}}[PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/bb96ddb1e504b586036e142bf55d6e5f211fe90b. [Broken] However, the term is also frequently (but mistakenly) used as a synonym for the observable universe; the latter is larger than the Hubble volume."

How can can the observable universe be larger (i.e. larger in diameter, right?) than the Hubble volume??

Last edited by a moderator: May 8, 2017
18. Jul 8, 2016

### Bandersnatch

Imagine you're living in a universe where the expansion rate (i.e. the Hubble constant) is constant not only in space (i.e., same everywhere at a given moment) but also in time (i.e., same everywhere and at any time).
In such a universe, if you draw a Hubble sphere ($H_R$) around yourself (observer), and then imagine a signal being sent from just outside $H_R$, then its approach velocity of $c$ will not be sufficient to ever get closer to the observer because the recession velocity at that distance is greater than $c$.
For example, a photon emitted just 1 metre outside the $H_R$ will have some very, very small net velocity away from you, that will only ever grow in time.

Now, make the Hubble constant change it time - specifically make it go down in time.

Look at the photon that was emitted 1 m outside the $H_R$. Now, providing sufficient time has passed, and the Hubble constant went down sufficiently (which is equivalent to growth of the $H_R$), the photon will find itself in a region where recession velocity is smaller than $c$, and as a result will have some net velocity towards the observer. From this point on, the photon is bound to eventually reach you to be observed. In other words, you will see the object that once emitted the photon - even though that object was outside the $H_R$!

This shows that you can observe objects located at distances higher than $H_R$ during emission.

Note that while the photon emitted just outside the $H_R$ is almost frozen in place (only from the observer's vantage point, locally it always moves at the speed of light), as its own velocity of c towards the observer combined with the recession velocity at that distance net something close to 0, the body that emitted the photon does not have such proper velocity and is instead carried away with full recession velocity. Hence while the photon of emitted light may sometime reach the observer, the object that emitted that photon will have been by that time carried away by the expansion much farther than where it had been when the light was emitted. This distance at which the emitter is NOW - at the time of reception of the once-emitted photon, is called proper distance. It is not where the emitter was when the light left it, but where it is now when the light reached the observer.
Proper distance of the farthest visible object (the CMBR) is called the Particle Horizon - or the radius of the observable universe.

The above paragraph works also for objects that emitted light from the inside of $H_R$. All you need for the proper distance to be larger than $H_R$ is for the recession to carry the emitting galaxy away fast enough to overtake the growing $H_R$ by the time its signal reaches the observer.

This shows that you can observe objects located at distances higher than $H_R$ during reception.

In all of the above, the particulars of how far an emission of light can be outside the $H_R$ to be eventually observable, depends on the exact way the Hubble constant goes down with time. The more rapid the reduction, the farther a photon can be emitted.
Since the way it changes in our universe looks like this:

the $H_R$ was growing much faster in the past, and is approaching 0 growth in the far future (i.e., at the limit of infinity), which means photons could have been emitted farther away from the $H_R$ in the past than today, and still be observable. In the far future no photons emitted outside the $H_R$ will be ever observable.

Last edited: Jul 8, 2016