# Observable boundary

1. Apr 27, 2010

### thecow99

If space is expanding then at at some point we must reach the point where it's expansion is faster than C. Do we know where that is?

Is the expansion uniform or is it dependent on something like Dark Matter clusters?

Is the expansion accelerating or decelerating at two fixed points?

In addition, is this why it takes so long for light from distant galaxies to reach us? If light from distant galaxies was actually emitted 13.8 billion years ago one would think 13.8 billion years ago those galaxies were MUCH closer, surely not 13.8 billion lightyears away.

Last edited: Apr 27, 2010
2. Apr 27, 2010

### AleLucca

First it should be said that nothing can travel faster than c. In addition, velocity is a quantity that is well-defined for a local observer only, so it is not clear in which sense we can talk about velocity for a very distant object.

Nevertheless, we can define the recessional velocity of an object as the rate of change of its "physical distance" (the istantaneous distance between two points) with time. This velocity increases with increasing physical distance according to the Hubble law:
v = H_0*d
Since H_0 = 70km/s/Mpc then you have v=c for d~4000Mpc.
We actually observe galaxies at this distance but they are not the observable boundary, since we can observe further out.

The expansion is uniform only on large (>100Mpc) scales. Locally there are departures from a uniform expansion: for example, our galaxy is moving towards the Andromeda galaxy, so they are getting closer.

The expansion is now accelerating, but it was decelerating until "not long ago" (cosmologically speaking).
Acceleration is not necessary to explain the long travel time of photons coming from distant galaxies.
But I'd better stop here because I don't want to say imprecise things.

3. Apr 27, 2010

### thecow99

Ok, so for every 1Mpc the rate of expansion is 70km/s

Then expansion would hit C at 4282Mpc or roughly 13.9 billion lightyears.

Is is not safe to say that the observable boundary is 13.9 billion lightyears since no electromagnetic wave beyond that can reach us?

Don't get me wrong, I'm not saying nothing exists beyond this point, I'm just saying nothing that we can observe in an electromagnetic wave format.

4. Apr 27, 2010

### AleLucca

No. The trick is the following:
The physical distance of a galaxy at redshift z~1 is now ~14 billion lightyears. However, this is the "istantaneous distance": the distance we would measure if the expansion stopped in this moment.
The photons that reach us today were emitted billions of years ago. At that time the Universe was much smaller and the galaxy was much closer to us. This allowed photons emitted from that galaxy and from more distant objects to reach us now.
In other words, the distance travelled by these photons is smaller than the "istantaneous distance"

Plus, the Hubble constant is not truly a constant but changes in time: at the time of emission of photons coming from sources at high redshift it was way much larger than 70 km/s/Mpc

5. Apr 27, 2010

### thecow99

Say there is a star 5000Mpc away. Are you saying light emitted from that object has the capability to reach us?

If no, then it must be outside the boundary of our observable universe.
If yes, how is this possible since the rate of expansion is greater than C?

6. Apr 27, 2010

### AleLucca

Yes.
1- The rate of expansion is greater than c now and it was even greater at the time of the emission of the signal.
But you should recall that this speed is not a physical speed, since in reality nothing can exceed the speed of light.
2- No matter what the recessional velocity of an object is, the photons emitted travel at c (the speed of light is c in any frame of reference), so the time it takes to reach the observer depends only on its distance.
3- The distance between such a source and the observer increases as the photon travels, but keeps finite, so that the photon will eventually reach the observer.

7. Apr 27, 2010

### AleLucca

The observable boundary is instead "the current position of a source for which we observe now the photons it emitted at the time of the Big Bang".
(you can convince yourself that this is the correct definition of the horizon by realizing that photons emitted later cannot overtake the photons emitted at the time of the Big Bang).

It should be clear that for an object located further away than the above source, the photons it emitted during his life haven't reached us yet.

8. Apr 27, 2010

### thecow99

I'm just having real trouble understanding how a photon traveling at C can overcome a gap in space that is expanding faster than C.

If you fire a bullet at 1200mph at a target receding away at 1201mph then the bullet is never going to reach the target no matter how much time elapses.

Would the electromagnetic wave not redshift into a flat line beyond the 13.9Bly point where spacial expansion is greater than C?

Unless you are saying that due to a deceleration in expansion that would in a sense allow the photon to gain ground in the distant future.

9. Apr 27, 2010

### thecow99

If I'm reading this correctly, I think we may be debating a moot point. I was looking for an observable boundary in a frozen frame.

I believe now you were telling me the observable boundary is not constant due to a deceleration in expansion. Yes?

10. Apr 27, 2010

### marcus

SciAm article explained that nicely. It happens because H changes over time. H used to be large like 100,000 and now much smaller, like as you said 70.
I have a princeton.edu link in my sig to the SciAm article by Charles Lineweaver that explains how the photon does it, by pictures.

Essentially what happens is that as H decreases the hubble distance c/H
which is the distance of something receding at c
grows larger.

Cosmos calculator also gives past values of H so you can realize how rapidly H has decreased over time (and it is still decreasing today but more slowly----someone who thinks that 'accelerating expansion' means that H should increase does not understand the meaning of accelerating expansion and should ask, its a separate question)

Any fraction with a decreasing denominator will grow.

So the growing hubble distance REACHES OUT TO THE STRUGGLING PHOTON.

Even tho the photon is being swept back by the current of expansion, the hubble distance reaches out to it (historically it has grown very fast)
and once the photon is within the hubble distance of us, it must reach us, because distances shorter than c/H grow slower than c.

You should learn to use the cosmos calculator. Google "cosmos calculator".
Ask anybody here how to use it. You put in the redshift z and it tells you how far away the galaxy is, and how fast it is receding.

11. Apr 28, 2010

### AleLucca

Thanks marcus, your explanation was way more convincing :)

Ok. So, what you mean by "observable boundary"?

12. Apr 28, 2010

### thecow99

I wanted to find out in a frozen frame what the maximum distance we could detect electromagnetic waves from, the "observable" boundary. With your help I was able to calculate ~13.9Bly. However, you were also correct that the boundary is not set due to a deceleration in the expansion rate which I was unaware of. You're answers just weren't exactly what I was looking for and I didn't quite understand because I was looking for a frozen frame reference and your answers explained the non frozen equivalent so I got a bit confused.

Thanks for supplying the link Marcus, it cleared up a couple things for me.

Till I think up something else that confuses me.... Cheers as always!

13. Apr 30, 2010

### AleLucca

But what do you mean with "frozen frame"?
Are you thinking about the hypothetical case of a constant expansion rate? Or really a "frozen" expansion (i.e. the galaxies are still relative to each other)?

How does this frozen frame relate with our capability of detecting photons from a given source?