# See light that is red-shifting z > 5.4

1. Sep 26, 2010

### niceboar

So we can see light that is red-shifting z > 5.4 which means it would be moving faster than light speed away in relation to us. How can we see this? We couldn't see the object's light while the distance between us is increasing at more than light speed right? I realize the light we are seeing is billions of years old but having a z of more than 5.4 doesn't make sense to me. The only thing I can figure is that the space fabric itself is expanding while the light is traveling through it elongating the light more than when it started. Could anyone shed light on this. I feel really bad for making that pun.

Last edited: Sep 26, 2010
2. Sep 26, 2010

### bcrowell

Staff Emeritus
Re: Redshifting

As the source's velocity approaches c, you get $z\rightarrow\infty$, not z=1. $z\approx v/c$ only holds for $v \ll c$.

It is possible to receive a photon from a source that is, and always has, been receding from us at v>c in some sense ("in some sense" because general relativity doesn't provide an unambiguous notion of the relative velocity of distant objects). However, this doesn't correspond to z>1.

3. Sep 28, 2010

### niceboar

Re: Redshifting

yeah I meant z>1.4 not 5.4

so eventually the object's light we see now will redshift to infinity and disappear? I might have misinterpreted something I read then that makes more sense. 1.4 of something we receive now would correspond to the value billions of years ago, now it would be moving faster than c away. Is that the idea?

4. Sep 28, 2010

### JesseM

Re: Redshifting

Last edited by a moderator: May 4, 2017
5. Sep 28, 2010

### niceboar

Re: Redshifting

So how would the Hubble constant work physically? The further away the light travels from the galaxy, the less expansion space undergoes until it reaches space expanding at less than c? I guess it makes sense considering locally we can't actually be moving at c so the light will eventually catch us. So if that is true what is responsible for the slower space expansion at further away from mass? AFAIK dark energy is treated as a sort of homogeneous constant in all space. Would the contraction caused by a galaxy moving and subsequent expansion as the galaxy moves away account for this?

edit
http://curious.astro.cornell.edu/question.php?number=575 [Broken]
So galaxies are the things accelerating, space expansion is actually decelerating.
Unrelated question but would that mean there would be a border to the universe then? A point where time and space itself can't exist?

Last edited by a moderator: May 4, 2017
6. Sep 28, 2010

### Calimero

Re: Redshifting

No, space expansion is not decelerating. Hubble constant is getting smaller, and Hubble constant defines Hubble radius (Hr=c/Ho) from where recession is superluminal. Read Jesse's post again carefully.

7. Sep 28, 2010

### niceboar

Re: Redshifting

That conflicts with the article I just read

8. Sep 28, 2010

### JesseM

Re: Redshifting

For someone who's more familiar with the math of cosmological coordinate systems, a quick question: how does the coordinate speed of light work in the type of "standard" coordinate system used in cosmology? In other words, if we have a coordinate system where a galaxy at a coordinate distance of D (and at rest relative to the local average rest frame of the cosmological microwave background radiation) will have a coordinate speed of v = HD (H=the value of the Hubble constant at that moment in coordinate time), what will be the coordinate speed of a photon passing next to that same galaxy? Will it just be c-v if the photon is moving toward us and and c+v if it's moving away from us, or is it more complicated?

9. Sep 28, 2010

### Calimero

Re: Redshifting

One simple way to understand why Hubble constant is decreasing: consider galaxy 1 Mpc away. It is receding from us at 71 km/s. Now, if value of Hubble constant remains the same, once it is 2 Mpc away it should be receding at 142 km/s, and so on. Our universe is accelerating in expansion, but not all that much.

10. Sep 28, 2010

### niceboar

Re: Redshifting

Okay but what would that have to do with space itself? That seems to me to have to do with the growing distances between galaxies.

What this means to me is the rate of spacetime expansion is slower. Galaxies themselves could be accelerating faster due to dark energy or something. Maybe it doesn't make a lot of sense but that is the only way I can interpret it.

11. Sep 28, 2010

### Calimero

Re: Redshifting

That is semantics issue. Distances = space. Anyway, how would you measure expansion of 'space itself' without objects embedded in it?

12. Sep 28, 2010

### JesseM

Re: Redshifting

Perhaps I am making a mistake when I assume that in the standard cosmological coordinate system, at coordinate time t the coordinate speed of a galaxy at coordinate distance D would be given by v = H(t)*D, where H(t) is the value of the Hubble parameter at that coordinate time? If this were true it would imply that if the Hubble parameter were decreasing with time, that would mean that the coordinate speed of objects passing by some fixed coordinate distance D decreases with time too, and I figured when cosmologists talk about the rate of expansion accelerating they meant the opposite of this (that the coordinate speed of objects passing by a fixed coordinate distance D increases with time). But on this thread George Jones writes:

13. Sep 28, 2010

### niceboar

Re: Redshifting

No idea and I'm not going to pretend I understand how you could do that. But if it is true then you can derive it somehow just don't ask me :)

How do you understand this analogy? To me this means the boat is the galaxy, so what would be the river? I think it would be spacetime.

In this context
Considering z=1.4 corresponds to a galaxy receding at light speed, he claims galaxies at z=1.69 light will still eventually reach us. How else is this possible? Mustn't the expansion of spacetime be decreasing?

Basically what does an isolated Hubble's constant mean physically?
e.g. what keeps the relationship nonlinear? I think it would be more or less linear given Newton's first law, at vast distances gravity's effects would go to 0.

wait is Hubble's constant a measure of the effect of dark energy then? but how would a galaxy light at a redshift of 1.69 reach us (billions of year from now)? I still can't reconcile both ideas.

Last edited: Sep 28, 2010
14. Sep 28, 2010

### niceboar

Re: Redshifting

This is what makes the most sense to me but it doesn't fit with the idea that we can't receive a photon after a certain point.

[PLAIN]http://img545.imageshack.us/img545/66/53429694.png [Broken]

right?

Last edited by a moderator: May 4, 2017
15. Sep 28, 2010

### JesseM

Re: Redshifting

niceboar, I'm pretty sure it's not true that light has a coordinate speed of c in the type of coordinate system used in cosmology to talk about velocities of distant galaxies. For example, the red lines in this diagram (from this section of Ned Wright's cosmology tutorial) represent the past light cone of a given event at our location:

Hopefully someone will answer the question I asked in post #8:

16. Sep 28, 2010

### niceboar

Re: Redshifting

Well I think this statement might answer that
I was wondering how you could reconcile an observation of an object moving faster than c with relativity.

Also if I had to guess I'd say it's more complicated than that. For one thing, the gravity of the galaxy would severely alter the spacetime around it I would think. So to an outside observer viewing the photon would see it going less than c? But I really don't know especially since x and t mean something different in this context. Still it doesn't make a whole lot of sense to me considering special relativity. The galaxy should be going less than c and the photon should be going c. So now I'm all confused I have no idea what this actually means.

So special relativity would say, no it is not expanding faster than c and an astronomical model would say it would?

Also this belongs in the astronomy section.

Last edited: Sep 28, 2010
17. Sep 28, 2010

### JesseM

Re: Redshifting

Yes, I understood that part already (it's impossible for any coordinate system covering a large region of curved spacetime to qualify as 'inertial', and the x and t which appear in the usual equations of SR only refer to coordinates in an inertial frame...note that even in the flat spacetime of SR you can pick a non-inertial frame like Rindler coordinates and the coordinate speed of light will not be c in these coordinates) but it doesn't answer my question about what the coordinate speed of light actually is in the standard cosmological coordinate system.
Cosmological models usually simplify things by assuming the universe is just filled with a homogeneous fluid with no local differences in density. My question was about what would be true in such a model.
What do you mean by "see"? Are you talking about visual appearances, or (change in coordinate position)/(change in coordinate time) in some coordinate system? In cosmology there's no notion that different observers each have their own "frame", so it's not like the speed of light would be different for an "outside observer" vs. any other observer.

18. Sep 28, 2010

### niceboar

Re: Redshifting

I was just saying I also think your hunch was right, that's all.

I thought you could translate non-inertial frames to inertial frames using extra psudoforces.

Obviously there is a lot I don't understand about the basics of cosmology.

19. Sep 28, 2010

### Calimero

Re: Redshifting

Hi Jesse, in the case of $$\Omega_{M}=\Omega_{\Lambda}=0$$ universe Hubble constant is just $$1/t$$, thus it necessarily must decrease over time. There is also exact solution from Friedmann's, which is: $$a(t)= H_{0}t$$

Last edited: Sep 28, 2010
20. Sep 28, 2010

### JesseM

Re: Redshifting

So is it true that a test particle at distance D from the origin at time t would have coordinate velocity D/t in such a universe? If so, would one say that the expansion rate is decreasing with time since the velocity of a particle at a given distance decreases over time, or when cosmologist talk about the expansion rate accelerating/decelerating do they mean something different?

21. Sep 29, 2010

### Ich

Re: Redshifting

Hi JesseM,

your#8:
you have
$$D= a r \dot D = \dot a r + a \dot r =H D + a \dot r$$
The term on the left is called "recession velocity"
The first term on the right side is called "recession velocity due to expansion" or something like that, the second term is called "peculiar velocity".
So yes, you simply add peculiar velocities to HD to get the "recession velocity" of something.

Yes.

They mean rather the velocity of some defined particle. In this case, H~1/t, but D~t, so v=const, as it should be in an empty universe. Essentially, we're talking about $\dot a$ being positive or not. See http://en.wikipedia.org/wiki/Deceleration_parameter" [Broken].

Last edited by a moderator: May 4, 2017
22. Sep 29, 2010

### JesseM

Re: Redshifting

Thanks Ich. Two more quick questions:
And I presume from your "yes" that it's also true that the cosmological coordinate system is constructed in such a way that the "peculiar velocity" of a light ray moving directly towards us or away from us is always c?
Ah, so when they talk about the expansion rate accelerating/decelerating they are talking about picking a test particle and seeing how its coordinate velocity changes with coordinate time in the cosmological coordinate system? But even in the standard FLRW cosmologies with cosmological constant set to zero this velocity v(t) would increase as the test particle moved further away from us, no? But then I'm confused about why cosmologists were surprised that the expansion rate was accelerating--does "expansion rate is accelerating" just mean the first derivative of v(t) for this test particle is positive (i.e. the velocity away from us increases with time) or does it mean something different?

Last edited by a moderator: May 5, 2017
23. Sep 29, 2010

### Calimero

Re: Redshifting

No, in empty universe velocity would remain constant and deceleration parameter is q=0. It is also good example to help understanding Hubble's parameter, because in such universe it is simply H=1/t.

In universe dominated with matter, velocity would decrease, and q>0. In cosmology constant dominated universe q is negative and velocities increase.

Why would you think that without cosmological constant, velocity should increase anyway?

They are not talking about v(t), they are talking about a(t). See http://en.wikipedia.org/wiki/File:Friedmann_universes.svg" [Broken].

Last edited by a moderator: May 5, 2017
24. Sep 29, 2010

### JesseM

Re: Redshifting

I don't know the physical meaning of "deceleration parameter"...
I suppose it was just the intuition that since velocity due to expansion increases with distance at any given moment, and the change in the Hubble constant occurs on very long cosmological timescales, I'd have expected the velocity of a given test particle (especially one very far away whose velocity is large and which therefore has a distance that's increasing rapidly) to increase as its distance increases. But of course there's no substitute for a more mathematical analysis, I guess my intuition was wrong here.
I said "the first derivative of v(t)", isn't that the same as a(t)? If so, does "expansion is accelerating" just mean a(t) is positive (assuming positive a means the velocity in the direction away from us is increasing) or something else?
The graph is dealing with the "average distance between galaxies", so presumably the slope of each line could be interpreted as v(t) for a test particle in each type of universe? So I can see from that that in each case where the cosmological constant is set to zero, it does look like the slope is continually decreasing, meaning a(t) is always negative.

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25. Sep 29, 2010

### Calimero

Re: Redshifting

It is really simple. If you look at the graph, ordinate represents average distance between galaxies, or scalefactor - a. Wherever slope of the curve is more then 45 deg relative to abcissa, universe is accelerating.

Edit: Stupid thing to say, I am surelly wrong. It comes to this equation: $$q=-\frac{\ddot{a}a}{\dot{a}^{2}}$$

As I said if q is negative then universe is accelerating.

Last edited: Sep 29, 2010