See light that is red-shifting z > 5.4

[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.

 

[bcrowell]

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.

 

[niceboar]

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.

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?

 

[JesseM]

http://www.scientificamerican.com/article.cfm?id=misconceptions-about-the-2005-03&page=4 says:

According to the usual Doppler formula, objects whose velocity through space approaches light speed have redshifts that approach infinity. Their wavelengths become too long to observe. If that were true for galaxies, the most distant visible objects in the sky would be receding at velocities just shy of the speed of light. But the cosmological redshift formula leads to a different conclusion. In the current standard model of cosmology, galaxies with a redshift of about 1.5--that is, whose light has a wavelength 150 percent longer than the laboratory reference value--are receding at the speed of light. Astronomers have observed about 1,000 galaxies with redshifts larger than 1.5. That is, they have observed about 1,000 objects receding from us faster than the speed of light. Equivalently, we are receding from those galaxies faster than the speed of light. The radiation of the cosmic microwave background has traveled even farther and has a redshift of about 1,000. When the hot plasma of the early universe emitted the radiation we now see, it was receding from our location at about 50 times the speed of light.

Running to Stay Still
The idea of seeing faster-than-light galaxies may sound mystical, but it is made possible by changes in the expansion rate. Imagine a light beam that is farther than the Hubble distance of 14 billion light-years and trying to travel in our direction. It is moving toward us at the speed of light with respect to its local space, but its local space is receding from us faster than the speed of light. Although the light beam is traveling toward us at the maximum speed possible, it cannot keep up with the stretching of space. It is a bit like a child trying to run the wrong way on a moving sidewalk. Photons at the Hubble distance are like the Red Queen and Alice, running as fast as they can just to stay in the same place.

One might conclude that the light beyond the Hubble distance would never reach us and that its source would be forever undetectable. But the Hubble distance is not fixed, because the Hubble constant, on which it depends, changes with time. In particular, the constant is proportional to the rate of increase in the distance between two galaxies, divided by that distance. (Any two galaxies can be used for this calculation.) In models of the universe that fit the observational data, the denominator increases faster than the numerator, so the Hubble constant decreases. In this way, the Hubble distance gets larger. As it does, light that was initially just outside the Hubble distance and receding from us can come within the Hubble distance. The photons then find themselves in a region of space that is receding slower than the speed of light. Thereafter they can approach us.

The galaxy they came from, though, may continue to recede superluminally. Thus, we can observe light from galaxies that have always been and will always be receding faster than the speed of light. Another way to put it is that the Hubble distance is not fixed and does not mark the edge of the observable universe.

 

[niceboar]

http://www.scientificamerican.com/article.cfm?id=misconceptions-about-the-2005-03&page=4 says:

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
looks like this article answered it
http://curious.astro.cornell.edu/question.php?number=575

You might be wondering how we could possibly see a galaxy that is moving away from us faster than the speed of light! The answer is that the motion of the galaxy now has no effect whatsoever on the light that it emitted billions of years ago. The light doesn't care what the galaxy is doing; it just cares about the stretching of space between its current location and us. So we can easily imagine a situation where the galaxy was not moving faster than the speed of light at the moment the light was emitted; therefore, the light was able to "outrun" the expansion of space and move towards us, while the galaxy moved away from us as the universe expanded. Keeping in mind what we learned above -- that farther objects recede faster in a proportionally stretching universe -- we can immediately see that right after the light is emitted, the galaxy is moving away from us faster than the point at which the light is located, and that this disparity will only increase as time goes on and the galaxy and light separate even more. Therefore, we can easily have a situation where the galaxy keeps on moving away faster and faster, eventually reaching or exceeding the speed of light relative to us, while the light which it emitted billions of years ago leisurely coasts on, never having to move across a region of space that was stretching faster than the speed of light, and therefore reaches us eventually.

You might also be wondering how a galaxy is ever able to surpass the speed of light barrier in the first place; for that, see our answer to a previous question.

The fact that galaxies we see now are moving away from us faster than the speed of light has some bleak consequences, however. Astronomers now have strong evidence that we live in an "accelerating universe," which means that the speed of each individual galaxy with respect to us will increase as time goes on. If we assume that this acceleration continues indefinitely, then galaxies which are currently moving away from us faster than the speed of light will always be moving away from us faster than the speed of light and will eventually reach a point where the space between us and them is stretching so rapidly that any light they emit after that point will never be able to reach us. As time goes by (billions of years in the future), we will see these galaxies freeze and fade, never to be heard from again. Furthermore, as more and more galaxies accelerate past the speed of light, any light that they emit after a certain point will also not be able to reach us, and they too will freeze and fade. Eventually, we will be left with a universe that is mostly invisible, with only the light from a few, very nearby galaxies (whose motions are strongly affected by local gravitational interaction) to keep us company. For more details, here is a technical paper on this topic.

Which galaxies are currently "saying their last goodbyes?" That is, if we imagine that there are aliens living in these galaxies who hope to make contact with us, which galaxies are running up against their deadline right at this moment? A reasonable guess would be that the galaxies which are currently moving at the speed of light with respect to us (at a distance of 4,200 megaparsecs and redshift of 1.4, as discussed above) are at the "critical point" where any light they emit after now will never be able to reach us. Roughly speaking, this is correct, but a detailed calculation (such as the one contained in this paper) shows that for the simplest viable model of the universe's acceleration, it is actually galaxies at a distance of 4,740 megaparsecs and redshift of 1.69 that are just now reaching the critical point, while galaxies at a redshift of 1.4 are still emitting light that will eventually reach us.

The difference is due to a rather subtle fact: Even though the universe is "accelerating" in the sense that each galaxy moves faster as time goes on, the Hubble constant is actually decreasing with time -- in other words, the rate at which space is expanding, measured at a point which is at a fixed distance from us, gets smaller as time goes on. If we keep our eyes on an individual galaxy as it moves away from us, we will see it accelerate, but if we keep our eyes on a fixed point in space and watch many different galaxies go past that point, each galaxy's speed will be slower than the one before it. (As a very rough analogy, the universe behaves like a river with rapids. If you put a boat in the river and allow it to be carried by the flow, it will accelerate as it moves downstream and enters the rapids. But if you sit on the bank and measure the speed of the water at one location, it changes based on an entirely different set of factors -- for example, the rate at which the supply of water from upstream is changing. It is possible for the water speed at your location to decrease with time, even though each boat that you release accelerates as it heads into the rapids.) Because of this effect, if light is able to "swim against the tide" and remain at a roughly constant distance with respect to us (as would happen if it is emitted from a galaxy moving away from us at the speed of light), then as time goes on and the Hubble constant decreases, it will eventually be able to gain ground, "swim upstream" and traverse the necessary distance of space to reach us.

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?

 

[Calimero]

So galaxies are the things accelerating, space expansion is actually decelerating.

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.

 

[niceboar]

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.

That conflicts with the article I just read

in other words, the rate at which space is expanding, measured at a point which is at a fixed distance from us, gets smaller as time goes on.

 

[JesseM]

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?

 

[Calimero]

That conflicts with the article I just read

in other words, the rate at which space is expanding, measured at a point which is at a fixed distance from us, gets smaller as time goes on.

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.

 

[niceboar]

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.

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.

in other words, the rate at which space is expanding, measured at a point which is at a fixed distance from us, gets smaller as time goes on.

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.

 

[Calimero]

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.

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

 

[JesseM]

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?

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:

A common misconception is that, since the expansion of the universe is at present accelerating, the value of the Hubble constant is presently increasing. Actually, it is presently decreasing.

To figure all this out, one needs the scale factor as a function of time, i.e, one needs to solve the Friedmann equation, so I think you want to know how observations lead to a particular best-fit solution to the Friedmann equation. But I could be wrong.

 

[niceboar]

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

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 :)

As a very rough analogy, the universe behaves like a river with rapids. If you put a boat in the river and allow it to be carried by the flow, it will accelerate as it moves downstream and enters the rapids. But if you sit on the bank and measure the speed of the water at one location, it changes based on an entirely different set of factors -- for example, the rate at which the supply of water from upstream is changing. It is possible for the water speed at your location to decrease with time, even though each boat that you release accelerates as it heads into the rapids.

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

Which galaxies are currently "saying their last goodbyes?" That is, if we imagine that there are aliens living in these galaxies who hope to make contact with us, which galaxies are running up against their deadline right at this moment? A reasonable guess would be that the galaxies which are currently moving at the speed of light with respect to us (at a distance of 4,200 megaparsecs and redshift of 1.4, as discussed above) are at the "critical point" where any light they emit after now will never be able to reach us. Roughly speaking, this is correct, but a detailed calculation (such as the one contained in this paper) shows that for the simplest viable model of the universe's acceleration, it is actually galaxies at a distance of 4,740 megaparsecs and redshift of 1.69 that are just now reaching the critical point, while galaxies at a redshift of 1.4 are still emitting light that will eventually reach us.

The difference is due to a rather subtle fact: Even though the universe is "accelerating" in the sense that each galaxy moves faster as time goes on, the Hubble constant is actually decreasing with time -- in other words, the rate at which space is expanding, measured at a point which is at a fixed distance from us, gets smaller as time goes on.

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?

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.

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.

 

[niceboar]

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

right?

 

[JesseM]

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:

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?

 

[niceboar]

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:

Well I think this statement might answer that

The time and distance used in the Hubble law are not the same as the x and t used in special relativity, and this often leads to confusion. In particular, galaxies that are far enough away from us necessarily have velocities greater than the speed of light:

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.

 

[JesseM]

Well I think this statement might answer that

The time and distance used in the Hubble law are not the same as the x and t used in special relativity, and this often leads to confusion. In particular, galaxies that are far enough away from us necessarily have velocities greater than the speed of light:

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.

Also if I had to guess I'd say it's more complicated than that. The gravity of the galaxy would severely alter the spacetime around it I would think.

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.

So to an outside observer viewing the photon would see it going less than c?

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.

 

[niceboar]

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.

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

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.

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.

 

[Calimero]

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).

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: $$a(t)= H_{0}t$$

 

[JesseM]

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: $$a(t)= H_{0}t$$

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?

 

[Ich]

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.

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?

Yes.

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?

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" .

 

[JesseM]

Thanks Ich. Two more quick questions:

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.

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?

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?

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" .

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?

 

[Calimero]

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?

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?

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? ?

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

 

[JesseM]

No, in empty universe velocity would remain constant and deceleration parameter is q=0. In universe dominated with matter, velocity would decrease, and q>0. In cosmology constant dominated universe q is negative and velocities increase.

I don't know the physical meaning of "deceleration parameter"...

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

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.

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?

They are not talking about v(t), they are talking about a(t).

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?

See http://en.wikipedia.org/wiki/File:Friedmann_universes.svg" .

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.

 

[Calimero]

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.

 

[JesseM]

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.

Ah, so "a" stands for the "scale factor" and not the acceleration--does that mean the vertical axis of the graph is "a", and that the scale factor at a given time is proportional to the distance of some randomly-selected test particle at that time? But if so, why would they use the term "accelerating" to mean that the slope (proportional to the velocity of a test particle if the above is correct) is greater than 45 degrees? This section of a wikipedia article seems to say that the terminology of the expansion "accelerating/decelerating" is based on the sign of the deceleration parameter which can be defined either in terms of the Hubble constant and its first derivative or the scale factor "a" and its first and second derivatives...are you saying there's some mathematical argument that shows that a positive deceleration parameter (decelerating expansion) implies a slope closer to horizontal than 45 degrees, while a negative deceleration parameter (accelerating expansion) implies a slope closer to vertical than 45 degrees? i.e. that for any FLRW solution, if $$\dot{a}$$ is less than 1 then $$-\frac{\ddot{a} a }{\dot{a}^2}$$ will always be positive, while if $$\dot{a}$$ is greater than 1 then $$-\frac{\ddot{a} a }{\dot{a}^2}$$ will always be negative?

edit: never mind, I see you corrected yourself. I guess the "deceleration parameter" has no simple intuitive physical meaning in terms of the motion of a test particle though, it's just defined that way because it simplifies various cosmological equations?

 

[JesseM]

Re: the non-intuitiveness of defining accelerating/decelerating expansion in terms of the "deceleration parameter"--looking on google books, I came across this textbook which denotes the scale factor as R(t), and indicates that the terminology of accelerating/decelerating expansion just refers to whether the second derivative $$\ddot{R}(t)$$ is positive or negative, but that if it's positive that always means the deceleration parameter is negative:

$$q(t) = \frac{-R(t)}{[\dot{R}(t)]^2} \ddot{R}(t)$$

Note the negative sign in this equation; this implies that if $$\ddot{R}(t)$$ is positive (i.e. if the expansion is accelerating) then the deceleration parameter will be negative.

I guess this does make sense in any universe that isn't collapsing, since in an expanding universe both $$R(t)$$ and $$\dot{R}(t)$$ should be positive. I do wonder whether cosmologists would say the expansion was "accelerating" or "decelerating" in a collapsing universe where both $$\ddot{R}(t)$$ and $$\dot{R}(t)$$ were negative so the deceleration parameter was negative too.

edit: never mind, I forgot that $$\dot{R}(t)$$ is squared in the equation for deceleration parameter, so it doesn't matter if it's positive or negative. And $$R(t)$$ is always positive by definition, so positive $$\ddot{R}(t)$$ always implies negative deceleration parameter and vice versa.

 

[Calimero]

I forgot that $$\dot{R}(t)$$ is squared in the equation for deceleration parameter, so it doesn't matter if it's positive or negative. And $$R(t)$$ is always positive by definition, so positive $$\ddot{R}(t)$$ always implies negative deceleration parameter and vice versa.

Yes, exactly. Basically, if you are looking for physical meaning of deceleration parametar, it is that when second derivative changes sign function changes from being concave to being convex, or so called inflection point.


That is what Ich was trying to say, he just swaped $\dot a$ and $\ddot a$ :

Essentially, we're talking about $\dot a$ being positive or not.

 

[Ich]

That is what Ich was trying to say, he just swaped $\dot a$ and $\ddot a$

Exactly, thanks for correcting that. I also didn't anticipate that "a" could be misinterpreted as acceleration, which of course it could.

Concerning the maths:
For a comoving particle, $\dot D = \dot a r$ and $\dot D = \ddot a r$. So the coordinate acceleration of a paricle is proportional to $\ddot a$. Accelerating/decelerating expansion = accelerating/decelerating particle.
For small distances (some 0.1 Hubble radii), this relation also holds for particles with peculiar motion. The reason is that the accelerating/decelerating expansion of the universe is exactly the same as accelerating/decelerating test particles, with exacly the same cause: the gravitational repulsion/attraction of the homogeneous fluid those particles are embedded in.

 

[JesseM]

Another question--does wikipedia have an incorrect definition of the scale factor in this article? They write the equation $$l_p = l_t a(t)$$ where l_p is the distance (say, to some galaxy) at present and l_t is the distance at some other arbitrary time t. This can be rearranged as $$a(t) = \frac{l_p}{l_t}$$. However, p. 363 of this textbook indicates that the scale factor is defined with the distance at time t in the numerator and the distance at present in the denominator:

Consider two galaxies at some particular reference time, say at the present moment $$t_0$$. Let their separation be $$d_0$$. At another time t they are separated by a distance d(t). We define the scale factor R to be the ratio of these distances

$$R = \frac{d(t)}{d_0}$$

 

[Ich]

Hmmm...

I really have problems understanding what they mean. The sentence

It relates the comoving distances for an expanding universe with the distances at a reference time arbitrarily taken to be the present.

$$l_p = l_t \; a(t)$$

where $\! l_t$ is the comoving distance at epoch $\! t$, $\! l_p$ is the distance at the present epoch $\! t_p$ and $\! a(t)$ is the scale factor.

could almost be correct if they chose nonstandard definitions.
If Lt is in fact a comoving distance taken at some reference time, then a(tp)*Lt is Lp, the distance now (tp).
However, they write a(t), which is wrong in this context. Further, you'd normally define Lp as the comoving distance, and relate it with the distance at a different time t -as you said.

I'll try to find a better wording for the article, or maybe you'd like to correct it?

 

[JesseM]

I'll try to find a better wording for the article, or maybe you'd like to correct it?

I had a go at correcting it, but feel free to modify further...

 

[Ich]

Better, but comoving distance is something different. The scale factor relates proper distance (Lp) with comoving distance (Lc): Lp= a*Lc.
Comoving distance is proper distance now.

 

[JesseM]

Better, but comoving distance is something different. The scale factor relates proper distance (Lp) with comoving distance (Lc): Lp= a*Lc.
Comoving distance is proper distance now.

Thanks, I edited it to distinguish between proper distance and comoving distance, and also edited the comoving distance article which discussed the proper distance but was missing that simple equation.

 

[JesseM]

Better, but comoving distance is something different. The scale factor relates proper distance (Lp) with comoving distance (Lc): Lp= a*Lc.
Comoving distance is proper distance now.

Hmm, but p. 263 of this book seems to say proper distance is something different from the scale factor times the comoving distance:

First it is important to remember that r in (10.23) is a comoving coordinate. If an observer here on Earth is at r=0 and a distant galaxy is at r=r_e, then the observer remains at r=0 and the distant galaxy remains at r=r_e. The term r is thus better thought of as a label than as a distance. The coordinate distance, d_C, is given by (10.20). If the light emitted by a galaxy with comoving radial coordinate r_e is observed by us at the present time t₀, then the present coordinate distance to the galaxy is given by

d_C(t₀) = R(t₀) r_e

where R(t₀) is the present value of the scale factor. The coordinate distance to the galaxy changes because R(t) changes, not because the galaxy has a large velocity through space away from us.

What is the actual distance to the galaxy? ... In the present context it is easiest to use the proper distance. To measure the proper distance to a galaxy, imagine that there is a chain of observers between us and the galaxy. Each observer measures the distance between himself and his immediate neighbor in the direction of the galaxy at the same cosmic time t. If we then add up all these small distance elements the result is the proper distance d_P to the galaxy at cosmic time t ... There are thus three expressions for the proper distance to an object, depending on the curvature of the universe:

d_P = R(t) sin^-1 r --- spherical
d_P = R(t) r --- flat
d_P = R(t) sinh^-1 r --- hyperbolic

Note that the proper distance is equal to the coordinate distance only in the case of a flat (i.e. k=0) space.

edit: On the other hand, p. 11-12 of this book distinguish between the "co-moving coordinate" r of a given galaxy and the function $$\chi(r)$$ which is multiplied by the scale factor to get the proper distance (defined at the top of p. 11 as the actual ruler distance), i.e. $$d_{proper}(r, t) = R(t) \chi(r)$$, with $$\chi(r)$$ working out to equal r when k=0 (flat universe), sin^-1(r) when k=1 (spherical) and sinh^-1(r) when k=-1 (hyperbolic). This mirrors the previous book but I am not sure whether r or $$\chi(r)$$ would normally be defined as the "comoving distance".