# Carried along by the Hubble flow.

1. Jun 15, 2012

### yuiop

Hi. If we fire a rocket into space at some significant fraction of the speed of light (say 0.1c) and then switch off the engines and let it cruise indefinitely, will it appear to accelerate away from us and in effect be carried along by the Hubble flow. Given a million or a billion years would it eventually appear to be receding at greater than the speed of light in the same way that distant galaxies appear to recede?

2. Jun 15, 2012

### Dickfore

Unless it falls into some local gravitational well by some nearby planet or a star and falls into oblivion.

3. Jun 15, 2012

### yuiop

So do you agree that in principle, an un-powered projectile with an initial local cruising velocity of 0.1c could (eventually) catch up with a distant galaxy receding at say 5c, if that galaxy is at rest with the local Hubble flow?

4. Jun 15, 2012

### Dickfore

no, if both the galaxy and the rocket are at rest with respect to the co-moving reference frame, they can never meet.

5. Jun 15, 2012

### yuiop

Yes, but the rocket is not at rest with the Hubble flow. Initially and locally it is moving at 0.1c relative to the local reference frame / CMBR / Hubble flow. Does it somehow get slowed down to come to rest with the Hubble flow?

P.S. Just thought. Isn't everything at rest with its own co-moving reference frame by definition?

6. Jun 15, 2012

### Dickfore

If the engines of the rocket get stopped, it starts moving along a geodesic, which, in FLRW metric, are the world lines of "Hubble flow".

7. Jun 15, 2012

### yuiop

Shouldn't then everything (un-powered) including the Moon going around the Earth, the Earth going around the Sun and the Sun going around the Galaxy all grind to a halt because they have peculiar velocities relative to the local Hubble flow?

8. Jun 15, 2012

### marcus

It's an intriguing discussion! Thanks both.

Y: So do you agree that in principle, an un-powered projectile with an initial local cruising velocity of 0.1c could (eventually) catch up with a distant galaxy receding at say 5c, if that galaxy is at rest with the local Hubble flow?[/QUOTE]

D: no, if both the galaxy and the rocket are at rest with respect to the co-moving reference frame, they can never meet.[/QUOTE]

Y: Yes, but the rocket is not at rest with the Hubble flow. Initially and locally it is moving at 0.1c relative to the local reference frame / CMBR / Hubble flow. Does it somehow get slowed down to come to rest with the Hubble flow?
P.S. Just thought. Isn't everything at rest with its own co-moving reference frame by definition?[/QUOTE]

D: If the engines of the rocket get stopped, it starts moving along a geodesic, which, in FLRW metric, are the world lines of "Hubble flow".[/QUOTE]
===============

I'm not sure how I would analyze or reply. Dick's intuition is absolutely right (I'm sure though not an expert myself.)
But I'm not sure how I would respond on a detailed level. There is something called the "cosmic event horizon" which is around 16 billion LY. If today you send a flash of light at a galaxy that is 16 Gly from here it will never get there. And those guys are not even receding at 5c. They might be receding at 1.2c or something modest like that. I will check.

And if they sent us a message today, it would never reach us. That is what the CEH is about. It is in some ways like a BH event horizon.

Last edited: Jun 15, 2012
9. Jun 15, 2012

### Dickfore

In GR, a reference frame has a different meaning than in Special Relativity. Namely, it is a set of (curvilinear) coordinates which uniquely determine an event. The space-time interval is given by the quadratic form:
$$ds^2 = g_{\mu \nu} \, dx^{\mu} \, dx^{\nu}$$
If we choose a system of coordinates in which:
$$g_{0i} = 0, \ (i = 1, 2, 3), \ g_{00} = 1$$
then $dx^{0} = d\tau$ the temproal coordinate coincides with the proper time, and the metric becomes:
$$ds^{2} = d\tau^{2} - \gamma_{i k} dx^{i} \, dx^{k}$$
This kind of coordinate system is what I mean by a co-moving frame.

10. Jun 15, 2012

### Dickfore

Well, the celestial bodies are actually "powered" by the extra gravitational pull from the local variations in the mean density of the Universe caused by nearby celestial bodies. That was my point about the rocket falling in a "local gravitational well".

The FLRW assumes a homogeneous and isotropic space. If you look at the night sky, you can easily confirm this is not the case. But, you have to bear in mind that the FLRW metric is "averaged out" over length scales of the order of magnitude of several intergalactic distances. Then, the "lumpiness" gets evened out.

Last edited: Jun 15, 2012
11. Jun 15, 2012

### yuiop

OK, I think I sort of get that and I assume it is something like the concept of a Momentarily Co-moving Inertial Reference Frame used to analyse accelerating systems in SR. The point is I don't get why inertial objects should come to rest with the local Hubble flow or CMBR if you prefer. Let's switch the scenario around. Let's say for the sake of argument that our galaxy is at rest with the local CMBR. Some distant freak Supernovae event hurls a huge star in our direction at 0.9c. Would it not be odd if we were saved because the speeding star came to a stop just short of colliding with us because it is un-powered and required to slow to the local Hubble flow/CMBR. What is that slows this projectile down, while other objects continue to orbit etc without any noticeable slow down?

12. Jun 15, 2012

### marcus

YUIOP, you asked an interesting question, it uncovers intriguing considerations. Counterintuitive things you might not expect. Here is something you should realize: expansion causes things to lose momentum relative to the CMB.
So if you send a rocket off at .1c (relative to ancient light) it will eventually slow down (relative to ancient light).
Not because of gravity, although if there is gravity pulling it back that would slow it also, but because of expansion.

Expansion bleeds momentum from a flash of light as well. It keeps going the same speed but its wavelength lengthens, which means its momentum is drained also. Just like happened to the rocket.

However this is a slow effect. Don't worry about it, neglect it. But do realize it is there.
=====================

Another thing. the CEH is not the same as the Hubble distance. The present Hubble distance is something like 13.8 Gly.
The CEH is something like 16 Gly.
Try dividing 13.8 by sqrt(0.73) and see what it comes to. We are working with standard model LCDM as usual with "dark energy fraction" 0.73.
=======================

Oooops! I see you guys are still busy conversing. I had better be quiet and come back later so as not to spam up the thread with extra posts.

13. Jun 15, 2012

### yuiop

OK, to eliminate the CEH issue from the analysis, I will settle for an un-powered projectile with a local peculiar velocity of 0.1c catching up with a distant galaxy receding at something more modest like 0.9c (but locally at rest with the Hubble flow/CMBR). Is that possible?

By the way, thanks Marcus (and Dickefore) for your detailed contributions, despite some of them being a bit too advanced for me. I am trying to establish if there is any drag effect (positive or negative) from the Hubble flow and expansion. So far the answer seems to be negative. Perhaps one way to get at the crux of the problem is put the question like this. If we launch an unpowered projectile at 0.1c and wait a long time, will it ever appear to be exceeding 0.1c as it recedes from us. (Please ignore any local gravity wells and assume the projectile is aimed at largely empty space).

P.s. Will the projectile even catch up with a not too far away Hubble flow object that appears to be receding at 0.1c now?

Last edited: Jun 15, 2012
14. Jun 15, 2012

### marcus

Just to complete CEH talk with rough estimates. A galaxy with a redshift of 1.8 is at the CEH and it is currently receding with speed 1.17 or 1.2 (I'm using Jorrie's calculatorhttp://www.einsteins-theory-of-relativity-4engineers.com/cosmocalc_2010.htm
with redshift z=1.8 plugged in.

Dickfore was showing you something, I just butted in. I'll wait a while and then see about your galaxy that is receding at 0.9c. "more modest" as you say [redshift in that case would be z=1.2, and distance would be 12.4 Gly at the present moment.]
I think Dick would say it still could not be reached by the rocket leaving today, and I would agree. Details later.

Over to you, D.

15. Jun 16, 2012

### marcus

Well, D. left, let's see what we can do with this new galaxy.
First of all be sure you know how to use:
http://www.einsteins-theory-of-relativity-4engineers.com/cosmocalc_2010.htm

It is very easy. Just put in 1.2 into the box labeled "redshift of source" or something like that.

You should get that the current recession speed is 0.9c. Or 0.896, but we can round up to 0.9.

You should also be getting that the current distance is 12.4 billion LY (Gly). that is the socalled proper distance now, which you would get if you could stop expansion, freeze it, so you'd have time to measure with some conventional means like radar or string. Of course that takes a long time 24.8 billion years for the radar round trip, but you don't care because you have frozen expansion in its tracks so you have plenty of time to do the measuring. That is what proper distance at a given moment in time means.

I'll see if you are still around. Say uh-huh or something.

Last edited: Jun 16, 2012
16. Jun 16, 2012

### yuiop

Hey! Its a public forum. Your contributions, knowledge and insights are always very welcome .

I will give it a break now and slow things down and hopefully come up with more considered responses. Hopefully you guys will also have time to consider my revised questions in post #13.

17. Jun 16, 2012

### marcus

Perhaps one way to get at the crux of the problem is put the question like this. If we launch an unpowered projectile at 0.1c and wait a long time, will it ever appear to be exceeding 0.1c as it recedes from us. (Please ignore any local gravity wells and assume the projectile is aimed at largely empty space).

P.s. Will the projectile even catch up with a not too far away Hubble flow object that appears to be receding at 0.1c now?

I think the answer to the blue question is YES! That should not be too hard to figure out. (There IS a drag effect in the CMB rest frame but as I said earlier you should neglect it. Too small to worry about.)

71/3.26 = about 22. Hubble constant is about 22 km/sec per Mly (per million LY)

So after your projectile has traveled a million years it will be more than 100,000 LY away and it will be receding at 2.2 km/second PLUS its projectile speed of .1c. So it will be departing from us at a little bit MORE than 0.1c. 2.2 km/s is not a lot more but it is something.
==============

your red question is harder. i am used to thinking about light signals, whether or not they reach the target. Not projectiles going 0.1c. But my intuitive hunch is NO! it does not ever reach the galaxy. You say the galaxy is receding 0.1c so it has redshift about z=0.102 which we can round to 0.1.

for small redshifts like 0.1 the redshift is almost the same as the recession speed as a fraction of light.

So put that in the calculator and find out how far away the darned thing is!
http://www.einsteins-theory-of-relativity-4engineers.com/cosmocalc_2010.htm
It comes out around 1.4 billion LY. And the projectile is only going .1c

It is going to take roughly 14 billion years for the projectile to get where the galaxy USED TO BE when the projectile started. And by that time the galaxy will have moved on. It does NOT look good for the projectile.

Someone else more wide awake than I am may step in and set up an equation and solve it. I'm just doing the approximate scribble on scrap paper.

[EDIT: I really was quite sleepy last night when i wrote that. My hunch was WRONG. The projectile sent at .1c towards a target receding at rate .1c actually does make it! For the same reason (linearly scaled down) that we could today send a flash of light to a galaxy at Hubble distance, i.e. receding at rate c, and it would eventually get there. See the post #21 further down the page.]

Last edited: Jun 16, 2012
18. Jun 16, 2012

### Chronos

Expansion never exceeds the speed of light in any local reference frame, so it appears to me photons eventually escape all regions of the universe, but, at a price - redshift. This appears to explain how we can still observe galaxies that were receeding at superluminal velocities when they emitted the photons we now observe.

19. Jun 16, 2012

### marcus

"Regions of the universe" are typically described as co-moving volumes. Consider a ball with comoving radius 16 billion LY, centered at our galaxy. Or just to be sure make it 17 billion LY.
If today you flash your flashlight up into the sky those photons are never going to escape from that region.

If you have a link to something online that says they escape all regions I'd be happy to look at it.

the fact that photons are reaching us today from galaxies which were receding faster than light when they emitted the light is due to the fact that the Hubble constant has been shrinking rapidly (and the Hubble radius growing) for most of the past 13.7 billion years.

That is not going to be happening so much in the future according to LCDM. H constant will tend to stabilize at around 60 km/s per Mpc. So I think its pretty clear you will NOT be seeing much of that. If you consider light emitted from now on, eventually we will not see any that was emitted by galaxies receding >c.

Last edited: Jun 16, 2012
20. Jun 16, 2012

### Dickfore

Just as a note, what I called a co-moving frame should be called a synchronous frame.

21. Jun 16, 2012

### marcus

Hi Dickfore!
just had an insight. The answer to the most recent question is YES the projectile CAN make it.
That was the case where the target is currently receding at .1c and the projectile is sent off towards it at .1c.

I had a mistaken hunch last night that the answer was no, for that case, so I want to correct my mistake. It just barely makes it.

What happens is for a long time it kind of hovers at the same proper distance from the target, just barely hanging on. Because its speed .1c just exactly cancels the rate the thing is getting farther way.

But the Hubble constant is still decreasing! According to standard LCDM it is now around 70.4 and destined to decline slowly and level out at (asymptotic) value of around 60 km/s per Mpc.

As soon as H gets appreciably below its present value the projectile will begin to slowly gain ground!

But for a long time it is just hanging in there, staying approximately 1.4 billion LY from the target. The target receding at .1c and it going towards at .1c, so no change. Then finally beause of declining H, the recession speed at that 1.4 billion LY distance will get slightly less, like .95c. And then the projectile will begin to creep towards it at 0.05c. From then on it is assured of getting there.

Interesting. if time permits I'll try to understand what you were saying earlier with that change, but so far have been distracted by other stuff and haven't tackled it
================

YUIOP, I think the fastest-receding galaxy which your .1c projectile could catch would be a galaxy that is currently receding at rate around .11c.
Then it would at first slowly lose ground (at around 0.01c) but eventually its determined persistence would be rewarded by Mother Nature who would graciously reduce H enough so that it would begin to shorten the distance between itself and target.

I'm pretty sure that it could NOT catch a galaxy receding at 0.12c. And I don't have an exact figure for the critical recession speed, just something around .11c

Last edited: Jun 16, 2012
22. Jun 16, 2012

### Chronos

I should have clarified I was speaking about the observable universe, which includes everything inside our particle horizon. We still receive photons emitted from the surface of last scattering, which I assume we all agree was receeding at superluminal velocity when the photons we now observe were emitted. Davis and Lineweaver explained this in http://arxiv.org/abs/astro-ph/0310808

A quote of interest from section 3.3. of the paper
"Our teardrop shaped past light cone in the top panel of Fig. 1 shows that any
photons we now observe that were emitted in the first ∼ five billion years were emitted
in regions that were receding superluminally, vrec > c. Thus their total velocity was
away from us. Only when the Hubble sphere expands past these photons do they move
into the region of subluminal recession and approach us. The most distant objects
that we can see now were outside the Hubble sphere when their comoving coordinates
intersected our past light cone. Thus, they were receding superluminally when they
emitted the photons we see now. Since their worldlines have always been beyond the
Hubble sphere these objects were, are, and always have been, receding from us faster
than the speed of light."

23. Jun 16, 2012

### marcus

You have just explained why what you said here in post#18 is wrong. The Hubble sphere is slated to expand much more slowly in future (in the usual LCDM model) and so photons will NOT range as freely as in in the past.
I just described the same mechanism facilitating photons getting here in the past, the mechanism you quote Lineweaver Davis explaining---H decreasing or the Hubble distance c/H increasing.

None of this is effected by whether you meant universe or currently observable portion, either is way larger than the regions that today's photons can NOT get out of. (Because the decline of H has slowed enormously and is slated to essentially stop.)

Last edited: Jun 16, 2012
24. Jun 17, 2012

### Chronos

I think this may be a case of semantical dissonance. Again, from the paper -
"... Most observationally viable cosmological models have event horizons and in
the CDM model of Fig. 1, galaxies with redshift z ∼ 1.8 are currently crossing our
event horizon. These are the most distant objects from which we will ever be able to
receive information about the present day. The particle horizon marks the size of our
observable universe. It is the distance to the most distant object we can see at any
particular time. The particle horizon can be larger than the event horizon because,
although we cannot see events that occur beyond our event horizon, we can still see
many galaxies that are beyond our current event horizon by light they emitted long
ago."

25. Jun 17, 2012

### marcus

Chronos, anybody can quote intelligent authors' stuff. It does not do any good if you don't understand the implication of what you are quoting. You say this
I've explained that photons emitted today are not going to be as free-ranging as those emitted during much of the past. So there are galaxies that we observe now, which are part of our observable universe, which today you could not send a flash of light that would get there. Most of the galaxies observable today, in fact, are OUTSIDE THE REACHABLE REGION.
We are surrounded by a region from which photons we emit today cannot escape.

You are quoting nice articles which I admire and know well and that's nice, but you don't seem to understand why what you said is wrong and inconsistent with the articles. So I'm inclined to give up and not try to explain further. Please excuse if I don't reply to you every time you quote Lineweaver and Davis.

The basic reason what you said is wrong is the positive cosmo constant---accelerated expansion. Its a feature of the standard LCDM model.