# Local effects of Hubble flow

1. Nov 9, 2005

### Jimmy Snyder

I did look through this forum for other threads on this topic, but didn't find any. If this has already been discussed, then please let me know where the thread is.

A common model for the expanding universe is an expanding ballon upon which some dots have been drawn. A well know flaw in this model is the fact that as the balloon expands, so do the dots. If the universe expanded like that, then we and our telescopes would expand as well and we wouldn't see the effect. A better model is proposed in which there are coins pasted to the balloon so that as the balloon expands, the coins stay the same size. In this model, the rubber of the balloon expands between the coins, but does not expand under the coins where the paste is. However, I would like to propose a somewhat idealized version of this model in which the paste only contacts the coin and the balloon at a single point of each, say the center of the coin tacked to a single point of the balloon. Then under the coin, except for that one point, the rubber is sliding as the balloon expands.

So how about the real world? Is space something that is stuck to us and our telescopes like the rubber is stuck to the coins, or does space slide through us as if attached at a single point? Or is there nothing attached and space is sliding through us at all points?

2. Nov 9, 2005

### hellfire

The expansion of space cannot be extrapolated to such small scales. The cosmological solution of Einstein equations is derived assuming homogeneity and isotropy of the distribution of matter in space. Therefore, the expanding solution must be valid at that scales where homogeneity and isotropy can be found. Observations tell us that this is 100 Mpc more or less. At smaler scales the geometry of spacetime must not be the cosmological one.

3. Nov 9, 2005

### Jimmy Snyder

But something is happening locally. What are the options?

4. Nov 9, 2005

### hellfire

What makes you think so?

5. Nov 9, 2005

### Jimmy Snyder

I know something is happening, but I don't know what it is. Do I?

6. Nov 9, 2005

### Staff: Mentor

Well, you can get around that by sticking them on with Scotch tape!

7. Nov 9, 2005

### Cosmo16

I like that analogy.

8. Nov 9, 2005

### Jimmy Snyder

I addressed that issue in my first post. I used paste, not tape.

9. Nov 9, 2005

### Jimmy Snyder

We see the distant galaxies slip sliding. What do they see when they look at us?

10. Nov 9, 2005

### pervect

Staff Emeritus
You might want to look at

http://xxx.lanl.gov/abs/astro-ph/9803097
http://arxiv.org/abs/astro-ph/0104349

and/or http://www.astro.ucla.edu/~wright/cosmology_faq.html#SS

This has been discussed quantitatively before here, too (without aid of these papers)

for example
works out a repulsive tidal force due to expansion (assuming a FRW metric) and the current cosmological values which include a cosmological constant.

Something worthwhile to know from one of the above published papers:

So, if we put all the pieces together, the conclusion I come to is this (a slightly more sophisticated view of what I wrote earlier) is this.

the cosmological constant terms will be very uniformly distributed over space, as they are due to space itself.

The cosmological constant terms currently give a repulsive force

If we imagine an empty sphere of space inside an FRW universe, there would be no forces.

But an empty sphere of space would be empty only if there were no cosmological constant. Because of the cosmological constant, an "empty" sphere of space actually contains negative mass. (I should clarify this - because the expansion of the universe is accelerating, an "empty" sphere of space must provide the forces that provide this acceleration. The energy density $\rho$ associated with a positive cosmological constant is postive, but gravity couples to $\rho + 3P$, i.e. to both energy and pressure, and the later quantity is negative.

This causes a net repulsive force between two objects that are far apart, via normal gravitational interactions. We can use the naive Newtonian viewpoint here about the gravity of a spherical shell of matter acting as if it were in the center if we assume spherical symmetry - at least as far as getting the sign correct. (Previously I resisted this, but now I think this naieve aproach does give the right answer).

If there is also matter inside the sphere, the repulsive force will be less. If there is sufficient matter inside the sphere, the repulsive force (due to the cosmological constant) will be overcome.

If the matter inside the sphere was exactly the "average" amount of matter that one would expect from the average matter density of the universe, you will get the number that I previously calculated for the "repulsive force" due to expansion, which assumed that the metric was the FRW metric. But it would probably be better to view the repulsive force as being due to the cosmological constant. You can then add in an observered matter density to get the sign of the resulting force assuming symmetry, or do some other calculations involving non-symmetrical matter and symmetrical "vacuum-energy".

Last edited: Nov 9, 2005
11. Nov 9, 2005

### hellfire

Sorry but I do not understand your question. From your initial question it seams that you are assuming that there must be some expansion effect at small scales (<< 100 Mpc). But, as I have already written, the cosmological solution is based on the assumption of homogeneity and isotropy and these are not given at small scales.

12. Nov 9, 2005

### Jimmy Snyder

Sorry if I was unclear. I mean that since there ISN'T any expansion effect at small scales, doesn't that mean that space must be slipping out the sides of things like the rubber of the balloon slips out of the edge of the coin. Because in the model, the rubber is space and the coins are matter.

Suppose you could bounce a beam of light around with mirrors in a small hollow contraption in such a way that the beam traveled 100 Mpsec from emiter to detector (that's an unreasonable number of bounces, I know). Because the contraption is small, it would not expand with the universe. What I want to know is if the space within which the beam travels is expanding so that the beam would eventually over time drift away from the detector.

You don't really need the beam to travel 100 Mpsec. The contraption should be as large as possible and still be able to maintain a rigid shape. It should be isolated from vibration. It should be given time for all of its mirrors to reach thermal equilibrium with the heat from the beam. It should bounce the beam as many times as it can and still have the beam be detectable. It should run for as long as it takes for the Hubble flow to move the end of the beam off of the detector.

My point is that at small scales, things don't expand, but does the space around those thing expand?

Last edited: Nov 9, 2005
13. Nov 9, 2005

### pervect

Staff Emeritus
The problem with answering this question is that it is not possible, even in principle, to perform a local measurement of "the velocity of space" - because the laws of physics are invariant with respect to velocity.

One thing we can say, though, is that scenarios such as Baez's "end"

http://math.ucr.edu/home/baez/end.html

eventually do wind up with a universe with a positive cosmological constant tearing apart matter, if you wait long enough. The current effects of expansion are not important, but if we wait long enough, this may no longer be the case.

Last edited: Nov 9, 2005
14. Nov 10, 2005

### Chronos

Simple analogy: Gravity is not strong enough to collapse the earth or the sun, but is strong enough to compel them to orbit one another. The cosmological constant is a puny force, but, put this into perspective: Archimedes swore he could move the earth itself given a long enough lever.

15. Nov 10, 2005

### Jimmy Snyder

If space is expanding, and light travels at c, then a local measurement of the velocity of space could be performed by measuring a delay in the arrival of a beam of light after it had traveled a long distance but within a small volume. That is the nature of my proposed experiment. When the experiment starts out, the beam reaches the detector, but after time and expansion, the beam may slip off of the detector.

Again, the contraption doesn't expand. We know that because our telescopes don't expand. At least if they do expand, they do so slower than Hubble flow. Otherwise we wouldn't be able to detect Hubble flow. But I wonder if space expands locally.

My guess is either that space does not expand locally as suggested by some. In that case the balloon model would be coins pasted to the balloon on the entire surface of the coin. The problem with that idea is that the only force between local galaxies is gravitational, and that same force attracts distance galaxies, just to a smaller degree. Or that space does expand locally in which case there may be some way to measure the effects of Hubble flow locally.

16. Nov 10, 2005

### Jimmy Snyder

My analogy would be that gravity is not strong enough to collapse the earth or sun, but is strong enough that we can measure how hard it tries.

17. Nov 10, 2005

### hellfire

According to my understanding it must not, but it could under some circumstances, which, I assume, are not given in our solar system.

Again, if you make some assumptions in order to find a solution to some equations, then the solution is only valid under that assumptions. This means that the expanding cosmological solution to Einstein equations is valid only where the cosmological principle aplies. In general this is condition is only given at large scales.

The local geometry is determined by the local distribution of matter in the solar system. This could be approximated by a point mass to give a Schwarzschild solution. This solution has to be matched "at infinity" with the geometry in which the sun is located. This geometry is not a homogeneous and isotropic expanding space, because the sun is located within the Milky Way. I assume that it sufficies to take flat spacetime as asymptotic condition.

Considering greater scales one would reach some state at wich it would be necessary to consider that the “local” geometry is embedded in a homogeneous and isotropic space and must match the expanding cosmological solution. My guess is that this occurs at the scale of galatic clusters and not before.

18. Nov 10, 2005

### Jimmy Snyder

I get some of my ideas from popularizations. Perhaps this is the problem. Everyone here seems to agree that there is some distance at which two different solutions of the metric tensor get glued together. This distance is defined by where space starts to look homogeneous, about 100 Mpsec. Further than that and the metric tensor exhibits expansion, closer and it doesn't. Well and good. But as far as I can remember, the popularizations don't speak of it. Here is an example. I added the bold emphasis, but left the misspelling of exapnsoin as is.

http://curious.astro.cornell.edu/question.php?number=75 [Broken]

Edit: I literally cannot emphasize enough "universal expnasion". Also, I note that the galaxy in Andromeda is blue-shifted and less than 1 Mpsec from Cornell. Surely the author of the article was aware of that.

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19. Nov 10, 2005

### Jimmy Snyder

Actually, here is a more scholarly article that seems to contradict what I have been hearing:

http://arxiv.org/PS_cache/astro-ph/pdf/9612/9612007.pdf [Broken]

The dynamics of Local Group and its environment provide a unique challenge to cosmological models. The velocity field within $5h^{-1}$Mpc of the Local Group (LG) is extremely "cold". The deviation from a pure Hubble flow, characterized by the observed radial peculiar velocity dispersion, is measured to be $~60km s^{-1}$.

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20. Nov 10, 2005

### hellfire

Yes, thank you for bringing this into our attention. I remember have read about this and I have not considered this in my postings here. It relates to the "Hubble-de Vaucouleurs paradox": the dynamics of the Local Group is extremely low compared to the peculiar velocities which would be expected due to the gravitational collapse in a Cold Dark Matter model. This has been explained here postulating that the action of dark energy becomes dominant at scales greater than 2 Mpc leading to a “cold” Hubble flow (a start of the expansion of space) which acts against gravitational collapse. In light of this, it might be that the expansion of space emerges actually at lower scales than the scales at which matter becomes homogeneous, due to the fact that dark energy permeates the whole space homogeneously even at low scales. (pervect was already talking about dark energy, but I was under the impression that it might not be relevant for this discussion, sorry). However, I assume that 2 Mpc is a lowest value for the matching between local geometries in clusters and the cosmological geometry and it is hard to believe that expansion could take place below this scale.

Last edited by a moderator: May 2, 2017