Does the expansion of the universe affect our observations in the real world?

[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 balloon 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?

 

[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.

 

[Jimmy Snyder]

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.

But something is happening locally. What are the options?

 

[hellfire]

But something is happening locally.

What makes you think so?

 

[Jimmy Snyder]

What makes you think so?

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

 

[russ_watters]

A common model for the expanding universe is an expanding balloon 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.

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

 

[Cosmo16]

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

I like that analogy.

 

[Jimmy Snyder]

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

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

 

[Jimmy Snyder]

The expansion of space cannot be extrapolated to such small scales.

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

 

[pervect]

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)

https://www.physicsforums.com/showthread.php?t=63805&page=2&highlight=force+expansion

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:

In the presence of spherical symmetry, the analysis of a spherical cavity embedded
in an FRW universe is well known: as a consequence of Birkhoff’s theorem, the metric
inside the spherical cavity is the Minkowski one, and the physics is the same as in flat
space (Einstein & Straus 1945; Sch¨ucking 1954; Dicke & Peebles 1964; Callan et al. 1965;
Bonnor 1996).

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 approach 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".

 

[hellfire]

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

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.

 

[Jimmy Snyder]

it seams that you are assuming that there must be some expansion effect at small scales.

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?

 

[pervect]

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.

 

[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.

 

[Jimmy Snyder]

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.

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.

 

[Jimmy Snyder]

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.

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.

 

[hellfire]

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

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 which 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.

 

[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.

Almost all galaxies are redshifted because of the Hubble expansion of the universe. Only a handful of the most nearby galaxies are blue-shifted. You see, in addition to the apparent motion imparted to galaxies due to universal expnasion, individual galaxies also have their own intrinsic, what we call "peculiar" motions. This is not because they are peculiar, as in strange, but rather because each galaxy is in motion irrespective of the universe's expansion, and each galaxy has its own unique velocity.

Generally, that velocity is some hundreds of kilometers per second. In regions close enough to our own galaxy where the Hubble expansion results in less outward expansion than this, the galaxies' peculiar velocities (if they are large enough and sufficiently towards us) can overcome that expansion, resulting in a blue-shift.

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

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.

 

[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

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}.

 

[hellfire]

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
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}.

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.

 

[pervect]

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.

Can you see that when you take a meter-stick, using the local defintion of a meter and the local defintion of a second, you always find the speed of light to be equal to 'c'?

Thus if you have a meter-wide container, it doesn't matter how many times back and forth the light beam bounces, you still get a speed of 'c'. Because the meter stick is defined to be of constant length, the time to traverse the meter does not vary as the universe ages.

When you say that we can measure "how hard it (gravity) tries", you are (hopefully!) agreeing that the important question is "are there any tidal forces on the meter stick", not "does the ideal meter stick expand or contract". The ideal meter-stick does not expand or contract as the universe ages.

Previous posts were about calculating the magnitude of said tidal forces, making various assumptions in the process. It turns out that the tidal forces are not correlated with the rate of expansion, but rather the deceleration parameter q.

http://www.site.uottawa.ca:4321/astronomy/index.html#decelerationparameter

Note: I have yet to read the paper you (Jimmysnyder) referenced, but I plan to do so in the near future.

 

[Chronos]

A fun exercise... reduce the Hubble constant to the Planck length [at which point it theoretically disappears] and see what number you get for the minimum distance over which it acts.

 

[Garth]

A fun exercise... reduce the Hubble constant to the Planck length [at which point it theoretically disappears] and see what number you get for the minimum distance over which it acts.

H = ~ 2 x 10 ^{- 18} sec^-1
P_L = ~ 10^-33 cm

H = v/d : over a Planck length the universe will expand at a present rate of

v = Hd ~ 2 x 10^-51 cm.sec^-1

or in 'pure' units

v/c ~ 2/3 x 10^-61 ~~ 10^-60: interesting! See my post #16 in the thread An infinitely old universe.

(Just putting the numbers a different way round)
Garth

 

[Chronos]

Well done Garth, you made my day! I admit an ulterior motive for raising that question. But it's a mighty peculiar number, wouldn't you say? SCC just might yield a better answer than most other explanations... which is why I found it attractive in the first place. I don't think GPB is the real test of the pudding. Do you not find it interesting the theoretical vacuum energy density just happens to be the inverse square of the particular number you just derived? I find it ... fascinating.

My point is [like I had one] anytime you can arrive at the same conclusion by different paths, they should be taken seriously.

 

[Jimmy Snyder]

Can you see that when you take a meter-stick, using the local defintion of a meter and the local defintion of a second, you always find the speed of light to be equal to 'c'?

First of all, thanks to you and everyone who is helping me think about this. I don't think I have been clear in my posts, perhaps because I am not clear in my own mind.

I started with a hypothesis that has not met with acceptance. Namely, that the Hubble flow occurs at all scales. I continue with that hypothesis in spite of opposition for the following reason (Please forgive me if I misrepresent what was said.) The objection was that Hubble flow can be derived only from the metric of a homogeneous and isotropic region and this region only begins at roughly 100 Mpc distance. For local geometry we use a different metric and that metric has no expansion in it. However, I have linked to some web sites that suggest that Hubble flow takes place at much smaller scales, roughly 1 or perhaps 2 Mpc. As far as I know, there is no problem at this distance of using different metrics, and so I conjecture that Hubble flow occurs at all scales. If you disagree, I would like to hear about it, but for this post, I will continue using the hypothesis.

We also know that the atoms in our telescopes are not participating in the Hubble flow, for if they did, then we wouldn't be able to detect the expansion. I would say that the forces between the atoms in the telescope, which are known to dwarf gravitational forces, are too much for the Hubble flow. In my estimation, the Hubble flow may cause a vanishingly small tug at the electrical bonds, but has no important effect (it obviously has no measureable effect because any effect would be shared by the measuring rod). But gravitational bonds are much weaker and if the Hubble flow occurs at all scales, then for instance, the moon must drift away from us at a rate predicted by H. Although gravity is much weaker than electrical forces, it still dwarfs the Hubble flow at that distance. I still think that Hubble flow would provide a vanishingly small tug at the moon.

My apparatus of mirrors should be small enough to be rigid. At least as rigid as a telescope which we already know does not expand Hubbley. Unlike the weak gravitational bond between Earth and moon, this thing is held by strong electrical bonds and won't expand. But what about the space within the apparatus? That space, by my hypothesis, is expanding. It slips out the sides of the apparatus like the rubber of the expanding balloon slips out of the rim of a coin tacked to it. The light beam touches the apparatus for short moments, but spends most of its time in that stretchy stuff between the mirrors. I wonder if we can measure the effect of that expansion by noting that the beam drifts away from the detector. I expect that if I could build a version of this thing 100 Mpc long, then the answer would be yes. But we know roughly what H is, we know how long the light beam is, and we know how much it would have to change in order to drift away from the detector. We should be able to calculate how long it will take before it does. If the time interval is on the order of decades or even centuries, then I think it is worth a try.

Note: I have yet to read the paper you (Jimmysnyder) referenced, but I plan to do so in the near future.

You don't really need to read the paper, though you can if you want. I quoted the part that I was concerned about. I linked to it just to show that I wasn't making it up.

 

[hellfire]

I started with a hypothesis that has not met with acceptance. Namely, that the Hubble flow occurs at all scales. I continue with that hypothesis in spite of opposition for the following reason (Please forgive me if I misrepresent what was said.) The objection was that Hubble flow can be derived only from the metric of a homogeneous and isotropic region and this region only begins at roughly 100 Mpc distance. For local geometry we use a different metric and that metric has no expansion in it. However, I have linked to some web sites that suggest that Hubble flow takes place at much smaller scales, roughly 1 or perhaps 2 Mpc. As far as I know, there is no problem at this distance of using different metrics, and so I conjecture that Hubble flow occurs at all scales. If you disagree, I would like to hear about it, but for this post, I will continue using the hypothesis.

If we would not have a homogeneous distribution of dark energy at all scales, then the expansion would take place only at scales of order of 100 Mpc. With a homogeneous distribution of dark energy at arbitary small scales things are different. According to the paper I referenced above, the critical distance of 1.5 or 2 Mpc in the Local Group is the distance at which the repulsive force of the dark energy starts to dominate over the gravity of matter. I would be very cautious to extrapolate the expansion to arbitrary small distances.

 

[Jimmy Snyder]

According to the paper I referenced above, the critical distance of 1.5 or 2 Mpc in the Local Group is the distance at which the repulsive force of the dark energy starts to dominate over the gravity of matter.

Does this mean that at distances greater than 100 Mpc, the universe expands Hubbley, that at distances between 2 and 100 Mpc, it expands dark energyly, and at distance less than 2 Mpc, it doesn't expand at all? All distances rough of course.

Edited to correct error.

 

[Jimmy Snyder]

hellfire, here is what is in the link you provided along with some self-serving comments of my own.

Our local environment at r < 10Mpc expands linearly and smoothly, as if ruled by a uniform matter distribution, while observations show the very clumpy local universe.

So the author agrees with you that since the local environment is not homogeneous, it shouldn't expand, at least not linearly and smoothly.

We argue that the recently discovered vacuum or quintessence (dark energy (DE) component ... may also manifest itself in the properties of the very local Hubble flow.

Here's the self-serving part. He doesn't say that there isn't a local Hubble flow, he says there is one.

at ... 1.5 Mpc the linear and very "cold" Hubble flow emerges, with about the global Hubble constant.

He's against me. He seems to say that Hubble expansion doesn't exist at distances below 1.5 Mpc. I'm clutching at a straw though. He doesn't say why. Is there a change in metric at that distance?

It explains why the Hubble law starts on the outskirts of the Local Group, with the same Hubble constant as globally

Same comment as above.

So why does the Hubble flow cut out at 1.5 Mpc? Or is it not that it disappears, but simply that it is overwhelmed by other effects. I don't even understand how you could know that Hubble flow was not occurring. For instance, look at the galaxy in Andromeda. Its distance is about .7 Mpc. Using H = 75 km/s/Mpc, it should have a Hubble speed of 50 km/s away from us. It is heading toward us at 120 km/s. So how do we know that it isn't heading toward us at 170 km/s and pulled away by Dr. Hubble at 50 km/s? Please forgive me if the numbers are not correct, I'm just asking about the principle.

 

[pervect]

First of all, thanks to you and everyone who is helping me think about this. I don't think I have been clear in my posts, perhaps because I am not clear in my own mind.
I started with a hypothesis that has not met with acceptance. Namely, that the Hubble flow occurs at all scales.

Here is my $.02

The "hubble flow" is a name we give to the motion of particles, all of which meet the following two conditions

1) The particles are all following geodesics in space-time
2) The particles are all at rest with respect to the CMB frame

This differs significantly from your viewpoint, you are thinking of space-time as being somehow like a substance, whose flow can be measured.

I do not think that this is a good approach. I've argued why before - but let me just suggest that we regard space-time as being described by a metric, and that this the standard way of describing space-time in GR.

This post is getting too long, and you'll fall asleep before the punchline, so let me state my conclusion up front - conditon 1) does not apply in genreal to a meter-stick, it does not in general follow a geodesic. This is why a meter-stick is not part of the Hubble flow.

Specifically, we can consdier a flat-space FRW metric with "co-moving" coordinates t,x,y,z such that

ds^2 = dt^2 - a(t)^2 (dx^2 + dy^2 + dz^2)

as the GR model of an expanding space-time.

The flat FRW metric will only apply in regions where space-time is not "lumpy", so we will have significant deviations in the real universe from the idealized flat FRW metric near lumps of matter like stars, galaxies, etc. Hopefully, however, the idealized problem of what happens when we are not near such lumps will illustrate the general point.

Now, given a flat FRW metric, we can ask a coupe of questions.

a) How do we represent a meter-stick that is at rest relative to the cosmos?

For a short meter-stick, it is only needed to to say that ends have coordinates \frac{\Delta x}{2a}, -\frac{\Delta x}{2a},
b) Do the endpoints of the meter-stick follow geodesics?For definiteness, let's say that our meter stick is oriented in the 'x' direction. Then tidal stretching forces in the x direction are zero if, and ONLY if

<br /> R^{x}{}_{txt} = 0<br />

Hopefully you picked the "why" of this up from your reading of Schutz. The detailed calculation of the Riemann gives

<br /> R^x{}_{txt} =-\frac{ \frac{d^2 a}{d t^2}}{a}<br />

which is zero if, and only if d^2 a/dt^2 = 0.

What I want to emphasize. The ends of the meter-stick (and hence the meter-stick itself) does *not* follow a geodesic except for a rare special case, one where the expansion of the universe is unaccelerated.

Thus a meter-stick is not part of the Hubble flow, because it does not follow a geodesic.

Actually this is somewhat oversimplified. If you really want to do this problem right, you introduce an orthonormal basis of one-forms

[1,0,0,0]
[0,a(t),0,0]
[0,0,a(t),0]
[0,0,0,a(t)]

with a diagonal inner product diag(1,-1,-1,-1)

You then plug this into a program like GrtensorII and get the result in terms of the orthonormal basis vectors

<br /> R^{\hat{x}}{}_{\hat{t}\hat{x}\hat{t}} = -\frac{\frac{d^2 a}{d t^2}}{a}<br />

which turns out to be the same answer.

c) I've separated this out as a non-critical but interesting question. What does a "long" meter-stick look like in FRW coordinates?

The fact that the meter-stick is "straight" implies that the meter-stick follows a space-like geodesic, i.e a spacelike solution of the differential equation

<br /> \frac{d^2 x^i}{d\tau^2} + \Gamma^i_{jk} \frac{dx^j}{d\tau} \frac{dx^k}{d\tau} = 0<br />

gives a parameterized equation for the shape of a curve representing a "straight" meterstick

This is most easily solved by looking for "killing vectors" which give rise to conserved quantites. Specifically if y=z=0, t=t(\tau), x=x(\tau)

then one of the geodesic equations reduces to

<br /> \frac{d}{d\tau} \left( a^2(t) \frac{dx}{d\tau} \right) = 0<br />

(This can be confirmed if you calculate the Christoffel symbols).

 

[Danger]

Archimedes swore he could move the Earth itself given a long enough lever.

I know that this is a tad off-topic, but I've always wondered just what he thought he would brace the fulcrum on. (Or was it just a throw-away line?)

 

[Garth]

I know that this is a tad off-topic, but I've always wondered just what he thought he would brace the fulcrum on. (Or was it just a throw-away line?)

The back of a turtle?
"GIVE ME A PLACE TO STAND AND I WILL MOVE THE EARTH"
That's the point of him having to have a place to stand.

Garth

 

[hellfire]

hellfire, here is what is in the link you provided along with some self-serving comments of my own...

Well, I am not sure I can tell you something new to what I have already written. The metic is determined by the distribution of masses and energy density. If you make the assumption of a homogeneous distribution in space, you obtain a spatially expanding metric. Disregarding dark energy, there is no homogeneity at small scales, therefore the metric must look different. Actually, if you consider for example the solar system, you could model it as a central mass with spherical symmetry. This leads to a very different, static, metric than the cosmological one. If one considers dark energy, one msut consider that there is homogeneity in its energy density even at small scales. The referenced paper seams to claim that the gravity of the local distribution of matter will "win" against the pressure of dark energy for scales under a critical value and this will lead to the emergence of an expanding metric at that scale and not before. This seams meaningful to me.

 

[hellfire]

pervect, I think you are right if your answer is related only to material objects. The question whether rods expand or not can be analized taking a look to the stress and therefore to the Riemann tensor. However, one may ask also whether two points in space separated by a "small" distance increase its distance or not. As far as I know, this is indeed measurable. This has nothing to do with stresses on rods and the Riemann tensor, but only with the increase of the scale factor: in a linearly expanding model with homogeneity at arbitrary scales and with zero Riemann tensor, the scale factor increases and two arbitrary points do always increase its distance. So, in my opinion, the question here reduces to know what metric is relevant at what scales.

 

[Jimmy Snyder]

This differs significantly from your viewpoint, you are thinking of space-time as being somehow like a substance, whose flow can be measured. I do not think that this is a good approach.

Most of the back and forth here derives from the fact that I don't really know what I am talking about and don't understand the math well enough to follow everything you have been saying. I have heard that one must think of Hubble expansion as the expansion of space because otherwise, you end up with distant objects traveling faster than light. Using 75 km/s/Mpc, this happens at 4000 Mpc. In other words, we shouldn't think of Hubble expansion as things moving away from us, but rather as space expanding.

Hopefully you picked the "why" of this up from your reading of Schutz.

You flatter me. But unfortunately, your hope dies unfulfilled. But don't tell me the answer, I am working on it as if it were an exercise from the book.

I have a question about the seam where the local (< 100 Mpc) metric is sewn to the global (> 100 Mpc) metric. How do you solve the problem of galaxy hopping. In other words, a galaxy 75 Mpc from here is not partaking of the Hubble flow in our metrics. Another galaxy 75 Mpc from it, and 150 Mpc from us is not partaking of the Hubble flow in its metrics. How can it do so in ours?

 

[Danger]

"GIVE ME A PLACE TO STAND AND I WILL MOVE THE EARTH"

Thanks, Garth. Believe it or not, I've never seen the full quote before.

 

[mccrone]

Isn't there some confusion in this thread between when the Hubble flow really kicks in (probably Planck scale) and when it is observeable (large scale)? The situation is much like classical vs relativity - relativity applies even at "Newtonian" scales, but is a correction so small it does not need to be considered.

Standard story is that atoms won't expand because nuclear forces far exceed any cosmological constant. And gravitationally bound systems like solar systems and probably galaxies, and even galaxy clusters, will also stay bound. So space between objects does grow over all scales, as Jimmy Snyder originally asked about, but the effect is too weak to muck up existing gravitational relationships except at very large scale.

If you google on big rip, you will see the speculation about what will happen if acceleration of the fabric of space picks up. At the end, even atomic forces would be overwhelmed and your molecules would get Hubble flowed!

Of course, much depends on what is doing the cosmological expansion. The above assumes a homogenous cosmological constant. If dark energy exists clumped between galaxies for some reason then our local space might not be expanding.

Anyway, for another source that takes local Hubble flow for granted, see...(any comments about the accuracy of the maths which seems to differ from that posted earlier in this thread?)...

http://hypertextbook.com/physics/mechanics/gravitational-energy/
Seeing the Hubble constant in inverse second form makes it a bit more accessible. The space around us is expanding at a rate of roughly one part in 1018 every second. Given that the diameter of a proton or neutron is roughly 10−15 m, and that 18 orders of magnitude greater than this 1000 meters, a good phrase to tell your family, friends, and neighbors is that one kilometer of space expands at a rate equivalent to the diameter of one proton every second.

 

[pervect]

Most of the back and forth here derives from the fact that I don't really know what I am talking about and don't understand the math well enough to follow everything you have been saying. I have heard that one must think of Hubble expansion as the expansion of space because otherwise, you end up with distant objects traveling faster than light.

Think of objects as following geodesics in the absence of non-gravitational forces - for every instant of proper time, tau, the object has a definite value of 3 space and 1 time coordiantes in whatever coordiante system one is using. This is the geodesic equation for the motion of that object.

Think of light as also following geodesics. Light doesn't have "proper time", but you can still express the geodesic of light in terms of an affine parameter lambda that serves essentially the same role.

In cosmology, we are mainly interested in objects which are stationary with respet to the CMB, which gives them very simple geodesics - x=y=z=constant.

This makes determining the geodesics of light the only hard part of the problem. The messy geodesic equations for light can be greatly simplified by a simple re-scaling of time known as "conformal time".

But this digresses totally from what I was saying earlier, and gives you a new set of problems.

What I was saying earlier was that the electromagnetic forces which hold a mete-stick together keep its ends from following geodesics. In short, there are in the general case, tidal forces on the meter stick.

A classic problem in GR is how one calculates the tidal forces near a black hole for an observer falling in. Hopefully Schutz solves this problem somewhere in his text, if not, you may have to find another text that does. (I know MTW does for a fact, but their notation is a bit old).

I am proposing that you solve a very similar problem, except that we want the tidal forces on an object in a FRW expanding space-time. And to make things a little simpler, this object has a velocity of zero, unlike the previous case.

The geodesic deviation equation is the main tool you'll need to solve this problem.

A messy issue which arises to make the calculation more difficult is the issue of how to deal with the fact that the tangent vectors in both the Schwarzschild coordinate system near a black hole, and in the FRW coordinate system, are not of unit length. For the purpose of calculating tidal forces, we really want an observer at that location who has a local coordiante system with orthonormal tangent vectors.

One approach to this issue is to imagine a set of one-forms, maps from vectors to scalars, which map the coordinate tangent vectors to a new coordinate system. The first one-form when applied to an arbitrary vector gives you the 't' coordinate of a local orthonormal cartesian coordiante system. The second one form, when applied to a vector, gives you the 'x' coordinate, the third the 'y', and the fourth the 'z'.

This is called an "orthonormal basis of one-forms", and is a powerful tool.

 

[Jimmy Snyder]

one kilometer of space expands at a rate equivalent to the diameter of one proton every second.

Thank's mccrone, for the support. You have said clearly what I have been saying in a muddled way. But these numbers seem like an embarrassment of riches. Why haven't we noticed light speeding up? It seems to me that a light beam traveling 1 km would traverse a measurably smaller amount of space in a matter of a few days. That is, at 86400 seconds per day, and 1 proton diameter per second, that's roughly 1 atomic diameter per day. How many days have to pass before we notice the effect?

 

[mccrone]

Speed of light stays the same but wavelengths get stretched or red-shifted. That is what we see at astronomical distances to get a measure on the Hubble flow. And it is what we see at every point of space as the CMB effectively - the way the hot glow aftermath of the Big Bang got stretched out to make a cold void filled with very long wavelength photons.

 

[pervect]

Meter sticks serve as the defintion of length, and do not change with time. A one kilometer meter stick does *not* change its length in a second, a minute, or 1000 years, because it serves as the standard by which distance is defined.

Do not confuse the expansion of space with the expansion of meter-sticks, please! Meter sticks do not expand.

What cosmology actually says is that if you have two points (particles following geodesics) that both have zero velocity with respect to the CMB, they will slowly drift apart.

This would theoretically be true even for two points as close as a kilometer. However, determining the doppler shift of the CMB to determine motion with respect to the CMB of such a small amount is well beyond our ability to observe or measure. In addition, the presence of neargy "lumps" of matter would probably disturb the theoretical, idealized "cosmological solution" if it were actually carried out over such a short distance. As hellfire has pointed out, the simple FRW cosmologies model the universe without any lumps - while we happen to live quite near some rather large lumps of matter (the Earth, the sun, the galaxy). So if we observed two points following geodesics near the Earth, we would see any cosmological terms totally swamped due to the gravitational effects of the Earth, moon, sun, etc.

A few more points:

Note that in a meter stick, both ends of the meter stick are at rest with respect *to each other*, not to the CMB!

Thus it is impossible for both ends of a meter stick to be at rest relative to the CMB. There can be at most one point on a meter stick which can be at total rest relative to the CMB. (It's probably convenient to take this point as the centerpoint of the meterstick).

We can go on from here to describe under what circumstances a meter stick experiences actual tidal forces. The answer is "almost always", specifically unless d^2 a/ dt^2 is zero, a(t) being the expansion factor of the universe, the meter stick will experience tidal forces.

I've mentioned this before, though, and get the feeling I'm not getting through. Sorry, but I don't know what the problem is, I've tried to explain things both informally and with the supporting mathematics, but I'm not getting any feedback that makes me believe I'm being understood :-(.

 

[mccrone]

Meter sticks serve as the defintion of length, and do not change with time.

Which does not address the point I made about red-shifting, just restates the fact that speed of light must always look the same to an embedded observer.

The essential question here is whether the Hubble flow expansion kicks in at a certain distance that is not the Planck distance - say at the scale of galaxies or galactic clusters.

The answer would be, as far as I have been able to learn, that it should start at the Planck scale. But it could hardly be observed as you would have to have two reference masses with no forces between them. So effectively nothing can be seen (or even happens due to gravitational clumping) until you get to cosmological scales.

In practice, the expansion does not even need to be considered at local scales. In theory, it does exist though. Or do you know opinion to the contrary?

Cheers - John McCrone.

 

[pervect]

As long as one accepts relativity, it is not possible, even in principle, to measure "the flow of space", because absolute velocities cannot be detected.

Therfore I am arguing for more precision in talking about "the Hubble flow". We cannot directly measure the flow of "space" with any sort of instrument because we cannot detect "absolute velocity". So what are we actually measuring when we measure the "Hubble flow"?

Because this is such a long thread, I'm going to re-iterate my proposal that this "flow" is measured by considering the paths of observers who are not experiencing any external forces and who are observing the CMB to be isotropic.

Certain answers to this question were given earlier that involve the speed of light or the length of meter sticks changing.

I re-read your response, you talk about how the CMB is red-shifting, and I have no disagreement that red-shifting occurs. Before we can say when the "flow" starts, we have to define what it is. I've already offered my defintion, if you have a different one than mine, please clarify it in more detail.

 

[mccrone]

Thanks, that is clear now.

We can observe certain things like redshift and we then have to assume that the reason is the expansion of space. Agreed. And the expansion of space would move two observers apart for "no reason". Although the two observers would have to be far enough apart for confounding forces like gravity to be overwhelmed.

So no argument. My point remains that IF there is a Hubble flow expansion, it is reasonable to assume it is scale invariant. The force or mechanism happens right down to the smallest scale.

I just noticed another confusion in Jimmy Snyder's posts which I don't think was picked up.

"I have heard that one must think of Hubble expansion as the expansion of space because otherwise, you end up with distant objects traveling faster than light. Using 75 km/s/Mpc, this happens at 4000 Mpc. In other words, we shouldn't think of Hubble expansion as things moving away from us, but rather as space expanding."

In fact all mass and space over the event horizon of the visible universe will be moving at supraluminal speed. It becomes lost from sight as it reaches the speed of light, and then keeps on going way past the speed of light...in effect. Relativity is preserved because there is no communication with stuff that has gone over the event horizon. Relativity is about the the speed of stuff in interaction.

Here is a good paper on this point...
http://arxiv.org/abs/astro-ph/0310808

Cheers - John McCrone.

 

[pervect]

pervect, I think you are right if your answer is related only to material objects. The question whether rods expand or not can be analized taking a look to the stress and therefore to the Riemann tensor. However, one may ask also whether two points in space separated by a "small" distance increase its distance or not. As far as I know, this is indeed measurable. This has nothing to do with stresses on rods and the Riemann tensor, but only with the increase of the scale factor: in a linearly expanding model with homogeneity at arbitrary scales and with zero Riemann tensor, the scale factor increases and two arbitrary points do always increase its distance. So, in my opinion, the question here reduces to know what metric is relevant at what scales.

The problem with using a metric to define the "expansion of space" is that we have the option of using many different metrics.

If we insist that the metric be homogeneous, isotropic, and spatially flat, and if we ignore "lumps", we pin the form of the metric down very closely, but one issue remains -defining the time scale for the 't' coordinate.

If we furthermore define the time scale 't' to be that of the proper time of an observer who observes the universe to be isotropic, we wind up with a unique metric. However, operationally, I believe that this definition is equivalent to my more physical defintion, as long as the universe is not a vacuum universe (more on this later). In a vacuum universe, there's nothing to look at to define the "preferred frame", so my proposal fails.

I agree that if a(t) is linear, there are no tidal forces on a meter stick. However, the Riemann is not zero everywhere in a FRW universe where a(t) is linear. Though the tidal force components of the Riemann are zero, other components, such as R^{x}_{yxy} are not zero. The Einstein tensor G is also not zero for a(t) linear, it reqires the presence of matter to have this solution.

[add]\rho is non-zero, P is non-zero, but \rho+3P is zero for a(t) linear - the solution requries negative pressures.

\rho = 3a_t/a^2 if you set the problem up with an orthonormal basis of one-forms.

The required presence of matter (a non-zero Einstein tensor which implies a non-zero stress-energy tensor) implies that a hypothetical observer has something to "look at" to define the "rest frame" of the linearly expanding universe, which is why I say my more physical definition is operationally equivalent to yours for this case.

Note that the observers at both ends of a meter-stick cannot both simultaneously be in frames which are "at rest" with respect to the universe, i.e. cannot be in the priveleged frame from which the universe appears to be isotropic. At most one obserer on a meter-stick can be in this special frame.

Thus there is no incosistency in saying that meter sticks do not expand and that space does expand. Usually there is a tidal force on the meter stick, but even when there is no such force, both ends of the meter stick are not and cannot be in the preferred "rest" frame which is how we are defining our distances to expand.

This is always true, even in the very special case (linear expansion) where there is no tidal force on a meter-stick.

I'll note that for the vacuum universe (where my defintion fails) we have an ambiguity as to what metric to use, the static, spatially flat, non-expanding universe, or the expanding non-spatially flat Milne universe. Thus, while my observational method fails to define an expansion factor when there is nothing to look at, your metric approach also has some ambiguities to deal with too. The Milne solution is the only expanding isotropic vacuum solution. Other solutions have the da/dt = 0.

The remaining issue is how to deal with "lumps". I'm not sure how best to do that, anyway this post is already too long.

 

[hellfire]

I agree that if a(t) is linear, there are no tidal forces on a meter stick. However, the Riemann is not zero everywhere in a FRW universe where a(t) is linear. Though the tidal force components of the Riemann are zero, other components, such as R^{x}_{yxy} are not zero. The Einstein tensor G is also not zero for a(t) linear, it reqires the presence of matter to have this solution.

I do not understand what you say here. If we are talking about a homogeneous and isotropic universe, then Friedmann's equations hold and then, with no matter i.e. \rho = p = 0, one has \ddot a = 0 and a ~ t (there may be however another fine tuning solution with matter to give a ~ t).

In such a case I cannot imagine that any of the components of the Riemann tensor are different from zero. Note that the Weyl tensor must be zero for the cosmological principle to hold and the Ricci tensor is zero: you can check this with equation (8.13) of http://arxiv.org/gr-qc/9712019 considering that \dot a^2 = - k and \ddot a = 0 in an empty universe without cosmological constant. The terms of the Riemann tensor which are contracted to give the Ricci and Weyl tensors must be all equal due to symmetry.

 

[Jimmy Snyder]

Do not confuse the expansion of space with the expansion of meter-sticks, please! Meter sticks do not expand.

This is the heart of my question. If space expands, and meter sticks don't and this all happens locally, then can't we perhaps measure it locally?

This would theoretically be true even for two points as close as a kilometer. However, determining the doppler shift of the CMB to determine motion with respect to the CMB of such a small amount is well beyond our ability to observe or measure.

OK, the answer is "no, not this way." How about some other way? Also, what about an experiment lasting many decades? Aren't the effects cumulative over time? We could magnify the effect by using a pair of highly reflective mirrors to bounce the light back and forth so that it travels much further than 1 km and yet remains within a 1 km apparatus.

I've mentioned this before, though, and get the feeling I'm not getting through. Sorry, but I don't know what the problem is, I've tried to explain things both informally and with the supporting mathematics, but I'm not getting any feedback that makes me believe I'm being understood :-(.

My fault entirely. Please don't think that I don't appreciate the effort. I have trouble seeing the big picture when it comes to physics. I can follow the text and work out the equations and still not understand what it all means.

 

[pervect]

I do not understand what you say here. If we are talking about a homogeneous and isotropic universe, then Friedmann's equations hold and then, with no matter i.e. \rho = p = 0, one has \ddot a = 0 and a ~ t (there may be however another fine tuning solution with matter to give a ~ t).
In such a case I cannot imagine that any of the components of the Riemann tensor are different from zero. Note that the Weyl tensor must be zero for the cosmological principle to hold and the Ricci tensor is zero: you can check this with equation (8.13) of http://arxiv.org/gr-qc/9712019 considering that \dot a^2 = - k and \ddot a = 0 in an empty universe without cosmological constant. The terms of the Riemann tensor which are contracted to give the Ricci and Weyl tensors must be all equal due to symmetry.

Sorry for the delay responding, I missed this somehow.

If you look at 8.14 of your reference, you'll see that the Ricci scalar is _not_ zero when \dot a = 1, \ddot a = 0 and k=0.

Therfore a(t)=t is not a vacuum solution for k=0 (flat space).

a(t)=t is a vacuum solution for k=-1, the Milne solution. As I mentioned in

https://www.physicsforums.com/showpost.php?p=754243&postcount=78

the expanding Milne solution (k=-1, a(t)=t) is equivalent to a static (non-expanding) flat space solution (k=0, a(t)=1) via a change of variables (that post gives the appropriate change of variables to demonstrate this).

Note that Sean Caroll is calculating the Riemann in a coordinate basis. My approach is very similar to that of MTW on pg 728 in that I calculate the Einstein tensor in an orthonormal basis of one forms.

Thus I calculate G_{\hat{t}\hat{t}}, not G_{00}.

Another way of saying this - I have introduced a local re-scaling of variables so that the local metric is diag(1,-1,-1,-1), i.e. Minkowskian.

The one-forms map a vector into a scalar. You can think of the first one-form as defining a the local 't' coordinate of a local observer - you feed the one-form a tangent vector, it spits out a scalar, which is the value of the local \hat{t} coordinate.

The second one form similarly defines the local \hat{x} coordinate, and the third and fourth define the local \hat{y} and \hat{z} coordinates. These local (\hat{t},\hat{x},\hat{y},\hat{z}) coordinates are Cartesian and have a Minkowskian metric.

Thus when I (or MTW) calculate the pressures and densities, I (we) calculate the pressures and densities that a local observer would observe with his local clocks and local meter-sticks in a locally Cartesian coordinate system.

So to recap - with flat space (k=0), a uniformly expanding universe requires a positive matter density and a negative pressure.

There are two sets of equivalent vacuum solutions, which can be interpreted as either a static, spatially flat universe, or an expanding non-spatially flat universe. There is no spatially flat expanding vacuum solution, however.

 

[hellfire]

I understand this that you agree that linearly expanding does not imply non-zero Riemann tensor. There exists a linearly expanding (fine tuned) solution in which the Riemann tensor is not zero, but in case of the Milne universe the Riemann tensor is zero.

 

[pervect]

I understand this that you agree that linearly expanding does not imply non-zero Riemann tensor. There exists a linearly expanding (fine tuned) solution in which the Riemann tensor is not zero, but in case of the Milne universe the Riemann tensor is zero.

Right.

I'm afraid I've lost track about what we were arguing about. Checking back, earlier I wrote:

I agree that if a(t) is linear, there are no tidal forces on a meter stick. However, the Riemann is not zero everywhere in a FRW universe where a(t) is linear. Though the tidal force components of the Riemann are zero, other components, such as R^{x}_{yxy} are not zero. The Einstein tensor G is also not zero for a(t) linear, it reqires the presence of matter to have this solution.

I guess I didn't make it clear that I was assuming that k=0, i.e. that space was flat, when making this remark.

 

[Jimmy Snyder]

In view of a new thread in the Astrophysics forum
https://www.physicsforums.com/showthread.php?p=1008842#post1008842,
I'm amazed that the Cooperstock paper was not mentioned in this thread.
I quote from near the end of the Cooperstock paper.

As a conclusion, it is reasonable to assume that the expansion of the universe affects all scales, but the magnitude of the effect is essentially negligible for local systems, even at the scale of galactic clusters.

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