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

[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

 

[pervect]

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.

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

I don't think I necessarily agree with the assumptions underlying Cooperstocks' approach as a "given". From what I can tell from Ned Wright's quote (which is where I found the Cooperstock reference in the first place), he also has similar reservations on Cooperstock's approach - i.e. Ned Wright feels that Cooperstocks approach could be justified only under certain particular conditions, such as a uniform background of "dark matter" in the universe.

Take a look at
http://www.astro.ucla.edu/~wright/cosmology_faq.html#SS

For the technically minded, Cooperstock et al. computes that the influence of the cosmological expansion on the Earth's orbit around the Sun amounts to a growth by only one part in a septillion over the age of the Solar System. This effect is caused by the cosmological background density within the Solar System going down as the Universe expands, which may or may not happen depending on the nature of the dark matter.

The thing that is probably the most important is that we can regard Cooperstocks' approach as being conservative, in the sense of giving us the biggest possible effect. And even being conservative in this manner, we can see that the effect would be very small and not measurable (parts in a septillion).To take one example, the loss of mass of the sun via radiation would have a much more significant effect on changing planetary orbits than the expansion of the universe - i.e. the sun is losing visible mass much faster than the rate at which hypothetical "dark matter" would be leaving the solar system due to cosmological expansion.