Does the steady-state model resolve Olbers' paradox?

[Ibix]

A couple of additional notes - as I understand it, $ds^2=dt^2-a(t)(\frac{1}{1-kr^2}dr^2+d\Omega^2)$ is the most general isotropic and homogenous pseudo-Riemannian metric, and this is argued on purely geometric grounds, independent of the EFEs. However, I think the form of $a(t)$ in Bonnor's paper is derived from requiring $\partial_tT^{tt}=0$ (no summation implied) in the EFEs.

So Bonnor's statement that this "is" the steady state metric appears to contain the EFEs as an unstated assumption. It does not appear to preclude other steady state metrics under other theories of gravity with different field equations.

 

[Ibix]

It turns out that Ben Crowell wrote an Insights article on the steady-state model:

https://www.physicsforums.com/insights/steady-state-model-no-longer-viable/

And that references section 8.4 (p. 363) in his GR book (downloadable from this page). That notes that Hoyle worked in a more-or-less GR framework, although his C field actually violates the equivalence principle because it picks a preferred frame, while Bondi and Gold proposed replacing GR entirely.

 

[Dale]

I disagree. A homogeneous and isotropic spacetime is necessarily described by the FLRW metric, regardless of whether or not it satisfies Einstein's field equations

This is clearly false. An obvious counterexample is Newtonian spacetime. It is homogenous and isotropic but is not described by any spacetime metric. It is described by a pair of degenerate metrics, one for space and one for time.

Similarly for any homogenous and isotropic spacetime with torsion. Homogeneity and isotropy of spacetime alone does not imply FLRW.

 

[Jaime Rudas]

My cosmology notes refer to the metric with $a(t)$ unconstrained by the Friedmann equations as the RW metric, and with the additional constraint of the Friedmann equations/EFEs as the FRW metric. Lemaitre isn't referenced. However, reading the Wiki article, I note that this particular distinction may reflect the personal preferences of my lecturer and/or whoever taught him.

I believe the issue stems from the historical sequence of events.

In 1922, Friedman found solutions to Einstein's field equations under the assumption of a homogeneous and isotropic universe. In this context, he employed the metric that is now known as the FLRW metric. In 1927, Lemaître independently derived dynamical solutions to Einstein's equations using essentially the same geometry. In 1935, Robertson demonstrated that the form of the metric does not depend on Einstein's field equations, but follows solely from the geometric assumptions of homogeneity and isotropy. In other words, he showed that it is the most general metric compatible with a space-time whose spatial hypersurfaces are homogeneous and isotropic. In 1937, Walker independently arrived at the same result through a more general and systematic mathematical treatment.

 

[Jaime Rudas]

I think we need to be careful with what we mean by "the steady state model", especially in light of what Bonnor shows in that article.

He certainly states that the metric is the metric of the steady state model, and McCrea clearly believed this too. However, Bonnor promptly shows that it is not a realistic model of our universe, being what we would now call pure dark energy.

Yes, when I mentioned the steady-state model in the original post, I assumed, perhaps mistakenly, that it would be understood that I was referring to a homogeneous and isotropic spacetime (without torsion!) that is expanding and has a positive density constant over time. Under these conditions, I assumed that the metric describing this spacetime is FLRW, where $\frac {\dot a}a$ is constant and, therefore, $a(t)$ is of the form $a_0 e^{Ht}$. That is, the form of the metric follows from the constancy of the density.

And, yes. As I believe I have also made clear, this model is not a realistic model of our universe.

 

[Ibix]

Yes, when I mentioned the steady-state model in the original post, I assumed, perhaps mistakenly, that it would be understood that I was referring to a homogeneous and isotropic spacetime (without torsion!) that is expanding and has a positive density constant over time. Under these conditions, I assumed that the metric describing this spacetime is FLRW, where $\frac {\dot a}a$ is constant and, therefore, $a(t)$ is of the form $a_0 e^{Ht}$. That is, the form of the metric follows from the constancy of the density.

And, yes. As I believe I have also made clear, this model is not a realistic model of our universe.

Well, in that case I think the answer is as given by Bonnor - the redshift is not determined by those constraints because there isn't any normal matter. Or to put it in Peter's more symmetry-based language, there isn't a unique congruence that you can identify as the co-moving family in order to define redshift between them.

 

[Roberto Pavani]

It seems to me that expansion alone does not necessarily prevent light from eventually reaching us.
For example, consider an expanding universe with scale factor $a(t)=t\log t$.
This is still an accelerating expansion because $a''(t)=\frac{1}{t}>0$.
However, $\int^\infty \frac{dt}{a(t)}=\int^\infty \frac{dt}{t\log t}$ diverges.
Therefore there is no future event horizon, even though the expansion is accelerating.
So, if I understand the OP's question correctly, the key issue for Olbers' paradox is not simply whether the universe expands, or even whether the expansion accelerates.
The question is whether redshift and time dilation suppress the contribution from arbitrarily distant sources enough that the total sky brightness remains finite, even though the received radiation may be highly redshifted.

 

[PeterDonis]

expansion alone does not necessarily prevent light from eventually reaching us.

Of course not. There are many expanding models that have no cosmological horizon.

However, de Sitter spacetime, which is the only candidate that has been discussed in this thread for a "steady-state" model (since the density and Hubble constant are truly constant everywhere, in space and in time), does have a cosmological horizon. So if the presence of a cosmological horizon is sufficient to resolve Olbers' paradox (i.e., to make the night sky dark), we would expect that in de Sitter spacetime.

Unfortunately, for reasons already discussed in this thread, de Sitter spacetime actually does not appear to describe what proponents of the steady-state model actually wanted to describe.

 

[Ibix]

It seems to me that expansion alone does not necessarily prevent light from eventually reaching us.
For example, consider an expanding universe with scale factor $a(t)=t\log t$.
This is still an accelerating expansion because $a''(t)=\frac{1}{t}>0$.
However, $\int^\infty \frac{dt}{a(t)}=\int^\infty \frac{dt}{t\log t}$ diverges.
Therefore there is no future event horizon, even though the expansion is accelerating.
So, if I understand the OP's question correctly, the key issue for Olbers' paradox is not simply whether the universe expands, or even whether the expansion accelerates.
The question is whether redshift and time dilation suppress the contribution from arbitrarily distant sources enough that the total sky brightness remains finite, even though the received radiation may be highly redshifted.

It wouldn't surprise me if it's possible to construct such a "bright sky" FLRW model - I already argued that the Einstein static universe was one such. When all's said and done, Olber's paradox simply declares such models implausible descriptions of our universe, not impossible.

 

[PeterDonis]

whether redshift and time dilation suppress the contribution from arbitrarily distant sources enough that the total sky brightness remains finite

Just knowing the spacetime geometry is not enough to assess this. You also need to know the worldlines of the observer (us here on Earth) and the sources.

In, for example, a matter-dominated FRW spacetime, there is a unique congruence of comoving worldlines that is the obvious choice for the worldlines of the sources and the observer.

But in de Sitter spacetime, as has already been said, there is no such unique congruence. There is no particular set of worldlines that is picked out. So there is no way to assess the redshift question.

 

[martinbn]

This is clearly false. An obvious counterexample is Newtonian spacetime. It is homogenous and isotropic but is not described by any spacetime metric. It is described by a pair of degenerate metrics, one for space and one for time.

Similarly for any homogenous and isotropic spacetime with torsion. Homogeneity and isotropy of spacetime alone does not imply FLRW.

I think the stament assumes that the manofold is Lorentzian to begin with. A manifold, with a Lorentzian metric, which is spacially homogenous and isotropic has a metric of the FRWL family.

 

[Dale]

I think the stament assumes that the manofold is Lorentzian to begin with. A manifold, with a Lorentzian metric, which is spacially homogenous and isotropic has a metric of the FRWL family.

That is kind of my point. If the EFE is not satisfied then the theory of gravity is not GR. So without specifying the alternative theory or class of alternative theories, you cannot just assume a metric theory of gravity or a Lorentzian metric or a torsion free connection.

 

[martinbn]

That is kind of my point. If the EFE is not satisfied then the theory of gravity is not GR. So without specifying the alternative theory or class of alternative theories, you cannot just assume a metric theory of gravity or a Lorentzian metric or a torsion free connection.

Any Lorentzian metric satisfies the EFE for some stress-energy tensor.

 

[PeterDonis]

If the EFE is not satisfied

Any Lorentzian metric satisfies the EFE for some stress-energy tensor.

See the last part of my post #19 for a little more detail about how we know this.

 

[Dale]

Any Lorentzian metric satisfies the EFE for some stress-energy tensor.

Any Lorentzian metric has an Einstein tensor. Whether the Einstein tensor is related to the stress-energy tensor depends on the theory of gravity.

 

[PeterDonis]

Whether the Einstein tensor is related to the stress-energy tensor depends on the theory of gravity.

Yes, but since you can always just divide the Einstein tensor by $8 \pi$ (and whatever other numerical factor is dictated by your choice of units) and call that the stress-energy tensor, as I said in post #19, you can always find a way for any Lorentzian metric to satisfy the EFE. So any Lorentzian metric will be a solution using standard GR with the EFE as the theory of gravity. Whether the stress-energy tensor implied by that solution has any reasonable physical interpretation on that basis is another question, but mathematically, it will be a solution for that theory of gravity.

 

[Dale]

Yes, but since you can always just divide the Einstein tensor by 8π (and whatever other numerical factor is dictated by your choice of units) and call that the stress-energy tensor, as I said in post #19, you can always find a way for any Lorentzian metric to satisfy the EFE.

As Abe Lincoln would say, calling a tail a leg doesn't give a calf five legs. You can call the Einstein tensor the stress-energy tensor, but that doesn't relate the physically measurable curvature of spacetime to the physically measurable stress-energy of matter.

Why don't we discuss physics instead of semantics.

 

[martinbn]

As Abe Lincoln would say, calling a tail a leg doesn't give a calf five legs. You can call the Einstein tensor the stress-energy tensor, but that doesn't relate the physically measurable curvature of spacetime to the physically measurable stress-energy of matter.

Why don't we discuss physics instead of semantics.

I might be wrong, but I think you are missing the point. Take any metric, calculate the Einstein tensor, divide it by the appropriate constant, use the result to define a stress-energy tensor. Then the metric statisfies the EFE for that stress-energy tensor.

 

[Ibix]

I might be wrong, but I think you are missing the point. Take any metric, calculate the Einstein tensor, divide it by the appropriate constant, use the result to define a stress-energy tensor. Then the metric statisfies the EFE for that stress-energy tensor.

I think @Dale's point is that this is true but meaningless unless you believe that the EFE is the correct theory of gravity. And at least @bcrowell seemed to believe that some Steady State proponents were considering alternate theories (arguably they all were, given that we know that the "steady state metric" in vanilla GR is de Sitter).

 

[Dale]

I think you are missing the point. Take any metric, calculate the Einstein tensor, divide it by the appropriate constant, use the result to define a stress-energy tensor

I get your point but it is a semantic point with no physics content.

 

[PeterDonis]

You can call the Einstein tensor the stress-energy tensor, but that doesn't relate the physically measurable curvature of spacetime to the physically measurable stress-energy of matter.

Yes, it does. The physical reasoning is "this is the kind of stress-energy that would be necessary to produce a metric with this Einstein tensor, given that we're using GR as our theory of gravity". And then you look at the properties of that tensor to see what its physical implications are. In the case of the metric shown in this thread that was claimed to be that of "the steady-state model", this process shows you that the stress-energy is dark energy aka a cosmological constant--a perfect fluid with equation of state $p = - \rho$.

 

[PeterDonis]

this is true but meaningless unless you believe that the EFE is the correct theory of gravity

It tells you what the implications are of assuming that the EFE is the correct theory of gravity and that the metric is the given ansatz.

some Steady State proponents were considering alternate theories

AFAIK Bondi and Gold explicitly said in at least one of their papers that they were assuming GR would need to be replaced by an alternate theory of gravity for their "steady-state model" to work. But they never actually proposed one. Without one, it's impossible to say what the physical implications would be of the metric ansatz that was given.

Hoyle's "C-field" was an attempt to construct a model within the framework of GR.

 

[Dale]

Yes, it does. The physical reasoning is "this is the kind of stress-energy that would be necessary to produce a metric with this Einstein tensor, given that we're using GR as our theory of gravity".

This is not even wrong.

 

[PeterDonis]

This is not even wrong.

The Bonnor paper that was referenced in this thread does not appear to think so--at least it expresses no issues with the approach it references in the McCrea paper it is analyzing:

"McCrea (1951) begins by noting that the field equations of general relativity, without cosmological term but with energy tensor T...do not restrict the Riemann space to which they are applied, so it must be possible to use them to describe the space (1.1). To adopt this procedure amounts to using (1.2) to define an energy tensor of matter which would, according to relativity, produce the space-time (1.1)."

In other words, compute the Einstein tensor from the metric and see what kind of stress-energy it implies according to the EFE, exactly as I said. I'm pretty sure that's not the only place in the GR literature where such an approach is taken.

 

[Dale]

When you propose something that cannot be false then it is not even wrong in the usual meaning of the phrase. If you choose to reduce the EFE to a tautology then it is not even wrong.

I think you are also misrepresenting Bonner. He is saying that given a metric you can calculate the stress-energy tensor of matter that would be required to satisfy the EFE. He is not saying that whatever you calculate from the curvature side of the EFE you can just call that the stress-energy tensor, as you are claiming.

and call that the stress-energy tensor

This right here is the issue. You are not saying that for any metric there exists a SET that would satisfy the EFE (that is Bonner’s statement). You are saying instead that from any metric you can simply call the thing that would satisfy the EFE the SET. You even italicized the word “call” to emphasize that this was a purely semantic exercise. That makes the EFE non-falsifiable and therefore not even wrong.