Using GR to predict the shape of the entire universe?

[Feynstein100]

So I've often heard that when GR is applied to the entire observable universe to calculate its curvature, we get a value of zero, meaning that the entire universe is flat. I've got 2 problems with this.
The first is that I thought GR was a local theory i.e. it only applies locally. Trying to apply it to the entire universe is taking the theory well beyond its range of application. This, of course, does not yield reliable results. Although, applying GR to the entire universe is what led to the prediction that the universe is expanding, which turned out to be true. So I'm probably wrong about this.
Second, it means that the shape of the entire universe depends on the average density of matter in it, which of course, is not constant because the universe is expanding. Meaning that just because the universe is flat now because the average matter density is 4 hydrogen atoms/m3 or whatever does not mean it was always this way. The density of matter was higher in the past. So what does that mean? The universe used to be closed in the past, but is now open/flat and eventually will be hyperbolic? And what happens in the far future of the universe, when there's no matter left? Does the hyperbolic universe go back to being flat? All this seems highly problematic to me.

 

[PeterDonis]

I've often heard

Where? Please give a specific reference.

when GR is applied to the entire observable universe to calculate its curvature, we get a value of zero

More precisely, when we build a model of our universe using GR and the Einstein Field Equation as our framework, and test it against observations, the best fit model has the universe being spatially flat.

The first is that I thought GR was a local theory i.e. it only applies locally.

This is not correct. GR is "local" in the sense that the Einstein Field Equation is a differential equation that must be valid at every point of a spacetime. But that in no way prevents you from constructing solutions that describe an entire spacetime.

Trying to apply it to the entire universe is taking the theory well beyond its range of application.

This is not correct either.

This, of course, does not yield reliable results.

Yes, it does. You are sadly misinformed.

applying GR to the entire universe is what led to the prediction that the universe is expanding, which turned out to be true. So I'm probably wrong about this.

And yet you are making confident statements--which are, in fact, wrong, as noted above. Perhaps it would be better to not make the confident statements in the first place.

Second, it means that the shape of the entire universe depends on the average density of matter in it

This is wrong as you state it. The correct statement is that the geometry of the entire universe depends on the distribution of stress-energy, via whatever particular solution of the Einstein Field Equation is valid for the universe.

The density of matter was higher in the past. So what does that mean? The universe used to be closed in the past, but is now open/flat and eventually will be hyperbolic?

No.

And what happens in the far future of the universe, when there's no matter left?

There will never be no stress-energy left. Stress-energy doesn't just disappear.

Does the hyperbolic universe go back to being flat?

No.

All this seems highly problematic to me.

That is because you are reasoning from incorrect premises. See above.

 

[Orodruin]

The first is that I thought GR was a local theory i.e. it only applies locally. Trying to apply it to the entire universe is taking the theory well beyond its range of application. This, of course, does not yield reliable results. Although, applying GR to the entire universe is what led to the prediction that the universe is expanding, which turned out to be true. So I'm probably wrong about this.

Being a local theory does not mean it applies only locally. It means things are described locally. What happens at an event only depends on things present at that event. Global geometry is certainly within its realm.

The universe used to be closed in the past, but is now open/flat and eventually will be hyperbolic?

No. It does not only depend on the density, but the density relative to the expansion rate. A flat universe stays flat, but the smallest of deviations from flatness will grow with time as long as the universe is dominated by matter of radiation. However, if dark energy dominates it will become flatter and flatter.

Also note that this is about spatial flatness, not flatness of spacetime.

 

[Feynstein100]

Where? Please give a specific reference.More precisely, when we build a model of our universe using GR and the Einstein Field Equation as our framework, and test it against observations, the best fit model has the universe being spatially flat.This is not correct. GR is "local" in the sense that the Einstein Field Equation is a differential equation that must be valid at every point of a spacetime. But that in no way prevents you from constructing solutions that describe an entire spacetime.This is not correct either.Yes, it does. You are sadly misinformed.And yet you are making confident statements--which are, in fact, wrong, as noted above. Perhaps it would be better to not make the confident statements in the first place.This is wrong as you state it. The correct statement is that the geometry of the entire universe depends on the distribution of stress-energy, via whatever particular solution of the Einstein Field Equation is valid for the universe.No.There will never be no stress-energy left. Stress-energy doesn't just disappear.No.That is because you are reasoning from incorrect premises. See above.

Thank you for the correction. It looks like I misunderstood what GR being local really means . It is applicable to the entire universe. I learned something new
However, the second point still kind of stands. Just replace matter density with stress-energy density. My problem is still that this density, and thus the curvature of the entire universe, is not constant over time as far as I know. Or is it?

 

[Feynstein100]

It does not only depend on the density, but the density relative to the expansion rate.

I'm sorry, I don't understand. What does that mean?

Also note that this is about spatial flatness, not flatness of spacetime.

Damn, I don't understand that either. I thought we were talking about the curvature of 4D spacetime? I mean, what other kinds of curvature are there?

 

[Orodruin]

I'm sorry, I don't understand. What does that mean?

Exactly what it says. Whether the universe is flat or not does not depend only on the density but also on the expansion rate (or Hubble rate $H$). The critical density for which the universe is flat is proportional to $H^2$. Since $H$ changes with time, so does the critical density.

I thought we were talking about the curvature of 4D spacetime? I mean, what other kinds of curvature are there?

No. When we talk about the curvature in cosmology it is usually about the spatial curvature, ie, the curvature of the spatial slices defined by constant cosmological time.

 

[Feynstein100]

Since changes with time, so does the critical density.

That's kind of what I'm having problem with. Does that mean the curvature of the universe changes with time too? i.e. It went from being closed/finite to being open/infinite? Which does not seem logically possible to me.

No. When we talk about the curvature in cosmology it is usually about the spatial curvature, ie, the curvature of the spatial slices defined by constant cosmological time.

I still don't get it

 

[vanhees71]

No. When we talk about the curvature in cosmology it is usually about the spatial curvature, ie, the curvature of the spatial slices defined by constant cosmological time.

To be pedantic: it's the coordinate time in the standard coordinates of the Friedmann-Lemaitre-Robertson-Walker solution, i.e., the proper time of an observer in the (local) rest frame of the cosmic microwave background.

 

[Orodruin]

Does that mean the curvature of the universe changes with time too?

No. See #3. It is unclear which part of that is unclear to you.

 

[Orodruin]

To be pedantic: it's the coordinate time in the standard coordinates of the Friedmann-Lemaitre-Robertson-Walker solution, i.e., the proper time of an observer in the (local) rest frame of the cosmic microwave background.

Not sure that is being pedantic, rather stating the definition of cosmological time.

 

[Ibix]

Does that mean the curvature of the universe changes with time too? i.e. It went from being closed/finite to being open/infinite? Which does not seem logically possible to me.

If the universe is not flat, the radius of curvature of space "now" changes over time, but does not change sign. So, for example, if the universe is closed then it is always closed, but the distance around the universe changes.

The point about the relationship between the Hubble constant and density is that the Hubble constant is not a constant over time - it's the same everywhere but it's not the same everywhen. So the density of the universe decreases over time, but the critical density needed for it to be flat/open also decreases.

 

[PeterDonis]

the second point still kind of stands.

No, it doesn't. See below.

Just replace matter density with stress-energy density. My problem is still that this density, and thus the curvature of the entire universe, is not constant over time as far as I know.

That is correct, but it does not mean what you are claiming it means. The universe can be spatially flat at all times while still having a stress-energy density that decreases with time. Post #6 by @Orodruin and post #11 by @Ibix have explained how.

 

[PeterDonis]

Does that mean the curvature of the universe changes with time too?

The local spacetime curvature along a particular comoving worldline changes with time, yes. This must be the case since the local spacetime curvature, or more specifically the local Einstein tensor, is directly proportional to the local stress-energy density as shown by the Einstein Field Equation.

i.e. It went from being closed/finite to being open/infinite?

This does not need to happen for the local spacetime curvature to change.

 

[Feynstein100]

If the universe is not flat, the radius of curvature of space "now" changes over time, but does not change sign. So, for example, if the universe is closed then it is always closed, but the distance around the universe changes.

The point about the relationship between the Hubble constant and density is that the Hubble constant is not a constant over time - it's the same everywhere but it's not the same everywhen. So the density of the universe decreases over time, but the critical density needed for it to be flat/open also decreases.

Ah I get it now. I thought the overall curvature of the universe was a function of only the stress-energy density i.e. C = f(ρ), hence the problem. But turns out, C = f(ρ,H) with H being the Hubble constant. And while ρ decreases with time, H increases, canceling out the overall effect and keeping the curvature constant. Or rather, preventing the curvature from changing sign i.e. going from flat to spherical/hyperbolic.
It seems a bit coincidental that the two variables of the function change just at the right rates to preserve the overall curvature. Is there a reason behind it perhaps or is it trivial?

 

[Bandersnatch]

H decreases too.

 

[PeterDonis]

I thought the overall curvature of the universe was a function of only the stress-energy density i.e. C = f(ρ), hence the problem. But turns out, C = f(ρ,H) with H being the Hubble constant.

No, that is not correct. $H$ is an effect of the curvature (the spacetime curvature, not the space curvature), not the source of it.

And while ρ decreases with time, H increases, canceling out the overall effect and keeping the curvature constant.

No, that is not correct. $H$ is not increasing with time in our universe; it is decreasing.

What you are missing here is that the stress-energy at a given event (point) in spacetime determines the spacetime curvature at that event (or more precisely the Einstein tensor, which is the only kind of spacetime curvature present in our models of the universe as a whole). The spatial curvature then depends on how you split up spacetime into space and time. The standard way of doing that in our model of the universe is to use "comoving" observers, which are observers who always see the universe as homogeneous and isotropic. There is a natural split of spacetime into space and time for those observers, and the "space" in this split is what is flat in our best current model. It stays flat because the stress-energy and $H$ both decrease with time in the right relationship.

 

[Feynstein100]

Whether the universe is flat or not does not depend only on the density but also on the expansion rate (or Hubble rate )

This seems to be in direct opposition to what you say here:

No, that is not correct. is an effect of the curvature (the spacetime curvature, not the space curvature), not the source of it.

This just keeps getting more confusing

and the "space" in this split is what is flat in our best current model

So it's the 3D portion of the 4D curvature that's actually flat? This seems unnecessarily confusing. Why not just refer to the 4D curvature as a whole to get the full picture?

 

[Orodruin]

Why not just refer to the 4D curvature as a whole to get the full picture?

Because it is actually less informative than the 4D curvature. Note that it is the curvature of a three-dimensional manifold, an isotropic and homogeneous spacelike surface. It is not a sub-part of the spacetime curvature. If we are interested in the structure and properties of the isotropic and homogeneous spatial slices, those are what we should consider.

The spacetime curvature also has its place, but not for the same purpose.

 

[Hill]

less informative than the 4D curvature

I think, you meant to say that the 4D curvature is less informative than the 3D curvature in this case.

Why not just refer to the 4D curvature

Because we want to know the geometry of the space we observe with our instruments.

 

[PeterDonis]

This seems to be in direct opposition to what you say here:

No, it isn't. (And the posts you quoted as seeming to be in direct opposition are not by the same person: the first is by @Orodruin, the second is by me.) We are just describing different aspects of the same thing.

So it's the 3D portion of the 4D curvature that's actually flat?

It's the 3D slices of constant comoving coordinate time (which equates to constant proper time since the Big Bang for comoving observers) that are flat.

 

[timmdeeg]

So it's the 3D portion of the 4D curvature that's actually flat? This seems unnecessarily confusing. Why not just refer to the 4D curvature as a whole to get the full picture?

Are you sure that you have the right understanding what spatial flatness means?

It just means that the sum of angles in a large enough triangle of a 3D slice is 180°.

 

[Feynstein100]

No, it isn't. (And the posts you quoted as seeming to be in direct opposition are not by the same person: the first is by @Orodruin, the second is by me.) We are just describing different aspects of the same thing.It's the 3D slices of constant comoving coordinate time (which equates to constant proper time since the Big Bang for comoving observers) that are flat.

I give up. This is too difficult

 

[Orodruin]

I give up. This is too difficult

I think part of the issue is trying to understand cosmology/GR at an I level. Creating a deeper understanding of what is actually going on and the math behind it is A level and requires actually studying GR from a relevant textbook source or taking a course that would typically be 4th or 5th year in a physics education.

 

[PeterDonis]

I give up. This is too difficult

The cautions @Orodruin gives about what background knowledge you need are worth considering. That said, you might want to try Sean Carroll's online lecture notes on GR:

https://arxiv.org/abs/gr-qc/9712019

Chapter 8 covers cosmology and addresses the sorts of questions you are asking here. However, working through the previous chapters that cover the basics of SR and GR might be helpful to increase your general understanding.

 

[Ibix]

I would say you need a solid understanding of special relativity before attempting Carroll's notes. The first chapter skims SR in highly abstract terms that are a good basis for developing GR, but probably not the most straightforward exposition of SR. Carroll himself says something of the sort, IIRC.