# Curvature of space; curvaure of spacetime

1. Mar 24, 2008

### jonmtkisco

I have a question about spacial curvature. Before asking the question, I will summarize my understanding of the distinction between spatial curvature and spacetime curvature. If I have this wrong, I would appreciate corrections.

Spacetime curvature occurs locally in the presence of gravity; i.e. mass-energy. Spacetime curvature causes timelike geodesics to be displaced from a "straight" line, towards a source of mass-energy, but only as function of the passage of time. At any given fixed instant in time, spacetime curvature does not imply any spacial curvature. For example, the spacelike coordinate geometry can be perfectly flat, even if located very close to an intense gravitational source such as a black hole. In that scenario, the circumference of a circle remains $$2\pi r$$.

On the other hand, spacial curvature occurs as a homogeneous geometry across the entire universe, depending on whether the universe is below, at, or above critical density $$\Omega$$ of mass-energy. Because spatial curvature is universe-wide, it is not possible to have positive curvature in one portion of the universe, flat in another portion, and negative curvature in yet another portion. (Or is it?) Spatial curvature (other than flat) causes the spacelike circumference of a circle $$\neq 2\pi r$$r at an instant in time.

Here's my question:

Is the heuristic of spatial curvature REQUIRED by general relativity, or is it merely a CONVENTION which is CONSISTENT WITH GR? I understand that mathematically, the curvature coefficient "k" provides the constant of integration for the Friedmann equations. However, the fact that a constant fits mathematically does not in itself require that constant to represent a physical reality.

So, it isn't clear to me why a universe with mass-energy > $$\Omega$$ is REQUIRED by GR to have 3-spherical positive spatial curvature. Why can't the geometry be flat?

If a universe were EXACTLY at $$\Omega$$, (being therefore geometrically flat), and also happened to be infinite, and then as a thought experiment a single extra "test" photon were "released" (inserted) into this universe, that universe would necessarily become above $$\Omega$$ ever so slightly, and therefore would gain positive curvature and become necessarily finite in extent. Such an instantaneous transformation from infinite to finite is a bit of a mental stretch.

My sense is that spatial curvature has been adopted as a "convention" in GR as a way to avoid the awkward discussion about whether the universe might be flat and finite, while simultaneously expanding or contracting. The discussion becomes awkward because this scenario implies the possibility of an "edge", or some kind of outer boundary to the universe. I understand why such a model would be less elegant than the spatial curve, "closed" approach.

So my question is, does GR REQUIRE spatial curvature if mass-energy density $$\neq \Omega,$$ or is spatial curvature more accurately described as a CONVENTION of GR?

Jon

Last edited: Mar 24, 2008
2. Mar 24, 2008

### marcus

Jon, the Friedmann equations do not describe the real universe.

they describe cases of an ideal universe which has matter distributed UNIFORMLY

So the various cases of the FRW model are just idealizations and the question is which one fits our universe best.

Our own universe is bumpy and warty and matter is distributed imperfectly and lumpy, so it has places where average curvature is zero or nearly and it has places where it is positive and negative etc.

We still don't know if you could somehow smooth it all out even with a giant rake
whether it would turn out this case or that----and which idea model it would fit.

Your question about adding a single particle does not make sense because you start with an ideal flat FRW case and if you add one particle it immediately is no longer ideally uniform. You are back in the real world. So you cant say that it jumps from the ideal flat infinite case to the ideal positive curved finite case. It doesn't jump from one clean ideal case to another, it stops being ideal.
It stays infinite, if that is the way it was assumed to be for starters, but it is just bumpy, with its newly uneven matter distribution.

as you realize the future is controlled by the lambda driven acceleration, so adding one particle doesnt change that either.

Last edited: Mar 24, 2008
3. Mar 24, 2008

### jonmtkisco

Marcus,

There is no doubt that the inhomogeneous gravitational clumping of matter causes the local spacetime curvature to vary significantly from place to place in the universe. However, I do not understand how localized spacetime curvature specifically causes localized spatial curvature. Is that outcome a requirement of GR, or rather an informed speculation (such as David Wiltshire has made in his theory, for example)? I view it as the latter.

Regarding my thought experiment of adding a single photon to a perfectly "flat" universe, I do not believe that such a small change in mass-energy could possibly be viewed as as affecting the overall homogeneity and isotropy of the universe. Surely local bumps and variations are far in excess of a single photon.

My point was simply that if the universe happened to actually be PERFECTLY flat because it was perfectly at critical density, then indeed a single photon would push it above critical density and (apparently) require the universe to gain positive curvature. And thereby change over from infinite to finite extent, in my example. Mathematically that seems to be a certainty if the Friedmann equations are valid even as a large-scale AVERAGE over a slightly inhomogeneous universe.

Jon

4. Mar 25, 2008

### Haelfix

Jon first of all, you are always free to choose local coordinates where the curvature is identically zero so I don't follow your question.

Globally, you also cannot simply add or subtract stress-energy content at will either, since that would violate causality. These things must always communicate through gravitational waves in a perfectly causal manner, so new entities must always be brought in from some sort of relevant asymptotia so that things rearrange smoothly.

Moreover two seperate segments of the universe (I use the word here referring to *everything*) that are out of causal contact need not and in general do not have the same matter density. Which is why you have some people postulating all sorts of various bubble universes with different physical properties.

5. Mar 25, 2008

### jonmtkisco

Haelfix,

I agree that my little thought experiment can't physically be performed; no doubt it violates additional principles besides causality. I intended it only to suggest how two hypothetical universes which differ by such a slight amount (1 photon) and are identical in all other respects must have such radically different internal geometries. And that in turn, the internal geometry dictates whether an infinite universe is ruled out. So let's not get fixated on the mechanics of the thought experiment itself.

I want to return to my question, which is, as a matter of formal modern GR theory, is spacial curvature a required element of GR, or merely a particularly elegant convention?

[EDIT: Or I can ask the question the other way around: Does formal modern GR theory rule out (mathematically or heuristically) the possibility of a (finite or infinite) spatially flat universe with $$\Omega \neq 1$$?]

The fact that spatial curvature cannot be defined by a single point in space coordinates does not imply that the reality or nonreality of spatial curvature is unimportant to understanding the physical functioning of the universe. Fundamental observations of the cosmos depend on whether circumference = $$2 \pi r$$ throughout at least the observable universe.

I agree that two regions of the universe which are out of causal contact with each other might have significantly different densities at the so-called "scale of homogeneity", but I think it is also quite plausible that they might not. [EDIT: I note that there are large regions within our observable universe which currently are out of causal contact with each other, and which do not appear to vary significantly from each other in terms of density or other large-scale characteristics.]

Wiltshire postulates that large voids are regions of our observable universe which can have locally negative spatial curvature even though surrounding filaments and clusters have locally flat curvature. I do not believe that mainstream cosmology has accepted that idea yet, although I'm not aware that mainstream cosmology has ruled it out. It seems to me to be a very important question. If it were true, then it raises the question why there are not postulated to be any regions in our observable universe with positive spatial curvature? The obvious candidates would be regions containing very substantial mass, such as black holes or rich galaxy clusters. But I am not aware of anything in GR [EDIT: e.g., in either an FLRW or Schwarzschild vacuum model] which says that local spacetime curvature specifically causes local positive spatial curvature. If one applied the FLRW equations to mass concentrations in the same way Wiltshire does to voids, I believe that such positive spatial curvature could be postulated. But of course that does not mean that it is an accurate description of physical reality. In any event, it seems logically inconsistent that the absence of local mass-energy would cause local negative spatial curvature if the presence of local mass-energy does not cause local positive spatial curvature. If there's an explanation for the distinction, I'd like to read it.

[EDIT: I don't want to get bogged down here in a discussion of the specific embedding mechanism underlying Wiltshire's type of model. The accepted swiss cheese model allows for certain kinds of different geometries to be theoretically embedded locally into a FLRW background, if the embedded geometry is spherically symmetrical, contains the same amount of mass-energy "carved" out of the background, and boundary conditions are met. Thus a Schwarzschild vacuum geometry and/or an FLRW geometry of arbitrary curvature "K" can be embedded locally into any particular FLRW background, such as a background with flat geometry.]

As far as I know, the physical reality of the entire concept of spatial curvature, on either a local or universal scale, has not been proved or even demonstrated to be likely. Again, if anyone is aware of research results on this topic, I would appreciate seeing them. Of course the WMAP analysis is trying to interpret the CMB anisotropy measurements to project whether the observable universe as a whole has a tiny amount of spatial curvature, but the error bars prevent any definitive projection at this time.

Jon

Last edited: Mar 25, 2008
6. Mar 26, 2008

### smallphi

Universes without gravitating matter (Minkowski) or filled with a cosmological constant density (DeSitter, AntiDesitter) can be foliated in 'constant time' hypersurfaces of constant spatial curvature of various sign. That's because such universes have time translation invariance (the matter density is the same constant at any space-time point) so nothing restrict how the homogeneous hypersurfaces must slice the manifold.

For example Minkowski in the usual inertial coordinates has spatial hypersurfaces 'of constant time' that have zero spatial curvature. It turns out a part of Minkowski can be reparametrized in coordinates that look like expanding FRW metric with spatial slices of negative spatial curvature - that parametrization is called Milne universe.

If the Universe is filled with matter whose density changes with time due to expansion, and one wants to use homogeneous and isotropic coordinates, the spatial slices are fixed to be the comoving matter slices. If you try to go away from those, the slices you get are not homogeneous anymore because the matter density at different points corresponds to different comoving time and will be different. For the homogeneous isotropic comoving slices, the spatial curvature is uniquely fixed by omega. You can't reparametrize such an universe into slices of constant spatial curvature of different sign.

Last edited: Mar 26, 2008
7. Mar 26, 2008

### jonmtkisco

Smallphi,

Thanks for the great answer. Now I want to persuade you to expand a bit on the two quoted sentences so I can better understand them. I interpret your bottom line to be that in an expanding FLRW universe with homogeneous isotropic matter distribution (which is the model I want to discuss here), if $$\Omega\neq 1$$, then it is mathematically impossible for the spatial geometry to be flat. I would like to understand why that is so.

Am I correct in interpreting you to mean that if I arbitrarily impose a constraint on the FRW curvature coefficient to require K=0 regardless of the value of omega, then the FRW metric is no longer an exact solution to the Einstein Field Equations?

I would appreciate if you could explain at more length why one can't reparameterize such a universe into slices of constant spatial curvature of different sign. For example, why is time translation invariance necessary in order to foliate with a different sign? And why doesn't the use of comoving coordinates itself circumvent that constraint? The answers aren't intuitive to me, so I hope you can help. If you can direct me to any source material that explains these specific details, that would be great also. Thanks!

Jon

Last edited: Mar 26, 2008
8. Mar 26, 2008

### jonmtkisco

It occurs to me that my question ("Is the paradigm of spatial curvature required by GR?") can be considered from another perspective.

In a simplified form of the Friedmann equation without Lambda or spatial curvature:

$$\dot{a}^{2} - \frac{8 \pi \rho}{3}{a^{2} = 0$$

it is obvious that there is no way to ever drive the expansion rate $$\dot{a}$$ negative at any positive value of density $$\rho$$. But we need to be able to generate a negative expansion rate when $$\rho$$ exceeds critical density.

When the curvature constant k = (+1, 0, or -1) is inserted into the Friedmann equation:

$$\dot{a}^{2} - \frac{8 \pi \rho}{3}{a^{2} = -k$$

it becomes possible for $$\dot{a}$$ to go negative when the density exceeds the expansion rate by a sufficient margin. This is an example of how the Friedmann equation can yield an incorrect answer if the curvature constant is arbitrarily constrained to k = 0.

As we've discussed previously, a simple way to characterize the Friedmann equation is that the spatial geometry is flat when the expansion rate of the specified region (e.g., the observable universe) exactly equals the Newtonian escape velocity of the total mass-energy contained in that region (with r as the radius of the region):

$$V_{esc} = \sqrt{\frac{2GM}{r}}$$

It seems to me that one underlying motivation for the Friedmann equation is to ensure that the kinetic energy of expansion always exactly equals the gravitational potential energy. Thus energy conservation of an Einstein-de Sitter model (without Lambda) is ensured. In a very loose sense, the energy of positive spatial curvature (k = +1) supplements the kinetic energy of expansion when the expansion rate is too slow ($$\Omega > 1$$), and the energy of negative spatial curvature (k = -1) supplements the gravitational potential energy when the expansion rate is too fast ($$\Omega < 1$$). The terms "too fast" and "too slow" mean as compared to escape velocity of course.

It is not clear to me why energy conservation must be ensured by an Einstein-de Sitter model. For example, it seems to me that if energy conservation is a necessary end state of an "initial condition" such as inflation, then inflation must be defined in a way that always results in a perfectly flat universe. (Happily inflation theory seems infinitely tuneable). Other than that, what fundamental principle requires that the kinetic energy of the Big Bang must be asymptotically exactly recaptured by the subsequent gravitational drag on expansion? Perhaps we are demanding one symmetry too many.

Or perhaps the nature of the "initial conditions" happens to be such that a Big Bang must inevitably be launched at precisely escape velocity.

In any event, it seems straightforward to me that the Friedmann equation can be modified so that it consistently generates the correct mathematical relationship between kinetic energy and gravitational drag without the need to insert an "artificial" curvature constant. I would hope such a modified version could still be an exact solution to the Einstein Field Equations, but I can't answer that question. A quick and dirty solution would be to leave the constant k = (+1, 0, -1) in the equation but simply interpret it to not represent physical curvature of space.

Jon

Last edited: Mar 26, 2008
9. Mar 26, 2008

### smallphi

'Newtonian cosmology' contrasted to GR cosmology, difference between 4-curvature and 3-curvature, Milne Universe, De Sitter Universe and their casting into different apparent spatial curvatures are discussed in "Physical foundations of cosmology" by V. Mukhanov, Chapter 1.

I have to warn that is a heavy calculational textbook, not a philosophical treatise.

10. Mar 26, 2008

### jonmtkisco

Smallphi,
Thanks. I am not looking to read an entire chapter of a treatise, which obviously covers far more than my narrow question, and may not directly answer my narrow question at all. Can you quote or summarize a particular excerpt which explains specifically why:

1. Time translation invariance is mathematically necessary in order to foliate with a different sign, for a FLRW universe; and/or

2. Why energy conservation is considered a necessary precondition for the Friedmann equations?

Jon

11. Mar 26, 2008

### smallphi

1. page 29: " Generally, one does not have a choice of foliation if it is to respect the homogeneity and isotropy of space. In particular, if the energy density is changing with time, the appropriate foliation is hypersurfaces of constant energy density. This choice is unique and has invariant physical meaning. Empty space, however, possesses extra time-translational invariance, so any space-like hypersurface has uniform 'energy density' equal to zero. The other example of a homogeneous and isotropic spacetime with extra time-translational invariance is de Sitter space. In the next section we will see that de Sitter space can be covered by three-dimensional hypersufaces of constant curvature with open, flat and closed geometry."

2. There's no such thing as potential gravitational energy in GR, so the problem is you are trying to transfer Newtonian intuition to GR. The Newtonian derivation of Friedman works only for a small local region in a spatially flat universe with presureless matter. The Newtonian value of k is interpreted as 'total energy' and has no space-curvature meaning - it doesn't influence the formulas for area or volume. In GR, k has a totally different meaning - it curves the spatial slices and has nothing to do with 'energy' of some expanding ball. The Newtonian and GR Friedman equations look exactly the same but the conceptual interpretation is very different so one can't transfer concepts from Newtonian physics to GR. The appearance of nonzero k in the GR Friedman equation does NOT violate the local energy conservation in GR: dE = - pdV. In fact Mukhanov derives the GR Friedman (with general nonzero k) using local energy conservation.

Last edited: Mar 26, 2008
12. Mar 26, 2008

### jonmtkisco

Smallphi,

Thanks! Despite your generosity in quoting the text, I went ahead and read the full section of Mukhanov you cited. It provides a clearer explanation of the relationship between Newtonian and GR expansion models than I've read before. I was surprised at how similar they are. At non-relativistic velocities, the only real differences between them are the role of pressure (not relevant to a dust model) and spatial curvature. So the Newtonian approximation really is useful. I understand that E in the Newtonian equation has a different meaning from k in the GR equation.

I agree that the GR Friedmann equation does not "violate" energy conservation. In fact, my question there was, does GR require the equation to conserve total energy in the first place, and if so, why? Why does the big bang need to conserve energy?

Mukhanov supplies two relevant Newtonian expansion equations which I've adapted slightly:

$$\dot{a}^{2} = \frac{2Gm}{a} - 2E$$

and

$$E = \frac{Gm_{cr}}{a} a^{2}\left[ 1 - \frac{m}{m_{cr}} \right]$$

where $$m_{cr}$$ means "critical total mass". When I substitute the 2nd equation into the first, I get:

$$\frac{\dot{a}^{2}}{a^{2}} = \frac{2Gm}{a^{2}} - 2Gm_{cr}\left( 1 - \frac{m}{m_{cr}}\right)$$

I offer this up as my attempt at defining the "Newtonian Friedmann Equation" without curvature or pressure. Again, my goal is to understand whether an FLRW model can be constructed with curvature constrained to k = 0, while maintaining the correct Newtonian relationship between kinetic energy of expansion and gravitational drag.

Jon

13. Mar 27, 2008

### Haelfix

Jon, GR does not respect global energy conservation, pretty much for any metric. You can think of it physically this way: You can sum up all the contributions from the stress energy tensor, but you hit a problem, namely there is no canonical way to include gravities contribution, since thats completely depends on what coordinates you want to use. Since there is no preffered way of doing this, even defining energy is arbitrary. People then make half solutions, and indeed you can make sense of certain notions of energy in some cases (for instance ADM energy), but this does not apply in full generality without imposing additional selection axioms on the classes of metrics you wish to study (for instance requiring asymptotic flatness, or certain killing vectors to be present).

Mathematically this is simply the statement that gauss's law doesn't go through like in the static case, or alternatively that there is no way to get Noethers law to apply without time translation symmetry (and that of course requires either an arbitrary foliation or isn't applicable for general metrics)

However, what GR does do is respect local energy conservation -always-, otherwise we'd be in trouble.

14. Mar 27, 2008

### jonmtkisco

Haelfix,

Makes sense to me. Do you think therefore that the Friedmann equation is NOT required by GR to ensure energy conservation? Its not intended to be a solution for a single point coordinate (unless you extend it back in time precisely to the big bang singularity).

Jon

Last edited: Mar 27, 2008
15. Mar 27, 2008

### smallphi

The full set of information in GR is contained in the Einstein equations.

The local energy conservation equation $$\nabla \mu T^{\mu \nu} = 0$$ , T = energy momentum tensor, contains part of that information but not all - it follows from Einstein but Einstein doesn't follow from it. Just like Einstein eq, the local energy conservation always applies - inhomogenous universe doesn't have Friedman equation yet it satisfies local energy conservation.

For the case of homogeneous isotropic universe filled with ideal fluid, Friedman is derived from Einstein. It can also be derived as Mukhanov did from the local energy conservation (which in this case boiled down to dE = - p dV) PLUS extra equation which he got heuristically by analogy with Newton, adding non zero pressure. The whole information of Einstein in this case is contained in any two equations from the set { Friedman, local energy conservation, the extra equation}. Choosing two of those, one can derive the third. That shows Friedman and local energy conservation are not the same equation, nor the contain the same information in GR.

Last edited: Mar 27, 2008
16. Mar 27, 2008

### Garth

Actually energy is NOT generally locally conserved in GR.

GR conserves energy-momentum, which is different, and it is this that is described by the expression:

$$\nabla_{\mu}T^{\mu \nu} = 0$$.

Note: $T^{\mu \nu}$ is the energy-momentum tensor, also often called the 'stress-energy' tensor.

The basic reason for this is that energy is a frame-dependent concept and the basis of GR is that it is frame-independent theory so that there is no preferred frame in which to measure an absolute value of energy.

In GR energy is only conserved in special cases, when there is a time-like Killing Vector, which is not the case if the gravitational field varies with time.

Thus you need to be an observer co-moving with a static gravitational system to observe energy conservation.

You can read more here: Is Energy Conserved in General Relativity?.

Garth

Last edited: Mar 27, 2008
17. Mar 27, 2008

### smallphi

People often call it loosely 'local energy conservation'. It means if you consider a constant local volume, any change of the energy inside that volume is accounted for by energy currents taking energy in or out of the volume, so energy doesn't appear suddenly without a flow from somewhere.

In the same way we say 'local charge conservation' when the pedantic way to say it is 'local charge-current conservation'.

18. Mar 27, 2008

### Garth

But the charge being considered here is that of four-momentum, and energy is only one component, the time component, of that charge.

Read MTW "Gravitation", 5.8 - pg 142 "Conservation of 4-Momentum, with figure 5.3, Box 5.3 - pg 147 "Volume Integrals, Surface Integrals and Gauss' Theorem in Component Notation", and Chapter 15 - pg 364 "Bianchi Identities and the Boundary of a Boundary".

You will find it is energy-momentum that is being conserved.

Garth

Last edited: Mar 27, 2008
19. Mar 27, 2008

### smallphi

Local conservation of 4-momentum implies local conservation of all it's components, the energy and the 3-momentum.

20. Mar 27, 2008

### Garth

This statement is incorrect.

Think of a vector of fixed length in 3-space, the x, y and z components can all vary depending on how the vector is rotating relative to the observer's frame of reference yet the total length remains constant.

In Minkowski four-space consider an accelerating particle. Its energy increases, its linear momentum increases and yet, because of the (_+++) signature of the metric, the total four-momentum remains the same, equal to its rest mass.

I hope this helps.

Garth