# Changing curvature of a manifold

1. May 18, 2014

### ChrisVer

Is it possible for a riemann manifold to change its curvature?
In practice could the universe in general change its curvature by time? (let's say in the past it was negative and today it's almost flat tending to positive);
If not which theorem disproves it?

2. May 18, 2014

### Staff: Mentor

Considered as a 4-dimensional manifold, spacetime does not "change" at all; it just is. Curvature can vary from event to event in spacetime (and in any physically realistic spacetime, it will), but that doesn't mean curvature is "changing"; it just means the 4-dimensional manifold is a manifold that does not have constant curvature, just as, for example, the surface of an ellipsoid is a 2-dimensional manifold that does not have constant curvature (unlike a 2-sphere, which does).

If we pick a particular slicing of a 4-dimensional spacetime into space and time, then yes, the curvature at a given point in space can certainly change with time; whether it does or not will, in general, depend on the spacetime and the slicing that we choose.

3. May 18, 2014

### ChrisVer

So could then the universe start with negative curvature and reach its value today?
I am not sure, because somewhere I read that in order to parametrize a surface you must impose that the vectors:
$x_{,1} \times x_{,2} \ne 0$
where with comma 1,2 I meant the derivative wrt the parameters $u^{1},u^{2}$ of the vector $x(u^{1},u^{2})$ characterizing the surface, i.e.:

$x_{,i}= \frac{\partial x}{\partial u^{i}}$

If I want by a continuous transformation to change the curvature's sign - that is equivelant to choosing $\bar{u}^{a}= \bar{u}^{a}(u^{b})$- wouldn't that mean that I'd had to reach a point where the above cross product would be equal to 0 (singularity)?

Last edited: May 18, 2014
4. May 18, 2014

### Bill_K

The cross product is the volume element, not the curvature. There's no reason it should be related to the curvature.

5. May 18, 2014

### ChrisVer

Isn't the cross product the normal vector on each point?
Since $x_{,i}$ is a vector tangent to the surface along the $u^{1}$ direction?

6. May 18, 2014

### Bill_K

There is something called The Equation of Gauss that relates the intrinsic curvature of a surface to its second fundamental form. The second fundamental form describes how the surface is embedded in a larger manifold, and in turn is related to the derivative of the unit normal vector.

For the second fundamental form, see here, the subsection "Physicist's notation". (Note that it's defined in terms of the second partial derivatives of the position vector, whereas the unit normal vector is determined by the first partials.) The second fundamental form is usually denoted bab, or B in index-free notation.

For The Equation of Gauss, see here, Eq.(3.1). It gives the Riemann tensor of the surface as a quadratic in the second fundamental form.

Last edited: May 18, 2014
7. May 18, 2014

### Staff: Mentor

First of all, it's important to distinguish the curvature of spacetime from the curvature of a particular spacelike slice, such as the universe at a given instant of time (where "time" here means time relative to a particular choice of slicing of spacetime into space and time). As best we can tell, the curvature of our current spacelike slice of the universe, i.e., the slice of constant time that corresponds to our "now", is zero--i.e., the universe right now is spatially flat. But the *spacetime* curvature of the universe in the region of spacetime we currently occupy is positive--more precisely, the dominant component, the one that is currently driving the expansion of the universe to accelerate, is positive. (There are also further complications due to the fact that the curvature of a 4-dimensional manifold--or, for that matter, a 3-dimensional manifold like a spacelike slice of the universe at constant time--can't be described by a single number, so strictly speaking you can't simply describe it as positive or negative. I won't open that can of worms unless you insist. )

As far as your question, in principle, yes, the curvature of the universe (in either sense, spacetime curvature or curvature of a spacelike slice) could have been negative in the far past and still have the values we observe now in the region of spacetime we currently occupy--in other words, such a model of the universe's evolution would be consistent with the laws of physics. However, as I understand it, our best current observations and theories indicate that in fact it wasn't.

8. May 19, 2014

### ChrisVer

In the Friedman equation, you have:
$\frac{\dot{a} ^{2}+ k^{2}}{a^{2}} = A \rho_{mat} + B \rho_{Λ}$
with A,B being some constants....
We can always evaluate though the value for $k^{2}$ at the present time ($t=t_{0}$ and insert it into the above equation... If that's true, then it shouldn't depend on time- shouldn't evolve that much...
On the other hand, I am still unable to understand the basic of it... If the curvature of the universe can change any time (changing the fate of the universe) then there's no predictivity in the theory- you can say that today the universe is almost flat ($Ω_{mat}=0.7, Ω_{Λ}=0.3$ approximately, so sum to 1=flat universe) , but tomorrow it may be closed instead of open... (check attachment)

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9. May 19, 2014

### Bill_K

In the Friedmann universe, the value of k is what determines whether the curvature of the space sections is positive, negative or zero. K is a constant. If it's positive today, it's positive tomorrow.

10. May 19, 2014

### ChrisVer

yes but that would mean the the curvature doesn't/cannot change by time (what I initially asked/proposed)... If curvature is to be changed (in sign) then the K would also need to change...

In other words, a universe started with positive curvature, can't change to negative and vice versa...

Initially I had this talk with a classmate of mine, who said that $Ω$ would depend on time...because $Ω=Ω_{mat}+Ω_{Λ}=Ω_{mat,0} (a)^{-3}+Ω_{Λ,0}$ (maybe it's +3 I don't remember exactly right now, but $a$ depends on time, and I wanted to show it's time dependence)...
Then we were talking whether Ω could change so that the curvature of universe would change (in the past or future)...I thought there cannot be a way to change the curvature of a given manifold, initially by the reasons I posted above..although they are lacking justification afterall (since I wasnt exactly correct hehehe)

However, I know this is getting into cosmology (wrong forum sector), but that's why I referred to a manifold instead.

Last edited: May 19, 2014
11. May 19, 2014

### WannabeNewton

Your original question was about the general Cauchy value problem in GR and not of the evolution of cosmological parameters in the FLRW universe or perturbations thereof. As such your original question has already been answered by Bill and Peter. If you give me a globally hyperbolic space-time and an initial Cauchy surface (an initial data surface) then I can tell you how this evolves in time from an appropriate 3+1 split of the Einstein equation through e.g. harmonic coordinates. The flow of time is given a priori by the hyperbolicity of the space-time. This then tells me how the curvature evolves in time as well.

I would suggest doing some readings on the Hamiltonian formulation of GR. A very instructive set of notes on the 3+1 formalism of GR are those by Eric Gourgoulhon: http://arxiv.org/pdf/gr-qc/0703035v1.pdf

If you have a question specific to time evolution of cosmological parameters then you should probably make a separate thread for that.

12. May 19, 2014

### Staff: Mentor

Is this really an absolute restriction? Yes, in a given FRW *model*, $k$ is a constant. But we don't use exactly the same FRW model for all phases of evolution of the universe; if nothing else, the effective equation of state changes depending on whether the evolution in the current phase is dominated by radiation, matter, or dark energy. Would it be possible to use models with different $k$ for different phases as well? Or is there no way of "patching together" models with different $k$ into a single overall spacetime?

13. May 19, 2014

### Staff: Mentor

The curvature, without qualification, is a very different thing from the *sign* of the curvature. The spacetime curvature of a given FRW model of the universe can be different at different events even if the value of $k$ in the metric is the same everywhere. (If $k$ is not zero, the spatial curvature of different spacelike slices can also change from slice to slice even though $k$ is constant.)

14. May 19, 2014

### Bill_K

The constancy of k doesn't depend on the equation of state. The only assumptions that go into it are that the space sections must be homogeneous and isotropic. (And Einstein's Equations!)

To patch two regions of spacetime together, a condition is that the intrinsic curvature of the "seam" must be the same when viewed from either side. So in the Friedmann case, the value of k would have to be the same in both patches.

15. May 19, 2014

### Staff: Mentor

Understood. I didn't mean to suggest that you need an equation of state to show that $k$ is constant.

Ah, got it.

16. May 19, 2014

### WannabeNewton

In standard cosmology the background FLRW metric is kept fixed (including $k$) and one does perturbation theory on this background. This leads to various perturbations of the evolution equations as well as the curvature and one studies the evolution resulting from these perturbations.