# Must the curvature of space be constant?

1. May 12, 2014

### Staff: Mentor

I was listening to one of Leonard Susskind's cosmology lectures.

He talks about the factor K having values of +1/0/-1 corresponding to positive/flat/negative curvature. We don't know what the real value of K is.

But then as he discussed K at the big bang and K now, it seemed that he was saying that whatever the answer, K was the same then and now.

My question: Is there any reason why the curvature of space should be constant in time?

2. May 12, 2014

### craigi

Positive curvature gives a closed universe. That is, it wraps around, in a similar way that coordinates of the earth wrap. Negative curvature gives an open universe, which doesn't wrap. It's hard to see how the transition from one to the other could be viable, for a whole universe.

3. May 13, 2014

### phsopher

FRW metric

$$\mathrm{d}s^2 = - \mathrm{d}t^2 + a^2(t)\left[\frac{\mathrm{d}r^2}{1-k r^2} + r^2 \mathrm{d}\theta^2 + \sin^2 \theta \, \mathrm{d}\phi^2\right]$$

is a solution to the Einstein equation for a homogeneous and isotropic universe. If you assume a matter distribution that is the same everywhere and in all directions you get a solution where k is constant. If you try to promote it to an arbitrary function of time and plug that into the Einstein equation you get an Einstein tensor component:

$$G^0_r = -\frac{r k'(t)}{1- k(t)r^2}$$

which, on the matter side, would correspond to non-vanishing momentum. And of course the equations become much more difficult. I'm not going as far as to say that no such solution exists, as I don't know, but if it did it would no longer describe a homogeneous universe.

EDIT: erroneous information removed.

Last edited: May 13, 2014
4. May 13, 2014

### Staff: Mentor

Thanks for an excellent answer PHSOPHER.

That is a very satisfying answer, that it would be inconsistent with an isotropic homogeneous universe and that we know that because the math tells us so.

5. May 13, 2014

### George Jones

Staff Emeritus
In some sense, $k$ characterizes the type of spatial curvature in a universe that is exactly FLRW, not the "amount" of spatial curvature. The "amount" of spatial curvature is time-dependent.

On any 3-dimensional hypersurface of constant $t$, the spacetime metric induces a spatial metric on the hypersurface, and the spatial metric is used to calculate the spatial curvature tensor for the hypersurce. Contracting the spatial curvature tensor to the constant and symmetric (in space) Ricci scalar curvature for the hypersurface gives $^\left(3\right)R = k/a^2$.

This makes sense: as cosmological time approaches zero, the magnitude of spatial curvature blows up for closed and non-flat open universes; as cosmological time approaches inifinity, spatial curvature goes to zero for non-flat open universes.

Lemaitre-Tolman-Bondi universes (FLRW are special cases of these) allow spatial curvature (including $k$) that varies with spatial position, somewhat along the lines that phsopher suggested (and that possibly varies with time, but I am not familiar enough with these universes to say).

Interesting treatments of Lemaitre-Tolman-Bondi universes are given in "An Introduction to General Relativity and Cosmology" by Plebanski and Krasinski, and "Relativistic Cosmology" by Ellis, Maartens, and MacCallum.

6. May 13, 2014

### Chalnoth

The way I like to think of it is in terms of the first Friedmann equation:

$$H^2 = {8 \pi G \over 3}\rho - {k c^2 \over a^2}$$

The parameter on the left, $H$, is the rate of expansion. Basically, it's how fast things are moving away one another (or towards, if negative).

The matter density, $\rho$, is how much stuff there is in the universe (note: this also includes exotic stuff like dark energy, so don't take the definition of "matter" too seriously or narrowly for the rest of what I'm saying here).

This is, in a way, the analog of what happens with a ball you throw from the Earth. If it's moving too slowly, it comes back and lands on the Earth again. If it's moving too quickly, it escapes the Earth's gravity and goes on forever.

Curvature is like this: it's a way of writing down the relationship between the rate of expansion and how much stuff is in the universe. It also happens to impact the shape of the universe, due to the way in which gravity behaves.

7. May 28, 2014

### Trenton

I too have contemplated this issue at length but too little avail. Firstly as I understand it, it is spacetime that is curved. Space is simply contracted under the stress-energy tensor. Secondly, there appears to be no way of measuring how contracted the space you are in is. Indeed this is the whole point of the word 'relativity' in both SR and GR.

The contraction does become more severe the nearer you are to the source of gravity of course. And as I understand it, this contraction sets up a radially inward 'flow' of space towards dense objects - and when this 'flow' hits the speed of light, we have a black hole event horizon. While nothing may move faster than c through space there is no limit to this flow in GR. So inside a BH, the flow gets faster and faster.

If we consider the case of two astronauts falling into a black hole, one a short time after the other, they can still see each other even when both are inside the EH. The first will be overtaken by photons from the second as they move inward at flow + c and the second will run into the photons emitted by the first as these move inward at flow - c. Only when the flow in the locale of the first exceeds the flow in the locale of the second by c does this cease to be the case (a horizon in its own right). In the meantime they each appear to the other as receeding and becoming redder, at an ever increasing rate.

If instead of two astronauts it were thousands, they would see something similar to what we see with the galaxies - ie the further away the faster the recession in an apparently linear relationship (ie space is flat)

In view of this I wonder if we can even determine that the universe is contracting or expanding - let alone at what rate. But assuming it is expanding, and that it is doing so from a disrupted singularity (big bang), is it not more reasonable to assume that a collapsing gravity field would cause an appently flat and accelerating expansion?

I note that if you devide c by the hubble constant you get 13.8 billion years. I find myself thinking like a policemen. I see something that looks like a coincidence and I smell a rat.