Answerer

But guys, sorry to be a skeptic, how can the total curvature of our universe's spacetime be flat when it, currently, expanding faster and faster?

Shadowy Man

There's nothing contradictory about 'flat' spacetime expanding.

Answerer

Can anyone elaborate more?

Friar Bellows

I think we all tend to get lazy in our use of the terms "space" and "spacetime". To be completely anal, I can't imagine spacetime (3 spatial dimensions + 1 temporal dimension) doing anything. It just is. It's the past, the present and future, as well as here, there and everywhere. On the other hand, I can imagine space (3 spatial dimensions) to be doing something, like expanding. When astronomers say that the universe is homogeneous (on the large scale) they're not talking about spacetime. After all, the universe is very different now to what it was before. For example, it was much denser and hotter at recombination all those billion of years ago. So the universe is not homogeneous in time, but only in the 3 spatial dimensions. Similarly, when astronomers say that the universe is flat, they are applying that concept to space, but not spacetime. Thus acceleration of the expansion is still compatible with the idea of a flat universe.

At least, that's my feeble understanding of the matter.

Answerer

Aren't things getting worse, my understanding is already slipping away at the speed of light.

Friar, I get your point that the flat universe is about three dimensional space. But I had found out that when physicists are talking about curvature, they are implying a four dimensional non-euclidean spacetime and since a 'flat' universe is being refered time and again by the physicists as an universe with zero curvature. Aren't they are implying that its 'spacetime', not the space, of the universe is homogeneous?

lpetrich

Actually, space-time can be curved with the space part being flat.

The way one describes curvature in General Relativity is with the help of a space-time metric, a variable-coefficient version of Pythagoras's Theorem.

In Cartesian coordinates x, y, and z, the familiar 3-space distance is

ds^2 = dx^2 + dy^2 + dz^2

where d means "differential of" (imagine the distance between two close points).

Now turn it into spherical coordinates:

x = r*sin(theta)*cos(phi)
y = r*sin(theta)*sin(phi)
z = r*cos(theta)

The distance formula above becomes

ds^2 = dr^2 + r^2*d(theta)^2 + r^2*sin(theta)^2*d(phi)^2

Variable-coefficient Pythagoras!

This can be generalized to

ds^2 = g(i,j)*dx(i)*dx(j)

summed over indices i and j, where g(i,j) is a function of the various x(i).

And for handling time, we simply treat it as an extra space variable, with a twist. In the flat-spacetime, Cartesian-coordinate case:

ds^2 = -c^2*dt^2 + dx1^2 + dx2^2 + dx3^2

Note the negative sign. The speed of light in a vacuum, c, is essentially a units factor, and is almost always set to 1 in relativistic theorizing.

Now how can one find the curvature, if the metric g(i,j) can vary from point to point? By calculating a rather complicated function called the "Riemann Tensor" (recommended only for those who know what "partial derivatives" are).

General relativity is a rather complicated subject, but a very simple summary is that the curvature of space-time is related in a simple way to the energy/momentum density/flux at each point in space-time.

But GR predicts that the metric produced by massive objects is, in an appropriate coordiante system and to first approximation:

ds^2 = - (1 + 2*V)*dt^2 + (1 - 2*V)*(dx1^2 + dx2^2 + dx3^2)

where V is the Newtonian gravitational potential.

And a flat, expanding Universe has a metric

ds^2 = - dt^2 + a(t)^2*(dx1^2 + dx2^2 + dx3^2)

where a(t) is the current "radius" of the Universe. The Hubble "constant" is simply (1/a(t))*(da(t)/dt) -- its rate of change over its value.

While 3-space is flat; notice the pure-Cartesian form of (dx1^2 + dx2^2 + dx3^2), space-time is not flat, as the presence of a(t) indicates. Only if a(t) = a0 + a1*t, where a0 and a1 are constants, can space-time be flat. That's the case of linear expansion.

But the Universe's expansion is nonlinear, meaning that space-time is curved.

Friar Bellows

Bah! Don't worry about that. A good portion of learning consists of unlearning.

When physicists talk about curvature in general terms, then yes, they are talking about the curvature of spacetime. But in practice, it often makes sense to split spacetime into its spatial and temporal parts (technically, foliate spacetime into spacelike slices), and then, for example, study the curvature of the spatial part alone. And that's what cosmologists are referring to when they talk about the homogeneity, isotropy and curvature of the universe.

But I urge you to re-read lpetrich's post: unlike me he really knows what he's talking about!

Shadowy Man

lpetrich:

Yeah, you are correct. I forgot that the a(t) part only modifies the dx part of the spacetime metric.

It's been a while since I took my gravitational physics course and I haven't used much of it since then.

lpetrich

What I've been discussing is the Friedmann-Robertson-Walker metric, which the Universe closely follows, to within localized clumpiness.

It has an interesting property: conformal flatness. Define a "conformal time", w, with

dt = a(t)*dw

w = Integral(dt/a(t))

Then the FRW metric turns into

ds^2 = a(t)^2*(-dw^2 + dx1^2 + dx2^2 + dx3^2)

Though the metric inside the ()'s is flat, the overall metric may be curved, since a(t) depends on w. In fact, the metric is only flat if a(t) = exp(k*w) where k is some constant.

Answerer

Thanks lpetrich and Friar, I think that most of the dark clouds in my head had been removed. Anyway, Friar, just curious, are you a physicist? You seems to know lots about physics, especially cosmology.