Does general relativity predict expansion?

1. Jan 18, 2006

Oxymoron

How does general relativity predict the expansion of the universe?

I know this is probably a very broad question, and perhaps it would be more useful to direct me somewhere else. Anything would be appreciated.

I need to know this because I want to find out if there is any link between gravitational redshift and expansion redshift.

2. Jan 18, 2006

George Jones

Staff Emeritus
Einstein's (tensor) equation is, in general, a set of 10 coupled, non-linear partial differential equations. Given how matter and energy distrbutes itself, i.e., given the stress-energy-momentum tensor $T$, the solution of Einstein's equation gives the spacetime metric.

We observe the universe and see large-scale symmtries, i.e., homogeneity,and isotropy (appropriately defined). When these symmetries are taken into account, Einstein's equation becomes a single ordinary differential equation called the Friedman equation.

The Friedman equation

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

governs the time-evolution of the spatial scale $a \left( t \right)$ of the universe.

The spacetime metric is

$$ds^{2} = - dt^{2} + a \left( t \right)^{2} \left[ \frac{dr^{2}}{1 - kr^{2}} + r^{2} \left( d \theta^{2} + \mathrm{sin}^{2} \theta d \phi^{2} \right) \right].$$

In the above, k = +1, 0, -1 if the universe is closed (compact spatial hypersurfaces), flat, or open.

Expansion redshift occurs because the scale-factor $a \left( t \right)$ is larger today than it was in the past.

Edit: In some (grossly oversimplified) sense: gravitational redshift in the Schwarzschild solution occurs because the time scale changes as we move from here to there in space (coefficient of dt depends on r); expansion redshift occurs because the distance scale changes as move from then to now in time.

Regards,
George

Regards,
George

Last edited: Jan 18, 2006
3. Jan 18, 2006

Ich

Actually, GR did not predict expansion. Einstein added the famous "cosmological constant" to get a static universe. As a result, you can fit the cosmological models to almost ervery observed behavior, so in this respect GR doesn´t predict anything.

4. Jan 18, 2006

Garth

Or, to specific about the zero cosmological constant case, in which the model does predict either expansion or contraction; if one takes the expanding regime, then that 'plain' GR model predicted a decelerating expansion.
The invocation of inflation, and now Dark Energy, to deliver two separate episodes of accelerated expansion to fit cosmological constraints is necessarily an anti-prediction of that standard GR model.
Garth

Last edited: Jan 18, 2006
5. Jan 18, 2006

JesseM

Even if you adjust the cosmological constant to prevent expansion/contraction, isn't this an unstable solution, so slight deviations from the precise static solution would eventually lead to rapid expansion or contraction?

6. Jan 18, 2006

RandallB

It what respect is that exactly?
As I understand Einstein’s problem with GR – it was giving him indications the universe was not static.
That’s why he was asking astronomers if things were looked like they were moving in relation to each other at great distances or not.
The answer he got based on observations at the time was NOT.
Therefore his GR to him was giving a false prediction of some kind that he attempted to correct or adjust with a constant.
Not that it’s real clear on exactly his "cosmological constant" was supposed to do.
I.E. did his insertion prevent the universe from collapsing back in on itself or from expanding forever to bring it into agreement with the static universe he wanted?

The Hubble constant and open vs. closed adjustments to GR are not the same as the "cosmological constant" he felt was needed to “correct” the implications of GR, so it’s hard to make direct comparisons or say what GR does without them applied.

So with respect to just raw GR without any of the adjustments; it seemed to imply something other than a static universe. Maybe someone else can make that part clear as to what it implies with zero adjustments. Or what is the "standard GR model" prediction?

Last edited: Jan 18, 2006
7. Jan 18, 2006

pervect

Staff Emeritus
Where did this form of the spacetime metric come from?

I usually see something that looks like:

$$ds^2 = -dt^2 + a \left( t \right) ^2 \left( d \chi^2 + f \left( \chi \right) ^2 \left( d \theta^2 + \mathrm{sin}^2 \theta d \phi^2 \right) \right)$$

with $f(\chi) = (sin(\chi), \chi, sinh(\chi))$ for k=(-1,0,1)

(Wald pg 95, also MTW, also http://en.wikipedia.org/wiki/FRW)

Your form looks simpler, assuming it's equivalent (I haven't checked).

Last edited: Jan 18, 2006
8. Jan 18, 2006

George Jones

Staff Emeritus
The 2 versions of the metric are equivalent, i.e., they're related by the coordinate transformations $r = \mathrm{sin} \chi$, $r = \chi$, $r = \mathrm{sinh} \chi$ for k = 1, 0, -1.

For example, if $r = \mathrm{sin} \chi$, then $dr = \mathrm{cos} \chi d \chi$ so that $dr^2 = \left( 1 - \mathrm{sin}^2 \chi \left) d \chi^2$.

Regards,
George

9. Jan 18, 2006

pervect

Staff Emeritus
To expand a bit on the Friedmann equation, the way the logic progresses is this:

First you find the metrics that represents a homogeneous isotropic universe.

From these metrics, g_ab, you calculate the Einstein tensor, G_ab.

G_ab gives you the stress-energy tensor T_ab by Einstien's equation

G_ab = 8 pi T_ab

(or, with a cosmological constant)

G_ab + $\Lambda$ g_ab = 8 pi T_ab

This then gives you two equations, for the density and pressure terms of T_ab. Only diagonal terms exist because of the homogeneous and isotropic assumptions. The pressure terms are equal by isotropy. So one has:

T_ab = diag(density, pressure, pressure, pressure)

This gives one equation for density, and one equation for pressure.

Finally you need to assume an "equation of state" or continuity conditions that relates density to pressure. A couple of simple but extreme assumptions are the "dust" model where pressure is zero, or the "radiation" model where the pressure is 1/3 the density. The radiation model is equivalent to assuming the universe is a "photon gas", an ideal fluid made out of radiation. (add: This would be a maximum pressure assumption.)

A very simple point that can be made is that without a cosmological constant, matter attracts itself. leading to an expansion that deaccelerates with time. If you imagine an expanding dust cloud, for instance, the velocity of the dust particles would be constant only if there were no force between dust grains. With a force between dust grains, you naievely expect the velocity to decrease with time. This naieve approach works well enough to explain why the expansion must deaccelerate with only normal matter.

In order for the expansion of the universe to accelerate (which is currently believed to be the case), you need some sort of exotic matter, or a cosmological constant. This is "dark energy".

Last edited: Jan 18, 2006
10. Jan 18, 2006

pervect

Staff Emeritus
Cool - your form is much simpler to write out. Thanks!

11. Jan 19, 2006

Ich

If Einstein had believed his original equations, he would have predicted a dynamic universe, ie either expanding or collapsing, according to George Jones´ formula. That´s why he called his "correction" with the cosmological constant his biggest blunder.
But now it seems that he rightly was cautious, because the cc seems to exist. So i think it´s more appropriate to say that GR has no specific prediction about the expansion of the universe than to say that Einstein failed to believe the correct prediction.

12. Jan 19, 2006

George Jones

Staff Emeritus
Not my form - I lifted it from Hartle! I really like the cosmology section of this book, which I find to be pedagogically very good.

As for being simple - I think the form of the metric that you gave gives more insight into the geometry of spatial sections.

Regards,
George

13. Jan 20, 2006

IACOB DUMITRU

Sorry, but for moment you wander about these inefficient study. The theoretical dada do not include a coherent scenario (as a theoretical solution) of the universe (as an emptiness or Euclidian space) before to any other elementary intruder (as energy, matter or particle). If you want to know more about the minimal spatial bit of information (theoretical possible), I can give you (free) my study “The space’s harmony”. anin1958@yahoo.com

14. Jan 20, 2006

Oxymoron

Q1) First a question on the most basic special relativity spacetime metric

$$s^2 = -c^2t^2 + x^2 + y^2 + z^2$$.

If we take this to be the metric in flat 4D Euclidean space, then this really isn't a Euclidean space at all. One of the metric axioms is not obeyed: The positive-definite one.

Q2) Is it true that a metric for flat space will have constant entries? As I understand it, if one considers flat space then the basis vectors are not dependent on the choice of coordinates. Is it true, then, in curved spacetime a metric would be coordinate dependent since entries in the matrix metric would change from one place to another?

Q3) In Minkowski space, are coordinates generally denoted by $x^{\mu}$ where mu ranges from zero to 3. My question is, I have seen

$$x^{\mu} = (ct,x,y,z)$$

and

$$x^{\mu} = (x,y,z,ct)$$

Is it simply a matter of opinion where the "ct" bit is written? Is this what is meant by the "signature" of the metric?

Q4) The infinitessimal length, $dx$. Is it a "4-vector"? I mean, in calculus we are taught that this is almost irrelevant, but now it is an actual vector (even a tensor?). Is this ok? Can you think of it as a infinitessimal vector?

Q5) $\bold{g}_{\mu\nu}$ is generally considered as the "invariant" metric tensor in Minkowski space, no? Are elements of the metric are actually "interior products" of the basis vectors? If the basis is orthonormal then the metric tensor is simpy a diagonal metric matrix?

Q6) When is $g_{\mu\nu} \equiv \eta_{\mu\nu}$?

15. Jan 20, 2006

George Jones

Staff Emeritus
The term metric has a different meaning in relativity than it does in a real analysis course.

I woudn't call this flat 4D Euclidean space, I would call it Minkowski spacetime. Any inertial observer can decompose spacetime into space and time, and the 3-dimensional space for any observer does have a metric in the positive-definite sense.

Emphatically no. None of this is true, for example, in spherical coordinates for Minkowski spacetime.

Yes. (Non-zero spacetime curvature) implies (there exits no global coordinate system in which the components of the metric tensor are all constant). Contrapositive: (there exits a global coordinate system in which the components of the metric tensor are all constant) implies (spacetime is flat).

You can see that these statements are true by looking at coordinate expressions for the components of the curvature tensor in terms of derivatives of the metric tensor. The converses of these statements are, however, false.

Even the true statements are a little subtle. Think of the 2-dimensional surface formed by a (loooong...) piece of paper roled up into a tube, i.e., S^1 x R. A curved surface, right? Well, no, not in the sense of curvature in relativity. As surface embedded in 3-dimesional space, this cylinder has non-zero (EDIT) extrinsic curvature. That accounts for all of the curvature we see, i.e., the surface of a cylinder has zero *intrinsic* curvature. Think of the paper as being graph paper. Before it's rolled, the piece of paper is flat, and the lines on it are geodesics. After it's rolled up, the graph paper is still flat, and the lines are still geodesics.

Similarly, we don't don't think of spacetime as being embedded in a higher dimensional space, and the curvature that we calculate is all intrinsic. Just as it's possible to have (intinsically) flat 2-dimesional spaces that aren't R^2, it's possible to have a flat spacetime that isn't Minkowski spacetime.

Suppose you're given a spacetime metric that has non-constant components. In general, there is no way to decide if the metric corresponds to a flat or a curved spaces. Who knows - you may have been given the metric for Minkowski spacetime expressed in an esoteric an unfamiliar set of curvilinear coordinates. To find out, by hook or by crook or by computer, calculate the connection coefficients. Again, look at the coordinate expressions for the components of the curvature tensor in terms of the connection coefficients.

Yes, it is purely a matter of convention where the "ct" bit is written. It's pretty standard these days to use the former convention, and I think this what everyone on this forum uses.

No, this not what is meant by the signature of the metric. This refers to whether x^0 y^0 - x^1 y^1 - x^2 y^2 - x^3 y^3 or -x^0 y^0 + x^1 y^1 + x^2 y^2 + x^3 y^3 is taken to the Minkowsk inner product. The generalizes to non-flat spacetimes. Most people on this forum (and most relativists?) use the latter; I (and the high-energy crowd) much prefer the latter.

Inuitively one thinks of dx as being an infinitesimal vector, but it is actually a one-form, i.e., a covector field.

I don't know what you mean, here. The g_{uv} are the components of the metric tensor with respect to a (fields of) basis vectors. As I have said before, these components are *defined* by g_{uv} = g(e_u , e_v). I would never bold the first g (scalars!) but I sometimes bold the second g. This is true in all spacetimes and for all (0,2) tensors, not just in Minkowski spacetime, and not just for the metric tensor.

I would say components, not elements, and it depends on what you mean by '"interior products" of the basis vectors.'

Look at the definition I gave for the components of the metric tensor. Do this definition answer this question?

Beats me; this is not a notation that I care to use. :tongue2:

The metric tensor tensor for Minkowki can be defined without using either basis vectors or coordinates. I did this in a post in one of your threads, but I don't remember which one.

eta_{uv} refers to the metric for Minkowski spacetime in a specific coordinate system. Generally, when changing kernel letter is used for a given tensor, e.g., T_{uv} and T_{u'v'}. What happens when you switch from cartesian to spherical coordinates? Do you switch from eta to g? I prefer to use g all the time and let context sort things out.

Regards,
George

Last edited: Jan 20, 2006
16. Jan 20, 2006

RandallB

I'm not clear on what your claiming here.
Are you saying that Einstein’s ”cc” (cosmological constant) eliminated the possibility of a “expanding or collapsing” universe.
But leaving it off allowed for Hubble expanding, plus more factors that may override the Hubble to switch back to collapsing or even flat.

Or are you saying that his “cc” is somehow a summary of all the modern cosmological factors currently in use.

17. Jan 20, 2006

Perturbation

Indeed, spaces with a negative metric component are referred to as "pseudo-Euclidean". Of course a time hypersurface of Minkowski space is Euclidean.

The entries in the matrix representing the metric tensor depend on the coordinate system. One has to transform between coordinates to get the different metric components in different coordinate systems, but the actual value of the metric tensor, its trace or whatever, is coordainte invariant, as coordinate transformations are orthogonal. Spherical polar coordinates are Euclidean, but the entries in the metric differ.

Yeah, $\mu =0,1,2,3$. It doesn't matter where you place the ct, as long as you keep your four-vector notation consistent. The signature is the sign of each metric component. The metric of space-times (curvature or not) has signature (+---). Ocasionally (-+++) is used.

It depends, $dx^{\mu}$ is an ifinitesimal four-vector displacement. $dx_{\mu}$ can be used as a coordinate basis for covariant tensors, also known as forms (the dual of vectors).

Yeah, the components of the metric tensor are generated from the inner product of the basis vectors of the space-time. The components of the tensor themselves are not the same in every coordinate frame, its trace etc. is invariant. In Minkowski space-time the components are the same in any Cartesian reference frame. If the coordinate frame is spanned by orthogonal basis vectors, the only non-zero components of the tensor are diagonal.

When we're dealing with Minkowski space-time in some way, either in SR as the actual metric or treating it as a background space-time. Some authors simply use $g_{\mu\nu}$ for the metric of Minkowski space-time. Quite often this is adopted in treatments of phenomena only in Minkowski space-time, so there's no need to distinguish between a general metric and the Minkowski metric.

Last edited: Jan 20, 2006
18. Jan 20, 2006

pervect

Staff Emeritus
As you may recall, the tangent space has been identified with derivative operators, thus $$\frac{\partial}{\partial x^\mu}$$ is a member of the tangent space. dx is, has been previously remarked, a one-form, because it provides a map from a derivative operator to a scalar (which is the defintion of a one-form), i.e.

$$\frac{\partial}{\partial x^\mu} {dx^\nu} = \delta^{\mu\nu}$$

Not in a general basis. Rather the interior (i.e. dot) product of $x^{\mu}$ and $x^{\nu}$ is

$$g_{\mu\nu} x^u x^v$$

These are different statements.

Yes - with an orthonormal basis at a point, you should have a Minkowskian metric at that same point.

You'll have to read the fine print of whoever uses it to be positive, but usually $\eta_{\mu\nu}$ is taken to be the "flat", Mikowskian metric used in SR. Typically one sees $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$, then equations are written in terms of h.

Last edited: Jan 20, 2006
19. Jan 23, 2006

Ich

No, the model with "cc" allow dynamic and static universes. If he had omitted it, he would have been forced to predict a dynamic universe, ie expanding or collapsing, before Hubble observed it.
It is one of these factors, the one associated with dark energy. Up to now, they have fitted GR to the data more ore less successfully (remember that more than 95% of the calculated energy density are unaccounted for), but afaik never acually predicted something significant.