# Is the Milne model compatible with Einstein's field equation?

1. Jun 29, 2012

### andrewkirk

I have just been reading about the Milne model of the universe, which is a version of the FLRW model with zero mass-energy everywhere and negative curvature.

This perplexes me a little as my understanding is that if there is no mass-energy the Einstein tensor will be uniformly zero, which I thought (but am not sure) implied that the Riemann curvature tensor would be uniformly zero. Hence if the Milne universe has negative curvature it would not be compatible with the field equation.

Am I misunderstanding something here?

2. Jun 30, 2012

### Orion1

Another problem about the Milne model is that it models the Universe as expanding at the speed of light without inflation, however the Universe is observed as inflating and expanding faster than the speed of light. And it is not matter that is expanding, but rather space-time.

Reference:
Milne model - Wikipedia

3. Jun 30, 2012

### nicksauce

No mass-energy implies the Einstein tensor is uniformly zero, but does not imply that the Riemann curvature tensor is uniformly zero. The obvious and hopefully familiar example is that in the Schwarzchild solution, the Einstein tensor is zero but the Riemann curvature tensor certainly is not.

4. Jun 30, 2012

### Chalnoth

The Milne cosmology is just another way of writing down flat Minkowski space-time. As such, it is only valid for an empty universe, and the expansion in such a cosmology is entirely illusory in the sense that it can be made to entirely disappear simply by changing the way you write your equations (in a universe with matter, you always get some sort of expansion or contraction, and while that expansion or contraction may manifest itself differently if you write your equations differently, it will always be there).

5. Jun 30, 2012

### andrewkirk

Thanks for that. This point about the Swarzschild exterior solution seems obvious once you say it: the stress-energy tensor is zero outside the star so the Einstein tensor must also be zero, yet the spacetime must be curved (nonzero Riemann tensor) in order for massive bodies to orbit the star.
Yet I had never noticed this before. My text - Schutz - omits to point this out in its derivation and in fact doesn't mention the Einstein tensor at all in its derivation of the exterior solution, and I was not alert enough to notice this for myself.

6. Jun 30, 2012

### andrewkirk

That is what I thought when I first came across the concept. Yet it seems to conflict with Wikipedia's claim that the Milne universe has negative curvature, as Minkowski space has zero curvature. How can this conflict be reconciled?
When you say 'write equations differently' are you just referring to choosing different coordinate systems, or is there more to it than that?

The 'Einstein universe' is static but contains matter. Is that a counterexample to the idea that some expansion or contraction will always be there, or does the Einstein universe not count because it assumes the matter is 'dust' which cannot really exist.

7. Jun 30, 2012

### jcsd

Spatial curvature doesn't mean there is curvature in spacetime.

It is a counterexample, but at the same time it can never be a realistic model as it is not stable: a small peturbation will cause a small amount of contraction/expansion and will grow over time.

8. Jun 30, 2012

### andrewkirk

That's an intriguing thought.
Of course in an open universe one can just put a Euclidean metric on the 3D manifold that is a hypersurface of constant time, then we have a 'flat' 3D hypersurface of a curved 4D manifold. But it's not very interesting because there's no relationship between the 3D and 4D metrics.
If we want to be realistic one assumes that the metric we apply to any 3D constant-time hypersurface is the metric inherited from the 4D manifold, based on the projections of the four-vectors on the tangent space of the hypersurface:

g(3)(u(3),v(3)) := g(4)(u(4),v(4))
where a (3) or (4) subscript refers to vectors and metrics in the tangent space to the 3D or 4D hypersurface/manifold respectively, and u(4) is u(3) with a zero time component.

Are you able to elaborate on the idea of a flat spacetime for which constant time hypersurfaces are curved, perhaps with an example?
Thanks

9. Jun 30, 2012

### Chalnoth

This has largely been answered, but I'll try to take another stab at it anyway.

As has already been mentioned, the curvature in the Milne cosmology is only spatial curvature. When I say the Milne cosmology is flat, I mean that its space-time is flat. What does this mean?

Well, the full curvature of space-time is given by the Ricci tensor:
$$R_{\mu\nu\sigma\tau}$$

I don't know how much you're comfortable with this sort of notation, but the main point to bear in mind here is that the indices $\mu\nu\sigma\tau$ cycle over the coordinates, so that this object depends critically upon which coordinate system you choose. In order to get a measure of what curvature means without any coordinates whatsoever, you have to factor these out. This is done rather simply by taking a sum:

$$R = \sum_{\mu\nu} R^{\mu\nu}_{\mu\nu}$$

This sum, that is taken from contracting over the coordinates, no longer has any indices, and is therefore independent of any coordinate system (the way in which tensors change when you change coordinates guarantees that this is the correct way to get a quantity which will be the same in any coordinate system*).

It is this value, $R$ which is independent of coordinates, that is identically zero in a Minkowski or Milne universe.

If you want, however, you can create a different value from not summing over all coordinates, but only summing over the spatial coordinates. This is the spatial curvature. Obviously this will depend precisely upon how your coordinate system divides between space and time. And this is why the spatial curvature is different between Minkowski and Milne space-times.

*The caveat here is that if the coordinate system has singularities in it, then those don't magically disappear: $R$ will have that same singularity. This is because once you've divided by zero, you can't get any sensible value back again.

10. Jun 30, 2012

### BillSaltLake

There may be a simpler way to say this. I think that some boundary conditions must exist such that GR gives a flat space (for all t) and a flat spacetime. This would be an empty space in which da/dt = 0. Normally, Milne has da/dt > 0, I believe.

11. Jun 30, 2012

### andrewkirk

Thanks Chalnoth: the fog is starting to lift a bit for me.

So if I am understanding correctly, the Milne universe is just Minkowski spacetime with a different coordinate system. Changing the coordinate system can't change the curvature of the spacetime, which is coordinate independent, being expressed by a tensor field (the Riemann tensor field) but it can change the curvature of the 3D space that is inherited from the 4D spacetime via the 3D metric that is, at any point, just the projection of the 4D metric at that point on the 3D tangent space at that point to the 3D manifold that is a constant-time hypersurface of the spacetime (the "induced metric"). (I am imagining the 3D space here as being embedded in the 4D spacetime).

Because the 3D hypersurface is not defined by valid tensor operations such as contractions but is rather coordinate dependent, its curvature depends on the choice of 4D coordinate system. Indeed, different coordinate systems will generate different constant-time hypersurfaces. The usual, Euclidean coordinate systems for Minkowski spacetime generate constant-time hypersurfaces that are flat under the induced metric. But the 'Milne' coordinate system generates constant-time hypersurfaces that are negatively curved under the induced metric.

Is that right?

12. Jul 1, 2012

Yup!