Milne vs. Minkowski metric

[PAllen]

[Moderator's note: Spin-off from another thread due to question change. Edited to remove content specific to the other thread.]

You can argue that Milne and Minkowski are different instances of manifold with metric, even though they are both flat, because there is no global isometry between them (and Milne is geodesically incomplete, while Minkowski is not).

(Note, there is, in fact, a homeomorphism between Milne and Minkowski, but there is no possible global isometry)

 

[PeterDonis]

there is no possible global isometry

Just to be clear: by this you mean that the Milne spacetime is only one "wedge" of Minkowski spacetime (the interior of the future light cone of a chosen "origin" event), correct? So while there is an isometry between the two within that "wedge", it's not global because it doesn't cover all of Minkowski spacetime, correct?

 

[PAllen]

Just to be clear: by this you mean that the Milne spacetime is only one "wedge" of Minkowski spacetime (the interior of the future light cone of a chosen "origin" event), correct? So while there is an isometry between the two within that "wedge", it's not global because it doesn't cover all of Minkowski spacetime, correct?

Correct.

It’s occurs to me that there is another way to look at this that strongly justifies treating the Milne spacetime as a completely different manifold with metric than Minkowski spacetime. Note that each is homeomorphic to $R^4$. Let me know if you need an argument to see this. So, starting with $R^4$ as a manifold without metric, we can place a flat metric on it in two completely different ways - one producing a geodesically complete manifold, the the other, a geodesically incomplete manifold.

 

[PeterDonis]

each is homeomorphic to $R^4$.

Yes, agreed.

starting with $R^4$ as a manifold without metric, we can place a flat metric on it in two completely different ways - one producing a geodesically complete manifold, the the other, a geodesically incomplete manifold.

I think this depends on how you define what a "metric" is. Is it the geometry itself? Or is it the specific form of the line element?

If you take the first view, then the Milne spacetime is not a different manifold with metric from Minkowski spacetime; it's just a "wedge" of Minkowski spacetime with a coordinate chart that obfuscates that fact. Note that you could define a "Milne chart" on the other wedges of Minkowski spacetime as well, if you don't mind switching which coordinate is timelike (in fact, if you do it on the "right wedge", you get the Rindler chart.)

If you take the second view, then yes, the Milne spacetime would be a different manifold with metric from Minkowski spacetime.

 

[PAllen]

If you take the first view, then the Milne spacetime is not a different manifold with metric from Minkowski spacetime; it's just a "wedge" of Minkowski spacetime with a coordinate chart that obfuscates that fact.

I think you are missing my point. We start with just topology, no metric. The topology we start will is all of $R^4$. In one case, we place the flat metric on it such that all geodesics are affine complete. In the other case, on the same starting topology, we place the flat metric such that no geodesics are affine complete. This has nothing to do with coordinates, per se. It is a difference in how we metrize the topology. Then, in the second case, you can cover this geodesically incomplete manifold with Milne coordinates.

 

[PAllen]

Also, in the other thread you thread, you said "No, that's not Minkowski coordinates, that's Milne coordinates. You can't put the Minkowski metric in Minkowski coordinates in the FLRW form, so you can't even define an a(t) for the Minkowski metric in Minkowski coordinates.", referring to a(t) = constant.

However, the standard FLRW metric with a(t)=constant, and the spatial slice Euclidean is exactly Minkowski coordinates covering the whole Minkowski spacetime. The Milne case of FLRW is a(t)=t, with hyperbolic slices. This case then turns out to have zero curvature, and can be isometrically mapped to the interior of the the future light cone of any point in Minkowski space.

 

[PeterDonis]

the standard FLRW metric with a(t)=constant, and the spatial slice Euclidean is exactly Minkowski coordinates covering the whole Minkowski spacetime.

Technically, yes, I suppose this is true. But this formulation lacks a key property of all other FLRW metrics: it's impossible to take a limit as $a \to 0$. I guess that isn't technically a required property to say that the metric is in the FLRW form, but I think it's an implicit assumption in a lot of discussions.

 

[PeterDonis]

I think you are missing my point.

I see what you're saying. I'm just not sure it's a slam dunk that the two cases you describe must count as "putting a different metric" on the same underlying topology. But that might be at least partly because I'm used to being satisfied with a physicist's level of rigor, which to a mathematician is hopelessly sloppy. To a physicist, or at least to this one, it seems obvious that "the Milne spacetime" is not a "different spacetime" from Minkowski, it's just one wedge of it with funny coordinates. But of course saying that leaves a lot out.

 

[PAllen]

I see what you're saying. I'm just not sure it's a slam dunk that the two cases you describe must count as "putting a different metric" on the same underlying topology. But that might be at least partly because I'm used to being satisfied with a physicist's level of rigor, which to a mathematician is hopelessly sloppy. To a physicist, or at least to this one, it seems obvious that "the Milne spacetime" is not a "different spacetime" from Minkowski, it's just one wedge of it with funny coordinates. But of course saying that leaves a lot out.

Of course you can look at it either way. Another statement is that Minkowski spacetime is the analytic continuation of the geodesically incomplete Milne spacetime, similar to Kruskal for the exterior Schwarzschild.

 

[Ibix]

Technically, yes, I suppose this is true. But this formulation lacks a key property of all other FLRW metrics: it's impossible to take a limit as $a \to 0$. I guess that isn't technically a required property to say that the metric is in the FLRW form, but I think it's an implicit assumption in a lot of discussions.

The Einstein static universe is also eternal, isn't it? It's another special case of primarily academic or historical interest, of course.

 

[PeterDonis]

The Einstein static universe is also eternal, isn't it?

Yes, it has $a(t)$ constant in FLRW coordinates. The difference is that the spatial part is a hypersphere, i.e., positive curvature, instead of Euclidean 3-space as with Minkowski spacetime, so the Ricci tensor is nonzero and there is matter present (as well as a positive cosmological constant).

 

[Ibix]

Yes, it has $a(t)$ constant in FLRW coordinates. The difference is that the spatial part is a hypersphere, i.e., positive curvature, instead of Euclidean 3-space as with Minkowski spacetime, so the Ricci tensor is nonzero and there is matter present (as well as a positive cosmological constant).

Sure - pretty much the only thing it has in common with Minkowski spacetime is satisfying the Friedmann equations and being eternal. But it's definitely FLRW and it's eternal.

I suspect this comment belongs on one of hedgehug's now closed threads, but I suspect the reason Minkowski spacetime isn't normally thought of as FLRW is that it's the "something tends to zero" limit of so many families of spacetimes (Kerr-Newman, de Sitter, Ellis wormhole - in fact, probably all of them given their locally Lorentzian nature) that we either include it in all of them or none. And it's often a very different place (e.g., compare the Penrose diagrams of Schwarzschild and Minkowski spacetimes), so seeing it as separate from the rest of every family doesn't seem unreasonable.

I'm not sure if that's an argument that one ought to see Milne as a distinct spacetime (because people do give it some kind of status as a zero-mass FLRW spacetime), or just that edge cases aren't necessarily neatly classifiable.

 

[Ibix]

What makes Ricci tensor nonzero, if the metric tensor is constant?

The spatial part is a hypersphere. It just doesn't change scale with time.

 

[rocco]

Sorry for the next dumb question: If only the spatial part is hypersphere, does it mean that time is not the 4th dimension of this hypersphere?

 

[renormalize]

Sorry for the next dumb question: If only the spatial part is hypersphere, does it mean that time is not the 4th dimension of this hypershere?

A hypersphere is a 3-dimensional space (a hypersurface embedded in 4D spacetime) and so it does not include the 4th dimension of time. See, e.g., https://en.wikipedia.org/wiki/3-sphere.

 

[rocco]

That makes me wonder how the flow of time of the observer can depend on the spatial geometry of the universe, since this geometry is always in the spatial part of the metric. For instance, there is gravitational time dilation in the Schwarzschild metric, but there is no time dilation due to the spatial geometry of the universe (flat or spherical or hyperbolic).

 

[Ibix]

For instance, there is gravitational time dilation in the Schwarzschild metric, but there is no time dilation due to the spatial geometry of the universe (flat or spherical or hyperbolic).

If you slice Schwarzschild spacetime into a stack of identical spacelike slices, the time between each successive slice depends on $r$ (that isn't stated rigorously, but a rigorous statement can be made). That's where gravitational time dilation comes from. However, in the Einstein static universe the time between identical spatial slices does not depend on any coordinate, so you have no time dilation effects.

In general, geometry is a feature of the 4d spacetime. In some specific spacetimes (called "stationary" spacetimes) you can pick a definition of time so tgat you can slice spacetime into sequences of identical spaces which have identical 3-geometries. That doesn't mean that the 4-geometry isn't a thing, though.

We're pretty off-topic here, too.

 

[PeterDonis]

there is no time dilation due to the spatial geometry of the universe

That's because the spatial geometry of the universe (including the Einstein static universe) is homogeneous--it's the same everywhere. So there can't be any time dilation due to it--that would require the geometry to be different in different places. In Schwarzschild spacetime, the geometry is different at different values of $r$.

 

[rocco]

If we were living in a spherical, expanding universe with a positive curvature, shouldn't this curvature change due to the expansion, resulting in geometry changing in time, but spatially homogeneous?

 

[PeterDonis]

If we were living in a spherical, expanding universe with a positive curvature, shouldn't this curvature change due to the expansion, resulting in geometry changing in time, but spatially homogeneous?

The spatial geometry changes with time, yes, but that doesn't produce any time dilation between one point in space and another, because, as you say, it remains spatially homogeneous. That means the geometry changes with time in the same way everywhere in space.

 

[Ibix]

If we were living in a spherical, expanding universe with a positive curvature, shouldn't this curvature change due to the expansion, resulting in geometry changing in time, but spatially homogeneous?

Yes, the curvature of the spatial slices changes (except in the Einstein static case), but it does so homogeneously so everywhere is the same.

A 1+1d analogy is a cone. Cosmological time is the direction along the surface away from the point. Every slice through the cone perpendicular to the axis is a circle (a 1d closed universe) of different size, but there's nothing to choose between any point on the circle, including its relationship to points on other such circles.

 

[rocco]

The spatial geometry changes with time, yes, but that doesn't produce any time dilation between one point in space and another, because, as you say, it remains spatially homogeneous. That means the geometry changes with time in the same way everywhere in space.

Does it produce time dilation between the observers living in different cosmological epochs?

 

[Ibix]

A 1+1d analogy is a cone

...or a rugby ball or American football if you want a closed universe with a Big Crunch too.

 

[Ibix]

Does it produce time dilation between the observers living in different cosmological epochs?

No, just redshift.

You will find occasional papers describing the redshift as time dilation, but this seems like a misuse of the term.

 

[rocco]

I conclude that changing, spatial geometry of the universe does not affect the flow of time of the observer. If so, then his flow of time should be the same whether he lives in a flat, spherical, or hyperbolic universe.

 

[Ibix]

I conclude that changing, spatial geometry of the universe does not affect the flow of time of the observer. If so, then his flow of time should be the same whether he lives in a flat, spherical, or hyperbolic universe.

Your flow of time is always one second per second, whatever the universe. The difference between different FLRW cosmologies (or other spacetimes) becomes apparent when you look around them. In different FLRW cosmologies the apparent size and brightness of objects varies differently with distance, as does the redshift-to-distance ratio (the Hubble "constant"). You might even see yourself in the far distance, if the mix of cosmological parameters works out.

 

[rocco]

@Ibix you can run two simulations of 1. Flat, 2. Spherical universe, both expanding. You can start them simultaneously and stop them simultaneously. Would you agree that simulated observers living in these two universes since the beginning will be the same age when you stop the simulations?

 

[Ibix]

@Ibix you can run two simulations of 1. Flat, 2. Spherical universe, both expanding. You can start them simultaneously and stop them simultaneously. Would you agree that simulated observers living in these two universes since the beginning will be the same age when you stop the simulations?

Depends what you mean by "simultaneously" and which observers you're talking about.

You probably mean to terminate the simulations at some equal cosmological time, in which case co-moving observers would be of equal ages, as this is a tautology. Non-co-moving observers will be different ages, in general.

 

[rocco]

I'm looking at it the other way around with the same result. Since the comoving observers in both universes will be the same age, the universes will be the same age regardless of the difference in their geometry.

 

[Ibix]

Since the comoving observers in both universes will be the same age, the universes will be the same age regardless of the difference in their geometry.

This is a tautology. The time experienced by a co-moving observer since the singularity is the definition of the term "age of the universe".