# Some simple minded aspects of gravity

1. Apr 23, 2007

### oldman

In the cosmology forum I have admitted to understanding GR only dimly despite, in a previous thread in this forum ("Metrics and Forces"), having been educated by Pervect to a slightly improved understanding. Here I would like to continue this education.

Consider a big observer inside an opaque box who is falling freely, in two extreme situations. First, let him be falling towards a nearby single massive object (say a neutron star). He will be unaware of gravity qua gravity since he is falling freely.

But he will be very aware of tidal forces, since he is big. He'll find himself settling into one orientation (head or heels toward the star, squashed sideways, stretched longways, feeling somewhat as if he were rotating head over heels at the orbital frequency appropriate to his distance from the star).

A General Relativist may explain to him that he is feeling dizzy because spacetime's spatial sections are curved by a nearby mass.

Consider a second extreme situation. Let the observer be surviving somehow in a flat inflating universe, where the Hubble constant is increasing exponentially fast. Here again he will experience strong tidal forces (as Pervect taught me to appreciate); in this case a symmetric dilating disruption.

Here the General Relativist can't attribute the disruption to curved space sections, because the inflating universe is flat.

Will he in this case invoke a curved time section?

Does this tell us that time sections would on a universal scale be almost flat? in a universe where H is almost constant: one that is hardly decelerating because of gravity, or accelerating due to dark energy, because the amount of mass/energy is well below the critical density, and where the Hubble flow is almost pure kinematic or coasting motion?

If so, one could invert this reasoning, and say that gravity and/or the antigravity of dark energy (both of which are attributed to the influence on the geometry of spacetime of mass/energy --- by means quite unknown) are simply manifestations of a curved time dimension. Too simple, I guess.

Last edited: Apr 23, 2007
2. Apr 23, 2007

### pervect

Staff Emeritus
One has to distinguish between "flat space" and "flat space-time".

The existence of tidal forces shows directly that the inflating universe does not have a flat space-time. To be more technical, it has a non-zero Riemann curvature tensor. Tidal forces can be identified with particular components of the Riemann tensor (one may have to add the further condition that the tidal forces are measured by a non-rotating observer following a geodesic to make this statement, but these conditions are not a problem).

Since our hypothetical observer meets these conditions, the existence of tidal forces shows that the space-time is not flat, because there is a non-zero Riemann (any non-zero component of the Riemann means that the tensor is non-zero).

This curvature of space-time is independent of the flatness of spatial slices. The flatness of the spatial slices depends to some extent on "how they are sliced" - i.e. one allows many different definitions of time in GR, thus one has to adopt some particular definition of time to be able to slice space-time into space and time. When one slices a curved space-time, it is perfectly possible to wind up with flat spatial slices. While one might informally call this a "curvature in time", this description will probably not be particularly clear and will cause a lot of arguments and discussion. It's probably better (though perhaps more cumbersome) to say a "curved space-time with flat spatial slices".

3. Apr 24, 2007

### oldman

Thanks for replying to my oversimplifications, Pervect. I agree with all you say, but have these comments:

I don’t know how to work the machinery of GR so as to identify which of the 20 independent components of the Riemann tensor are non-zero in scenarios like the two I described, using non-rotating reference frames moving along geodesics.

I assumed the slicing used in the FRW standard model universe, where it seems permissible to talk of “flat” space-sections and a "universal" time -- I suppose “flat” is used in the sense that parallel-transportation is path-independent in the space dimensions.

If it is OK to talk colloquially of “flat space sections”, I feel it should be OK (even if unconventional) to talk of a “curved time-dimension” during inflation, which is hardly a conventional process. In its later stages, where people talk of it flattening out any space-curvature, inflation’s chief peculiarity seems to me to be “curved time”, whatever that may be like. I find it interesting that gravity manifests itself to freely falling observers in this odd way during inflation.

Do you have any thoughts on what "curved time" might be like?

4. Apr 24, 2007

### George Jones

Staff Emeritus
One has to be somewhat careful here; there are two different metrics in play here, and the notions of flat/curved and parallel transport are metric dependent.

Let me give an example in flat R^3. Consider the 2-dimensional surface (sphere) picked out by the equation of constraint x^2 + y^2 +z^2 - R^2 = 0. The metric for R^3, when restricted to the surface, induces a metric on the surface. So we have two metrics, the standard flat 3x3 (after a basis is introduce) on R^3, and the induced 2x2 metric on the 2-dimensional surface.

Consider parallel transport on the surface. With respect to the R^3 metric, parallel transport on the surface is path-independent; with respect to the induced metric, parallel transport is path dependent. The curvature tensor tensor obtained from the flat R^3 metric is zero everywhere, including on the surface. The curvature tensor obtained from the metric for the surface is non-zero.

The cosmological situation is, in some sense, the reverse of the above example. Setting cosmological time equal to a constant picks out a 3-dimensional spatial hypersurface in spacetime. The (4x4) spacetime metric restricted to the hypersurface can be used to induce a definite (3x3) metric on the hypersurface.

Consider curves in the hypersurface. With respect to the spacetime metric, parallel transport on the hypersurface is path-dependent; with respect to the induced metric, parallel transport is path-independent. The curvature tensor obtained from spacetime metric is non-zero on the hypersurface. The curvature tensor obtained from the metric for the hypersurface is zero.

Edit: instead of talking about "curved time", maybe it's better to talk about non-zero extrinsic curvature. There exist cosmological models in which spacetime can be foliated by spatial hypersurfaces that have zero intrinsic curvature and non-zero extrinsic curvature. These universes are sometimes called flat, but flat doesn't refer the spacetime curvature tensor.

This is similar to another R^3 example, the surface of a cylinder, which has zero intrinsic curvature and non-zero extrinsic curvature.

Last edited: Apr 24, 2007
5. Apr 24, 2007

### oldman

I take it that the metric for R^3 is what you call "intrinsic" below, whereas the 2x2 metric is what you call "extrinsic". If this is so, I follow you. But in the end there is a matter of terminology preference, and I can't rid myself of a hankering after "curved time" in the specific case I considered in my OP, even if this is technically confusing, as Pervect warned it might be.

Thanks for explaining this to me. I need all the help I can get!

6. Apr 24, 2007

### pervect

Staff Emeritus
I don't know if this will help, but

$$R_{\hat{x}\hat{t}\hat{x}\hat{t}}$$ is the component of the Riemann that is equivalent to a tidal stretching force along the x axis.

This picks out the exact component of the Riemann that is equivalent to a tidal force. It uses the concept of a "frame field", which basically means that we set up some local coordinates so that the metric is locally Minkowskian, i.e. so that g_ab = diag(-1,1,1,1) near the origin. You know it's a frame field because of the "hats" that x and t are wearing.

Of course we can't make g_ab Minkowskian everywhere, but we can make it happen at one point, and in a small region around that point it won't change much. This is basically what a frame field is, it's the Riemann in some locally Minkowskian tangent space on a manifold.

Similarly, if you replace $$R_{\hat{y}\hat{t}\hat{y}\hat{t}}$$ is a tidal stretching force along the y axis, and $$R_{\hat{z}\hat{t}\hat{z}\hat{t}}$$ is, you guessed it, a stretching force along the z axis.

Terms like $$R_{\hat{x}\hat{t}\hat{y}\hat{t}}$$ can be eliminated by a proper orientation of spatial axis, i.e. via a principal axis coordinate transformation.

Other terms in the Riemann represent things other than tidal stretching. Some terms represent gravitomagnetic types of forces associated with frame dragging, other terms contribute to the Ricci tensor which can be regarded as the source of gravity - by Einstein's equation, the Ricci will be determined by the stress-energy at the point, so that in a vacuum region, for instance, the Ricci is always zero.

I assumed that's the slicing that you used. See George's notes about parallel transport.

Maybe it is, but in my experience it has confused at least some people to talk about "curved time", even though it doesn't confuse many people to talk about "flat spatial slices". Go figure. I don't have any particular problem with you thinking about it as "curved time", it's just that if you try to talk with someone else, they might not know what you were talking about.

If I could explain "curved time" without using so many words, I'd use it more, but it takes a while to explain and is rather involved, so for both operational and aesthetic reasons I don't use it much.

7. Apr 25, 2007

### oldman

Yes, it does clarify things for me. Although I am not proficient in manipulating the tensors of GR, I am familiar with the second-rank stress and strain tensors of ordinary elasticity. Seeing repeated spatial indices in the Riemann components that you say are responsible for the dilation strains produced during inflation was therefore reassuring. I take it that when different spatial indices occur they represent shears that, as you say, can be transformed away by a suitable choice of axes. As in ordinary elasticity.

The rest of your post was equally illuminating.

You went on to say:

I agree.

It seems to me that the physical manifestation of gravity during inflation must (for purely hypothetical free-falling observers) include phenomena like an extreme redshift, dilation stresses, and I wonder what else?

Maybe "curved time" causes clocks to tick at space or time-varying rates. If the latter, I can't begin to think how this would be detected.

The help yourself and George Jones have given me is much appreciated.

8. Apr 26, 2007

### MeJennifer

So do I understand correctly here that a FRW model with a positive cosmological constant, which is a conformally flat spacetime, and thus has a vanishing Weyl tensor, has tidal forces?
If Ricci curvature is meant, how standard is it to call Ricci curvature a tidal force?

Last edited: Apr 27, 2007
9. Apr 27, 2007

### oldman

But, if H is not constant, (if that is what happens in a universe with a cosmological constant) the answer becomes Yes. In a slowly decelerating (or accelerating) FRW universe (as cosmologists think ours is), the answer is yes. A tiny and unimportant effect.

In an inflating universe, in which acceleration is exponentially huge, the answer is YES^ !!!!.... (raised to some large power). I suspect that neither cosmologists nor relativists have concerned themselves with this aspect of inflation, though.

This is a technical matter which I'm not able to answer. But I don't like using "tidal force" for a spherically symmetric dilation. I prefer saying that it's the result of the "time dimension being curved", which is both non-standard and, I'm told, confusing. But hey, this is a hitherto obscure effect ... so what's in a name!

Last edited: Apr 27, 2007
10. Apr 27, 2007

### MeJennifer

Thanks oldman (sorry I don't know your name and I hope oldman is ok)

So then the Petrov classification does not cover it? Or these are not FRW type universes?
It seems I miss something here.

Last edited: Apr 27, 2007
11. Apr 27, 2007

### oldman

But: as far as I'm aware all universes considered in the standard model of cosmology (with or without its inflationary scenario) are FRW models. Amazing that nothing is thought to modify this scaffolding over something like 10^64 plus orders of magnitude change in the scale factor, but that's cosmology for you!)

12. Apr 27, 2007

### MeJennifer

That's what I thought as well. And according to the Petrov classification such spaces are conformally flat and have a vanishing Weyl tensor.

13. Apr 27, 2007

### oldman

Though I understand what you are telling me, one must remember that here the hypersurface is a space section. And, as my clever wife (younglady) has just pointed out to me, parallel transporting a vector around a circuit in any space section must in practice be a process that takes time. Therefore it is not practically possible to explore the consequence of there being an "extrinsic" metric of a flat hypersurface in the particular case of an accelerating FRW universe.

"Curved time", as I like to call it (despite your gentle persuasion below), cannot be concealed when measuring space curvature or its absence.

Spatial slices like those nested potato crisps sold by Pringles, I assume?

14. Apr 27, 2007

### pervect

Staff Emeritus
An interesting point. The issue is whether if we have the appropriate non-zero terms in the Riemann, but zero terms in the Weyl, whether we really have tidal forces.

If we consider the case of a pair of test particle in a De-sitter space-time, I think it's fairly clear that the answer should be yes.

To be more specific, use a standard flat FRW chart

ds^2 = -dt^2 + a(t)^2*(dx^2 + dy^2 + dz^2)

where a(t) = exp(K t)

We can trace geodesics in this spacetime, (x,y,z=constant are the simplest standard examples) and see that they separate at an accelerating rate.

This means that we have to apply forces to particles to keep them from separating.

We can also see that the Weyl is zero, and the Riemann is non-zero.

15. Apr 27, 2007

### MeJennifer

FRW models allow only co-moving objects.
That non co-moving objects in such a model with a positive cosmological constant need to apply forces to stay together seems to me a bit of a questionable physical interpretation.

Last edited: Apr 28, 2007
16. Apr 27, 2007

### Chris Hillman

Definition of tidal tensor

Hi, pervect,

Haven't had time to read this thread carefully, but this is the reason why the tidal tensor, aka "electrogravitic tensor", aka "time-time curvature components", is defined (with respect to a timelike congruence $\vec{X}$) as
$$E[\vec{X}]_{ab} = R_{a m b n} \, X^m \, X^n$$
not
$$C_{a m b n} \, X^m \, X^n$$
Similarly, the magnetogravitic tensor, aka "space-time curvature components", is defined as
$$B[\vec{X}]_{ab} = {R^\star}_{a m b n} \, X^m \, X^n$$
not
$${C^\star}_{a m b n} \, X^m \, X^n$$
The cognoscenti can ponder left versus right dual (an issue for Riemann but not for Weyl).

Last edited: Apr 27, 2007
17. May 1, 2007

### pervect

Staff Emeritus
I have no idea where you got that idea.

18. May 2, 2007

### MeJennifer

Pervect, known solutions in general relativity are very specific and idealized. For instance the Schwarzschild solution is a vacuum solution of one and only point mass, the Kerr solution a vacuum solution of a rotating point mass, the FRW solution is a solution of point masses that are all co-moving (plus a few of more assumptions).

You simply cannot introduce all kind of ad hoc point masses into an existing solution and then claim that "forces" are at work there.

Last edited: May 2, 2007
19. May 2, 2007

### pervect

Staff Emeritus
You're rather missing the point. We can add zero mass test particles to any geometry. If no forces act on them, they follow geodesics. There is nothing wrong with doing this.

We can consider a pair (or more generally a family, sometimes called a congruence) of geodesics in the FRW space-time. This family separate from each other. There's a technical way of describing this, the "expansion scalar".

The geometrical view is an expansion of the geodesics. The forces are just an interpretation. If we have a pair of particles that are following geodesics in Earth orbit, and the particles diverge, because the geodesics diverge, we interpret this in familiar Newtonian language as a "tidal force".

For instance, in our example, one particle is closer to the Earth than the other, so the "force" of gravity in Newtonian gravity is different, and we call this a tidal force. From the geometric point of view, we call it something else. The point is, it's basically the same entity, described in geometrical terms and described in more familiar Newtonian language.

This is not my invention, this is discussed for instance in MTW in the section on "tidal gravitational forces and the Riemann curvature", pg 270. (But it's rather technical).