Natural Warping

1. Nov 21, 2005

-Job-

I was wondering if in the GR model for gravity it is possible that space-time be naturally warped (by which i mean warping not caused by matter). In such a region would one experience gravity, even without any matter around?

2. Nov 21, 2005

pervect

Staff Emeritus
This could be caused by a negative cosmological constant. The current cosmological constant is expected to have the opposite effect, i.e. it makes the universe accelerate its expansion, rather than slowing it down (see also "dark energy").

http://www.astro.ucla.edu/~wright/cosmo_constant.html

3. Nov 21, 2005

robphy

The presence of matter, via the Field Equations with zero cosmological constant, is associated with a nonzero Einstein [Curvature] tensor. However, a zero Einstein [Curvature] tensor does not imply a zero [Riemann] Curvature tensor.

Technically speaking,...
the Schwarzschild solution is a vacuum solution (a solution with no matter in the spacetime). http://www.pma.caltech.edu/Courses/ph136/yr2002/chap25/0225.1.pdf

4. Nov 21, 2005

pervect

Staff Emeritus
The Schwarzschild solution may not have matter, but it has a singularty at r=0 that gives the solution a total mass. Usually this mass is thought of as being concentrated at a singularity at the origin of the coordinate system.

5. Nov 21, 2005

JesseM

The answer to the OP depends on what you define as "matter"--energy also contributes to the warping of spacetime in GR, so I would think (though I'm not sure) that an electromagnetic wave packet travelling through a region empty of matter would locally curve spacetime around it as it moved. Likewise, how about a spacetime empty of matter but containing gravitational waves? And if you do count energy as "matter", then the cosmological constant can itself be considered as a form of energy filling all of space, no? Physicists seem to think that the vacuum energy of quantum field theory and the cosmological constant are the same thing...

6. Nov 22, 2005

-Job-

So in Einstein's model, the "warping" of space at any point is just a visual interpretation of the distribution of energy in space (energy density) if we choose to see matter as "denser energy". This seems to follow since:
More Energy/Mass -> Bigger Gravity -> Bigger Warping.
So in the model, the warping of space is equivalent to a graph where for each point x,y,z (t?) there is associated some energy density level. If we had the energy density distribution in the solar system at a time t, then a graph of x, y, z, d (d = energy density) is a 4D graph that shows the warping of space at the region our solar system in precisely the same way as Einstein's model.

7. Nov 22, 2005

Ich

Interesting Question - I stumbled across it in a SF novel by Benford. There people live in folded spacetime orbiting a black hole.
Is this equivalent to static electric fields w/o charge, or does GR allow for self sustaining warps in spacetime (other than moving at c)?

8. Nov 22, 2005

robphy

Following up on my first comment,
at each event in spacetime, the [Riemann] Curvature has two parts, a part determined by the Ricci Curvature (which, via Einstein's equations, is related to the matter density at that event) and the remaining part called the Weyl Curvature. (See, for example, https://www.physicsforums.com/showthread.php?t=93396&highlight=riemann+weyl and http://math.ucr.edu/home/baez/gr/ricci.weyl.html)

So, where there is matter, there is curvature. However, where there is curvature, matter need not be.

In the Schwarzschild solution, there is curvature at events where the matter density is zero. [The Earth (in the presence of the distant Sun) experiences the curvature of spacetime even though there is no matter encountered by the Earth (ignoring the Earth itself, which is regarded as a test body).] This is an example of a "Vacuum Solution" http://en.wikipedia.org/wiki/Exact_solutions_of_Einstein's_field_equations .

Last edited: Nov 22, 2005
9. Nov 22, 2005

pervect

Staff Emeritus
Very close, but actually at every point (x,y,z,t) we find we really need a quantity known as the stress-energy tensor, which consists of 16 numbes.

The reason is this - without relativity, energy / unit volume is the same in all reference frames. With relativity, this is no longer true, and if you know only the energy/unit volume in one frame, you cannot compute it in another.

What you need to compute the energy/unit volume in all frames is the stress-energy tensor. This consists of the energy density at a point, the "direction of flow" of the energy at that point (i.e. the momentum density at that point), and the pressure at that point. The pressure is not always a single number, BTW, but depends on direction. Perfect fluids have the property that pressure is a single number the same in all directions, solids and more general fluids do not have this property.

All of these quantities in the stress-energy tensor wind up contributiting to the gravitational field, though one could single out energy and pressure as being "source" terms.

10. Nov 22, 2005

-Job-

That's a good point, I completely missed that.
One thing that confuses me, how reasonable is it to expect that the "vaccum" around an object like the moon be more energetic than "vaccum" farther away? I have to ask this because if pure vaccum has some energy level (possibly the lowest of all possible values) and matter is very dense energy (i don't mean that literally, but by E=mc^2 matter seems to pack alot of energy), then it's hard not to expect a smooth transition in between these two.

11. Nov 22, 2005

pervect

Staff Emeritus
It is necessary that the vacuum appear to be the same for all observers, no matter how they are moving, if relativity is to be correct.

The transition is not an issue for a classical theory like GR. I'm not really sure if there are any quantum or semi-classical issues here. I suppose that since the position of a planet isn't perfectly well definied, quantum mechanically the boundary of the planet must be "fuzzy" on the scale of the uncertanity of the position of the planet. Which is of course, very very small.

To deal with this sort of issue, you'd have to talk to someone who knows a lot more about quantum gravity than I do. But in practical terms, such issues aren't going to have any major effects on large masses, which is where GR is applicable. For very small masses, gravity isn't usually very important, and if it does need to be analyzed, Newtonian theory is sufficient.