Can Weyl Tensor Look Like Negative Mass?

[gnnmartin]

The smoothed Weyl tensor can look like space that contains a non-zero Einstein tensor. To verify this, consider that gravitational waves carry mass away from (say) a rotating binary, so the apparent mass at infinity of a large sphere containing a radiating binary will be greater than the mass at infinity of a much smaller sphere. The difference (roughly speaking) will be the non-local mass carried by the radiation in the space between the two spheres.

Thus the surface of a sphere containing only the Weyl tensor can have similar curvature to that of a sphere containing positive mass (say, dust). Is it mathematically (as opposed to physically) possible for the surface of a sphere containing only some different mix of Weyl tensor to have a curvature similar to that of the surface of a sphere containing negative mass? If not, is there a neat proof?

 

[PeterDonis]

consider that gravitational waves carry mass away from (say) a rotating binary, so the apparent mass at infinity of a large sphere containing a radiating binary will be greater than the mass at infinity of a much smaller sphere.

No, this is not correct. "Mass at infinity" is not a property of spheres that can vary with the size of the sphere. It is a property of the spacetime as a whole. You can't carve out just a portion of the spacetime and apply the concept of "mass at infinity" to it, leaving the rest out.

Also, there isn't just one "mass at infinity" in GR. There are at least two: the ADM mass and the Bondi mass. A system that is emitting radiation (gravitational or otherwise) will have a Bondi mass that is smaller than its ADM mass, because the ADM mass includes the energy carried away by the radiation, while the Bondi mass does not. In the case of a system emitting gravitational radiation, the radiation itself has no stress-energy associated with it, but the Bondi mass is still less than the ADM mass, which is what we mean when we say the gravitational radiation carries away energy from the system.

the surface of a sphere containing only the Weyl tensor can have similar curvature to that of a sphere containing positive mass

If this claim is based on the reasoning in your previous paragraph, it is incorrect, for the reason given above.

If you are getting it from some reference (textbook or peer-reviewed paper), please provide a reference. (I don't think there is one, because I don't think the claim is true.)

 

[gnnmartin]

Thanks for the response. I will rephrase my question. Imagine a universe where the Einstein tensor is zero everywhere, but where the Weyl tensor is non-zero.
1: Is it possible to add a condition that (on a large scale) the universe is roughly the same everywhere, or can one prove that if the only curvature is Weyl curvature, then large scale near self similarity is not possible?
2: If it is possible, is it possible that the universe is closed? In other words, is it possible that the universe has the same topology as a dust filled universe?
3: If it is possible that the universe is closed, is it possible that it is not, in other words, that it is a perturbation of flat space time.
4: If it is possible that such a universe is flat, is it possible that it could be negatively curved?

My gut feel (that I tried to justify in my original post) is that the answers are 1: yes, 2: yes, 3: no, 4: not applicable. I am interested to know if anyone has looked at this question.

 

[PeterDonis]

Is it possible to add a condition that (on a large scale) the universe is roughly the same everywhere

What does "roughly the same everywhere" mean? Can you formulate that condition mathematically?

We can't address the rest of your questions until we get this one resolved.

 

[PeterDonis]

@gnnmartin , you might also want to look into the Petrov classification of metrics, described briefly here:

https://en.wikipedia.org/wiki/Petrov_classification

Advanced GR textbooks will give more detailed information. This classification includes looking at things like what the implications are of particular forms for the Weyl tensor, Ricci tensor, etc.

 

[gnnmartin]

What does "roughly the same everywhere" mean?

I mean that the metric is a an arbitrarily small perturbation of the metric of a space of constant curvature.

 

[PeterDonis]

I mean that the metric is a an arbitrarily small perturbation of the metric of a space of constant curvature.

Ok. Then it seems like the answer is obviously no, since a spacetime with constant curvature has a zero Weyl tensor, and any arbitrarily small perturbation of the metric of such a spacetime would also have an arbitrarily small Weyl tensor, meaning the Weyl tensor could not have any significant effects.

 

[gnnmartin]

Then it seems like the answer is obviously no, since a spacetime with constant curvature has a zero Weyl tensor, and any arbitrarily small perturbation of the metric of such a spacetime would also have an arbitrarily small Weyl tensor,

I do not think that follows. That metric A is an arbitrarily small perturbation of metric B does not imply (does it?) that the curvature over the space/time described by metric A is an arbitrarily small perturbation of the curvature over the space/time described by metric B.

The corollary certainly is not true. Working in 2 dimensions for example, the surface of the Earth can be described as a small perturbation of a surface of constant curvature, but the fine detail of the curvature includes near singularities (for example, where a wall meets the ground).

 

[gnnmartin]

@gnnmartin , you might also want to look into the Petrov classification of metrics, described briefly here:

https://en.wikipedia.org/wiki/Petrov_classification

Thanks for the suggestion. I don't see how the Petrov classification helps, since the Petrov classification classifies a point, so does not tell us anything about a space/time unless we assume (for example) that the classification is the same everywhere.

 

[PeterDonis]

That metric A is an arbitrarily small perturbation of metric B does not imply (does it?) that the curvature over the space/time described by metric A is an arbitrarily small perturbation of the curvature over the space/time described by metric B.

If you just mean in a small local neighborhood, then you are correct, the implication does not hold. (This is obvious from the equivalence principle, since in a small enough neighborhood every spacetime, regardless of its curvature, looks flat.)

However, you appear to be making a stronger claim, which is that the metric that has a significantly nonzero Weyl tensor can be everywhere an arbitrarily small perturbation of a metric with constant curvature. That is the claim I am questioning. Globally speaking, I think my implication is valid: if metric A is everywhere an arbitrarily small perturbation of metric B, then the same must hold for the curvature tensors, because if metric A is everywhere an arbitrarily small perturbation of metric B, that constrains the derivatives (to any order) of metric A as well as its values, and the curvature tensor is built from derivatives of the metric.

the surface of the Earth can be described as a small perturbation of a surface of constant curvature, but the fine detail of the curvature includes near singularities

Please show your work. I think you will find that in the neighborhood of these "near singularities", the metric can no longer be described as a small perturbation of a surface of constant curvature. You are basically looking at the metric of the Earth's surface averaged over a large scale, and then treating small-scale near singularities as though they shared the same metric properties as the large scale average.

 

[gnnmartin]

if metric A is everywhere an arbitrarily small perturbation of metric B, then the same must hold for the curvature tensors, because if metric A is everywhere an arbitrarily small perturbation of metric B, that constrains the derivatives (to any order) of metric A as well as its values

I don't think that is correct. An assertion that A is a small perturbation of B does not imply that dA is a small perturbation of dB. Still, I won't try to push this line.

Thank you for your help. I think I can't usefully argue my point any further.