Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.
up vote 2 down vote favorite

Suppose I have two point/line singularities in spacetime (what is important to me is that they are localized). Also suppose I have some fields in spacetime and that the two singularities interact with each other via the fields (fields are non-minimally coupled to the curvatures). Now if I denote the Ricci scalar of the entire system by $R$ and for each of the singularities by $R_1$ and $R_2$, is $R=R_1+R_2$? I am not able to make up my mind. I don't think so and I think only to some order in perturbation this is true. But your help/comments will be greatly appreciated.

share|improve this question
I doubt that a singularity has a finite Ricci scalar. – C.R. Feb 8 '12 at 7:38
You can have a delta function Ricci scalar for such localized singularities. That's for example what happens for a cosmic string. So, the spacetime just outside is flat. – user1349 Feb 8 '12 at 13:29
1  
When you refer to Ricci scalars "for" the singularities, do you really mean, for example the "Ricci scalar $R_1$ of the spacetime with matter field distribution number 1, $T^1_{\mu\nu}$, which has a singularity", and ditto for number 2 etc ? If so, I've no idea how to talk about singularities interacting with each other, since almost by definition, we don't know how to mathematically model them. – twistor59 Feb 8 '12 at 13:37
1  
Yea, this is ill-posed and demonstrates a lack of understanding of mathematical curvature. – Chris Gerig Feb 8 '12 at 21:46
1  
GR doesn't have any notion of superposition of fields. If you have one spacetime and another spacetime, there is no meaningful way to define how to add them, because there is no meaningful way to even say which points correspond to one another. You could pick coordinates and identify points that had the same coordinates, but the coordinates aren't built in to the spacetime. Someone else could have picked different coordinates. – Ben Crowell May 3 at 21:17
show 1 more comment

closed as not a real question by Qmechanic♦ May 3 at 22:03

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.

1 Answer

up vote 0 down vote

A few things:

1) If you have a spacetime containing an isolated object like a black hole or a cosmic string, or any matter distribution whatsoever, $R$ is zero outside the matter distribution, since $T_{ab}$ is zero.

2) Also, consider the Schwarzschild solution in asymptotically Cartesian coordinates. Place the $r=0$ singularity at $(x_{1},y_{1},z_{1})$. If you try and write the metric according to the ansatz $g_{ab}={}^{1}g_{ab}+{}^{2}g_{ab}$, where ${}^{1}g_{ab}$ is the Schwarzschild solution with the hole at $(x_{1},y_{1},z_{1})$, you will find that the solution you get is only an approximation of a vacuum solution of Einstein's equation.

3) There are terms in $R_{abcd}$ that have to do with things like the boundary charges and initial conditions in the spacetime. In particular, if you two singularities are moving with respect to each other, or have appreciable gravitational binding energy, or emit gravitational radiation, then you will have boundary terms that don't satisfy $M_{adm}=M_{1}+M_{2}$, and so, you wouldn't expect that $R_{abcd}={}^{1}R_{abcd}+{}^{2}R_{abcd}$

share|improve this answer
But in the case OP considers--- two cosmic strings--- the 2d Gauss curvature adds up between the different strings in a straightforward way. – Ron Maimon Jun 7 '12 at 19:30

Not the answer you're looking for? Browse other questions tagged or ask your own question.