# Two-mass Schwarzschild metric instead of Kerr metric?

1. Jan 21, 2016

### Rlam90

Just a thought...

Would there be any implicit differences between (A) a two-body metric where the two central masses are drawn ever further together, with angular momentum included, and (B) the Kerr metric? Angular momentum would still be part of the system, but it would be explained by a more intuitive form rather than the assumption that mass inherently has angular momentum. It would also allow for symmetry breaking if one desired, as the closer to the center of the coordinate system you become, the more apparent other subsystems could be. This would allow a sort of binary-splitting capability into the metric, allowing one to explore the events near the center in more detail than simply having everything fall towards some mysterious rotating mass at the center.

Of course, it would be extremely difficult to formulate this explicitly in all possible details, but that's always the case with models of the universe. It's a step up from the Kerr metric though.

If I were to start deriving the metric, I would base it in bispherical coordinates and have the focal distance parameter "a" evolve with time, as well as the relative azimuthal angle as the central bodies orbit each other. What wouldn't be included is the effect that the test mass would have on the orbit of the central bodies, but we could assume that's so small as to not be relevant.

2. Jan 21, 2016

### Staff: Mentor

Most certainly. There is no exact solution known for (A); we only know how to analyze it numerically, so we can't enumerate all the specific differences mathematically. But just looking at (A), there are at least two obvious differences from (B):

First, (A) is not a vacuum solution; (B) is.

Second, (A) is not axially symmetric (the two masses have particular angular locations, which breaks the symmetry); (B) is.

That's not what the Kerr solution says. What it does say will seem even more outlandish to you (but is nevertheless true): it says that spacetime inherently has angular momentum. The angular momentum in Kerr spacetime is a property of the spacetime geometry; it is not connected to any "mass" in the sense of a nonzero stress-energy tensor, because the Kerr solution is a vacuum solution (no stress-energy anywhere), as noted above. (Yes, there is a "mass" that appears in the metric, but it's not associated with any matter or energy; it's also a property of the spacetime geometry.)

If you think you can, please do so. As noted above, nobody currently knows of an exact solution for this case; you would be a good candidate for a Nobel prize if you came up with one.

3. Jan 21, 2016

### Rlam90

What I've got so far is the metric tensor for flat-space in bispherical coordinates with a stationary focal distance term "a".

$(1, 2, 3, 4) = (σ, τ, ϕ, t)$

$g = \begin{pmatrix} L^2 & 0 & 0 & 0\\ 0 & L^2 & 0 & 0\\ 0 & 0 & L^2 sin^2 σ & 0\\ 0 & 0 & 0 & c^2 \end{pmatrix}$

Where

$L = a / (cosh τ - cos σ)$

Working on the Christoffel symbols for fixed "a" right now.

I'm using tau for now because of habit. I'm going to need to change it, probably to xi, when I actually get to working with the line element.

4. Jan 21, 2016

### Staff: Mentor

Can you give more details or a reference for these coordinates? I'm not familiar with them.

5. Jan 21, 2016

### Rlam90

Sigma is the angle <F1 P F2

Tau is the logarithm of the ratio of the distances d1 and d2 from the foci.

Phi is the standard azimuthal angle.

The foci are located at +a and -a from origin, on the z axis.

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

*update* I'm switching over to toroidal coordinates because it will be easier to keep track of the movement of the coordinates, limiting the movement in coordinates due to the rotation of the foci to just the phi coordinate.

Last edited: Jan 21, 2016
6. Jan 21, 2016

### pervect

Staff Emeritus
I'm currently not finding that this line element is flat, though it's possible I'm making some mistake (possibly it's just a matter of incomplete simplification). Basically, I put your metric into grtensor - after changing the sign of dt^2 to make it Lorentzian. Because L^2 should be positive, the metric as written has 4 positive diagonal terms - which isn't Lorentzian. So I changed the sign of the dt^2 term.

The constraint equations I used were:

$${\frac {\partial }{\partial \tau}}L \left( \tau, \sigma \right) =-{\frac { \left( L \left( \tau,\sigma \right) \right) ^{2}\sinh \left( \tau \right) }{a}} \quad {\frac {\partial }{ \partial \sigma}}L \left( \tau,\sigma \right) =-{\frac { \left( L \left( \tau,\sigma \right) \right) ^{2}\sinh \left( \tau \right) }{a }}$$
and got several apparently non-zero components of the Riemann, one of which is, for instance
$$R_{\sigma\tau\sigma\tau}={\frac { \left( L \left( \tau,\sigma \right) \right) ^{3} \left( -L \left( \tau,\sigma \right) \left( \sinh \left( \tau \right) \right) ^{2}+\cosh \left( \tau \right) a-L \left( \tau,\sigma \right) \left( \sin \left( \sigma \right) \right) ^{2}+\cos \left( \sigma \right) a \right) }{{a}^{2}} }$$

That doesn't look like it will simplify to zero. I can't say a lot more at the current time - I did try substituting $\tau = \sigma = 0$ into the above expression to try and prove it wasn't just a matter of something that evaluated to zero even though it didn't appear to. However, those particular values make L "blow up" so it may not be a good test.

I'm a bit skeptical that the line element actually is a dual body solution to GR, so my first thoughts was to test the flatness, the second thought is to understand why it would be considered a "dual body" metric at all.

7. Jan 21, 2016

### Staff: Mentor

If it's flat, it can't be, because a dual body solution has nonzero stress-energy in it, and flat spacetime is only a solution to the EFE if the stress-energy tensor is zero (as well as the cosmological constant).

If it's not actually flat, it could be a dual body solution (though I think that's highly unlikely); but I don't think that was his intent. I had interpreted that metric as an attempt to describe flat spacetime in the coordinates he was proposing to use for the (curved) spacetime of his dual body solution, as a warmup exercise. (But it looks as if he's now going to use different coordinates anyway.)

8. Jan 21, 2016

### Rlam90

Thanks for the input.

I opted for toroidal coordinates because of the symmetry along the "orbital" plane of the foci. In this case the sin^2 in the metric tensor is sinh^2, otherwise the same.

I probably shouldn't have marked this as "advanced", but I didn't know what level to choose. What does "experimental" count as?

Also: out of curiosity, does that same program give odd solutions for the Schwarzschild metric at r=0?

I know there's something interesting in this once I dig deep enough, but at this point it's not much more than a hunch. I'm sure if I want to achieve what I have in my head I'm going to need to include some fractal concepts and take this outside the regular scope of relativity. What I'm seeing in this model might be more of an algorithm to decide when to switch to a new frame based on the shape of spacetime in the section being examined. A more natural, iterative or stochastic model is probably going to be the end result. My main goal right now is to create a Schwarzschild-like metric with two stationary singularities based on these coordinates, and then to examine the geometry of the resulting model. Gotta start somewhere, right?

It might actually be easier to achieve that in bispherical coordinate (two actual foci as opposed to a circular focal ring), but formulating them side-by-side shouldn't be too much more difficult given their similarities, and in case one becomes much more obviously appropriate it would make sense to do both.

9. Jan 21, 2016

### Staff: Mentor

Not sure what "program" you are talking about here. One key thing you should keep in mind is that physics is independent of coordinates; a particular choice of coordinates might make it easier to do calculations or describe their results, but it doesn't change any physics. So whatever "odd" thing is happening at $r = 0$ in the Schwarzschild metric, physically speaking, it's happening regardless of what coordinates you choose.

10. Jan 21, 2016

### Staff: Mentor

That won't be a solution unless you also include something to hold the singularities apart (otherwise they will fall together), which is going to be difficult if by "singularities" you mean black holes, since black holes are vacuum--you can't push on them or attach anything to them.

11. Jan 22, 2016

### Markus Hanke

Also, (A) has a non-vanishing quadruple moment associated with it, unlike (B), meaning this binary system will gravitationally radiate. This is a geometry which is really completely different from the Kerr case; the Kerr space-time is Petrov type D, whereas the radiating binary would be type N, if I am not mistaken - correct me if I'm wrong.

12. Jan 22, 2016

### Rlam90

That's kind of my point of researching this. It doesn't make sense to me that there would be some infinitesimal parts of spacetime that wouldn't radiate at least a minute amount of gravitational radiation.

Also: the coordinate system itself would be what holds gravity back between the foci. I see it as making all space intervals relative to the interval between the foci of the coordinate system. Imagine focus 1 (F1) as a Schwarzschild black hole, and focus 2 (F2) as an object around the Schwarzschild. Now, we know that gravity attracts more strongly towards a center of mass. So, an object closer to F2 than F1 would be attracted more strongly to it, and visa versa. From an observer positioned so that F1 and F2 always maintain a steady distance, this would make it appear that anything close enough to F2 would be drawn more strongly towards it and the same for F1. Let's disregard any angular movement between F1 and F2 for simplicity right now. What would happen is that anything far from both foci would appear to be drawn to the center of the system. Anything between the two foci would be drawn to either one of them. Everything between the two would appear to move away from the center of mass of the foci in this point of view, while from an infinitely distant observer it would appear that the distance between the foci would be removed. This system of coordinates would make even an infinitely situated observer move, to maintain the set distance between the foci. In fact, if the geometry of this coordinate system was anything other than flat it would force a physically stationary point to move with regard to the coordinate system. Does this make sense?

13. Jan 22, 2016

### Staff: Mentor

This can't be correct. Coordinates aren't physical things; they can't have physical effects.

Do the hole and the object have the same mass? It seems like you are assuming they do, but I can't be sure.

From the observer or from each other? I assume the former, but again I can't be sure. As I pointed out before, F1 and F2 cannot maintain a steady distance from each other unless there is something holding them apart (otherwise they will fall together), and I don't see how you can have anything pushing on a black hole to hold it apart from something else.

I can't make sense of the rest of your post, but that might be because I'm misunderstanding the conditions you are specifying. (It might also be because the conditions you are specifying are inconsistent.)

14. Jan 22, 2016

### Markus Hanke

Why doesn't that make sense to you ? Generally speaking, in order to get gravitational radiation, you need a source with a non-vanishing quadrupole ( or higher multipole ) moment. It is not a general feature of just any arbitrary source.

15. Jan 22, 2016

### PAllen

I believe I've seen exact solutions of two non-rotating BH (with singularity) held statically apart by a 'strut' that is a conical singularity. If I remember right, the whole construction is all vacuum within the manifold.