Schwarzschild metric in 3-dimensional space-time

[Zinger0]

I'm trying to find Schwarzschild solution for 3-dimensional space-time (i.e. time\otimes space^2). The problem is, I can't take the 4-dimensional solution

\[ds^2=\left(1-\frac{r_g}{r}\right) dt^2-\left(1-\frac{r_g}{r}\right)^{-1} dr^2-r^2\left(d\theta^2+sin^2\theta d\phi^2\right)\]​

and substitute \theta = \frac{\pi}{2}, which would yield a tidy result. I have to solve this explicitly, starting with 2 spatial dimensions instead of 3, calculating Ricci tensor and equating its components to zero: R_{\mu\nu}=0. Now, that does not seem to be very hard, but the results obtained are not satisfactory.

Starting with metric in the general form (see http://en.wikipedia.org/wiki/Deriving_the_Schwarzschild_solution" for defining analogous 4-dimensional metric) \[ds^2=e^{\nu(r)} dt^2-e^{\lambda(r)} dr^2-r^2d\phi^2, \]
I calculate the Ricci tensor (prime denoting differentiation with respect to r)

R_{00}=e^{\nu-\lambda} \left(\frac{1}{2}\nu''+\frac{1}{4}\nu'^2-\frac{1}{4}\nu'\lambda'+\frac{1}{2}\frac{\nu'}{r} \right)=0
<br /> R_{11}=-\frac{1}{2}\nu&#039;&#039;-\frac{1}{4} \nu&#039;^2 +\frac{1}{4}\nu&#039;\lambda&#039;+\frac{1}{2}\frac{\lambda&#039;}{r}=0 \nonumber
R_{22}=\frac{1}{2}re^{-\lambda}(\lambda&#039;-\nu&#039;)=0​

Solving this gives

\lambda&#039;+\nu&#039;=0 \nonumber
\lambda&#039;-\nu&#039;=0 \nonumber​

or \lambda&#039;=\nu&#039;=0. Applying the condition for r \to \infty: \; e^\nu \to 1, \ e^\lambda \to 1, the metric turns out to be Euclidean:

\[ds^2= dt^2- dr^2-r^2d\phi^2\]​

which does not seem right at all.

Note that the Ricci tensor is completely determined by metric tensor. Thus, once the metric in general form is defined - with e^\nu and e^\lambda, one inevitably finds himself at differential equations resulting in flat space-time (or so it seems); and I don't see any way to define metric differently.

I've created a text with a little more details (in attached file), comparing the 4- and 3-dimensional derivations in parallel. Hope it would somehow help.

I'm sort of stuck. Any ideas are greatly appreciated.

 

[bcrowell]

For insight, you can automatically write down the weak-field version based on the Newtonian theory. For Newtonian gravity in 2 spatial dimensions, the field lines spread out radially in a plane, so the density of field lines falls off like 1/r. That means that Newton's law of gravity becomes F=GMm/r. In the weak-field limit, the time part of the line element has to be 1+\Phi, where \Phi is the gravitational potential. Let G=M=m=1 for convenience. Then \Phi=\ln r. That means your metric should have the form (1+\ln r)dt^2+\ldots in the limit of large r.

 

[bcrowell]

Here is a discussion of the Schwarzschild solution in an arbitrary number of dimensions d: http://relativity.livingreviews.org/open?pubNo=lrr-2008-6&page=articlese5.html They're only interested in d=4 or greater. It's pretty clear that the form of their solution becomes incorrect for d=3.

If you look at section 1 of the same article, I think I see the answer to your problem. If black hole solutions exist in dimension d, then "black string" solutions exist in dimension d+1. Your calculation seems to prove that there aren't black hole solutions in d=2. This seems right to me, because I've never heard of black string solutions in d=3.

 

[bcrowell]

This is a fun example -- thanks for posting it!

I've tried to confirm the nonexistence result by a different calculation, using the open-source symbolic math program Maxima. The following Maxima code calculates the Ricci tensor for a metric with g_tt=h and g_rr=k.

load(ctensor);
dim:3;
ct_coords:[t,r,phi];
depends(h,r);
depends(k,r);
lg:matrix([h,0,0],
          [0,-k,0],
          [0,0,-r^2]);
cmetric();
ricci(true);

Inspecting R_phiphi, we see that we need k'/k=h'/h. Since the logarithmic derivatives of h and k are they same, the two functions can differ by at most a constant factor c. So now we do a second iteration of the calculation:

load(ctensor);
dim:3;
ct_coords:[t,r,phi];
depends(h,r);
lg:matrix([h,0,0],
          [0,-c*h,0],
          [0,0,-r^2]);
cmetric();
ricci(true);

The result for R_rr is independent of c, and is positive for large r, so there's no way we can get a vacuum solution.

 

[Zinger0]

For insight, you can automatically write down the weak-field version based on the Newtonian theory. For Newtonian gravity in 2 spatial dimensions, the field lines spread out radially in a plane, so the density of field lines falls off like 1/r. That means that Newton's law of gravity becomes F=GMm/r. In the weak-field limit, the time part of the line element has to be 1+\Phi, where \Phi is the gravitational potential. Let G=M=m=1 for convenience. Then \Phi=\ln r. That means your metric should have the form (1+\ln r)dt^2+\ldots in the limit of large r.

Thanks for suggestion!

But I have some questions. First, I'm rather convinced that Newton's law of gravity should be F=GMm/r^2 in a plane, same as in 3-dim. space. That is because: 1). what matters is the distance from the source, which is r in either case. 2). just the units analysis. Force should have the same units disregarding the dimensionality of space.
The density of the field lines indeed falls off like 1/r in flat space; that is if we define the density as \frac{N}{2\pi r}, where N is the number of field lines piercing the circle of radius r. In 3-dim. it falls like 1/r^2 because it's defined as \frac{N}{4\pi r^2}. If we consider 3-dim. gravity and restrict ourselves to some plane within it, we should obtain the same result as in 2-dim.

Nevertheless your idea is very helpful: if Newton's law is really the same in 2 as in 3 space dimensions, than gravitational potential \Phi(r) is proportional to -1/r: \Phi(r) \propto -\frac{1}{r} in both instances. Therefore the metric in 2-dim. space would be
(1-\frac{const}{r})dt^2+\dots, just as in regular Schwarzschild, which seems to be reasonable to me. The problem is to prove it.

Here is a discussion of the Schwarzschild solution in an arbitrary number of dimensions d: http://relativity.livingreviews.org/open?pubNo=lrr-2008-6&page=articlese5.html They're only interested in d=4 or greater. It's pretty clear that the form of their solution becomes incorrect for d=3.

If you look at section 1 of the same article, I think I see the answer to your problem. If black hole solutions exist in dimension d, then "black string" solutions exist in dimension d+1. Your calculation seems to prove that there aren't black hole solutions in d=2. This seems right to me, because I've never heard of black string solutions in d=3.

Seems to be highly relevant link (thanks again!), but quite a lengthy read, so I'll post comments a bit later. However, what I can say now is that Schwarzschild solution pertains not only to black holes, but any metric of massive spherically (cylindrically) symmetric body in vacuum. No one has heard of black strings (or, rather, disks) or start just because we live (apparently) in 3-dim space. But I don't see how anything should change if we consider a plane in that space. What I obtain with my computations is Euclidean flat space-time metric; not likely if anything massive is present.

 

[PAllen]

Isn't the BTZ black hole a 2+1 black hole? Search in arxiv and there are hundreds of papers on this.

 

[Zinger0]

Isn't the BTZ black hole a 2+1 black hole? Search in arxiv and there are hundreds of papers on this.

This is GREAT! Just what I was looking for! Did not manage to google it myself, not for the lack of trying.

Just to be absolutely sure: do I understand correctly that Schwarzschild solution in 3-dimensional space-time without cosmological constant should be flat (just as I've obtained)?

 

[PAllen]

This is GREAT! Just what I was looking for! Did not manage to google it myself, not for the lack of trying.

Just to be absolutely sure: do I understand correctly that Schwarzschild solution in 3-dimensional space-time without cosmological constant should be flat (just as I've obtained)?

I don't know, I've never tried to work it out, or research this specifically.

 

[Dickfore]

What are the Einstein's equations in 3-dimensional space-time?

 

[bcrowell]

Just to be absolutely sure: do I understand correctly that Schwarzschild solution in 3-dimensional space-time without cosmological constant should be flat (just as I've obtained)?

Yes. You and I both proved it, by two different methods. You can also confirm it by looking here: http://en.wikipedia.org/wiki/BTZ_black_hole

But I have some questions. First, I'm rather convinced that Newton's law of gravity should be F=GMm/r^2 in a plane, same as in 3-dim. space. That is because: 1). what matters is the distance from the source, which is r in either case. 2). just the units analysis. Force should have the same units disregarding the dimensionality of space.

G would have different units in 2+1 dimensions. Force would have the same units. Gravitational forces definitely cannot fall off as 1/r^2 in d=3 or d>4. This is totally accepted and well established. That's why, for example, you can get microscopic black holes at the LHC if there are large extra dimensions: Giddings and Mangano, "Comments on claimed risk from metastable black holes," http://arxiv.org/abs/0808.4087 You can also see this from the Schwarzschild–Tangherlini metric for d>=4 in the Living Reviews article I linked to. They have g_{tt}=(1-r^{-(d-3)}) (taking GM=1 and converting to the +--- metric that you and I have both been using). In the weak field limit, this is essentially the gravitational potential. Differentiation gives the gravitational field, which varies as r^{-(d-2)}. This shows that the radial variation of the field strenth has a different exponent depending on d. (It doesn't work for d=3, because there aren't black-hole solutions for d=3.)

If we consider 3-dim. gravity and restrict ourselves to some plane within it, we should obtain the same result as in 2-dim.

This is incorrect. The only way to match up results in d and d+1 dimensions is if the spherically symmetric result in d dimensions is a cylindrically symmetric result in d+1 dimensions. Otherwise you don't have a solution to the vacuum field equations in both cases.

Nevertheless your idea is very helpful: if Newton's law is really the same in 2 as in 3 space dimensions, than gravitational potential \Phi(r) is proportional to -1/r: \Phi(r) \propto -\frac{1}{r} in both instances. Therefore the metric in 2-dim. space would be
(1-\frac{const}{r})dt^2+\dots, just as in regular Schwarzschild, which seems to be reasonable to me. The problem is to prove it.

We've established conclusively (your calculation, my calculation, and WP) that this is *not* the case. We know that there is no such solution for d=3.

 

[Zinger0]

I don't know, I've never tried to work it out, or research this specifically.

Thanks anyway! Well, it seems to be flat metric after all.

 

[Zinger0]

What are the Einstein's equations in 3-dimensional space-time?

They are still R_{\mu \nu} - {1 \over 2}g_{\mu \nu}\,R = {8 \pi G \over c^4} T_{\mu \nu} (if we ignore cosmological constant), only \mu, \nu are now restricted to 0..2 instead of 0..3. In Schwarzschild's case we consider vacuum solution, so T_{\mu \nu}=0 and, therefore,
R_{\mu \nu}=0.

 

[Passionflower]

They are still R_{\mu \nu} - {1 \over 2}g_{\mu \nu}\,R = {8 \pi G \over c^4} T_{\mu \nu} (if we ignore cosmological constant), only \mu, \nu are now restricted to 0..2 instead of 0..3. In Schwarzschild's case we consider vacuum solution, so T_{\mu \nu}=0 and, therefore,
R_{\mu \nu}=0.

Sorry, but how can you state what the Einstein equations are for 3-dimensional spacetime? 3 dimensional spacetime does not exist.

Of course one can do pure mathematics but I don't think it makes any sense to call something Einstein's equations for situations that are clearly not physical.

After all Einstein was a theoretical physicist not a mathematician.

 

[Zinger0]

Just to be absolutely sure: do I understand correctly that Schwarzschild solution in 3-dimensional space-time without cosmological constant should be flat (just as I've obtained)?

Yes. You and I both proved it, by two different methods. You can also confirm it by looking here: http://en.wikipedia.org/wiki/BTZ_black_hole

Indeed we've proved it, but I can't quite get the hold of it.

First, how can the metric be flat, when there is a massive body, forming the gravitational field and therefore curving the metric?

Second, is the gravitational force in a plane really different from it in 3d space? I mean, if we consider some classical gravitational problem involving two bodies, which could be taken as points, moving in a plane, what force law should we choose - F~1/r^2 or F~1/r?
The metric in Newtonian limit for those force laws would be ds^2=(1-\frac{const}{r})dt^2+\dots and ds^2=(1+\ln r)dt^2+ \dots respectively. In any event, this is not flat ds^2=1\cdot dt^2 + \dots. How can we explain this?

 

[Passionflower]

Indeed we've proved it, but I can't quite get the hold of it.

First, how can the metric be flat, when there is a massive body, forming the gravitational field and therefore curving the metric?

You asking physics questions about models that are non-physical and then you try to get a hold of it?

A massive body in two dimensions?
You are simply not doing physics.

 

[Ben Niehoff]

The solution looks correct. Gravity is trivial in 2+1 dimensions. All singularities are conical singularities. While your metric does look flat, remember that it might have a conical point if you choose the range of the angular variable to be something other than [0,2pi).

 

[Zinger0]

Sorry, but how can you state what the Einstein equations are for 3-dimensional spacetime? 3 dimensional spacetime does not exist.

Of course one can do pure mathematics but I don't think it makes any sense to call something Einstein's equations for situations that are clearly not physical.

After all Einstein was a theoretical physicist not a mathematician.

Let me first state that I don't claim this treatment to be anything other than pure math.

That being said, however, given approach has a certain value. I give a citation from http://www.math.columbia.edu/~woit/wordpress/?p=555":

[the motivation to consider such model is] ... to study fully quantum black holes in an exactly solvable toy model. There’s no exactly solvable model in 3+1d, and 1+1d is too simple, so that leaves 2+1d.

But I can't really defend the model, because I've just learned about it. For serious discussion see the papers in the link to Wiki page on http://en.wikipedia.org/wiki/BTZ_black_hole" and others mentioned above.

 

[bcrowell]

First, how can the metric be flat, when there is a massive body, forming the gravitational field and therefore curving the metric?

What you've proved is that massive singularities can't exist in 2+1 dimensions.

 

[bcrowell]

Sorry, but how can you state what the Einstein equations are for 3-dimensional spacetime? 3 dimensional spacetime does not exist.

By the same logic, we can't state the Einstein field equations for a universe with a zero cosmological constant, since our own universe has a nonzero one.

BTW, we actually don't know the dimensionality of our universe. It could be greater than 4: http://en.wikipedia.org/wiki/Large_extra_dimensions

Zinger0 has correctly stated the vacuum Einstein field equations in 2+1 dimensions, in the sense that this is what every physicist working on this sort of thing takes them to be, and also in the sense that they are the only sensible generalization.

 

[Zinger0]

The solution looks correct. Gravity is trivial in 2+1 dimensions. All singularities are conical singularities. While your metric does look flat, remember that it might have a conical point if you choose the range of the angular variable to be something other than [0,2pi).

That sounds a bit too advanced for me. You mean if I restrict the range of \phi to something less than the whole interval [0,2pi)?

What you've proved is that massive singularities can't exist in 2+1 dimensions.

But the Schwarzshcild solution should apply not only to singularities, but to any symmetrical bodies in vacuum. Probably the flatness of metrics tells us there could be no massive bodies in 2d? That would also solve the problem of Newtonian limit.

 

[bcrowell]

That sounds a bit too advanced for me. You mean if I restrict the range of \phi to something less than the whole interval [0,2pi)?

It's not a hard idea. If you draw the diagram showing Euclid's proof of the Pythagorean theorem on a piece of paper, the proof is just as valid when you roll the paper up into a cone or a cylinder. That means that cones and cylinders don't have intrinsic curvature. An ant can't detect the curvature by measuring lengths and angles on the paper. But at the tip of the cone, the ant will notice that if he walks around the tip, his orientation comes out funny. If you take a piece of circular paper and cut out a wedge, you can then bend it into a cone.

But the Schwarzshcild solution should apply not only to singularities, but to any symmetrical bodies in vacuum. Probably the flatness of metrics tells us there could be no massive bodies in 2d? That would also solve the problem of Newtonian limit.

This kind of stuff is discussed in the two review articles by Carlip linked to from the WP article http://en.wikipedia.org/wiki/BTZ_black_hole . I've only skimmed them so far, so I'm not sure if your generalization of my statement is right, but I'd guess that it is.

 

[Dickfore]

They are still R_{\mu \nu} - {1 \over 2}g_{\mu \nu}\,R = {8 \pi G \over c^4} T_{\mu \nu} (if we ignore cosmological constant), only \mu, \nu are now restricted to 0..2 instead of 0..3. In Schwarzschild's case we consider vacuum solution, so T_{\mu \nu}=0 and, therefore,
R_{\mu \nu}=0.

It would be more parenthetic to start from 4-d spacetime, but nothing should depend on one of the spatial coordinates, say x₃. This would correspond to everything being translationally invariant along the x₃ axis. See if you can find the general form of a metric in this case. Then, if you write down the (4-d) Einstein's equations, you might get something different than by the method you were using.

Another option is to place x_{3} = 0 everywhere. This would correspond to the existence of 1/r^{2}-law even in 3 space-time dimension, as someone was already suggesting, but it is a completely different physical situation.

In the first case, as circular distribution of matter is equivalent to a uniform cylinder in space, whereas in the second case it is truly a disc.

 

[Zinger0]

If you take a piece of circular paper and cut out a wedge, you can then bend it into a cone.

I think I'm starting to get the general idea.

This kind of stuff is discussed in the two review articles by Carlip linked to from the WP article http://en.wikipedia.org/wiki/BTZ_black_hole . I've only skimmed them so far, so I'm not sure if your generalization of my statement is right, but I'd guess that it is.

I'll have to do some research on interpretation, but at least the solution itself is correct.

Thanks for everybody's help! I would never consider flat metric anything but a blunder.

 

[Zinger0]

It would be more parenthetic to start from 4-d spacetime, but nothing should depend on one of the spatial coordinates, say x₃. This would correspond to everything being translationally invariant along the x₃ axis. See if you can find the general form of a metric in this case. Then, if you write down the (4-d) Einstein's equations, you might get something different than by the method you were using.

Another option is to place x_{3} = 0 everywhere. This would correspond to the existence of 1/r^{2}-law even in 3 space-time dimension, as someone was already suggesting, but it is a completely different physical situation.

In the first case, as circular distribution of matter is equivalent to a uniform cylinder in space, whereas in the second case it is truly a disc.

If I understand correctly, you suggest solving 4-dimensional spacetime equations omitting one of spherical coordinates. But that would not exactly yield a required (2+1) metric. Apparently, the "physics" in general is different when applied to space of different number of dimensions. Thus, dealing with such spaces one can not rely on any solutions for spaces of different dimensionality: all the quantities should be calculated from scratch using the most fundamental equations (EFE in the form written above) and general assumptions.

As it turns out, physics on a plane within 3d space is not the same is physics on a plane in 2d.

 

[Dickfore]

If I understand correctly, you suggest solving 4-dimensional spacetime equations omitting one of spherical coordinates.

No. When you solve for the Schwartzshild metric in 4-d spacetime, you actually omit 2 of the 3 spherical coordinates, namely the polar and azimuthal angles.

 

[Zinger0]

No. When you solve for the Schwartzshild metric in 4-d spacetime, you actually omit 2 of the 3 spherical coordinates, namely the polar and azimuthal angles.

By omitting coordiante I've meant your suggestion

Another option is to place x_{3} = 0

.
In my reply I was referring to that approach.

 

[Dickfore]

In my reply I was referring to that approach.

Yes, but I gave another option as well.

 

[Zinger0]

Yes, but I gave another option as well.

You mean, to make x_3 invariant? But both setting it to zero and making it invariant would still keep us in 3d space (where we no longer want to be). So, I rather was calling 'omitting' either of those approaches.
Actually, I've been trying to do the same at first. But I was told that such derivation would be incorrect.

 

[Dickfore]

Ok, I said what I had to say. Play with (1+2)-spacetime. However, it is a mathematical exercise with no connection to Physical Reality.

Just as a sidenote. If the Ricci tensor:

<br /> R_{\mu \nu} = 0<br />

is zero, it still does not necessarily mean space-time is flat. There is something called Riemann curvature tensor R^{\mu}{}_{\nu \rho \pi}, which is a 4-fold tensor. This one has to be equal to zero in order for space-time to be flat.

As you vary the dimensions, the number of independent components of the two change. In 2-d (total number of dimensions), they have the same number and are uniquely expressible between each other. There, it would mean that space-time is flat. However, in higher number of dimensions, I think there are more components to the Riemann tensor, then there are to the Ricci tensor and no such unique relationship exists.

 

[Zinger0]

Ok, I said what I had to say. Play with (1+2)-spacetime. However, it is a mathematical exercise with no connection to Physical Reality.

Just as a sidenote. If the Ricci tensor: R_{\mu \nu} = 0 is zero, it still does not necessarily mean space-time is flat.

Well, I've already made a disclaimer about its real-world value. Nevertheless, it might prove helpful as an analogy for a more complex 3d world.
As for the Ricci tensor, you are right: in an ordinary (4-d spacetime) Schwarzschild solution R_{\mu \nu} = 0 too, but the metric is not flat.

 

[Stingray]

As Dickfore points out, the 3D Einstein equation has a completely different physical character than the 4D one. All degrees of freedom in the curvature are locally determined by the matter distribution. This means, for example, that there can be no gravitational radiation. And there can be no analog of the Schwarzschild solution.

What might be more interesting is to look at a 2+1 slice of a static, vacuum, cylindrically symmetric 3+1 solution. This describes the Levi-Civita metric:
<br /> ds^2 = - \rho^{4 \alpha} d t^2 + \rho^{4 \alpha (2 \alpha-1)} ( d \rho^2 + d z^2 ) + \beta^{-2} \rho^{2-4\alpha} d \phi^2 .<br />
\alpha can be interpreted as a mass/length and \beta is an angular defect parameter.