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Einstein field equations have vacuum solutions that (probably) assumes the source of curvature (either energy-momentum tensor or the cosmological constant term or both) is elsewhere. Like, in Schwarzschild solutions we assume that the source is a star or a planet or even black hole singularity - but it is there somewhere. Now - can we prove that there must be a source somewhere in 4D space-time? In 2d the Einstein tensor is zero - thought we can have non-zero Riemann and therefore curvature - and a zero Einstein Tensor necessarily means no source is allowed, the sum of energy-momentum tensor and cosmological constant term must be zero.

Qmechanic
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Nayeem1
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The fact that we have Ricci-flat solutions means we have solutions to the field equations $G_{ab}=8\pi T_{ab}$ with $T_{ab}$ identically vanishing. In other words, we truly have non-trivial vacuum solutions to the field equations (even if we assume all the nice initial and boundary conditions, like a smooth initial hypersurface, asymptotically flat etc). This partly has to do with the non-linear dependence of the Ricci tensor on the metric and its derivatives. I’m not sure what you mean by “assumes the source of curvature is elsewhere”.

For example, thinking of the curvature in Schwarzschild as being sourced by the blackhole region doesn’t make sense, because the blackhole region is still perfectly smooth and the Ricci tensor vanishes identically there. Also, trying to ascribe the non-zero curvature to the $r=0$ singularity, doesn’t really seem to be a good solution. Note that Schwarzschild solutions are a 1-parameter family labelled by $M$; which we interpret as mass. This is a geometric feature of the spacetime itself, and this is the reason why it is curved. Note that when $M=0$ we get Minkowski, where $r=0$ is suddenly no longer a singularity of spacetime (it’s simply a breakdown of polar coordinates) and this is a nice perfectly flat spacetime.

Another reason why trying to ascribe the source of curvature to $r=0$ is bad is that if for example you look at the (maximal globally hyperbolic developments, not the maximal analytic extensions of) Kerr spacetimes, then they do not possess such a singularity. They have so-called Cauchy-horizons which is the boundary of the region where the spacetime is globally hyperbolic (roughly speaking, the region of spacetime that is uniquely determined from the initial data), but as a Lorentzian manifold itself, one can smooth extend beyond these Cauchy-horizons. Again, the Kerr-spacetimes are Ricci-flat everywhere, and sicne there are no singularities like in Schwarzschild, we now have to really confront the fact that the curvature is due to the spacetime itself (particularly the two parameters $M,a$ of mass and specific angular momentum).

peek-a-boo
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    Does that imply it’s not necessary to have a source of gravity? – Nayeem1 Oct 29 '23 at 05:18
  • @Nayeem1 I haven't read the question or the answer, but in a way, in modified theories of gravity $T=0$ does not imply $r=0$. So there are subtleties, but GR is the more viable approach -- so I would expect a vacuum solution to necessarily imply $R=0$. – VaibhavK Oct 29 '23 at 07:09
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    @Nayeem1 yes. Typically we take ‘source’ to mean $T_{ab}$. So, (as mentioned in the other answrr) unless you mean ‘source’ to mean something completely different. Maybe I should have emphasized more in my answer, but you can think of the non-trivial solutions as being ascribed to the non-trivial initial data as well (even if the initial data is nice and smoothly behaved). As a very trivial example, consider the ODE $f’’=0$. If the initial conditions are $f(0)=f’(0)=0$, then $f=0$. However, if we require $f(0)=0,f’(0)=1$, then the solution is $f(t)=t$ for all $t$. – peek-a-boo Oct 29 '23 at 21:21
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    So, we have a non-trivial (smooth) solution to our evolution problem even though the ODE were solving is $f’’=0$ (i.e the RHS is $0$, i.e ‘no sources’). Maybe this is too simple, so you might want to think about the linear wave equation $\partial_t^2\psi-\Delta\psi=0$ on $\Bbb{R}^{1+n}$, and ask about solving this source-free PDE with prescribed initial conditions $\psi(0,\cdot)=\psi_0$ and $\partial_t\psi(0,\cdot)=\psi_1$. These are of course very naive and simple remarks, and GR has other technicalities, but I think this serves as a decent initial mental model. – peek-a-boo Oct 29 '23 at 21:22
  • Schwarzchild metric uses M - but in another universe (following EFE) M can just be a constant - having nothing to do with real mass. And we can definitely find many other vacuum EFE solutions with non-zero Riemann. So, does that mean the source (stress-energy tensor and/or cosmological constant) is not a necessity? After all, we can have a universe with 0 energy, 0 stress-energy tensor, 0 lambda and still that universe can have non zero Riemann. – Nayeem1 Oct 30 '23 at 03:20
  • Yes exactly; we have Ricci-flat metrics which are not Riemann flat. And my point is that if you want to think ‘dynamically’ of where this is coming from, then it’s in the initial conditions, that’s what the previous two comments are about. – peek-a-boo Oct 30 '23 at 03:22
  • @Nayeem1 I think it's helpful to draw a comparison to Maxwell's equations in basic E&M. These relate electromagnetic fields to sources (charges/current), but they also admit nontrivial eternal vacuum solutions (e.g. plane waves). That is, Maxwell's equations in principle allow for a universe with propagating light waves that never came from any sources, and GR is much the same in this respect. In our practical universe, however, we have pretty much always been able to trace back every electromagnetic wave we've seen to a source, and this has been our experience with gravity as well. – jawheele Oct 30 '23 at 14:43
  • @jawheele I think vacuum solutions are valid only in some part of spacetime. Gravitational waves must have somewhere their source, means somewhere there is a non zero energy stress tensor. – JanG Oct 31 '23 at 19:04
  • @JanG "Gravitational waves must have a source" is certainly a reasonable expectation for the universe's "real" spacetime manifold based on our experience, but that does not change the fact that the Einstein equations admit global nontrivial solutions that have no stress-energy anywhere, just like Maxwell's equations admit solutions with nontrivial E&M fields that have no charges/currents anywhere. Such solutions can be useful idealizations, just like plane waves are useful idealizations in studying EM waves. – jawheele Oct 31 '23 at 19:46
  • @jawheele I understand your argumentation but we should first agree to what is solution of Einstein field equations (EFE). If you just mean a metric that components satisfy EFE without satisfying the corresponding physical boundary conditions then in my view it is not a true solution of EFE. – JanG Nov 01 '23 at 18:59
  • @JanG Though perhaps not standard parlance, that's a reasonable perspective to take. I'd be surprised if your idea of physical boundary conditions was so stringent that it ruled out (say) the vacuum evolution of well-behaved perturbations to Minkowski initial data, though. In any event, I've used "solution" here in the same sense that we call a plane wave a solution to Maxwell's equations (namely, in spite of the boundary conditions' being clearly nonphysical). – jawheele Nov 01 '23 at 21:00
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Is source of space-time curvature necessary?

This is largely a matter of definitions: What kind of “sources” do we admit and where exactly is this “somewhere” can be?

When we are looking at the metric of a star, the stress–energy tensor of the star's matter would be the “source” of curvature. But what about spacetimes containing only gravitational waves? An example of such spacetime would have zero stress–energy tensor everywhere inside the spacetime yet such a spacetime would satisfy nontrivial boundary conditions at infinity. Can we call these boundary conditions “source of curvature”? If we answer this question in the affirmative we can then recast specific boundary conditions as a form of “boundary matter” and associate some boundary stress–energy tensor to them.

This program is outlined in paper:

Note, that when we have such boundary matter as “sources” then we would realize a version of Mach's principle very close to original formulation by Einstein, where metric (and thus curvature) is uniquely specified by the stress–energy tensor, now understood as containing both “bulk” and “boundary” parts.

While “boundary matter” was quite uncommon during the first half-century of GR development there are many examples nowadays: black hole membrane paradigm, Brown–York boundary stress–energy, brane–world models, Hořava–Witten end-of-the-world brane, holographically dual matter …

A.V.S.
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  • Upvote from me. A physical solution of Einstein field equations is always a solution to the physical boundary value problem, i.e. solution to the differential equation that also satisfies the imposed physical boundary conditions. The last are defined by the right side of Einstein equation in defining value of the energy stress tensor on some boundary. – JanG Oct 29 '23 at 15:50
  • +1 This is very interesting! My GR is a bit rusty. I'm not sure I yet fully believe in treating boundary conditions as boundary sources in the case of an arbitrarily far away boundary. In the case of asymptotically flat spacetimes, if you take the boundary arbitrarily far out, will the corresponding boundary stress energy tensor go to zero? What about the case of gravitational plane waves, where the metric is not asymptotically flat? – user196574 Oct 30 '23 at 04:53
  • Also, from other questions you've answered, you might be able to comment on my question about asymptotically flat metrics without curvature singularities, as well as on John Rennie's interesting answer. There I'm nervous that there is some singularity still lurking in the geon solutions as some appropriate time coordinate goes to plus or minus infinity. My GR knowledge is lacking, so I worry my understanding is not even wrong. – user196574 Oct 30 '23 at 06:32
  • @user196574 Just a short remark. At least for the case of static perfect fluid sphere the asymptotically flatness of spacetime results from the boundary condition $p(R)=0$ on the star surface - see the equations in https://physics.stackexchange.com/a/679431/281096. – JanG Oct 30 '23 at 16:25
  • @user196574 if you take the boundary arbitrarily far out, will the corresponding boundary stress energy tensor go to zero? Yes, but some charges (obtained from integrating over “sphere at infinity”) might be diverging, so we would need to keep regularization parameter, ($r_0$ the size of sphere) and then do some kind if renormalization to obtain finite physically meaningful quantities. – A.V.S. Oct 31 '23 at 07:53
  • @A.V.S. "Goes to zero" means "never zero" like the function ${e}^{-x}$ for $x\rightarrow \infty$, right? – JanG Oct 31 '23 at 18:55
  • @JanG Right, only polynomially in inverse radius, like $\mathcal{O} (R_0^{-1})$ rather than exponentially. – A.V.S. Nov 01 '23 at 18:16
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There must be a "source," i.e. a mass or the metric reduces to the 3+1 flat-space metric.

LiveProton
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  • Can you prove that? – Nayeem1 Nov 04 '23 at 09:35
  • Your comment leads me to believe you cannot do math. Without mass in the Schwarzschild metric, it reduces to the flat-space metric. You should demonstrate this to yourself – LiveProton Nov 04 '23 at 16:57
  • Your comment leads me to believe that you didn’t understand exactly how Schwarzchild metric is derived. M in that metric can be thought of just a constant - nothing in it demands it to be actual mass. And you can find countless other solutions without M. – Nayeem1 Nov 11 '23 at 16:05