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Schwarschild found his solution to Einstein's field equations for vacuum ($T_{\mu \nu}=0$) by placing a point-mass in the center of origin.

Since the Ricci tensor $R_{\mu \nu}$ and the Einstein tensor $G_{\mu \nu}$ are trace-reversed, meaning they vanish identically:

Which point of a universe with a mass in it would meet this equation: $R_{\mu \nu}=0$ (and therefore $G_{\mu \nu}=0$)?

(to make sure, infinity is not a point, i.e. if gravity is infinite than you can't find a particular point, where it vanishes down to zero due to distance)

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    $\begingroup$ Not really sure what you're asking. The Einstein tensor is zero everywhere except for the worldline of the point mass. $\endgroup$ – Muphrid Mar 29 '14 at 20:18
  • $\begingroup$ The Einstein tensor denotes gravitational curvature. What you say means that there is no gravity in the universe except within matter. $\endgroup$ – bright magus Mar 30 '14 at 7:48
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    $\begingroup$ I see you have this misconception that I will address with an answer. $\endgroup$ – Muphrid Mar 30 '14 at 8:09
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Gravitational curvature is not fully described by the Einstein or stress-energy tensors. It is only fully described by the Riemann tensor.

Consider, by analogy, a simple example from electromagnetism: $\nabla \cdot E =\rho/\epsilon_0$, Gauss's law for the electric field. The function $\rho$ describes the distribution of charge density. The LHS is zero wherever there is no charge density, but that does not mean the electric field is zero.

Similarly, in regions of vacuum, the stress-energy tensor is zero, and thus so is the Einstein tensor, but that does not mean the Riemann tensor is zero!

Indeed, the Riemann tensor can be decomposed into two parts: a traceful part that corresponds to the Ricci or stress-energy tensors but also a traceless, self-dual part, the Weyl tensor, which can be nonzero even in empty space (space without energy-momentum).

So Einstein's equations in vacuum mean exactly that: that $G_{\mu \nu} = 8\pi T_{\mu \nu} = 0$ in a region without mass-energy. That is far from saying that there is no gravity, just as it would be silly to say there is no electric field in the exterior of a charged ball.

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  • $\begingroup$ In our case, when I say "gravity" I mean "attraction" between particular mass bodies. If there is no Einstein tensor (resulting from energy-momentum) outside of mass, how do material bodies interact with each other gravitationally? Only "through" the Weyl tensor? How come it's not included in Einstein field equations? $\endgroup$ – bright magus Mar 30 '14 at 8:33
  • $\begingroup$ Because the Einstein field equations are not a complete system. You also need the Bianchi identity. $\endgroup$ – Robin Ekman Mar 30 '14 at 8:55
  • $\begingroup$ OK, I would like to have a clear picture of gravitational interaction between bodies, so if you could clarify the Weyl tensor issue in brief, please? Is the Weyl tensor (the only one) responsible for the space curvature (non-zero Riemann tensor) that locally pulls one body toward another? Or the one that makes light bend? And what is the physical meaning of Bianch identity that also plays a role in here, as you say? $\endgroup$ – bright magus Mar 30 '14 at 9:19
  • $\begingroup$ And, by the way - if you say that the Einstein tensor is confined to matter and therefore should depend on matter "generated" stress-energy only, than the cosmological constant (along with a spacetime metric) included in it looks like a total mess to me. Seems that the cosmological constant should be outside of this equation as much as the Weyl tensor, right? But again, I must be missing something ... (I'm not sure of the forum rules - perhaps I should pose a separate question/thread concerning this? $\endgroup$ – bright magus Mar 30 '14 at 9:41
  • $\begingroup$ Personally, I don't regard the Einstein equation as the full equation that describes the curvature of spacetime. To me, that equation is the equation for the Riemann tensor, but there are good reasons it's seldom used: working with the full Riemann tensor means you have to worry about 20 components instead of 10, after all, and since most people just work with the metric and its 10 components, working with the Einstein tensor is often just easier. - As far as the cosmological constant goes, it's regard it as a source of "dark energy" and just move it to the RHS with the stress-energy. $\endgroup$ – Muphrid Mar 30 '14 at 16:21
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The vacuum equation $G_{\mu\nu} = 0$ holds wherever there is vacuum. This is tautological so I don't really see what the content of the question is.

If you are hesiant about point masses you can consider the Schwarschild interior solution, which describes a spherically symmetric non-rotating star with uniform density. The Schwarschild interior solution, matched with the exterior solution, has $R_{\mu\nu} = 0$ everywhere outside the star. I don't have nearby right now so I can't give a precise page number, but the standard reference here is Exact Solutions to Einstein's Field Equations by Stephani et al.

To deal with a point mass you need to weaken your demands on smoothness and such, but this is not unique to general relativity, it's the same in Newtonian gravitation and classical electrodynamics.

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  • $\begingroup$ "The vacuum equation Gμν=0 holds wherever there is vacuum." You mean that matter doesn't make (vaccum) space curve? (There are 2 vacuums actually: one is understood as vacuum at particular point of space, and the other as the whole universe devoid of matter - I'm not sure which one you mean) $\endgroup$ – bright magus Mar 29 '14 at 17:32
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In 3+1 dimensions, the Ricci tensor vanishes when the stress-energy tensor vanishes indeed. Which means that, whenever there's a vacuum, the Ricci tensor vanishes as well.

But it also happens that, in that number of dimension, the metric is not entirely determined by the Ricci tensor. The full curvature tensor (the Riemann tensor) is a mixture of the Ricci tensor and the Weyl tensor. The Weyl tensor roughly describes how gravity propagates : if you have some energy, it will influence the metric of the vacuum regions around it.

I emphasize the 3+1 dimensions because that is not the case in 2+1 dimensions (gravity stops in a vacuum, because the Riemann tensor is $\propto$ the Ricci tensor) and even less so in 1+1 dimensions (the Einstein tensor is a topological invariant, and does not vary with the stress energy tensor).

In the case of Schwarzschild, what you have is a distribution of energy in the middle. There's several ways to describe it. You can use distributions, but Schwarze distributions handle non-linear equations poorly, and non-linear distributions are a rather thorny topic. But there's still some ressources talking about how to describe a singularity as a Dirac distribution. It has the benefit of being rather clear : The stress energy tensor is a Dirac, and so is the Ricci tensor.

You can show it indirectly, as well, by calculating the Komar mass. You find that, in an arbitrarily small region around the singularity, there's a mass M. The Ricci tensor is, as usual, 0 outside of the singularity and undefined at r = 0.

Of course you can also just have a plain old spherical body. In that case, the metric is just Schwarzschild outside (with energy density 0) and whatever may be inside, with, hopefully, a continuity in the metric between the inside and the outside.

A good way to illustrate how the curvature of space does not depend entirely on the Ricci tensor is to look at empty space. The simplest vacuum spacetime is Minkowski space, but it is not the only one. You can freely add sourceless gravitational radiations (using for instance pp-wave metrics), which will also be a vacuum solution.

In the case of the Schwarzschild spacetime, the stress energy tensor is not enough to get the full metric because you are not looking at all the spacetime. You are only looking at its vacuum regions without looking at its energy distributions, which can influence other regions. In the case of gravitational waves, it's a boundary value problem : solving the Einstein equation still requires, like all PDEs, some boundary value of the field.

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  • $\begingroup$ "You can show it indirectly, as well, by calculating the Komar mass. You find that, in an arbitrarily small region around the singularity, there's a mass M. The Ricci tensor is, as usual, 0 outside of the singularity and undefined at r = 0." - You mean that this mass M does is not a source of any stress energy? $\endgroup$ – bright magus Mar 30 '14 at 9:59
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There is no per se 'point'. The entire 'mass' of the universe defines itself, no matter how we might perceive it, as an infinite density with no locality. In other words, the entire universe is, always has been and always will be a pointless so-called 'singularity'. We can't perceive it or measure it as such, because we are part of it, so it appears to us to have dimensional extensiveness.

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