Einstein's field equations are:

$R_{ab} - {1 \over 2}g_{ab}\,R + g_{ab} \Lambda = {8 \pi G \over c^4} T_{ab}$

And since the Ricci curvature tensor is "less information" than the Riemann curvature tensor because:

$R_{ab} = R^c{}_{acb}$

and the Riccie curvature scalar is even "less information" than the Ricci curvature tensor because:

$R = R^a{}_{a}$

This appears to indicate the GR field equations don't contain enough information to specify how the full Riemann curvature tensor evolves. So it looks like something missing. Is it even possible to say how the full Riemann curvature tensor evolves in GR?

Since the Riemann curvature tensor can be obtained from the metric, if we can solve for how the metric evolves using the Ricci curvature tensor, we can then get the Riemann curvature tensor evolution. So the equation is equivalent to the title question, what all do we need to solve for the metric in GR?

In books where they obtain the solution outside a static spherical object, they seem to always refer back to the Newtonian limit and compare to Newtonian gravity to fully fix the answer. This seems a bit scary, as it seems to suggest indeed that GR needs some other equations specified to get the answer.

If I choose a coordinate system and specify the metric everywhere at some initial time, is that initial metric + the GR field equations enough to solve for the metric everywhere in spacetime? Or is there some way to use GR to get the metric without any prior geometry put in?

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    $\begingroup$ Your question appears to be riddled with misunderstandings about basic notions in GR such as gauge invariance and the counting of degrees of freedom, and is likely to confuse a newcomer. So -1. $\endgroup$
    – user346
    Mar 22, 2011 at 5:08
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    $\begingroup$ I probably am misunderstanding basic things. Can you suggest how to improve the question? (Or edit it yourself?) $\endgroup$
    – John
    Mar 22, 2011 at 5:22
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    $\begingroup$ @John: first, talk about Riemann tensor is completely off topic. You should know that you can determine it from the metric, so it's not an independent information. Also, look at Einstein's equations: how many of them there are and how many degrees of freedom do you have in metric. Do the numbers match? $\endgroup$
    – Marek
    Mar 22, 2011 at 8:41
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    $\begingroup$ I have Carroll and the large Gravitation book. They both start out nice and slow, and build upon step after step. But eventually I get to a point where I stop and look around and don't understand what the physics content is anymore. It is not clear to me what it even means to take a measurement in such broad notions, so likewise it is not clear to me what is even measurable anymore. The "exploring black holes" book by Taylor is much easier and just static spacetime, but stops to debate any bizarre details along the way. I wish there was a simpler bridge from that to full GR. Any suggestions? $\endgroup$
    – John
    Mar 22, 2011 at 10:51
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    $\begingroup$ For those reasons, I have sympathy for the people historically that thought gravitational waves may just be a coordinate effect and have no physical meaning. For me, it's just plain hard to grab onto what is "physical" in GR. So I can easily see how they got confused too. $\endgroup$
    – John
    Mar 22, 2011 at 10:56

1 Answer 1


Dear John, let me post the same thing that Marek has said as a standard answer.

Einstein's equations are not equations for the Riemann tensor because the Riemann tensor's components are not independent fields. Instead, Einstein's equations are differential equations for the metric tensor.

In 4 dimensions, the metric tensor has 10 components - a symmetric tensor - and Einstein's equations have 10 components - a symmetric tensor - too. It doesn't matter that the Riemann tensor has 20 components because these 20 functions of space and time are calculated from the 10 component functions of the metric tensor and its (first and second) derivatives.

In fact, the 10-10 counting is oversimplified. Four "differential combinations" of Einstein's equations vanish identically because $\nabla_\mu R^{\mu\nu}=0$ is an identity (that always holds, even if the equations of motion are not satisfied). The same identity holds for the corresponding other tensors that are added to the Ricci tensor in Einstein's equations.

So instead of 10 equations, the Riemann equations are, in some sense, just 6 independent equations. That means that they don't determine the metric completely: they leave 4 functions undetermined and these are exactly the 4 functions that you may choose arbitrarily to specify a diffeomorphism, mapping one solution into another (equivalent) solution.

Up to the coordinate transformations which are always allowed to be made, initial conditions for the metric and its first derivative determine the metric tensor - and therefore the whole Riemann tensor - everywhere in the future. One doesn't need any Newtonian equations as a "mandatory supplement" in general relativity. That doesn't mean that the Newtonian limit is unimportant: of course, it is one of the most important approximate consequences of general relativity.

  • $\begingroup$ I'm not sure I'm entirely understanding. Let me try to summarize to see if I am getting it. You are saying that the field equations allow one to find $R^{\mu\nu}$ everywhere, and this can be used to obtain a unique physical solution of $g^{\mu\nu}$ (which is actually a class of physically equivalent solutions related by 4 functions to specify a diffeomorphism). Is that correct? So this means the Riemann curvature tensor doesn't contain more physical information, just Ricci + specific choice out of diffeomorphism related solutions? $\endgroup$
    – John
    Mar 22, 2011 at 9:29
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    $\begingroup$ Dear John, sorry but I think you're still not getting the basic point. Einstein's equations are differential equations for the metric tensor, not algebraic equations for the curvature tensor. If you don't understand this point, you shouldn't be using shortcuts such as $R_{ab}$ at all. Instead, you should substitute the right definition of the Ricci tensor into the equations. The equations say something like $\nabla^a \nabla_a g_{cd} + {\rm other\,\,terms} = K T_{cd}$: second-order (second derivatives) partial differential equations for the metric tensor $g_{ab}$. $\endgroup$ Mar 22, 2011 at 12:16
  • $\begingroup$ Otherwise, the Riemann curvature tensor at one point contains 20 independent components. But if you ask about the Riemann curvature tensor in the whole spacetime, it's 20 functions of spacetime coordinates that are not not independent; they satisfy lots of identities proving that they're not independent and these identities reflect the fact that the 20 functions $R_{abcd}(x,y,z,t)$ may be written in terms of 10 functions $g_{ab}(x,y,z,t)$. The Riemann tensor is not an independent variable, and it doesn't appear as such in Einstein's equations at all. $\endgroup$ Mar 22, 2011 at 12:18
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    $\begingroup$ @Lubos -- Am I confused, or do you have a typo in your answer? The thing that's covariantly conserved is the Einstein tensor, not the Ricci tensor. That is, I think you mean $\nabla_\mu G^{\mu\nu}=0$, instead of $\nabla_\mu R^{\mu\nu}=0$. $\endgroup$
    – Ted Bunn
    Mar 22, 2011 at 13:26
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    $\begingroup$ @John, there is a further complication here in that the Einstein equations (and such hyperbolic differential equns in general) do not have a unique solution ($g^{uv}$) in general. In the $T^{uv}=\Lambda=0$ case we have Minkowski $\eta$, Schwarzchild, Kerr, etc solutions - all physically different metrics. $\endgroup$ Mar 22, 2011 at 14:31

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