Ricci Tensor in Empty Space

1. Nov 9, 2009

MaxwellsDemon

I've been wondering two things lately. Why did Einstein make the assumption that the Ricci tensor is 0 in empty space. Is there a physical/mathematical reason? I know later he set it equal to another tensor...which leads to all the cosmological constant business, but I'm just curious why he favored 0 initially. What was his motivation? The other thing I don't understand is, what sorts of restrictions does setting the Ricci tensor place on the form of the metric? Any insights would be greatly appreciated.

2. Nov 9, 2009

clamtrox

Einsteins first guess to the form of general relativity was $$R^{\mu \nu} \sim T^{\mu \nu}$$. This is a natural guess, since both are symmetric second order tensors. However this is simply wrong; it's not compatible with energy-momentum conservation. Instead, the correct form for the equation is $$R^{\mu \nu} - R g^{\mu \nu}/2 \sim T^{\mu \nu}$$.

Ricci tensor depends on the affine connection (which in turn depends on metric in the usual GR formalism). It's not a pretty sight though, since you sum over a lot of different permutations of 2nd order derivatives of the metric tensor.

3. Nov 9, 2009

waht

I suppose Einstein noticed that in a weak limit the zero components of the Ricci tensor simplify to a laplacian of a potential, and if equated to a energy density - it simplies to a Poisson's equation used in Newtonian theory of gravity.

$$R_{00} \sim \nabla^2$$

And:

$$R_{00} \sim T_{00}$$

implies,

$$\nabla^2 \phi = 4\pi G\rho$$

4. Nov 9, 2009

ramparts

Perhaps I'm misreading your question, but the most sensible thing for Einstein to believe, so shortly after the formulation of special relativity, is that empty space is Minkowski, which is certainly Ricci flat. So that's a limit that would be very reasonable for his field equations to satisfy.

5. Nov 10, 2009

MaxwellsDemon

I think I'm starting to understand the reasoning behind this assumption now, but I still have some questions. Waht and ramparts' suggestion that the correspondence principle would be a strong motivation makes sense to me. He would certainly want to reproduce Newton's results and Special Relativity in the appropriate limits. Since the Ricci tensor is dependant on derivatives of the metric, don't you need boundary conditions to obtain Minkowski space?...couldn't I reproduce Minkowski space in an empty space where the Ricci tensor is non zero by choosing appropriate boundary conditions?...or is that mathematically absurd and illogical? I guess what I'm having difficulty with is that, sure setting the Ricci tensor equal to zero makes sense under the conditions within our solar system to reproduce classic results, but might it not be different really close to a black hole or way out there in between galaxies? Would Special Relativty have to apply locally everywhere in empty space? 4-momentum conservation would be a strong arguement in favor of setting it zero everywhere as clamtrox suggests...but in GR isn't it only strictly conserved locally? (I ask this because of the second term with the Christoffel symbol when taking the covariant derivative of the stress tensor) Is setting the Ricci tensor equal to zero just a mathematical way of saying that there is always a local patch of spacetime that "looks flat" when I zoom in? Because I thought that assumption only came into play when defining derivatives and parallel transport... Sorry if I'm confused or asking strange questions... I've been trying to teach myself GR and have been a little hung up on the motivation for Einstein's equation.

6. Nov 10, 2009

ramparts

The Ricci tensor is constructed from the metric and its derivatives, so if you have a Minkowski metric, there's only one possible Ricci tensor (and all of its components are 0). Of course, it doesn't work the other way around - there are plenty of other possible metrics with a 0 Ricci tensor (like Schwarzschild and Kerr).

As for your next question - might it be non-zero near, say, a black hole - well, not in GR. Einstein's equations show pretty clearly that if the stress-energy tensor is zero at a point, so is the Ricci tensor.

7. Nov 10, 2009

waht

Minkowski's metric is just a flat Euclidean space, whose Ricci tensor is zero.

But, there are also other metric solutions that have a zero Ricci tensor. If the stress tensor is zero, the solution to:

$$R_{uv} = 0$$

implies that an empty space can be curved without the presence of matter. If you do some approximations, you can simplify the Ricci tensor or the Einstein tensor to a wave equation which describes a propagating gravitational wave far away from a source.

8. Jan 13, 2010

utku

In General relativity,there arent any forces,there are curvature because of the matter and energy.
In special relativity valid only inertial frames.So it doenst include any gravitational force or acceleration.After,Special relativity was generalized by einstein with constracting the general relativity.In general relativity all physics laws are valid in all frames including inertian and non inertial frames.So you have to use tensor for describing physical laws covariantly and independent from referance frames.
Einstein firsly tried for propotunion of Curveture tensor and energy and momentum tensor.But energy and momentum tensor has 2 rank ,simmetric and divergence free so after that he tried to solve the which tensor he must use for describing spacetime curvature.So he found that combination of ricci tensor diergence free and it has other properties.It is einstein tensor.
In tensor analysis we know that reimann tensor describing the curvature of the space.
And ricci tensor too.When ricci tensor goes zero this is interpretation of the flat spacetime.But our aim is not that.Question is why ricci tensor is equal to zero in vacuum.
In above we mention about proportion of energy- momentum tensor and ricci or einstein tensor.So in vacuum energy and momentum tensor is equal to zero and after this stuation ricci and einstein tensor are equal to zero.
Because there isnt any matter in vacuum only exception is source of gravitational field.

9. Jan 13, 2010

bcrowell

Staff Emeritus
The Riemann tensor includes both tidal effects from distant sources and effects from local sources. The motivation for constructing the Ricci tensor from the Riemann tensor is that it averages out the tidal effects, leaving only the effects of local sources. When there are no local sources, you expect that the only curvature will be tidal, hence the Ricci tensor should vanish. Here http://www.lightandmatter.com/html_books/genrel/ch05/ch05.html#Section5.1 [Broken] is a more detailed discussion.

One way of seeing that you really need the Einstein tensor rather than the Ricci tensor, when sources are present, is by counting degrees of freedom: http://www.lightandmatter.com/html_books/genrel/ch07/ch07.html#Section7.1 [Broken]

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