# Why Riemann tensor?

1. May 15, 2006

### eljose

A doubt..why einstein Chose Riemann Tensor for GR?..i know its covariant derivative is zero and all that..but Why Riemann tensor?...was not other tensor avaliable or simpler than that?..i studied that and found that for Geodesic deviation ( i didn,t understand that concept..sorry) the Riemann tensor was involved....:uhh: :uhh: :uhh:

2. May 15, 2006

### pervect

Staff Emeritus
We can break this down into parts

Part 1: Why use a tensor? The answer is that using tensors make the underlying theory independent of the coordinate system used. All tensors transform under a change of coordinates in a specific way (in the case of relativity and most of modern physics they transform via the Lorentz transform, the transform that Einstein originally derived in his first paper on special relativity). Because the transformation properties are known, a tensor equation is good in any coordinate system one chooses to use.

Part 2: What tensor(s) should one use? This is not an easy question, and it took Einstein a very long time to derive GR. The short version of the answer is that the choices Einstein made Einstein's field equations automatically conserve momentum and energy in almost the same way that Maxwell's equations automatically conserve charge.

The tensor that Einstein chose to represent matter was the stress-energy tensor, T_ij. This choice was made fairly early on, and can be regardes as "reasonably obvious" - at least, to someone familiar with tensors. The tensor that Einstein chose to represent "curvature" was not the Riemann, but a modified version of the Ricci tensor, a contraction of the Riemann.

The Riemann is R_abcd, the Ricci is formed by a process known as contraction which reduces the number of indices (the rank) of the tensor to two. The Ricci is usually written using the same symbol as the Riemann, but it can be distinguised by the fact that the Ricci only has two indices, i.e. R_ab.

Given that matter is represented by a rank-2 tensor, it is a necessity for a tensor equation that the left hand side of the equations also be a rank 2 tensor, so the Ricci is an obvious candidate.

Einstein originally tried using the Ricci on the left hand side, but he realized that the resulting theory had problems with the conservation of energy and momentum

I.e. R_ab = (some constant) T_ab does NOT work

Einstein eventually realiesed that what he needed on the left hand side was not the Ricci tensor, but a different tensor, known as the Einstein tensor, G_ab

where G_ab = R_ab - (1/2) g_ab R

and R is the contraction of the Ricci, which reduces the number of indices to zero, making it a scalar.

The result of this is Einsteins field equation:

G_ab = 8 pi T_ab does work properly.

Not only does it avoid the problems with momentum and energy conservation that the first formula had, it automatically implies the differential form of the conservation of energy and momentum of T_ab from the geometric properties of G_ab.

Note the form of the equation: the left-hand side contains information about the curvature of space-time, the right hand side contains information about matter - the density of energy and momentum.

This can be summed up neatly by the expression

"Space tells matter how to move, and matter tells space how to curve".

For more detailed online reading, I'd recommend

http://math.ucr.edu/home/baez/einstein/

Much of the information in my post is derived from MTW's textbook "Gravitation", but this is not a popular read, it's a graduate-level textbook. While large portions of the book are written in an informal style that might be accessible to the lay reader, equally large portions of the book are not so accessible.

3. May 15, 2006

### masudr

I'd argue that the first part is given by the geodesic equation, and the second part by the Einstein Field equations (and not just the latter).

EDIT: even better. the first part by extremizing proper time with respect to neighbouring paths and the second part by extremizing the spacetime integral of the Ricci scalar with respect to the metric.

4. May 15, 2006

### pervect

Staff Emeritus
I'm pretty sure that geodesic motion follows from Einstein's field equations alone - it does not need a separate postulate.

Google finds, for instance

http://www.mathpages.com/home/kmath588/kmath588.htm

The author goes on to point out that the differential conservation laws "built into" GR imply that particles follow geodesics in the lack of any external forces.

Unfortunately, I'm not quite sure who the author of the above actually is, which makes it a less than perfect source to resolve a debate, but it does agree with my recollection that we don't need to postulate geodesic motion separately.

5. May 16, 2006

### masudr

That is a very appealing argument. I wonder if anyone has worked through the problem and derived the geodesic equation...

6. May 16, 2006

### robphy

The Gravitational Equations and the Problem of Motion
A. Einstein, L. Infeld, B. Hoffmann
Annals of Mathematics, 2nd Ser., Vol. 39, No. 1 (Jan., 1938) , pp. 65-100
"In this paper we investigate the fundamentally simple question of the extent to which the relativistic equations of gravitation determine the motion of ponderable bodies...."

http://prola.aps.org/abstract/RMP/v21/i3/p408_1
On the Motion of Test Particles in General Relativity
L. Infeld and A. Schild
Rev. Mod. Phys. 21, 408–413 (1949)
"In this paper we give a simple derivation of the geodesic motion of test particles from Einstein's gravitational equations for empty space. The history of this problem is connected with the development of basic physical concepts...."

http://www.artsci.wustl.edu/~jashiffl/einstein_infeld.html [Broken]

Last edited by a moderator: May 2, 2017
7. May 16, 2006

### robphy

8. May 16, 2006

### Stingray

Yes, to most peoples' satisfaction, this is has been done. The simplest derivations are present in most textbooks, actually. These have a rather limited scope, though, which led a number of (picky) people to call the principle that objects move along geodesics the "geodesic hypothesis."

Some particularly picky people still call it this, but its precise applicability is much better understood now. Some work to this effect is discussed in http://www.arxiv.org/abs/gr-qc/0309074" [Broken]. The two authors (Ehlers and Geroch) have made many important contributions to the rigorous understanding of relativistic mechanics.

Anyway, the fact the Einstein's equation determines mechanics (to a large extent) was historically considered one of GR's most attractive features.

Last edited by a moderator: May 2, 2017
9. May 16, 2006

### dextercioby

See Dirac's 70-page textbook.

Daniel.

10. May 18, 2006

### samalkhaiat

Hi,

The following does not undermine answers given by others.

1) The importance of the Riemann tensor for general relativity can be understood solely from differential geometry (tensor analysis) which shows that a space is flat if and only if the riemann tensor vanishes.

2) Einstein was looking for generally covariant (i.e tensorial) geometrodynamical field theory (i.e the dynamical variables are intrinsic geometrical objects) which has causal solutions(i.e the field equations are 2nd-order differential equations).
To arrive at such theory, one needs a scalar density(Lagrangian) that satisfies;

(i) it is constructed out of the geometry of spacetime alone (i.e the metric tensor)
(ii) it contains no higher derivatives (of the metric tensor) than the second.
(iii) it is linear in the 2nd derivatives of the metric so that it can be put in the form;

$$(g)^{1/2}\mathcal{L} = (g)^{1/2} G(g^{\mu\nu},\partial^{\rho}g^{\mu\nu}) + \partial_{\mu}[(g)^{1/2} J^{\mu}(g^{\mu\nu}, \partial^{\rho}g^{\mu\nu})]$$

(iv) it vanishes in flat spacetime (the geometry of flat spacetime is not dynamical)

3) From the symmetries and the antisymmetries of $R^{\mu\nu\rho\sigma}$ one can show that the scalar curvature is the only scalar that exists that satisfies (i)-(iv).

4) If one takes

$$\mathcal{L}=R=g^{\mu\nu}R_{\mu\nu}=g^{\mu\nu}R^{\rho} {}_{\mu\rho\nu}$$

one arrives (through action principle) at Einstein equation in free space;

$$R^{\mu\nu} - 1/2 g^{\mu\nu} R = 0$$

regards

sam

Last edited: May 18, 2006
11. May 18, 2006

### clj4

He is professor Ken Brown from Cornell. See here:

http://www.math.cornell.edu/People/Faculty/brownk.html
.

Last edited: May 18, 2006