A General Relativity Question

[Ken Whight]

Given the excellent replies I got to a previous forum posting on “A possible Cosmological Paradigm?”, I thought I would try another more answerable question.
As stated in my last posting I am making another attempt to understand General Relativity using the book “Covarient Physics” by Moataz H. Emam that I mentioned in my previous posting and it’s going quite well. However in many books it is stated that the 4th rank Riemann curvature tensor in 4 dimensions has 256 (i.e 4^4) elements of which only 20 are independent due to symmetries. I’ve got the total down to 21 as follows (excuse the latin rather than Greek indices!).
The fact that Rijkl is antisymmetric in the first and last pair of indicies means that there are only 4*3 ways of choosing ij that doesn’t result in a zero element and of these only 6 are independent as Rijkl = -Rjikl. The same holds for choosing kl, so the number of independent elements Rijkl is reduced (so far) to 6*6 = 36. they can be written as a 6*6 array:-

0101 0102 0103 0112 0113 0123
0201 0202 0203 0212 0213 0223
0301 0302 0303 0312 0313 0323
1201 1202 1203 1212 1213 1223
1301 1302 1303 1312 1313 1323
2301 2302 2303 2312 2313 2323

Now as Rijkl is symmetric in the first and last index pairs R1201 = R0112 etc and the number of independent elements further reduces to 6+5+4+3+2+1 = 21.
So I still need another relationship, presumably there is some normalisation factor determined by the basis vectors that finally reduces the number of independent elements to 20. Is this correct?

[Ken Whight]

I realised that R2301, R1302 and R1203 are linked by a Bianchi identity such that we know R2301+R1203=R1302 so the total reduces to 20 independent elements QED.

[Paul G. Abel]

There are many subtleties to this but it sounds like you haven’t included the Bianchi identities, in particular the first and perhaps the second Bianchi identities. There are reasons why this tensor has these properties. There are also the anti-skew symmetries and one of them is to do with defining a suitable inner product on the tangent space T(M) which is induced by the metric tensor g_ab.

If a curvature tensor satisfies the skew symmetry, anti skew symmetry and first Bianchi identities then it is trivial to show that there exists a curvature tensor at each point p in M (a pseudo-Riemmannian manifold) which has n^2(n^2-1)/12 components.

It’s all in two standard texts:
-Introducing Einstein’s Relativity by D’Inverno
-The Mathematical theory of Black Holes by Chandrasekhar.

• This reply was modified 1 year, 3 months ago by Paul G. Abel.

[Ken Whight]

So the curvature tensor has 20 independent elements whilst a solution of Einstein’s field equations determines the 10 independent elements of the Ricci tensor (a contraction of the 4 index curvature tensor first&last indices). This means that given a solution of the field equations the full curvature tensor is not uniquely deternined (pairs of Rijkl independent elements can be related to single or combinations of Rjk elements), does this mean anything physically?

[Dr Paul Leyland]

I strongly recommend Gravitation, aka MTW, for a comprehensive and relatively (pun intended) accessible course in classical geometrodynamics, despite it being wrotten 50 years ago.

That said, I can’t answer your question off the top of my head. Let me think about it and/or consult MTW.

[Ken Whight]

But… the GR curvature tensor is fully calculable from the metric tensor, so does this mean that a metrical connection reduces the independent components of the 4 index curvature tensor to just 10?

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