# Ricci tensor problem

1. May 4, 2007

### Mentz114

Starting with this definition of the Reimann tensor

$$R^a_{mbn}=\Gamma ^{a}_{mn,b}-\Gamma ^{a}_{mb,n}+\Gamma ^{a}_{rb}\Gamma ^{r}_{mn}-\Gamma ^{a}_{rn}\Gamma ^{r}_{mb}$$

Can I contract on indices a,b and r to get $$R_{mn}$$ ?

It bothers me that the expression on the right is not symmetric in m,n. I worked out

$$R_{12}=\Gamma ^{2}_{12}\Gamma ^{3}_{23}-\Gamma ^{3}_{13,2}-\Gamma ^{3}_{13,2}-\Gamma ^{3}_{13,2}-\Gamma ^{3}_{13,2}-\Gamma ^{3}_{13}\Gamma ^{3}_{32}$$
and
$$R_{21}=\Gamma ^{2}_{21}\Gamma ^{3}_{23}-\Gamma ^{3}_{23,1}-\Gamma ^{3}_{23,1}-\Gamma ^{3}_{23,1}-\Gamma ^{3}_{23,1}-\Gamma ^{3}_{23}\Gamma ^{3}_{31}$$.
I thought they should be equal. I'm Using the Schwarzschild metric, with signature (-1, 1, 1, 1), x0 = t, x1=r, x2=theta, x3=phi.

Last edited: May 4, 2007
2. May 4, 2007

### Norman Albers

Check your opening formulation. I think I'm reading differently, and I'll see if I can present it.$${R^a}_{mbn}={\Gamma^a}_{mn|b} -{\Gamma^a}_{bm|n}+{\Gamma^a}_{kb}{\Gamma^k}_{nm} - {\Gamma^a}_{kn}{\Gamma^k}_{bm}.$$ What do you need to do to see LaTex edits?

Last edited: May 4, 2007
3. May 4, 2007

### AKG

A priori, you can't contract on a,b, since the Ricci curvature is the (1,4) contraction of the Riemann curvature, not the (1,3) contraction. But the right hand side of your first equation is anti-symmetric in n and b, so $R^a_{mbn} = -R^a_{mnb}$. So if you were to contract the right side, you'd simply get the negative of the Ricci curvature.

4. May 4, 2007

### Mentz114

Thanks AKG, I'll try the (1,4) contraction - at least it looks symmetric.

Norman, I'm not sure what your notation means.

5. May 4, 2007

### Norman Albers

One vertical line means simple differentiation; two mean covariant. Maybe you're used to comma and semicolon? I read that the both the first pair of indices and the second, are antisymmetric, and that the pairs themselves are symmetric upon exchange.

Last edited: May 4, 2007
6. May 4, 2007

### Mentz114

Hi Norman, yes, I'm accustomed to the comma and semicolon. I'm going to check out the (1,4) contraction a little later. I'll be glad to get past this one, I'm going cross-eyed doing index gymnastics.

7. May 4, 2007

### Norman Albers

Yah, this demands more coffee or woodchopping breaks than calculus. I am slowly but steadily putting together the pieces to do what I mention in the 'tarpits' thread.

8. May 4, 2007

### Mentz114

AKG, according to my book, Stephani, the 1,3 and 1,4 contractions of that expression give R_mn and -R_mn. It doesn't matter how I do it, it won't come out symmetrical. In fact I don't see how it possibly can. The original expression is not symmetrical, so there has to be some cancellation of terms presumably.

9. May 4, 2007

### AKG

That's exactly what I said, but with the signs reversed (i.e. I said that the 1,4 contraction gives R_mn, and the 1,3 gives -R_mn). This is not surprising because there 7 different sign conventions in current use.
I know, that's why I said it's anti-symmetrical.
It all depends on the definitions and conventions in your book. Why haven't you bothered to post the definitions (esp. the definition of the Ricci curvature) you're working with? You should always do this when asking for help with problems in the future.

10. May 4, 2007

### Mentz114

AKG, I was affirming what you told me, not trying to contradict you. I am grateful for your advice, believe me.

Sorry, I thought there was enough information in my original post. The thing is I'm contracting the tensor I've been given and ending up with an expression that has two lower indices, m,n which is supposed to be R_mn, a symmetric tensor. But I don't see how a symmetric tensor can arise if I start with something that is unsymmetric in m,n ( neither symmetric nor antisymmetric ).

11. May 5, 2007

### Mentz114

Yes.

It shouldn't bother you, the Ricci tensor is not always symmetric or anti-symmetric. Work out the algebra.

Easy isn't it ?

Don't mention it. By the way you'd better stop answering your own posts ...

12. May 5, 2007

### Norman Albers

symmetries

There are the symmetries I described.

13. May 5, 2007

### robphy

The Ricci tensor is symmetric with a symmetric Christoffel symbol.

Write
$$R^a_{mbn}=\Gamma ^{a}_{mn,b}-\Gamma ^{a}_{mb,n}+\Gamma ^{a}_{rb}\Gamma ^{r}_{mn}-\Gamma ^{a}_{rn}\Gamma ^{r}_{mb}$$
and, by index substitution,
$$R^a_{nbm}=\Gamma ^{a}_{nm,b}-\Gamma ^{a}_{nb,m}+\Gamma ^{a}_{rb}\Gamma ^{r}_{nm}-\Gamma ^{a}_{rm}\Gamma ^{r}_{nb}$$
Subtract the two... forming the combination that is antisymmetric in m and n. This is not identically zero, although its contraction with a and b is zero [using the symmetry of the Christoffel symbols and of the metric.. and probably some dummy-index-substitutions].... meaning that this contracted tensor is equal to its symmetric part.

$$R^a_{man}-R^a_{nam} = ( \Gamma ^{a}_{mn,a}-\Gamma ^{a}_{nm,a} ) -( \Gamma ^{a}_{ma,n} -\Gamma ^{a}_{na,m} ) +( \Gamma ^{a}_{ra}\Gamma ^{r}_{mn}-\Gamma ^{a}_{ra}\Gamma ^{r}_{nm} ) -( \Gamma ^{a}_{rn}\Gamma ^{r}_{ma} -\Gamma ^{a}_{rm}\Gamma ^{r}_{na} )$$

The first and third terms are obviously zero [by symmetry of the Christoffel symbols].
The last term is zero by dummy-index-substitution (swapping all occurrences of the dummy indices a and r) on the second term :
\begin{align*} ( \Gamma ^{a}_{rn}\Gamma ^{r}_{ma} -\Gamma ^{a}_{rm}\Gamma ^{r}_{na} ) &=( \Gamma ^{a}_{rn}\Gamma ^{r}_{ma} -\Gamma ^{r}_{am}\Gamma ^{a}_{nr} )\\ &=( \Gamma ^{a}_{rn}\Gamma ^{r}_{ma} -\Gamma ^{r}_{ma}\Gamma ^{a}_{rn} )\\ &=( \Gamma ^{a}_{rn}\Gamma ^{r}_{ma} -\Gamma ^{a}_{rn}\Gamma ^{r}_{ma} ) \end{align*}
where we use the symmetry of the Christoffel symbol in the second-to-the-last line and commutativity in the last line.

The second term will probably turn out to be zero by writing out the Christoffel symbol in terms of the [symmetric] metric and its derivatives.

I hope I didn't make any errors.

Last edited: May 5, 2007
14. May 5, 2007

### Mentz114

Thanks, robphy, I understand what you're saying. After finding out more about the Reimann tensor, i.e. anti-symmetric under 1<->2 and 3<->4 interchange, and symmetric under (12)<->(34), it occured to me that something like this needs doing because we're left with 2,4 after contraction. To be symmetric 2,4 must be acompanied by a 1,3 interchange, but since we're summing over 1,3, extra symmetrization is needed (?).

It's becoming clearer, but it's more important to me right now that I can get the text-book answers from my own calculations.

15. May 7, 2007

### Mentz114

I wrote out the components of this
$$\Gamma ^{a}_{ma,n} -\Gamma ^{a}_{na,m}$$

in 3 dimensions and compared components for m<>n. They are not equal...

16. May 7, 2007

### robphy

Note that there is an identity involving $$\Gamma ^{a}_{ma}$$:

See "The contracting relations on the Christoffel symbols" near the bottom of http://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry

In our symbols, $$\Gamma ^{a}_{ma} = (\log{\sqrt{\det g}})_{,m}$$, that is the partial derivative of a scalar function.

So,
\begin{align*} \Gamma ^{a}_{ma,n} - \Gamma ^{a}_{na,m} &= (\log{\sqrt{\det g}})_{,mn} - (\log{\sqrt{\det g}})_{,nm}\\ &=0 \end{align*}
since partial derivatives commute on a scalar function.

I hope I didn't make any errors.

17. May 8, 2007

### Mentz114

Food for thought. Thanks for pointing that out, robphy. My doubts about the symmetry of the Ricci tensor are getting very weak.

 I did the full subtraction as you suggest in your earlier post. All the residual terms have the form -

$$\Gamma ^{k}_{mk,n}-\Gamma ^{k}_{nk,m}$$

which is zero thanks to the identity you quoted. So all doubt gone, and much learnt. Thanks again.

Last edited: May 8, 2007