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 Excersise [14.30] 
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Joined: 04 Feb 2009, 15:52
Posts: 8
Post Excersise [14.30]
Regards to all you!

I´m not sure if the formula \frac{1}{12}n^2(n^2-1) about the quantity of independents components of Riemann tensor is right. I think this value must be \frac{1}{12}n^2(n-1)^2

I say this because:

R_{abcd}=-R_{bacd}\Rightarrow R_{aacd}=-R_{aacd}\Rightarrow R_{aacd}=0 (1)
Moreover a=1,..,n and c=1,..,n and d=1,..,n then, there are n^3 components whose value is 0.

The second antisymmetry tells us that R_{abcd}=-R_{abdc}\Rightarrow R_{abcc}=-R_{abcc}\Rightarrow R_{abcc}=0
So, there are n^3 components whose vale is 0, but the components R_{aacc}=0 were already counted in (1); and a=1,..,n and c=1,..,n. Then there are n^3-n^2 more null components.

So, R has n^4-(2n^3-n^2)=n^2(n^2-2n+1)=n^2(n-1)(n-1)=n^2(n-1)^2 not-null components (that is, independent components)

But the first antisymmetry R_{abcd}=-R_{bacd} tells us that half the components of R are dependent of another half, so the independent componets are \frac{1}{2}n^2(n-1)^2

And the second antisymmetry R_{abcd}=-R_{abdc} tells us the same, so remain \frac{1}{4}n^2(n-1)^2 elements. The interchange symmetry R_{abcd}=-R_{cdab} tells the same, so remain \frac{1}{8}n^2(n-1)^2 elements.

The Bianchi symmetry R_{abcd}+R_{bcad}+R_{cabd}=0 implies that each component depends of another two elements, so only 2/3 of remaining indepependent components are really independent.

Then, finally, there are \frac{2}{3}\frac{1}{8}n^2(n-1)^2=\frac{1}{12}n^2(n-1)^2 independent components in Riemann tensor R.

This is my reasoning. Is there misprint in the book? Am I right or I'm wrong? Please I wish someone answer me.
See you soon!!

07 Jun 2009, 09:07

Joined: 05 Jan 2009, 15:24
Posts: 4
Post Re: Excersise [14.30]
Hi there, don't have time to check through your derivation carefully, but your expression is certainly wrong, and Penroses is correct.

See for instance http://mathworld.wolfram.com/RiemannTensor.html.

Might have a look at it a bit later.


09 Jun 2009, 09:53
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