- #1

- 18

- 0

2[itex]\frac{R_1212}{det (g_μ√)}[/itex]

?

- Thread starter ssamsymn
- Start date

- #1

- 18

- 0

2[itex]\frac{R_1212}{det (g_μ√)}[/itex]

?

- #2

- 18

- 0

I am asked to show it by the symmetries of the Riemann tensor by the way.

- #3

- 18

- 0

Ok here is my thoughts, I try to stay away from the connection coefficients. So, I don't write the R tensors in form of [itex]\Gamma[/itex]'s. So I am trying:

R= g[itex]^{αβ}[/itex] R[itex]_{αβ}[/itex]

= g[itex]^{αβ}[/itex] R[itex]^{c}[/itex][itex]_{αcβ}[/itex]

= g[itex]^{αβ}[/itex] g[itex]^{αb}[/itex] R[itex]_{bαcβ}[/itex]

but now I can't have the R[itex]_{αβαβ}[/itex] form. Since it is 2-d, I put α=1 and β=2, but the c and b contractions doesn't give me what I want.

How can get that 2R[itex]_{1212}[/itex] ?

R= g[itex]^{αβ}[/itex] R[itex]_{αβ}[/itex]

= g[itex]^{αβ}[/itex] R[itex]^{c}[/itex][itex]_{αcβ}[/itex]

= g[itex]^{αβ}[/itex] g[itex]^{αb}[/itex] R[itex]_{bαcβ}[/itex]

but now I can't have the R[itex]_{αβαβ}[/itex] form. Since it is 2-d, I put α=1 and β=2, but the c and b contractions doesn't give me what I want.

How can get that 2R[itex]_{1212}[/itex] ?

Last edited:

- #4

Bill_K

Science Advisor

- 4,155

- 195

Write out completely the individual components of g

- #5

- 9,853

- 1,049

[tex]

\left[ \begin{array}{cc}

0 & R \\

-R & 0 \\

\end{array} \right]

[/tex]

Now we know that R_abcd = 0 if a=b or c=d by the anti-symmetry properties, and also that R_abcd = R_bacd and that R_abcd = -R_abdc

This, and a little thought, gives us the value of all the components of R, which can be described as an anti-symmetric 2d array of two-forms, i.e. it looks like the array above, but the members of the array are the anti-symmetric two-forms.

Next we just have to compute the contractions to get the Riemann tensor and scalar, which I'm too lazy to do by hand.

- #6

- 18

- 0

Yes I rewrited everything with g lower index.

I am not sure about that symmetry:

R[itex]^{αβ}[/itex][itex]_{αβ}[/itex]= - R[itex]^{βα}[/itex][itex]_{βα}[/itex]

But I feel I am close to it. Thank you again.

- Last Post

- Replies
- 11

- Views
- 5K

- Replies
- 13

- Views
- 3K

- Last Post

- Replies
- 6

- Views
- 3K

- Last Post

- Replies
- 40

- Views
- 7K

- Replies
- 1

- Views
- 1K

- Replies
- 13

- Views
- 3K

- Replies
- 4

- Views
- 3K

- Replies
- 2

- Views
- 947

- Replies
- 2

- Views
- 3K

- Replies
- 8

- Views
- 7K