Solving Curvature Math Problems with Christoffel Symbols

[Feynman]

Hello
I have a small pb:
Let \displaystyle{M=\mathbb{R^{2}\{(x,y);x=0\quad ou \quad y=-1}
Let D such that Christoffel symboles different from 0 are\\\\
\displaystyle\Gamma_{11}^{1}(x,y)=-\frac{1}{x}\\\\\\\\<br /> \Gamma_{22}^{2}(x,y)=\frac{1}{1+y}\\\\\\
How calculate curvurture?\\

 

[Timbuqtu]

The Riemann-curvature is given by:

R^a_{b c d} = \frac{\partial \Gamma^a_{b d}}{\partial x^c} - \frac{\partial \Gamma^a_{b c}}{\partial x^d} + \Gamma^a_{c e} \Gamma^e_{b d} - \Gamma^a_{d e} \Gamma^e_{b c}

 

[Feynman]

excuse me \displaystyle\Gamma_{11}^{1}(x,y)=-\frac{1}{x}\quad and \quad Gamma_{22}^{2}(x,y)=\frac{1}{1+y}
And
\displaystyle{M={R^{2}\{(x,y);x=0\quad or \quad y=-1}}
So How we use this formula in this case
and how we obtain this formula?

 

[Timbuqtu]

a, b, c, d and e are indices, so in this case they can take the values 1 or 2: x = x^1 and y = x^2. Repeates indices are summed over (Einstein summation convention).

So for example:

R^1_{1 2 2} = \frac{\partial \Gamma^1_{1 2}}{\partial x^2} - \frac{\partial \Gamma^1_{1 2}}{\partial x^2} + \sum_{e=1}^2 (\Gamma^1_{2 e} \Gamma^e_{1 2} - \Gamma^1_{2 e} \Gamma^e_{1 2})

which in this case is 0 because only \Gamma^1_{1 1} and \Gamma^2_{2 2} are different from 0.

If you want to see a derivation of this formula, you could have a look at (chapter 3 of) Sean M. Carroll's lecture notes: http://pancake.uchicago.edu/~carroll/notes/

 

[Feynman]

But in my case
R=0?

 

[dextercioby]

Are u looking for the tensor,or for its contractions (Ricci tensor,Ricci scalar)...?

Daniel.

 

[PBRMEASAP]

What are those hieroglyphic looking symbols in the first post? I clicked on it and the latex code doesn't look anything like it.

 

[dextercioby]

U didn't close \mathbb function right after R...

Daniel.

 

[PBRMEASAP]

\displaystyle{M=\mathbb{R}^{2}\{(x,y);x=0\quad ou \quad y=-1}

Ah, there we go. I still don't know what that means, but it looks more recognizable. I was hoping those were tensor diagrams like Penrose uses. I don't get how they work, but they sure look nifty.