GRstudent said:
Your example is very good; however, I didn't understand why have you not included the Riemann tensor into your equation. And secondly, I am still waiting to have some clarification about having the products of Gammas in the Riemann tensor definition. Also, how I can derive Riemann tensor?
Since reading your question, I spent about two days (not full-time, I do have a life) trying to derive the Riemann tensor in terms of the connection coefficients \Gamma^{\mu}_{\nu\lambda}, and I managed to convince myself that I know how to do it, but it is a mess.
Conceptually, it works this way:
- Start with three vectors U, V and W.
Let U_1 = U, V_1 = V and W_1 = W.
- Parallel-transport all three vectors along the vector U_1. This gives you new vectors U_2, V_2 and W_2.
- Parallel-transport all three vectors along the vector V_2. This gives you new vectors U_3, V_3 and W_3.
- Parallel-transport V_3 and W_3 (we don't need to bother with U_3) along the vector -U_3. This gives you new vectors V_4 and W_4.
- Parallel-transport W_4 (we don't need the other ones) along the vector -V_4. This gives you a new vector W_5.
- Now define a final vector Q to be the vector that gets you back to where you started: Q = -U_1 - V_2 + U_3 + V_4
- Now, finally parallel-transport W_5 along Q to get a final vector W_6.
- Let \delta W = W_6 - W_1
- Then, to second order in the vectors U and V, the Riemann tensor R is defined by R(U,V,W) = \delta W. Or in terms of components, R^{\alpha}_{\beta\gamma\mu} U^{\beta} V^{\gamma}W^{\mu} = \delta W^{\alpha} (where \beta, \gamma and \mu are summed over).
This prescription still doesn't tell you how to compute Riemann in terms of connection coefficients. I don't know the most elegant way to do it, but the following approach works (although it's a bit of a mess).
Here's a fact about parallel transport that I have not seen written down anywhere, but I've convinced myself is true. Suppose you have two vectors A and B and you want to parallel-transport A along B to get a new vector \tilde{A}. How do the components of \tilde{A} relate to the components of A? This is a coordinate-dependent question, so it involves connection coefficients. To first order, the answer is:
\tilde{A}^{\mu} = A^{\mu} - \Gamma^{\mu}_{\nu \lambda} B^{\nu} A^{\lambda}
The following is bad notation, but since it's possible to deduce what the indices have to be, we can write this more simply as
\tilde{A} = A - \Gamma B A
That doesn't look too bad. Unfortunately, for computing the Riemann tensor, we need \tilde{A}^{\mu} to second-order in B. I'm pretty sure the answer is, in terms of components:
\tilde{A}^{\mu} = A^{\mu} - \Gamma^{\mu}_{\nu \lambda} B^{\nu} A^{\lambda} - \frac{1}{2} \partial_{\tau}\Gamma^{\mu}_{\nu \lambda} B^{\tau} B^{\nu} A^{\lambda} + \frac{1}{2} \Gamma^{\mu}_{\tau \nu} B^{\tau} \Gamma^{\nu}_{\lambda \alpha} B^{\lambda}A^{\alpha}
Using my bad, compact notation:
\tilde{A} = A - \Gamma B A - \frac{1}{2} (\partial\Gamma) B B A + \frac{1}{2}\Gamma B (\Gamma B A)
Also, the connection coefficients themselves change after moving along B. Afterwards, the new value of the connection coefficients \tilde{\Gamma}^{\mu}_{\nu \lambda} is given by:
\tilde{\Gamma}^{\mu}_{\nu \lambda} = \Gamma^{\mu}_{\nu \lambda} + \partial_{\alpha} \Gamma^{\mu}_{\nu \lambda} B^{\alpha}
In my bad compact notation:
\tilde{\Gamma} = \Gamma + (\partial \Gamma) BIf you use these formulas repeatedly to compute how the components of U, V, W and \Gamma change around the loop, only keeping the first- and second-order terms, then you will get an expression for \delta W^\mu in terms of U^\alpha, V^\beta, W^\gamma, \Gamma^{\mu}_{\nu \tau} and \partial_{\lambda}\Gamma^{\mu}_{\nu \tau}
