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Christoffel symbols in exact plane waves

Physics Asked by G.Tos on January 5, 2021

In the book "A first course of General Relativity" by Schutz I am stuck in trying to calculate Christoffel’s symbols for an exact plane wave. I have the metric:
$$ds^2 = -dudv+f^2(u)dx^2+g^2(u)dy^2$$
from which I think I get $g_{uv}=g_{vu}=-1/2$ , $g_{xx}=f^2(u)$ and $g_{yy}=g^2(u)$.

So, in order to calculate $Gamma^x_{xu}$ and$ Gamma^x_{xu}$ , I thought to use the formula $Gamma^k_{li}=frac{1}{2}g^{jm}(frac{partial g_{ij}}{partial q^l}+frac{partial g_{lj}}{partial q^i}-frac{partial g_{il}}{partial q^j})$ .

So I get $ Gamma^x_{xu} = frac{1}{2} g^{xx}2dot{f}(u)/f(u)=frac{dot{f}(u)}{f(u)}$ which is equal to the result of the book.
Though, when calculating in this way $Gamma^v_{xx}$ what I get is $frac{1}{2} g^{uv}(-2dot{f}(u)f(u))=2dot{f(u)}f(u)$. The problem is that the correct answer would be $2frac{dot{f}(u)}{f(u)}$. Where have I gone wrong here?

EDIT:

Schutz’s book also reports the value of $R^x_{uxu} = -frac{ddot{f}}{f}$ . Calculating it through the formula : $R^{alpha}_{beta mu nu}=frac{1}{2} g ^{alpha sigma}(g_{sigma nu , beta mu} – g_{sigma mu , beta nu }+g_{beta mu , sigma nu}-g_{beta nu , sigma mu})$
I get $R^x_{uxu} = -frac{ddot{f}}{f} – frac{dot{f}^2}{f^2} $ which is in agreement with my previous result because (given $R^l_{ijk} = frac{partial}{partial x^j} Gamma^l_{ik} – frac{partial}{partial x^k} Gamma^l_{ij} + Gamma^s_{ik}Gamma^l_{sj} – Gamma^s_{ij}Gamma^l_{sk}$) $R^x_{uxu}= – frac {d}{du}(frac{dot{f}}{f})$

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