Help with Kaluza Klein Christoffel symbols

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To calculate the Kaluza-Klein Christoffel symbols, the expression for ##\tilde{\Gamma}^\lambda_{\mu 5}## is derived using the metric tensor components and their derivatives. The discussion confirms that the simplified form of the expression can be written as ##\tilde{\Gamma}^\lambda_{\mu 5}=\frac{1}{2} \tilde{g}^{\lambda\sigma} \left(\partial_\mu \tilde{g}_{5\sigma} - \partial_\sigma \tilde{g}_{\mu 5}\right)##. It is clarified that the indices range from 0 to 5, consistent with the 5-dimensional nature of Kaluza-Klein theory. The participants emphasize the importance of understanding the dimensionality in the context of general relativity. The conversation highlights the intricacies involved in calculating Christoffel symbols in higher-dimensional theories.
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Homework Statement
I want to calculate ##\tilde{\Gamma}^\lambda_{\mu 5}##.
Relevant Equations
\begin{align}
\tilde{\Gamma}^\lambda_{\mu\nu} & = \frac{1}{2} \tilde{g}^{\lambda X} \left(\partial_\mu \tilde{g}_{\nu X} + \partial_\nu \tilde{g}_{\mu X} - \partial_X \tilde{g}_{\mu\nu}\right) \\
& =\frac{1}{2} \tilde{g}^{\lambda\sigma} \left(\partial_\mu \tilde{g}_{\nu\sigma} + \partial_\nu \tilde{g}_{\mu\sigma} - \partial_\sigma \tilde{g}_{\mu\nu}\right) + \frac{1}{2} \tilde{g}^{\lambda5} \left(\partial_\mu \tilde{g}_{\nu5} + \partial_\nu \tilde{g}_{\mu 5} - \partial_5 \tilde{g}_{\mu\nu}\right)
\end{align}

where

\begin{cases}
\tilde{g}_{\mu\nu} = g_{\mu\nu} + k A_\mu A_\nu \\
\tilde{g}_{\mu5} = k A_\mu \\
\tilde{g}_{55} = k\,(\mathrm{constant})
\end{cases}

and

\begin{cases}
\tilde{g}^{\mu\nu} = g^{\mu\nu} \\
\tilde{g}^{\mu5} = -A_\mu \\
\tilde{g}^{55} = \frac{1}{k} + A_\mu A^\mu.
\end{cases}
If I want to calculate ##\tilde{\Gamma}^\lambda_{\mu 5}##, I will write

\begin{align}
\tilde{\Gamma}^\lambda_{\mu 5} & = \frac{1}{2} \tilde{g}^{\lambda X} \left(\partial_\mu \tilde{g}_{5 X} + \partial_5 \tilde{g}_{\mu X} - \partial_X \tilde{g}_{\mu 5}\right) \\
& =\frac{1}{2} \tilde{g}^{\lambda\sigma} \left(\partial_\mu \tilde{g}_{5\sigma} + \partial_5 \tilde{g}_{\mu\sigma} - \partial_\sigma \tilde{g}_{\mu 5}\right) + \frac{1}{2} \tilde{g}^{\lambda5} \left(\partial_\mu \tilde{g}_{55} + \partial_5 \tilde{g}_{\mu5} - \partial_5 \tilde{g}_{\mu 5}\right)
\end{align}

Is it then correct to write that the above reduces to

$$\tilde{\Gamma}^\lambda_{\mu 5}=\frac{1}{2} \tilde{g}^{\lambda\sigma} \left(\partial_\mu \tilde{g}_{5\sigma} - \partial_\sigma \tilde{g}_{\mu 5}\right)?$$
 
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Do you mean all the indexes take number of 0,1,2,3,4,5 though usually 0,1,2,3?
 
Kaluza-Klein is a 5-dimentional theory, if I remember my Relativity correctly.
 
I want to find the solution to the integral ##\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}## I can see that ##\frac{d^2u}{d\theta^2} = A +Bu+Cu^2## is a Weierstrass elliptic function, which can be generated from ##\Large(\normalsize\frac{du}{d\theta}\Large)\normalsize^2 = c-u^2 +2u^3## (A = 0, B=-1, C=3) So does this make my integral an elliptic integral? I haven't been able to find a table of integrals anywhere which contains an integral of this form so I'm a bit stuck. TerryW

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