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I have a question regarding the connection coefficients as they pertain to the following paper: http://www.weylmann.com/kaluza.pdf . When I try to calculate the 4D Christoffel symbols from the 4D part of the 5D metric $\tilde{g}_{\mu \nu}=g_{\mu \nu}+kA_\mu A_\nu$ I get an extra term that shouldn't be there. I get:

$$\tilde{\Gamma}^\lambda_{\mu \nu }=\frac{1}{2}\tilde{g}^{\lambda \sigma}(\partial_\mu \tilde{g}_{\nu \sigma}+\partial_\nu \tilde{g}_{\mu \sigma}-\partial_\sigma \tilde{g}_{\mu \nu})$$

$$=\Gamma^\lambda_{\mu \nu }+\frac{k}{2} g^{\lambda \sigma}(\partial_\mu (A_\nu A_\sigma)+\partial_\nu (A_\mu A_\sigma)-\partial_\sigma (A_\mu A_\nu))$$

$$=\Gamma^\lambda_{\mu \nu }+\frac{k}{2} g^{\lambda \sigma}(A_\mu (\partial_\nu A_\sigma -\partial_\sigma A_\nu)+A_\nu (\partial_\mu A_\sigma -\partial_\sigma A_\mu)+A_\sigma (\partial_\mu A_\nu +\partial_\nu A_\mu ))$$

$$=\Gamma^\lambda_{\mu \nu }+\frac{k}{2} g^{\lambda \sigma}(A_\mu F_{\nu \sigma }+A_\nu F_{\mu \sigma}+A_\sigma (\partial_\mu A_\nu +\partial_\nu A_\mu ))$$

According to the paper the $A_\sigma$ term shouldn't be there, but I can't figure out how to make it go away. Anybody have any ideas?

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I should note that there appear to be a number of errors in the paper, but I'm unsure as to whether or not this is an error as well. For example, in the Lorentz force law he derives there is a factor 1/2 in front of the second term on the RHS which keeps the equation from being gauge-invariant (so the 1/2 should just be 1). He also says second term "has no classical correspondence" although a quick check reveals that $k(u^5 + A_\mu u^\mu)$ is the Noether conserved momentum about the 5th dimension which can be associated with charge. So the second term is just an illusion. –  jld Feb 10 '13 at 8:11

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You missed a term in expanding the upper-indexed metric. The full version is below: \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}_{\nu5} - \partial_5 \tilde{g}_{\mu\nu}\right) \\ & = \frac{1}{2} g^{\lambda\sigma} \left(\partial_\mu \left(g_{\nu\sigma} + k A_\nu A_\sigma\right) + \partial_\nu \left(g_{\mu\sigma} + k A_\mu A_\sigma\right) - \partial_\sigma \left(g_{\mu\nu} + k A_\mu A_\nu\right)\right) \\ & \quad \qquad + \frac{1}{2} \left(-A^\lambda\right) \left(\partial_\mu \left(k A_\nu\right) + \partial_\nu \left(k A_\mu\right) - \partial_5 \left(g_{\mu\nu} + k A_\mu A_\nu\right)\right) \\ & = \frac{1}{2} g^{\lambda\sigma} \left(\partial_\mu g_{\nu\sigma} + \partial_\nu g_{\mu\sigma} - \partial_\sigma g_{\mu\nu}\right) + \frac{k}{2} g^{\lambda\sigma} \left(\partial_\mu \left(A_\nu A_\sigma\right) \partial_\nu \left(A_\mu A_\sigma\right) - \partial_\sigma \left(A_\mu A_\nu\right)\right) \\ & \quad \qquad - \frac{k}{2} A^\lambda \left(\partial_\mu A_\nu + \partial_\nu A_\mu\right) \\ & = \Gamma^\lambda_{\mu\nu} + \frac{k}{2} g^{\lambda\sigma} \left(A_\mu \left(\partial_\nu A_\sigma - \partial_\sigma A_\nu\right) + A_\nu \left(\partial_\mu A_\sigma - \partial_\sigma A_\mu\right) + A_\sigma \left(\partial_\mu A_\nu + \partial_\nu A_\mu\right)\right) \\ & \quad \qquad - \frac{k}{2} g^{\lambda\sigma} A_\sigma \left(\partial_\mu A_\nu + \partial_\nu A_\mu\right) \\ & = \Gamma^\lambda_{\mu\nu} + \frac{k}{2} \left(A_\mu F_{\nu\sigma} + A_\nu F_{\mu\sigma}\right). \end{align} This is because we have \begin{cases} \tilde{g}_{\mu\nu} = g_{\mu\nu} + k A_\mu A_\nu \\ \tilde{g}_{\mu5} = k A_\mu \\ \tilde{g}_{55} = k \end{cases} and also \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}

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Perfect! Silly mistake on my part, thanks for your help. –  jld Feb 10 '13 at 9:36

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