Kaluza Klein Equations of Motion I have found a derivation of the Kaluza-Klein equations of motion on this webpage: http://www.konfluence.org/Williams_31Mar2012.pdf
As I understand it, he starts with the 5d geodesic equation of motion
$$\tilde{U}^a \tilde{\nabla}_a \tilde{U}^b=0$$ (tildes and roman indices refer to 5d).
He asks how a 4d geodesic would move (greek indices for 4d) and finds (eqn 34):
$$\tilde{U}^a \tilde{\nabla}_a \tilde{U}^\mu=0 \Rightarrow \frac{d U^\mu}{d \tau} + \tilde{\Gamma}^\mu_{\alpha \beta} \tilde{U}^\alpha \tilde{U}^\beta + 2\tilde{\Gamma}^\mu_{5 \alpha} U^\alpha U^5 + \tilde{\Gamma}^\mu_{55} U^5U^5 + U^\mu \frac{d}{d \tau} \ln{ \left( \frac{c d \tau}{ds} \right)}=0$$ with $$\tilde{U}^a = \frac{dx^a}{ds}. U^a=\frac{dx^a}{d \tau}.$$
Question 1: where does the last term (with log) in the above equation of motion come from?
He then says that he wants to match this with standard equation of motion for particle in presence of EM field and so identifies $kU^5=\frac{q}{mc}$. Somehow this can then be rearranged to give his eqn 2:
$$\frac{d U^\mu}{d \tau} + \Gamma^\mu_{\alpha \beta} U^\alpha U^\beta =\frac{q}{mc} F^{\mu \nu} U_\nu$$
Question 2: How on earth does this substitution help bring it to this form? How do I change the $\tilde{\Gamma}$'s to $\Gamma$'s? How do I get $F^{\mu \nu}$? Where does the crazy log term go?
 A: *

*Observe that
$$ \frac{\mathrm{d}\tilde{U}^\mu}{\mathrm{d}s} = \frac{\mathrm{d}}{\mathrm{d}s}\frac{\mathrm{d}x^\mu}{\mathrm{d}s} = \frac{\mathrm{d}}{\mathrm{d}s}\left(U^\mu\frac{\mathrm{d}\tau}{\mathrm{d}s}\right) = \frac{\mathrm{d}U^\mu}{\mathrm{d}\tau}\left(\frac{\mathrm{d}\tau}{\mathrm{d}s}\right)^2 + U^\mu\frac{\mathrm{d}^2\tau}{\mathrm{d}s^2}$$
and
$$ \frac{\mathrm{d}}{\mathrm{d}\tau}\ln\left(c\frac{\mathrm{d}\tau}{\mathrm{d}s}\right) = \frac{\mathrm{d}s}{\mathrm{d}\tau}\frac{\mathrm{d}^2\tau}{\mathrm{d}s^2}$$
and therefore
$$ \frac{\mathrm{d}\tilde{U}^\mu}{\mathrm{d}\tau} = \frac{\mathrm{d}U^\mu}{\mathrm{d}s} + U^\mu\frac{\mathrm{d}}{\mathrm{d}\tau}\ln\left(c\frac{\mathrm{d}\tau}{\mathrm{d}s}\right)$$

*To change the $\tilde{\Gamma}$ to $\Gamma$ you have to actually calculate the Christoffels for the 5D and 4D metric and express them in term of each other. It's tedious. The $F$ is, as usual, the curvature of the form $A$, i.e. $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$, and it appears when you explicitly calculate the Christoffels (plausibility argument: The Christoffels are certain symmetric combinations of derivatives of the metric, in which $A$ appears, and $F$ is also a special (anti-)symmetric combination of derivatives of $A$).
