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?

  • $\begingroup$ Minor comment to the post (v1): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$ – Qmechanic Jan 28 '16 at 22:43
  1. 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 $$ \frac{\mathrm{d}\tilde{U}^\mu}{\mathrm{d}\tau} = \frac{\mathrm{d}U^\mu}{\mathrm{d}s} + U^\mu\frac{\mathrm{d}}{\mathrm{d}\tau}\ln\left(c\frac{\mathrm{d}\tau}{\mathrm{d}s}\right)$$

  2. 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 $A$).

  • $\begingroup$ So I'm reading that if you get the 4d EoM from the 5d EoM, in order to make it look like the Lorentz Law, you should identify $\dot{x}^5+A_\mu \dot{x}^\mu =\frac{q}{m}$ (see eqn 43.28 in blau.itp.unibe.ch/newlecturesGR.pdf). But, like you pointed out in a different post of mine, we really want to identify $U^5:=m \dot{x}^5=\frac{q}{m}$ (which makes sense - motion in 5th direction gives charge). How do I get rid of the $A_\mu \dot{x}^\mu$ term? Thanks. $\endgroup$ – user11128 Jan 29 '16 at 18:09
  • $\begingroup$ @user11128: The dot in those notes refers to the derivative w.r.t. $s$, while the $U^5$ is the derivative of $x$ w.r.t. $\tau$. $\endgroup$ – ACuriousMind Jan 29 '16 at 18:22
  • $\begingroup$ Does that help to get rid of $A_\mu \dot{x}^\mu$? I see what you want to do: rewrite $\dot{x}^5+A_\mu \dot{x}^\mu = U^5$. But to my eyes, $U^5=\frac{dx^\mu}{d \tau}$ and I don't see how I can split that into the two terms above? $\endgroup$ – user11128 Jan 29 '16 at 18:56
  • $\begingroup$ Am I understanding you correctly? I'm assuming $\dot{x}^5+A_\mu \dot{x}^\mu$ need to give a momentum in 5th direction (with respect to 5d proper time?) because the normal Kaluza Klein mantra is that momentum in extra direction becomes the electric charge. Is this the right idea? Thanks $\endgroup$ – user11128 Jan 30 '16 at 11:23
  • $\begingroup$ @user11128: Divide the KK $\mathrm{d}s^2$ by $\mathrm{d}\tau^2$. You find $\frac{\mathrm{d}s^2}{\mathrm{d}\tau} = c^2 + \phi^2(A_\mu\dot{x}^\mu + \dot{x}^5)^2$. Recognize that this is $U_i U^i + U_5 U^5$, but from the viewpoint of the 4D metric, $U_5 = U^5$, so you get $U^5 = \dot{x}^5 + A_\mu\mathrm{d}x^\mu$. $\endgroup$ – ACuriousMind Jan 30 '16 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.