# Kaluza Klein charge

If I take a $(d+1)$ dimensional Einstein Hilbert Lagrangian $L_{d+1}=\sqrt{-\hat{g}} \hat{R}$ and perform a standard Kaluza Klein dimensional reduction by periodically identifying one direction, let's say $z$, by $z \sim z + 2 \pi r$, I arrive at a $d$ dimensional Lagrangian $L_d=\sqrt{-g}(R - \frac{1}{2} (\partial \phi)^2 - \frac{1}{4} e^{-2(d-1) \alpha \phi} F^2)$.

We can see that the Kaluza Klein vector in the $(d+1)$ dimensional metric manifests itself as a $d$ dimensional gauge field in the lower dimensional system. This gauge field has some associated electric charge and I would like to know how, and why, this gets quantized as a result of the identification $z \sim z + 2 \pi r$.

Thanks very much.

The Kaluza-Klein equations of motion (the geodesic equations) for a particle moving in the 5D spacetime contain the equations of motion of a particle in 4D spacetime under influence of electromagnetism if and only if one identifies $p^5 = mU^5 = \frac{1}{\sqrt{G}}cq$, i.e. relates the momentum in the fifth dimension $p^5$ to electric charge $q$. (And yes, the momentum in the fifth dimension is a constant of motion, so this is allowed.)
Now, for the Kaluza-Klein theory on a 5D cylinder $\mathbb{R}^{3,1}\times S^1$, the fifth coordinate is that of a circle, and since position and momentum are Fourier transforms of each other in the quantum theory, this means the allowed momenta $p^5$ are discrete, yielding discretization of the electric charge.
• Thanks for the reply. Can you explain why the momentum is to be identified with the specific quantity $\frac{1}{\sqrt{G}} cq$? Or possibly provide a reference? Intuitively, shouldn't the charge have some dependence on the size of the circle? – user11128 Jan 20 '16 at 21:14
• @user11128: Yes, the charge has someting to do with the size of the circle. Use the deBroglie relation $p = \hbar / \lambda$ to get the allowed momenta in terms of the circle length (the allowed wavefunctions are standing waves on the circle with circumference $L$). This gives a naive bound on the size of the fifth dimension $\sim 10^{-30}\mathrm{m}$, iirc. – ACuriousMind Jan 21 '16 at 2:27
• Thanks for the notes. Usually in KK reductions we truncate to the massless sector ($S^1$ needs to be small enough so as not to be visible forcing the masses to be very high - too high to detect). If we truncate the massive modes then 5.47 is telling me $k^2=-\frac{m^2}{h^2}$. What is this $m$? and how can $k^2<0$ since above 5.46 he just required $k^2>0$? Thanks. – user11128 Jan 21 '16 at 10:01