Differentiation on orbifolds

1. Dec 18, 2008

jdstokes

I'm trying to understand how Randall and Sundrum go from Eq. (9) to Eq. (10) in their RS1 paper:

http://arxiv.org/abs/hep-ph/9905221

I understand that since the extra dimension $\phi$ is periodic, we must have

$\frac{d^2}{d\phi^2}|\phi|\propto \delta(0) - \delta(\phi - \pi)$.

However, I'm not entirely sure why the proportionality constant is 2, i.e, why

$\frac{d^2}{d\phi^2}|\phi|= 2[\delta(0) - \delta(\phi - \pi)]$.

I'm assuming that it's related to the $\mathbb{Z}_2$ symmetry of the $S^1/\mathbb{Z}_2$ orbifold, but I'm not sure how to show his.

Thanks.

2. Dec 18, 2008

Hurkyl

Staff Emeritus
Well, if we lift the absolute value function back to be a function on R, we get a triangle wave; specifically, the function defined on $[-\pi, \pi]$ by $f(x) = |x|$, and extended periodically by $f(x + 2\pi) = f(x)$.

The first derivative of this function is a square wave (defined almost everywhere), alternating between 1 and -1. The second (distributional) derivative is therefore

$$f''(x) = \sum_{k = -\infty}^{+\infty} (-1)^k 2 \delta(x - k \pi)$$

This distribution has the appropriate symmetry to be a distribution on your orbifold, being even and periodic, so I would think it ought to be the correct result for the derivative.

(Disclaimer: I haven't actually worked with orbifolds before)

3. Dec 19, 2008

jdstokes

Yes, but where did the factor of 2 come from??

Edit: Oh I get it. Thanks

$|z|'' = sgn(z)' = [2\theta(z)-1]' = 2 \delta(z)$.

Last edited: Dec 19, 2008