1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differentiation on orbifolds

  1. Dec 18, 2008 #1
    I'm trying to understand how Randall and Sundrum go from Eq. (9) to Eq. (10) in their RS1 paper:


    I understand that since the extra dimension [itex]\phi[/itex] is periodic, we must have

    [itex]\frac{d^2}{d\phi^2}|\phi|\propto \delta(0) - \delta(\phi - \pi)[/itex].

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

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

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

  2. jcsd
  3. Dec 18, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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 [itex][-\pi, \pi][/itex] by [itex]f(x) = |x|[/itex], and extended periodically by [itex]f(x + 2\pi) = f(x)[/itex].

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

    [tex]f''(x) = \sum_{k = -\infty}^{+\infty} (-1)^k 2 \delta(x - k \pi)[/tex]

    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)
  4. Dec 19, 2008 #3
    Yes, but where did the factor of 2 come from??

    Edit: Oh I get it. Thanks

    [itex]|z|'' = sgn(z)' = [2\theta(z)-1]' = 2 \delta(z)[/itex].
    Last edited: Dec 19, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?