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Volume of a simplex using delta function

  1. Feb 15, 2010 #1
    I tried to calculate the volume of a simplex, but got an integral I couldn't do.

    For simplicity take a 2-simplex (the volume of a 2-simplex is 1/6)

    [tex]V=\int da \int dx \int dy \int dz \mbox{ } \delta(1-a-x-y-z) [/tex]

    where the integration limits are over the 4-cube.

    My reasoning for this formula is that as (1-a) varies from 0 to 1, the integral counts the volume when x+y+z=(1-a), and gives zero otherwise, so you end up measuring the volume 0<x+y+z<1.

    I then wrote the delta function as a Fourier integral:

    [tex]V=\frac{1}{2\pi}\int dk \int da \int dx \int dy \int dz \mbox{ }e^{ik(1-a-x-y-z)}=
    \frac{1}{2\pi}\int dk \mbox{ }e^{ik} \int da \mbox{ }e^{-ika} \int dx \mbox{ }e^{-ikx}\int dy \mbox{ }e^{-iky}\int dz \mbox{ }e^{-ikz}[/tex]

    The integral of a,x,y,z all give the same value of [tex] \frac{1}{-ik}\left(e^{-ik}-1 \right) [/tex] since the limits are from 0 to 1. There are (d+2) of these terms for a d-simplex. So you end up with the integral:

    [tex]\frac{i^{d+2}}{2\pi} \int dk \mbox{ } e^{ik} \left(\frac{1}{k}\right)^{d+2}\left(e^{-ik}-1 \right)^{d+2} [/tex]

    I'm aware there are easier ways to calculate the volume of a d-simplex.

    But I want to know if in principle this method works, and why is the integral so hard when the result should be easy, since the volume of a d-simplex is a simple expression.
  2. jcsd
  3. Feb 19, 2010 #2
    I figured it out. The method I outlined is correct.

    So we had this:

    \frac{i^{d+2}}{2\pi} \int dk \mbox{ } e^{ik} \left(\frac{1}{k}\right)^{d+2}\left(e^{-ik}-1 \right)^{d+2}


    [tex]\frac{2^{d+1}}{\pi} \int dk \left(\frac{1}{k}\right)^{d+2}sin^{d+2}\left(\frac{k}{2}\right)\left[cos\left(\frac{dk}{2}\right)-isin\left(\frac{dk}{2}\right) \right][/tex]

    For some reason Mathematica can't handle this expression (or maybe just the online free version can't handle this). So what I did was that I inserted by hand several different values of d, and thankfully, the imaginary part vanished (as it should), and you get:


    which is the correct answer.

    I'm quite interested in seeing how the T.A. solves this problem next Monday in class...because a T.A. ought to spend more time on the homework than even the students, because the T.A. has a responsibility to be even more prepared, so his answer ought to be really good too.
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