Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Integral over a region in spacetime

  1. Nov 10, 2018 #1
    Hello, can anyone show me if this integral can be evaluated?

    ##\frac{1}{a_0^2}\int_\Sigma\frac{dy'dz'}{\bigg(y'^2+z'^2+\tfrac{1}{(2a_0)^2}\bigg)^2}##
     
  2. jcsd
  3. Nov 10, 2018 #2

    Svein

    User Avatar
    Science Advisor

    I am thinking polar coordinates - but what does the summation sign stand for?
     
  4. Nov 10, 2018 #3
    I have been thinking of polar form too. The summation sign represent a spacelike 2 surface in spacetime.

    Initially, I want to evaluate this integral in spacetime.
    $$\int_{\Sigma} \frac{dydz}{[a_{o}(y^{2}+z^{2})+2f_{o}y+2g_{o}z+c_{o}]^{2}}$$ where $$a_{o}c_{o}-f_{o}^{2}-g_{0}^{2}=\frac{1}{4}.$$

    My way is to define
    $$y':=y+\frac{f_0}{a_0},\,z':=z+\frac{g_0}{a_0}$$
    Then the quadratic becomes $$a_0(y'^2+z'^2)+c_0-\frac{f_0^2+g_0^2}{a_0}=a_0\bigg(y'^2+z'^2+\tfrac{1}{(2a_0)^2}\bigg).$$

    So now, I will need to evaluate$$\frac{1}{a_0^2}\int_\Sigma\frac{dy'dz'}{\bigg(y'^2+z'^2+\tfrac{1}{(2a_0)^2}\bigg)^2}\cdot$$ But this seems complicated. So I am thinking of using polar coordinate and if I do so, it becomes:
    $$\frac{1}{a_0^2}\int_\Sigma\frac{dy'dz'}{\bigg(y'^2+z'^2+\tfrac{1}{(2a_0)^2}\bigg)^2} = \frac{1}{a_0^2}\int_\Sigma\frac{rd\phi dr}{\bigg(r^2+\tfrac{1}{(2a_0)^2}\bigg)^2}$$

    Since this is in reference to spacetime, I am stuck as to how I should continue. If it is about a disk with radius R then I know what to do but for a spacelike 2 surface, I am still wandering.
     
  5. Nov 10, 2018 #4

    Svein

    User Avatar
    Science Advisor

    Hm. This is far outside my area (complex function algebras), but I can throw in some random thoughts and hope it will inspire you.
    • I don't know why, but Stokes' theorem keeps swirling around in my head (https://en.wikipedia.org/wiki/Stokes'_theorem)
    • I have a tendency to think complex variables - have you considered that
    • If I haven't done a blunder, I think [itex]\frac{d}{dr}\frac{1}{r^{2}+C}=\frac{-2r}{(r^{2}+C)^{2}} [/itex]
     
  6. Nov 10, 2018 #5

    mathman

    User Avatar
    Science Advisor

    It looks like your main problem is defining the domain of integration in purely geometric terms. You need to explicitly define ##\Sigma##.
     
  7. Nov 10, 2018 #6
    Thanks Svein for the ideas. I will check these out.

    Yes, mathman. The summation sign represent a spacelike 2 surface in spacetime. Its mentioned in the third post. Sorry about that.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted