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A bothersomeIntegral

  1. Apr 14, 2010 #1
    1. The problem statement, all variables and given/known data

    Hello

    The following integral turns out to be a bit nerve-wracking for me as I’ve tried almost any possible way to get the answer but all my efforts got blocked up and fell flat in their face somehow. The integral which may sound familiar is

    [tex]\tau=\int_{0}^{2m}{\frac{1}{\sqrt{2m/r-1}}}dr[/tex],

    and shows the lifetime of an observer swallowed up by the Schwarzschild BH from the time he enters the BH through the hypersurface [tex]r=2m[/tex] to the time he falls into the spacetime singularity at [tex]r=0[/tex] where he is doomed to die dreadfully.

    Any help will be highly appreciated.

    AB
     
  2. jcsd
  3. Apr 14, 2010 #2

    Mark44

    Staff: Mentor

    The integrand can be written as
    [tex]\frac{\sqrt{r}}{\sqrt{2m - r}}[/tex]

    I would start first with an ordinary substitution such as u = sqrt(2m - r) and see if that got me somewhere.

    Also, you integrand is undefined at r = 0 and r = 2m, so you have an improper integral than you'll need to use limits to find. Since there are discontinuities at both endpoints, you'll need to split the integral into two separate integrals, with limits for each.
     
  4. Apr 14, 2010 #3
    Unfortunately the substitution you offer doesn't work because the integrand then turns again into another indocile function that can't be integrated by any method. I mean the modified integrand

    [tex]-2\sqrt{2m - u^2}[/tex]

    is not reducible to a simple function which would be capable of being written in terms of elementary functions if one integrated it wrt the new variable u! So what now?
     
  5. Apr 14, 2010 #4
    Can we put [tex]u^2/2m=\cos^2(k)[/tex] and keep going?
     
  6. Apr 14, 2010 #5
    Why not put 2m/r - 1 = u^2 in the original integral?
     
  7. Apr 14, 2010 #6

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    You might begin with the sub [itex]u=\frac{2m}{r}[/itex], and follow that with an appropriate trig sub.
     
  8. Apr 14, 2010 #7
    I tried a simple substitution x=1/r and seemed to get an answer very quickly using a table for the new integral. It's possible I worked too fast and made a mistake, but you can check it out quickly enough.
     
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