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Differential cross section/simple integral

  1. Nov 11, 2008 #1
    At least it should be a simple integral...

    1. The problem statement, all variables and given/known data

    The whole text is here- http://i35.tinypic.com/2nisnp.jpg

    Basically (I think) I need to integrate the differential over all angles theta and phi, and get sigma(naught) back out.

    2. Relevant equations

    given in pic

    3. The attempt at a solution

    If I use the given substitution and integrate I get log(tan(pi/2)) which is log(0) which is broken... I don't really know where to go other than that.

    Thanks! Been reading the forum for a long time but this is my first question :)
  2. jcsd
  3. Nov 11, 2008 #2


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    You shouldn't be getting any Log terms when you integrate; if you show me your work, I can tell you where you are going wrong.
  4. Nov 11, 2008 #3
    I substitute sin(theta) for d(omega), and the integral for 1/sin(theta) is log(tan(theta/2)) at least as given by mathematica and the back of my book.

    I kind of figured I shouldn't be getting that, which is what makes me think I'm approaching the problem all out of whack.
  5. Nov 11, 2008 #4


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    You should be substituting [itex]sin(\theta) d \theta d \phi[/itex] for [itex]d \Omega[/itex] not just [itex]sin(\theta)[/itex]...[itex]d \Omega[/itex] is a differential, [itex]sin(\theta)[/itex] is just a function...you should have:

    [tex]\frac{d \sigma}{d \Omega}=\frac{d \sigma}{sin(\theta) d \theta d \phi}=\frac{\sigma_0}{4 \pi}[/tex]

    [tex]\Rightarrow d \sigma=\frac{\sigma_0}{4 \pi} sin(\theta) d \theta d \phi[/tex]

    ...do you follow?

    Then just integrate both sides of the equation.
  6. Nov 11, 2008 #5

    I figured it was something simple. I follow perfectly, thank you very much :)

    [...I need to figure out that LaTeX code stuff, that's pretty cool.]
  7. Nov 11, 2008 #6


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    Your welcome. There's an introduction to LaTeX here :smile:
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