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Calculating Cross Section

  1. Jul 25, 2016 #1
    1. The problem statement, all variables and given/known data
    I have a problem in calculating cross-section in elektron-positron -> muon-antimuon scattering.

    2. Relevant equations
    In the relativistic limit, we find the differential cross-section of {e}+{e^-} -> {μ}+{μ^-} is

    \frac{d\sigma}{d\Omega}=\frac{\alpha^2}{4s}*(1+cos^2{\theta}).

    When I integrating over \theta, i get an answer:

    \sigma=\frac{3*\pi*\alpha^2}{8*s},

    but that is not the correct answer. Due to my reference (Halzen and Martin), the answer should be

    \sigma=\frac{4*\pi*\alpha^2}{3*s}.

    3. The attempt at a solution
    Can someone help me through this problem? My intuition said I have mistaken the integration. But I've tried to re-integral-ing it, still not come up with correct solution.
     
  2. jcsd
  3. Jul 25, 2016 #2

    BvU

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    Hello,

    Is ##d\Omega## just ##d\theta## or has ##\phi## something to do with it too ?

    Anyway, show your steps in detail, please. That makaes it easier to follow and point out where things may go wrong...
     
  4. Jul 26, 2016 #3

    Hello Mr., thank you for your respond.

    In my first reference (Halzen and Martin), the instructor is to integrate over ##\theta## and ##\phi##. Here is my doodling:

    $$ \sigma = \frac{\alpha^2}{4s}\int{(1+cos^2{\theta})}d\theta d\phi $$

    Substituting the trigonometry identity,

    $$ \sigma = \frac{\alpha^2}{4s} \int {\frac{(3+cos{2\theta})}{2}} d\theta d\phi $$.

    Dividing the integration,

    $$ \int 3 d\theta d\phi , \quad \int {cos{2\theta}}d\theta d\phi $$

    and integrating from ## 0 ## to ## \pi ## for ##\theta##, ##0## to ##2\pi## fpr ##\phi##. The second term is vanish, so I have

    $$ \int 3 d\theta d\phi=3(\pi)(2\pi)$$.

    Thus, when inserting to ## \sigma ##, I have

    $$ \sigma = \frac{3\pi^2\alpha^2}{4s} $$.

    Using my second ref (Schwartz), the integration just applied over ##\theta## and I get

    $$ \sigma = \frac{3\pi\alpha^2}{8s}$$.

    Which step do I miss?
     
  5. Jul 26, 2016 #4

    BvU

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    If I remember well, ##d\Omega \ne d\phi d\theta ## because then you would end up with ##2\pi^2## where you expect ##4\pi## . Wasn't it ##d\Omega = \sin\theta \;d\theta d\phi ## ?
     
  6. Jul 26, 2016 #5
    Speechless. That's true Mr. Thank you very much. (shy)
     
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