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Yukawa Potential derivation

  1. Feb 17, 2015 #1
  2. jcsd
  3. Feb 17, 2015 #2

    ChrisVer

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    1st step: go to "spherical coordinates" and integrating out the [itex]\phi[/itex]... [itex] u = \cos \theta [/itex]
    2nd step: Integrating wrt to [itex]u[/itex], will give a difference of [itex]e^{i~something} - e^{-i~something} \propto \sin (something) [/itex]. In fact I wouldn't ever write it in terms of sin...
    3rd step: uses that he has two integrals one with the exponential with + and the other with the exponential with - ... then changes the integration variable of the one from k to -k, and gets this result.
    4th step: factorizes the denominator.
    5th step and then final: solves the (complex) integral

    If you want to see a step in more details, you can ask for a specific one.
     
    Last edited: Feb 17, 2015
  4. Feb 17, 2015 #3
    I actually get it with your points! Thank you very much!
     
  5. Feb 26, 2016 #4
    hello sir,
    i can't get the 3rd step. is change the integration variable, change the value from k to -k in that term, including the dk -> d(-k)?
     
  6. Feb 27, 2016 #5

    ChrisVer

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    like everytime you change your integration variable from [itex]x[/itex] to [itex]y(x)[/itex] what has to change is of course the differential, the integrand and the limits of the integral.
     
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