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Peskin page 23

  1. Jan 14, 2013 #1
    Hello everybody!

    Looking on solving equation 2.34 I am stuck at the fourth line. What is the calculation to do to go from:

    [tex] \delta^{(3)} (p'- q') \gamma (1 + \beta \frac{dE}{dp_3}) [/tex]

    to

    [tex] \delta^{(3)} (p'- q') \frac{\gamma}{E} (E + \beta p_3). [/tex]

    Many thanks for your help!!
     
  2. jcsd
  3. Jan 14, 2013 #2

    jedishrfu

    Staff: Mentor

    Can the function of E(p3) be determined?

    I can see how you can mix in a E by multipying by E / E making the last factor become:

    β E dE / dp3

    hence the step implies E dE / dp3 = p3 which would work if E = p3
     
  4. Jan 14, 2013 #3
    Well I had the same guess! :)

    BUT, let's set p^2= E^2 -|p|^2=m^2 equal to 0 for a massless particle, which would be actually ok as we're looking at the KG eq, well then E^2= |p|^2. But then I should set p_1 = p_2 = 0.. that I don't understand..
     
  5. Jan 14, 2013 #4

    Avodyne

    User Avatar
    Science Advisor

    [tex]E=(p_1^2+p_2^2+p_3^2+m^2)^{1/2}[/tex]
    [tex]{dE\over dp_3}={1\over2}(p_1^2+p_2^2+p_3^2+m^2)^{-1/2}(2p_3)={p_3\over E}[/tex]
    Faster:
    [tex]E^2=p_1^2+p_2^2+p_3^2+m^2[/tex]
    [tex]E\,dE = p_1\,dp_1 + p_2\,dp_2 + p_3\,dp_3[/tex]
    Set [itex]dp_1=dp_2=0[/itex], and rearrange to get
    [tex]{dE\over dp_3}={p_3\over E}[/tex]
     
  6. Jan 15, 2013 #5
    Got it!! It's that easy...

    Many many Thanks! :)
     
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