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Peskin Equation 12.66

  1. Jul 11, 2015 #1
    I have a problem with that equation. I understand (dont know if I'm right) that ##p = - M##. But than, isn't ##g\left( { - {{{p^2}} \over {{M^2}}}} \right)## just equal ##g\left( { - 1} \right)##?

    And my bigest problem: in 12.66

    ##\left[ {p{\partial \over {\partial p}} - \beta \left( \lambda \right){\partial \over {\partial \lambda }} + 2 - 2\gamma \left( \lambda \right)} \right]{G^{\left( 2 \right)}}\left( p \right) = 0##

    Where dose the free 2 (after the ##\beta ## term) comes from?

    From looking at 12.78 and 12.84 I realize it's the inverse mass dimantionalty of ##G##, But how did it got ther?

    Any help would be apriciated :)
     
  2. jcsd
  3. Jul 13, 2015 #2

    ChrisVer

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    there is a transition I think from the 4-momentum ##p## to a 3-momentum magnitude ##p## in 12.66?
     
  4. Jul 13, 2015 #3

    vanhees71

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    No, by definition Peskin/Schroeder sets ##p=\sqrt{-p^2}## for space-like ##p##, i.e., ##p^2<0##, where you evaluate the Green's functions (corresponding to Euclidean field theory) first. The key is (12.65), which makes use of dimensional analysis: For a massless theory the mass-dimension (-2) quantitiy ##G^{(2)}## must be proportional to ##1/p^2## times a dimensionless function of ##p##. Since for a massless theory the only way to get a dimensionless function like this, it must be a function of ##p^2/M^2##, where ##M## is the renormalization scale. Thus the two-point Green's function must be of the form (12.65):
    $$G^{(2)}(p^2,M^2)=\frac{\mathrm{i}}{p^2} g(-p^2/M^2).$$
    Now you can indeed express the derivative wrt. ##M## by a derivative with respect to ##p##, just knowing this functional form. In the original Callan-Symanzik equation you need the derivative
    $$M \partial_M G^{(2)}=\frac{2\mathrm{i}}{M^2} g'(-p^2/M^2).$$
    In the last step, I used (12.65). On the other hand from this ansatz you get
    $$p \partial_p G^{(2)}=-\frac{2 \mathrm{i}}{p^2} g(-p^2/M^2)-\frac{2 \mathrm{i}}{M^2} g'(p^2/M^2)=-2 G^{(2)}-M \partial_M G^{(2)}.$$
    In the last step used again (12.65) and our result for ##M \partial_M G^{(2)}##. From this you get
    $$M \partial_M G^{(2)}=-p\partial_p G^{(2)}-2 G^{(2)}.$$
    Now substituting this into the Callan-Symanzik equation and flip the sign on the left-hand side gives (12.66).

    That's a fine trick to get the behavior of the Green's function by solving this RG flow equation, given the functions ##\beta## and ##\gamma##. This goes beyond perturbation theory despite the fact that ##\beta## and ##\gamma## are given only perturbatively. It's a kind of leading-log resummation (see Weinberg, The quantum theory of fields, vol. 2 for a nice explanation of this issue).
     
  5. Jul 16, 2015 #4
    My Hero :)

    Just so I understand, I was supposed to get to that by myself?
     
  6. Jul 16, 2015 #5

    ChrisVer

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    Ehmmm, is that a question...? you will either have to derive the things that seem non-trivial to you [in order to understand them], or you will have to just accept them as they are [if you don't care about understanding them]
     
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