Solving Peskin Equation 12.66 Problem

[physichu]

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 :)

 

[ChrisVer]

there is a transition I think from the 4-momentum $p$ to a 3-momentum magnitude $p$ in 12.66?

 

[vanhees71]

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).

 

[physichu]

My Hero :)

Just so I understand, I was supposed to get to that by myself?

 

[ChrisVer]

I was supposed to get to that by myself?

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]