Solving the Callan-Symanzik Equation

  • Thread starter latentcorpse
  • Start date
In summary, the person is trying to solve for the massless case in a paper from 2009 and is stuck on two parts. They are unsure of what the Taylor expansion for the integral is and are also struggling to get Feynman rules from it.
  • #36
fzero said:
The energy of a massive particle is never zero!

Nevertheless, you can use conservation of energy and momentum to show that [tex]E_1 = E_{1'}[/tex], so the rest of your calculation works out.



Yex.

So I have been trying another CS question. I am asked at the end of q3 in this paper:
http://www.maths.cam.ac.uk/postgrad/mathiii/pastpapers/2006/Paper49.pdf
to explain why [itex]\mu \frac{dF}{d \mu}=0[/itex]

For this I wrote that we can substitute [itex]h=\mu^{\frac{1}{2}\epsilon} Z^{\frac{1}{2}}h_0[/itex] (where a subscript 0 denotes a bare quantity) to rewrite the definition of F all in terms of bare quantities

i.e. [itex]e^{iF(\lambda,m^2,h;\mu)}=\frac{1}{N_0} \int d[\phi] e^{iS_0[\phi]+h_0 \int d^dx \phi_0}[/itex]
[itex]\Rightarrow {iF(\lambda,m^2,h;\mu)}=\ln{\frac{1}{N_0} \int d[\phi] e^{iS_0[\phi]+h_0 \int d^dx \phi_0}}[/itex]
Now since bare quantities are by construction independent of the RG scale [itex]\mu[/itex], the RHS of this equality will be [itex]\mu[/itex] independent and therefore by equality so will the LHS.
Therefore [itex]\frac{dF}{d \mu}=0 \Rightarrow \mu \frac{dF}{d \mu}=0[/itex]

Does that look ok?

What about the next part though? We need to derive the CS equation. At first glance it looks like you just bash out the same chain rule procedure as we used previously but it turns out to be a bit more complicated. Previously, we defined the [itex]\gamma_\phi[/itex] term to go with the [itex]\frac{\partial}{\partial \sqrt{Z}}[/itex] term so that when we acted it on [itex]G_n^B=Z^{n/2}G_n[/itex] we pulled down a factor of n by the chain rule. However, this was because we were able to write the bare green's function in terms of the renormalised greens function with that factor of [itex]Z^{n/2}[/itex] to operate on.

But now we have the bare quantity F (which we aren't able to write in terms of any renormalised quantity) so I don't understand how we will get the h next to the [itex]\gamma_h[/itex] term?

Do we just define [itex]\gamma_h[/itex] differently from [itex]\gamma_\phi[/itex]? I guess I could just write [itex]\gamma_h=\frac{\mu}{h} \frac{dh}{d \mu}[/itex]. Does this work ok?

Thanks again!
 
Last edited by a moderator:
Physics news on Phys.org
  • #37
latentcorpse said:
Do we just define [itex]\gamma_h[/itex] differently from [itex]\gamma_\phi[/itex]? I guess I could just write [itex]\gamma_h=\frac{\mu}{h} \frac{dh}{d \mu}[/itex]. Does this work ok?

That's the definition they're using. It's very natural for anomalous dimensions to be related to logarithmic derivatives, since they naturally appear as factors like [tex]\mu^\gamma[/tex] in scaling relations.
 
Back
Top