• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

174 submissions , 137 unreviewed
4,308 questions , 1,640 unanswered
5,089 answers , 21,602 comments
1,470 users with positive rep
635 active unimported users
More ...

  What is the physical meaning of this simplification to calculate the effective coupling constants for a Gaussian model with quartic interactions?

+ 3 like - 0 dislike

To calculate the effective coupling constants $u'_2(q)$ and $u'_4(q)$ of the effective Hamiltinian eq (4.9) of this paper

$$ H' = -\frac{1}{2}\int\limits_q u'_2(q)\sigma'_q\sigma'_{-q} - \int\limits_{q_1}\int\limits_{q_2}\int\limits_{q_3}u'_4(q_1,q_2,q_3,-q_1-q_2-q_3) \sigma'_{q_1}\sigma'_{q_2}\sigma'_{q_3}\sigma'_{q_-q_1-q_2-q_3}$$

The following simplifications are introduced into eq (4.20) and (4.21) to calculate $u'_2(q)$ and $u'_4(q)$ respectively

  1. $u'_2(q)$ is only evaluated to order u, which means only tree level diagrams are considered

  2. Higher order than quartic interactions are neglected

  3. $\int\limits_{\frac{1}{2} < ¦p¦ < 1} \frac{1}{p^2+r} \rightarrow \frac{1}{1+r}\int\limits_{\frac{1}{2} < ¦p¦ < 1} 1 $

  4. $\int\limits_{\frac{1}{2} < ¦p¦ < 1} \frac{1}{p^2+r}\frac{1}{[(\frac{1}{2}q_1+\frac{1}{2}q_2-p)^2 +r]} \rightarrow \frac{1}{(1+r)^2}\int\limits_{\frac{1}{2} < ¦p¦ < 1} 1 $

What is the physical meaning of 3. and 4. ? Is there an "intuitive explanation for the physical meaning of these two simplifications?

PS: here is an alternative link to the paper that maybe works better.

asked Feb 10, 2013 in Theoretical Physics by Dilaton (5,240 points) [ revision history ]
retagged Mar 12, 2014
As they are written down, these equations are nonsensical, as they don't even have the right units. $r$ cannot have dimension mass^2 and 0 at the same time.

This post imported from StackExchange Physics at 2014-03-12 15:25 (UCT), posted by SE-user Vibert

1 Answer

+ 2 like - 0 dislike
  1. $\int\limits_{\frac{1}{2} < ¦p¦ < 1} \frac{1}{p^2+r} \rightarrow \frac{1}{1+r}\int\limits_{\frac{1}{2} < ¦p¦ < 1} 1 $

  2. $\int\limits_{\frac{1}{2} < ¦p¦ < 1} \frac{1}{p^2+r}\frac{1}{[(\frac{1}{2}q_1+\frac{1}{2}q_2-p)^2 +r]} \rightarrow \frac{1}{(1+r)^2}\int\limits_{\frac{1}{2} < ¦p¦ < 1} 1 $

Since the referenced document is behind a paywall, it is a little difficult to know for sure of the context, but when I see these two equations, it would appear they are making a simplification about the amplitudes associated with propagators (internal lines of Feynman diagrams). It appears to be saying the upper half of the range of values associate with momentum follow the rules of geometric progression or more appropriately a geometric series and the region of integration has a constant slope (e.g. there is a uniform accumulation in the identified region of integration) Represented as. $$ \int\limits_{\frac{1}{2} < ¦p¦ < 1} 1 $$

I would understand it as a statement about the behavior of the propagators at high momentum, e.g. they are well behaved, and the contribution of more complicated diagrams diminishes at higher orders.

Since the paper is Wilson's paper on renormalization, this interpretation would at least be consistent.

If anyone would like to further clarify, please do.

This post imported from StackExchange Physics at 2014-03-12 15:25 (UCT), posted by SE-user Hal Swyers
answered Feb 10, 2013 by Hal Swyers (20 points) [ no revision ]

Please log in or register to answer this question.

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights