# Renormalization Group for dummies

by waterfall
Tags: dummies, renormalization
P: 381
 Quote by atyy One way to think about the changing constants is to realise you are just writing and "effective theory". In the Box-Jenkins philosophy: all models are wrong, but some are useful. Say you have a curve. Every point on the curve can be approximated by a straight line. Depending on which part of the curve you are approximating, the slope of the line will change. The straight line is your "effective theory" and the changing slope like your changing coupling constant. This example is not a detailed comparison, but it's the general philosophy of the renormalization group. As for detailed mathematical correspondence, apart from quantum field theory, the renormalization group has been applied in classical statistical mechanics and classical mechanics.
I read in wiki that "Geometric series are used throughout mathematics, and they have important applications in physics, engineering, biology, economics, computer science, queueing theory, and finance."

So you are saying that Renormalization Group concepts and regulator thing are also used in biology, economics, finance and not just in QFT? So in the calculations in biology. The coupling constant equivalent can become infinite in the second term but if one makes a cut-off at first term. it is finite?
P: 8,800
 Quote by waterfall I read in wiki that "Geometric series are used throughout mathematics, and they have important applications in physics, engineering, biology, economics, computer science, queueing theory, and finance." So you are saying that Renormalization Group concepts and regulator thing are also used in biology, economics, finance and not just in QFT? So in the calculations in biology. The coupling constant equivalent can become infinite in the second term but if one makes a cut-off at first term. it is finite?
Renormalization has nothing to do with infinities. QED is renormalizable and it has a cut-off - it is not a true theory valide at all energies, it is only an effective theory like gravity, valid below the Planck scale. Once you have a cut-off, there are no infinities. Sometimes you are lucky and you get a theory where you can remove the cut-off, like QCD. But in QED, as far as we know, the cut-off probably cannot be removed.
P: 381
 Quote by atyy Renormalization has nothing to do with infinities. QED is renormalizable and it has a cut-off - it is not a true theory valide at all energies, it is only an effective theory like gravity, valid below the Planck scale. Once you have a cut-off, there are no infinities. Sometimes you are lucky and you get a theory where you can remove the cut-off, like QCD. But in QED, as far as we know, the cut-off probably cannot be removed.
I can't believe why no one is answer my simple question. I know in power series, one expand the terms. I'm asking if the techniques in Renormalizaton Group can be applied to Finance or Biology problem, not just in QFT. If one knows the answer to this. Please let me know. Thanks.
P: 1,940
 Quote by waterfall I'm asking if the techniques in Renormalizaton Group can be applied to Finance or Biology problem
Then perhaps you should post the question in the Biology forum.
P: 8,800
 Quote by waterfall I can't believe why no one is answer my simple question. I know in power series, one expand the terms. I'm asking if the techniques in Renormalizaton Group can be applied to Finance or Biology problem, not just in QFT. If one knows the answer to this. Please let me know. Thanks.
In very, very broad terms - yes - everyone uses effective theories.

Less facetiously, the philosophy of the central limit theorem is like the renormalization group. Take the heights of people. That's due to hundreds of genes, whose interactions we hardly understand. Yet when you plot a distribution of heights of populations, very often you get a Gaussian distribution characterized fully by two parameters: mean and variance. So if you are looking coarsely on a population level, you don't need the theory of all the genes with hundreds of parameters - you just need the Gaussian distribution with two parameters.

Here the Gaussian distribution is derived using the renormalization group:
http://www.math.princeton.edu/facult...gorovLec07.pdf
P: 381
 Quote by atyy In very, very broad terms - yes - everyone uses effective theories. Less facetiously, the philosophy of the central limit theorem is like the renormalization group. Take the heights of people. That's due to hundreds of genes, whose interactions we hardly understand. Yet when you plot a distribution of heights of populations, very often you get a Gaussian distribution characterized fully by two parameters: mean and variance. So if you are looking coarsely on a population level, you don't need the theory of all the genes with hundreds of parameters - you just need the Gaussian distribution with two parameters. Here the Gaussian distribution is derived using the renormalization group: http://www.math.princeton.edu/facult...gorovLec07.pdf
Thanks. So it applies elsewhere too.

Going back to QED and the coupling constant. When higher-order terms in the perturbation series is used, the length scale becomes smaller. I wonder if this is the reason why the coupling constant becomes larger when more terms are used. Bill Hobba kept saying this but he didn't explain the physical reason why. Maybe it's because as more virtual particles or more terms are used, the length scale get smaller and smaller and hence the coupling constant secretly being dependent on how many terms means the coupling constant is equal to the sum of the feynmann vertex in the virtual particles interactions in 1st, 2nd, 3rd terms (I'm aware virtual particles are just the terms in the perturbation expansion so I'll use them interchangeably)? How do you understand this physically.
P: 381

Thanks for the updated descriptions. You'd make a great PF science advisor someday.
Say, in the magnetic moment of the electron with measured value of 1.00115965219 and calculated value of 1.0011596522 in the fourth term in the power series. Did they use Renormalization Group there? Maybe the reason the second term to fourth term didn't produce an infinite coupling constant is because the calculation replaces it with 1/137 as you described. What then is the original coupling constant when no Renormalization Group procedure was used?
PF Gold
P: 2,912
 Quote by waterfall Thanks for the updated descriptions. You'd make a great PF science advisor someday. Say, in the magnetic moment of the electron with measured value of 1.00115965219 and calculated value of 1.0011596522 in the fourth term in the power series. Did they use Renormalization Group there? Maybe the reason the second term to fourth term didn't produce an infinite coupling constant is because the calculation replaces it with 1/137 as you described. What then is the original coupling constant when no Renormalization Group procedure was used?
They used re-normalisation to calculate it - the re-normalisation group is simply a description of how quantities that are regulator (ie cutoff) dependant such as the coupling constant behave as the value of the regulator changes.

Thanks
Bill
P: 381
note the interesting comments by Science Advisor tom.stoer that the Renormalization Group is just a DIRTY TRICK.

 Renormalization inorer to remove infinities is just a dirty trick. The renormalization group itself says that you can look at a theory at different energy scales E and E' (or at different length scales which is the same) and renromalize masses and coupling constants g(E) and g'(E') such that the predictions of both theories are the same. Essentially they are THE SAME theory. This can e.g. be interpreted as "integrating out" degrees of freedom and hiding their contribution in the renormalization of parameters. Have a look at http://en.wikipedia.org/wiki/Renorma...lization_group

Renormalization in perturbative QFT is constantly teached to be "removing divergences". Unfortunately this is only a dirty trick! Renormalization (non-perturbative renormalization) is something totally different - and it may even appear in systems w/o any divergences.

Have a look at http://en.wikipedia.org/wiki/Renormalization_group

 Suppose you have a theory with certain coupling constants ga, gb, ...; suppose you study this theory at some energy scale E. Then you will observe that you can re-express the theory in terms of different coupling constants g*a, g*b,... at a different energy scale E* (and different interactions, i.e. changing E may turn on new couplings). It is interesting that these two 'theories' defined at different energy scales E and E* are essentially the same theory! The way how to find the description at scale E* given the description at E is via studying trajectories ga(E), gb(E) in the space of all coupling constants. Different points on the same trajectory correspond to different scales E, E*, ... of the same theory; different theories are defined by points not connected by a trajectory. These trajectories define something called renormalization group flow. The renormalization group of a certain theory is defined via a set of coupled differential equation defining this flow of coupling constants. Perturbative QFT is defined by studying the flow near the point ga = gb = ... = 0; in some case (QCD) it can be shown that this is reasonable in certain regimes; in other cases (QG) there are strong indications that G=0 may never be a valid starting point, neither in the UV (not asymptotically free), nor in the IR (coupling constant of Newtonian gravity G is not zero).
PF Gold
P: 2,912
 Quote by waterfall Bill Hobba kept saying this but he didn't explain the physical reason why
The physical reason the coupling constant gets larger is the the shielding effect of the virtual particles around an electron. Landau showed that because of that for any finite charge at any distance the charge would measure zero. This created the famous zero charge problem. The only way around it is for the charge to actually be infinite and to grow to an infinite value the closer and closer you get to it - this is called the Landau pole. That is the reason for the infinities - the closer you get the higher the energy you must use. If I remember correctly this is not just theory at 90gv it measured 1/127 (OK to avoid egg on my face I checked it)

As one other person posted the fact it blows up to infinity means the theory is sick and incorrect - but we already know that because it gets replaced by the electroweak theory which will probably also get replaced by something else someday and that something else will hopefully be free of these problems.

Thanks
Bill
PF Gold
P: 2,912
 Quote by waterfall note the interesting comments by Science Advisor tom.stoer that the Renormalization Group is just a DIRTY TRICK.
I do not agree with that. The effective field theory approach, as he sort of says, avoids this blowing up to infinity business:
http://en.wikipedia.org/wiki/Effective_field_theory

The only issue is theories like QED etc have a domain of applicability beyond which is does things like blow up to infinity - stick to that domain and everything is fine. Its hardly flabbergasting news and a DIRTY TRICK theories may not be valid to all energies.

Thanks
Bill
P: 381
 Quote by bhobba The physical reason the coupling constant gets larger is the the shielding effect of the virtual particles around an electron. Landau showed that because of that for any finite charge at any distance the charge would measure zero. This created the famous zero charge problem. The only way around it is for the charge to actually be infinite and to grow to an infinite value the closer and closer you get to it - this is called the Landau pole. That is the reason for the infinities - the closer you get the higher the energy you must use. If I remember correctly this is not just theory at 90gv it measured 1/127. As one other person posted the fact it blows up to infinity means the theory is sick and incorrect - but we already know that because it gets replaced by the electroweak theory which will probably also get replaced by something else someday and that something else will hopefully be free of these problems. Thanks Bill
Thanks for the facts. Last question for this thread before I sit back and read all the papers as well as study deeper aspects of calculus with all this nice background in mind.

My last question is this (I saw a similar question asked in the archives but unanswered).

In the Dirac Equation. The magnetic moment of the electron is calculated as 1. In the 4th term in the power series, it's equal to 1.0011596522. The interacting fields are the electron self magnetic field and the electron. What about the interactions of say two electrons, what would be the Dirac Equation counterpart of 1.0 in the magnetic moment of the electron calculation? Do you calculate the dirac equation of each electron by adding them or calculate both of them combined? And if the result is for example 3.0. After the fourth term in the power series, would the result only be 3.0111 or would it be 5.0 (I don't think it would just be a small 3.0111 isn't it because the electric field strength is bigger). Thanks.
PF Gold
P: 2,912
 Quote by waterfall In the Dirac Equation. The magnetic moment of the electron is calculated as 1. In the 4th term in the power series, it's equal to 1.0011596522. The interacting fields are the electron self magnetic field and the electron. What about the interactions of say two electrons, what would be the Dirac Equation counterpart of 1.0 in the magnetic moment of the electron calculation? Do you calculate the dirac equation of each electron by adding them or calculate both of them combined? And if the result is for example 3.0. After the fourth term in the power series, would the result only be 3.0111 or would it be 5.0 (I don't think it would just be a small 3.0111 isn't it because the electric field strength is bigger). Thanks.
Don't know the answer to that one - sorry.

Thanks
Bill
P: 381
 Quote by bhobba Don't know the answer to that one - sorry. Thanks Bill
Anyone else know? atyy?
P: 1,943
 Quote by Ken G Do you mean that the full integral has a closed-form expression (involving modified Bessel functions) for any value of lambda, but the series expression (involving Gamma functions) has terms that only converge absolutely when lambda<1? So if we had lambda>1, and all we had was the series form, we might worry the integral doesn't exist, when in fact it does?
The series for this integral diverges for _any_ nonzero value of lambda. That's why it is called an asymptotic series.