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Renormalization Group for dummies

  1. Feb 19, 2012 #1
    Renormalization Group concept is rarely given in laymen book on QM and QFT and even Quantum Gravity book like Lisa Randall Warped Passages. They mostly described about
    infinity minus infinity and left it from there. So if you were to write about QFT for Dummies. How would you share it such that common folks can understand them? I'll share what I know and some questions.


    In the convensional popularization on the infinity problem, it is often said that:

    M_correction = infinity
    m_bare = m- infinity = - infinity

    And in renormalization group, I understood it simply that instead of it, one simply assume m_bare is some definite value? Is that correct?
    How about the M_correction. How did the value gets lower to finite?

    But I went to many references. In the book The Story of Light. It was mentioned:

    "With the bare mass also taken to be of infinite value, the two infinities - the infinities coming out of the perturbation calculations and the infinity of the bare mass - cancel each other out leaving us with a finite value for the actual, physical mass of an electron".

    So as more detailed accounts or Renormalization. It is not just m_bare = m - infinity, but the perturbation calculation infinity minus the - m_bare = m_observed. Do you agree?

    Now in Renormalization Group calculations. According to http://fds.oup.com/www.oup.co.uk/pdf/0-19-922719-5.pdf [Broken] the fine structure constant for example is altered and this altered value is entered into the perturbation equation as well as mass and charge.. but how do you make a power series with an altered fine structure constant no longer diverge?? Landa pole is still landa pole whatever is the fine structure constant values.

    Also someone said said "What's really happening is that your approximate theory is incomplete, and at some high energy, new physical processes show up, and change how the effective mass (charge, etc) varies with energy, so that the "bare" quantities are more reasonable.".

    What is this example of new physical processes showing up at high energy that can affect or make effective mass varies with energy. I have a rough idea of Renormalization Group. Checked out many references for hours but want to get the essence and gist of it. I think this details of the nature of how new physical process showing up at high energy that can affect or make effective mass varies with energy (as well as fine structure constant varies with energy) can give the heart of the understanding.

    Thanks.
     
    Last edited by a moderator: May 5, 2017
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  3. Feb 19, 2012 #2

    Ken G

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    You want a better "dumbed down" version such that you can understand the answers to your questions. I don't really understand renormalization group physics either, but I fear that the only real answer to your question is, if you want a "dumbed down" version to work for you, you have to try to be dumber, and not ask those questions. If you insist on not being dumb, and ask those questions, then no "dumbed down" version is going to work for you, you'll need the real deal. Your choice!
     
  4. Feb 20, 2012 #3

    Of course not further dumbing down. I just want answers in terms of power series, coupling constant, higher energies and those terms which a mere conceptual description is enough without getting into deep rigorous mathematics which most introductory sites on Renormalization Group contain that repel the laymen from understanding its essence. Of course one has to understand some basics of calculus like the infinite series, divergences and other basic which I have.

    So anyone can share what the heck is the Renormalization Group in terms of my original questions?
     
  5. Feb 20, 2012 #4

    bhobba

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    Unfortunately the jig is up on this one - you need some math. Here is the simplest explanation I know:
    http://arxiv.org/pdf/hep-th/0212049.pdf

    There is a trick in applied math called perturbation theory. The idea is you expand your solutions in a power series about a parameter that is small and you can calculate your solution term by term getting better accuracy with each term. The issue is the coupling constant is thought to be small so you expand about it. The first term is fine. You then calculate the second term - oh oh - its infinite - bummer. Whats wrong? It turns out the coupling constant in fact is not small - but rather is itself infinite so its a really bad choice. Ok how to get around it. What you do about it is what is called regularize the equations so the equations are of the form of a limit depending on a parameter called the regulator. You then choose a different parameter to expand about called the renormalized parameter and you fix its value by saying its the value you would get from measurement so you know its finite when you take the limit. If you do that you immediately see the original problem - the coupling constant secretly depends on the regulator so when you take the limit it blows up to infinity. The infinity minus infinity thing is really historical before they worked out exactly what was going on and resolved by what is known as the effective field theory approach.

    Thanks
    Bill
     
    Last edited: Feb 20, 2012
  6. Feb 20, 2012 #5
    Thanks. I kinda got the concept now. Anyway. In a power series, [itex]y= y_0+ \epsilon y_1+ \epsilon^2 y_2+ \cdot\cdot\cdot [/itex], is "[itex]\epsilon[/itex]" equivalent to the coupling constant which must be very small like 1/137 and present in each series (although I know it is in more complex form)?
     
  7. Feb 21, 2012 #6

    bhobba

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    That's it. If you have an energy cutoff (that is one kind or regularization you can do) then the above trick of using perpetuation theory works because epsilon is small and as you raise it to higher and higher powers in the series it gets smaller ans smaller so the trick works. However when you take the limit as the cutoff goes to infinity ie remove the cutoff you find that epsilon secretly depends on the cutoff and goes to infinity, so instead of getting smaller and smaller for large values of the cutoff it gets bigger and bigger (in fact when it goes to infinity its infinite) and the method fails. To get around it you use a different epsilon to expand about called the renormalized quantity that due to the way you chose it by insisting it is something you measure then you have no problems when the cutoff is taken to infinity.

    Thanks
    Bill
     
  8. Feb 21, 2012 #7
    I'm studying Power Series. What specific concept is it called where the epsilon getting larger in value if there is no cutoff? Does this apply to all Power Series or selected ones like Taylor Series or others? Thanks.
     
  9. Feb 21, 2012 #8

    bhobba

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    Sorry if what I posted wasn't clear - its a good idea to read the link I gave. Without a cutoff the value of the power series parameter you expand about turns out to be infinite so it obviously will not work. For the variable in the power series you expand about (that's epsilon in the equation you posted) substitute infinity and the result is infinity. However if you impose a cutoff and choose a low enough value then it is small and epsilon to some power gets smaller and smaller as the power you raise it to gets bigger so the method works - each term gets smaller and smaller. That's because it secretly depends on the cutoff. As the cutoff is made larger and larger the coupling constant gets larger and larger until in the limit it is infinite. That's why you need to expand about something better - that something is called the renormalised value. When this is done it does not blow up as you take the cutoff to infinity so the method now works. All this is made clear in the paper I linked to - its a bit heavy going - but persevere.

    Thanks
    Bill
     
    Last edited: Feb 21, 2012
  10. Feb 21, 2012 #9

    Vanadium 50

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    If you are just now studying Power Series, you are by my count about 23 courses prior to where renormalization will be discussed. I think you're going to have to accept that the answers you get will be kind of hand-wavy.
     
  11. Feb 21, 2012 #10
    I tried to read the paper for more than 30 minutes and see some web references and calculus book and thinking all this for more than an hour already. But I still can't understand the very basic question whether it applies to all power series. To know what I'm asking. Let us forget about Renormalization first. In a power series like

    The p-series rule:

    (infinity)
    sum sign 1/n^p
    n = 1

    for p-series p=2

    1 + (1/2^2) + (1/3^2) + (1/4^2) + (1/5^2)....


    So is the coupling constant equivalent to the p or n in the above equation, or the terms 2^2, 3^2, etc.?

    Also about the coupling constant getting larger for longer series without cutoff and it getting normal in value or smaller when there is cutoff. Do you also apply this to nonQED thing like trajectory of a ball thrown or is it only in QED?

    Just answer whether it is only in QED or present in all power series. This is all I need to know now. If it is only in QED.. then it has to do with the quantum nature or probability amplitude and all those path-integrals, etc. thing which I already understood and can relate and I will continue with the paper you gave. But if it is present in all power series.. i can't find it in a basic calculus book about power series where the equivalent of coupling constant gets infinite depending on whether you make a cut-off and will need to find it in other calculus book about power series. Again don't mention about renormalization first. Thanks.
     
  12. Feb 21, 2012 #11

    Physics Monkey

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    Typically the power series for some physical quantity would be of the form
    [tex]
    \sum_{n=0}^\infty c_n g^n .
    [/tex]
    We would call g the coupling constant and the coefficients c are what you compute e.g. from feynman diagrams. The coefficients c are typically computed by doing various integrals, and the integrals sometimes diverge if the range of integration is not cut off.

    The procedure of "subtracting infinities" can then sometimes be used to render the sum above finite term by term. That is, each individual [itex] c_n [/itex] is finite (as is g). However, the series may still diverge.

    Examples:
    If [itex] c_n = 1/n! [/itex] then the radius of convergence in g is infinite.
    If [itex] c_n = 1/g^n_0 [/itex] then the radius of convergence in g is [itex] g_0 [/itex].
    If [itex] c_n = n! [/itex] then the radius of convergence is zero. The series still gives infinity if g is different from zero even though every term is finite. This situation often happens in qft and is related to the concept of an asymptotic series.

    For a simple example, try doing the integral
    [tex]
    \int_{-\infty}^{\infty} dx \,\exp{\left(-x^2 - \lambda x^4\right)}
    [/tex]
    by first expanding the exponential as a power series in [itex] \lambda [/itex] and then exchanging the order of summation and integration. Such gaussian integrals are extremely common in qft.
     
  13. Feb 21, 2012 #12
    Thanks for the above. I've been looking for the summation sign for days.
    I'm presently reading Ryan's "Calculus for Dummies".

    About the coupling constant getting larger for longer series without cutoff and it getting normal in value or smaller when there is cutoff. Do you also apply this to nonQED thing like trajectory of a ball thrown or is it only in QFT due to the peculiar nature of the quantum amplitude thing? This is what I need to know for now. Thanks.
     
  14. Feb 21, 2012 #13

    Ken G

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    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?
     
  15. Feb 21, 2012 #14

    atyy

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    Here is an example from rainbows where a divergent, but "asymptotic series" is useful.

    http://www.ams.org/samplings/feature-column/fcarc-rainbows
    "There were two major contributions by Stokes. The first was that the Airy integral could be approximated for large values of |m| by asymptotic series. The one for m > 0 approximates A(m) by a slowly decreasing oscillation, and the one for m < 0 approximates it by an exponentially decreasing function. These series are expansions in negative powers of m. ...... These series do not converge, but the initial terms decrease reasonably rapidly, and the series give fair approximations to A(m) if one breaks off calculation when the terms start to grow."
     
  16. Feb 21, 2012 #15
    This question has trouble me enough to lose 4 hours of sleep thinking about it and I had to take Ambien just to sleep so hope someone can settle it before another night comes.

    Bill Hobba is saying that the coupling constant of 1/137 in the first term of the power series can become 1/50 depending on how many terms in the series you have and whether there is cutoff?

    If there is cutoff. It's like the fine structure constant is 1/137 in the first term and when none. It's 1/infinity in the first term?

    I have not heard of this before.

    Now I just want to know if this also occurs in normal power series like calculating for trajectory of a spacecraft or just in QED where all paths were taken. So just answer 1 or 2:

    1. this occurs in all power series like calculating for trajectory of a spacecraft
    2. just in QED/QFT where "any thing that can happen, does" as Brian Cox put it.

    Well?
     
  17. Feb 21, 2012 #16

    bhobba

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    Its not quite like that. You write out the power series and you calculate the terms term by term using perturbation theory. The first term is finite - no problem. Second and higher terms turn out to be infinite. It took people a long time to figure out why this happened but the answer turned out to be the thing you expand the power series in, the coupling constant, was infinite and not 1/137 like they thought. Substitute infinity in any power series and its infinite or undefined. To get around this problem you impose a cutoff (this can be looked on as taking the term you are expanding about as finite and later taking its limit to infinity) redo your perturbation procedure and you find it is all OK. The reason is the coupling constant secretly depends on the cutoff - which is rather trivial the way I explained it - but it took people a long time to realize this is what is going on. The value of 1/137 they used was the value measured at a certain energy scale which in effect was measuring the value with a cutoff. But the equations they used had no cutoff so you were really using the value as the cutoff goes to infinity ie infinity. As you take the cutoff to infinity the coupling constant goes from 1/137 to infinity which is why without the cutoff terms in the power series are infinite. Now what you do is assume the coupling constant is a function of what is called the renormalised coupling constant (which is the value from experiment ie 1/137) so you know it will not blow up. You assume it is a function of the un-renormalised parameter ie the value that does blow up to infinity, expand it in a power series, substitute into the original power series, collect terms so you now have a power series in the renormalised parameter. But you have chosen it so it is the value found from experiment so does not blow up. Carry out your calculations, take the cutoff to infinity and low and behold you find the answer is finite.

    The infinity minus infinity thing comes from when you analyse the behavior of the series when you use the renormalised value and take the limit - you find a term that is the original un-renormalised coupling constant and a term that is a function of the renormalised coupling constant - they in fact both blow up to infinity as you take the limit - but are subtracted from each other so the answer is finite.

    If you are at the level of Calculus For Dummies its probably going to be difficult to understand the paper I linked to. I have a degree in applied math and I found it tough going. So don't feel bad you are finding it tough - I congratulate you for trying.

    If you want to get your math up to the level you can understand that paper you will have to a study a more advanced textbook. The one I recommend is Boas - Mathematical Methods
    https://www.amazon.com/Mathematical-Methods-Physical-Sciences-Mary/dp/0471198269/ref=ntt_at_ep_dpi_1

    Unfortunately otherwise you will have to accept the hand-wavey arguments. As I said in my original post the jig is up with this one - you need to do the math.

    To give a specific answer to the questions you raised and how to relate it to renormalisation I will see what I can do. If you substitute infinity into any power series it will give either infinity or terms like infinity minus infinity that are undefined. An example of the first would be the power series e^x where each term is positive and an example of the second would be sine x which has positive and negative terms. Now one way to try and get around this is let x be finite and take the limit. Before you take the limit everything is fine - its finite and perfectly OK. Now what you do is assume the variable in the power series is a function of another variable (in this case called the re-normalized variable) that you hope does not blow up to infinity as you take the limit. You expand that out as a power series and you collect terms so you have a new power series in that variable. Now you take the limit and low and behold, for the case of what are called re-normalizable theories, everything is finite. You look deeper into why this occurred and you find changing to this new variable introduced another term in your equations that also blows up to infinity but is subtracted from the original variable that blows up to infinity - as you take the limit they cancel and you are left with finite answers.

    Normally when you calculate the terms in a power series using perturbation theory it does not blow up to infinity. That's because it is very unusual to chose a variable to expand the power series in that is infinity. The only reason it was done is they did not understand the physics well enough then - they did not understand the measurement of the constant they thought was small at 1/137, and was a good thing to expand in a power series about since as it is raised to a power it gets smaller and smaller, was a measurement made with a cutoff basically in effect. The equations they used had no cutoff and it all went pair shaped. When this happened it left some of the greatest physicists and mathematicians in the world totally flummoxed - these are guys like Dirac with awesome mathematical talent. It was a long hard struggle over many years to sort out what was going on. The thing that fooled them was the parameter you expanded about as a power series secretly depended on the regulator or cutoff and as you took its limit to infinity it went to infinity. When you expanded about a different one that didn't blow up to infinity everything worked OK.

    As I was penning this I remembered John Baez wrote an interesting article about re-normalisation that may be of help:
    http://math.ucr.edu/home/baez/renormalization.html

    Thanks
    Bill
     
    Last edited by a moderator: May 5, 2017
  18. Feb 21, 2012 #17
    I actually understood most everything you were saying.. but I just want to know if you can apply this coupling constant getting bigger dependent on terms in the series too to non-QFT problems like calculating for the trajectory of a ball. This is simply what I want to know. Thanks.
     
    Last edited by a moderator: May 5, 2017
  19. Feb 21, 2012 #18

    atyy

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    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.
     
  20. Feb 21, 2012 #19
    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?
     
  21. Feb 21, 2012 #20

    atyy

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    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.
     
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