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Why do couplings run with log of energy?

  1. Dec 13, 2008 #1
    Coupling constants run with the logarithm of energy.
    This is a result of renormalization, its equations, and the
    change of the effect of the virtual particle cloud with energy.

    Is there a simple way to understand why this happens with the
    log of energy, instead of with another function?

    Logarithm is the integral of 1 over r; does this enter somehow?

  2. jcsd
  3. Dec 13, 2008 #2


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    It comes from "the equations of renormalization", you calculate a lot of loop integrals which gives you Log's

    And that is why in fact you "can do" renormalization, the log's are "slow functions"
  4. Dec 13, 2008 #3
    Well, I know that, as I wrote in my post. But why log? Why not something else?
    Can one explain this in simple terms?

  5. Dec 13, 2008 #4


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    why is not this "simple"??

    The log just comes from math.
  6. Dec 13, 2008 #5

    Vanadium 50

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    Yes, but that by itself doesn't really tell you anything. You can always take an equation with a logarithm in it and express it as a differential equation with a 1/x in it instead. That DE may or may not provide additional insight.

    As Malawi_Glenn points out, the key is that the coupling constant varies slowly. If the coupling constant varied quickly, we wouldn't call it a constant and wouldn't use the mathematical machinery that we do to describe it. We'd pick something else.
  7. Dec 16, 2008 #6
    In fact, it is the other way round:

    You know for sure, that the theories we construct suffer from being well-behaved either at low or high energy limits. As a consequence, many of the integrals are divergent (either IR or UV).

    To get a proper meaningful theory out of these divergent quantities (a theory which learns to live with the problems and still remains meaningful), one redefines the physical quantities, what is called re-normalization. And hence the coupling constants, for example, run (with energy).

    Now, it is not that the quantities always run as log of energy... depending upon how you constructed your Lagrangian they can run more violently .... but, we select ONLY those theories which show a log dependance behavior: the logarithmic function is the slowest changing function and confirms the least sensitivity of your theory to the cut off energy (value of which you are not generally aware)

    So, in short, it is not that the couplings run always logarithmically with energy, but it is your choice which decides the dependence and we (generally) chose theories which are renormalizable (and hence showing a log dependence)
  8. Dec 21, 2008 #7


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    Expanding somewhat on what njoshi3 said:

    The reason has to do with the phenomenon of "scaling." When you renormalize a coupling you are forced to introduce an energy scale (this is standard renormalization theory - see your favorite textbook), called the "subtraction scale" (M). Then by ordinary dimensional analysis, without any other energy scales in the problem (let's imagine that all the masses are zero for the moment) all dimensionless functions can only depend on the dimensionless scale E/M, where E is the energy, since this is the only dimensionless quantity left! Adding masses or dimensionful couplings means that there are more ratios you can construct, but that's fine.

    It is an old trick that whenever you have a dimensionless ratio that can cover the entire positive real line, it is often very useful to take the log of that ratio, which maintains the dimensionless-ness of the ratio and is bijective, so it has an entire inverse (the exponential function). Find me another function that does this.

    But it is NOT necessarily true that operators only run logarithmically with energy. They could run as a power law as well.

    For example, QCD with a vanishing beta function runs power law, not logarithmically. This happens, for example, in supersymmetric QCD with Dirac gluinos (something I've been researching recently!). So the running can be more general, it's just that in the Standard Model, all the Lagrangian parameters have logarithmic running. But that's not true in the most general case.
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