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I Two loop beta function result

  1. Nov 17, 2017 #1

    CAF123

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    A one loop expression for the strong coupling can be calculated with the result $$\alpha_s^{(1)}(\mu^2) = \frac{4\pi}{b_o \ln(\mu^2/\Lambda^2)}\,\,\,\,\,(1), $$see e.g in the first few pages of the attached file, where $$Q^2 \frac{\partial \alpha_s(Q^2)}{\partial Q^2} = \beta(\alpha_s) = -\frac{b_o}{4\pi} \alpha_s^2 -\frac{b_1}{8\pi^2}\alpha_s^3 + \dots$$

    My first question is,
    1) how is (1) derived? I don't see how to get this expression without an initial condition ##\alpha_s(\mu_o^2)## or something after integration.

    2) Can I principle use the above to get a two loop running equation for alpha strong? Would it just be solving $$Q^2 \frac{\partial \alpha_s(Q^2)}{\partial Q^2} = -\frac{b_1}{8\pi^2}\alpha_s^3$$

    Thanks!
     

    Attached Files:

  2. jcsd
  3. Nov 18, 2017 #2
    1) There is an initial condition, if you like of the form you wrote. It is absorbed into the definition of ##\Lambda##.

    2) The above (your second equation) is already the equation for two loop running, i.e. you have to solve the equation including ##b_0## and ##b_1##. I think there are only approximate solutions (or numerical codes). The QCD review by the particle data group http://pdg.lbl.gov/2015/reviews/rpp2015-rev-qcd.pdf list one that can be used for four loops, if I am not mistaken the first two terms should correspond to a possible two loop result, at least that is how it looks schematically.
     
  4. Nov 18, 2017 #3

    CAF123

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    Thanks, I've got it now.

    I see. So is it correct to say that the ##\Lambda## I get depends on the the order at which I truncate the perturbative expansion for the strong coupling? And because of this, it is preferable to solve equations numerically without introducing ##\Lambda## and such a truncation but instead dealing with an ##\alpha_s(\mu_o^2)## with ##\mu_0=M_Z## the typically chosen value?
     
  5. Nov 18, 2017 #4
    For a fixed value of ##\alpha_s(\mu^2)##. Of course it would reversably be true that the value of ##\alpha_s(\mu^2)## depends on the order at which you truncate, given a fixed value of ##\Lambda##. It just is the case that experiments typically measure ##\alpha_s(\mu^2)## rather than ##\Lambda##. And specifically LEP measured a lot of data at a CME of the ##Z## mass, I think this is why the value is usually quoted as ##\alpha_s(M_Z^2)## (maybe more senior members can shed more light on this).

    As far as I am aware, most codes (like muti purpose monte carlo simulations) use the approximate solutions as quoted in the pdg review (or corresponding equations at lower loop level, I think for most purposes two loop running is sufficient). Theoretically it does of course not make a difference, just the value of ##\Lambda## is not very well known experimentally, while the value of ##\alpha_s(M_Z^2)## is.

    Also, the value of ##\Lambda##, for a fixes value of ##\alpha_s(M_Z^2)## depends on details of for example how you handle the masses of ##b## and ##c## quarks.
     
  6. Nov 18, 2017 #5

    CAF123

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    Ok, so for the latter statement do you mean I can truncate at first order, extract the value of ##\Lambda##, then use this value of ##\Lambda## in the two loop expression for the running coupling (ie the first two terms in the PDG four loop result you mentioned in the paper linked).

    Is this dependence of Lambda on the masses you mention through the choice of ##n_f## in the beta function coefficients?
     
  7. Nov 18, 2017 #6
    No, I just mean that you can specify your (perturbative) theory by either quoting ##\Lambda=\text{some value}## or by quoting ##\alpha_s(M_Z^2)=\text{some value}##. The relation between the two will only be known up to some higher order term. I do not think the prescription you give is consistent, i.e. you have to specify your theory by giving either ##\Lambda## or the value of the strong coupling at some mass scale. Then calculate ##\alpha_s## at the scale you need it with the loop accuracy you need it. I should be clear that you have to use the first plus second term in any case.

    You can for example choose to run from ##\mu=M_Z## to ##\mu=m_b## with ##n_f=5## and then continue with ##n_f=4##, which is a reasonable choice, if that is what you mean.
     
  8. Nov 18, 2017 #7

    CAF123

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    I see. By truncating at first order and rearranging the solution for ##\alpha_s## into the form I wanted in the OP I find I must choose ##\Lambda^2 = \mu_0^2 \exp(-1/\beta_0/\alpha_s(\mu_0^2))##. As this was derived using a first order truncation, then the value of Lambda I get can only be used at the one loop accuracy for determinations of alpha_s at some other mu^2. Is that correct?

    If so, I’m just trying to understand why the theory can be specified with only a single Lambda because the value of Lambda I get above may not be used at second order alpha_s accuracy etc, if I understood your last comment.

    Thanks!
     
  9. Nov 20, 2017 #8
    What I meant to say is that you can either say "##\Lambda=\text{number}## and I use two loop running" or you say "##\alpha(M_Z^2)=\text{number}## and I use two loop running", i.e. you always need just one inital condition, no matter on how many loops you calculate, and you can either specify if as ##\alpha## at some scale, or as ##\Lambda##, i.e. as the point where ##\alpha## formally diverges. Have a look at the four loop solution, it also involves ##\Lambda## so if you tell me its value I can tell you ##\alpha## at any scale (and reversely).
     
  10. Nov 20, 2017 #9

    CAF123

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    I see, thanks! And if I'm interested in, for example, the one loop running for alpha the value of ##\Lambda## I get through solving the equation in #7 is different from the ##\Lambda## I would get in using the fourth loop running. E.g in book by Schwartz, the value of lambda in the one loop approximation is around 89MeV and that for the fourth loop is around 213MeV. But I guess these numbers too depend on the number of flavours one uses too in the beta function coefficients.

    When I see the equation ##\alpha(M_Z^2) = 0.118## is this referring to the fourth loop value of ##\alpha(M_Z^2)## that I'd get by using some ##\Lambda## derived at fourth order?
     
  11. Nov 20, 2017 #10
    Experiments usually directly measure ##\alpha## at some scale, rather than ##\Lambda##. This is why all values of ##\Lambda## have relatively large errors. It is then conventional to quote the value of ##\alpha## at the scale of the Z mass. Usually, what is used at the end is a fit to several experiments performed at different scales. I think the pdg also discusses that.
     
  12. Nov 21, 2017 #11

    CAF123

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    Ok thanks, so is the value of ##\Lambda## usually quoted to be around 200 MeV the value one gets through using the input value of 0.118 for ##\alpha_s(M_Z^2)##(determined experimentally) in the relation between alphas(M_z) and Lambda at fourth loop order assuming some number of flavours? (I think this is what Schwartz is saying about the 213MeV quoted above)
     
  13. Nov 25, 2017 #12
    It would be common to have some matching at the b and c quark thresholds, and calculate with less flavours below them. I dont know how Schwartz derives the exact values, and I dont have his book. You get "around 200 MeV" already from the above description, I dont think it matters too much how many loops you use (at least beyond two loop results). The experimental error is exponentiated, so "around 200 MeV" means something like "about 100 to 400 MeV" here anyway, as far as I am aware.
     
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