What is the dependence of the beta function of pure $SU(N_\text{c})$ Yang-Mills theory on the number of colors? I guess $$\mu\frac{dg_\text{YM}}{d\mu}=-\beta_0N_\text{c}g_\text{YM}^3-\beta_1N_\text{c}^2g_\text{YM}^5-\beta_2N_\text{c}^3g_\text{YM}^7-\cdots\,,$$ with the appropriate constants $\beta_0,\beta_1,...$, as has been computed in QCD (including quarks) at four loops [arXiv:hep-ph/9701390]. Is this correct? How could it be proved (to all orders)?

Then the 't Hooft coupling $\lambda=g^2_\text{YM}N_\text{c}$ runs independently of $N_\text{c}$: $$\frac{\mu}{2}\frac{d\lambda}{d\mu}=-\beta_0\lambda^2-\beta_1\lambda^3-\beta_2\lambda^4-\cdots\,.$$ The goal is to assure that in the large $N_\text{c}$ limit the coupling runs independently of $N_\text{c}$, so that the confinement scale is held fixed. So the series in $\lambda$ could be truncated for $N_\text{c}\rightarrow\infty$ (there could be negative powers of $N_\text{c}$).

  • 5
    $\begingroup$ Some suggestions for improving this question: 1) Indicate why you are interested in this problem. 2) Tell us how you guessed this expression. (Why should the leading term vanish?) 3) Tell us what you've tried as far as proving it goes. $\endgroup$
    – user1504
    Commented Apr 18, 2013 at 2:54
  • $\begingroup$ Looks like an expansion in the 't Hooft coupling $g_{YM}^2 N_c$ so you should probably look into the large-$N_c$ approximation literature. Maybe try this relevant sounding thing. (I haven't read it.) $\endgroup$
    – Michael
    Commented Apr 18, 2013 at 3:03

1 Answer 1


Based on the exact result for a supersymmetric Yang-Mills

$$ \beta(\alpha_s)=-\frac{\alpha_s^2}{4\pi}\frac{3N}{1-\frac{N\alpha_s}{2\pi}} $$

these guys arXiv:0711.3745 postulated the following exact result for a non-supersymmetric Yang-Mills

$$ \beta(\alpha_s)=-\alpha_s^2\frac{11N}{12\pi}\frac{1}{1-\frac{17N}{11}\frac{\alpha_s}{2\pi}} $$

Here the beta-function is defined as $$ \mu^2\frac{d\alpha_s}{d\mu^2}=\beta(\alpha_s) $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.