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 $$Q^2 \frac{\partial \alpha_s(Q^2)}{\partial Q^2} = \beta(\alpha_s) = -\frac{b_o}{4\pi} \alpha_s^2 -\frac{b_1}{8\pi^2}\alpha_s^3 + \dots$$(adsbygoogle = window.adsbygoogle || []).push({});

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!

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# I Two loop beta function result

Have something to add?

Draft saved
Draft deleted

Loading...

Similar Threads - loop beta function | Date |
---|---|

I Particle physics -- The collision of two Beta particles... | Mar 17, 2018 |

I Relativistic particle in non-uniform magnetic field (math) | Jan 18, 2018 |

A Is it possible that a particle is much heavier through a loop correction | Oct 16, 2017 |

A Highest loop order of experimental relevance? | Sep 9, 2017 |

I Correct Feynman rules for one loop diagram? | Jun 5, 2017 |

**Physics Forums - The Fusion of Science and Community**