The beta function for dimensional reguarlization in MS scheme looks like:(adsbygoogle = window.adsbygoogle || []).push({});

β(ε,λ)=-ελ+f(λ)

But does this mean that the total derivative of β with respect to ε is nonzero, namely equal to -λ ?

λ should be invariant to total changes in ε?

But doesn't β also have to be invariant with respect to changes in either λ_{0}or ε?

If you were using a cutoff scheme, then I think β has to be independent of the cutoff (or I guess what happens is the cutoff changes, but the bare couplings change, but the renormalized coupling should not be affected by any of that). The analogy should be the same, except now ε is your bare cutoff and λ_{0}= λ_{0}(ε) should adjust as you change ε so that β doesn't depend on ε at all.

**Physics Forums | Science Articles, Homework Help, Discussion**

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

# Invariance of beta function in dimensional regularization?

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

Loading...

Similar Threads for Invariance beta function |
---|

I Difference between Schrodinger's equation and wave function? |

I Find the energy from the graph of the wave function |

I Probabilities Associated with Sudden Changes in Potential |

I Question about gauge invariance and the A-B effect |

**Physics Forums | Science Articles, Homework Help, Discussion**