Hi Florian,
There are loop functions have the term Log (m^2/mu^2), where mu is the renormalization scale, e.g. C00 which is defined in SPheno as
C00m = ((-m1 + m3)*((m1 - m2)*(m2 - m3)*(-3 + 2*Log(m1)) - &
& 2*m2**2*Log(m2/m1)) + 2*(-m1 + m2)*m3**2*Log(m3/m1))/ &
& (8.*(m1 - m2)*(m1 - m3)*(m2 - m3))
It seems that mu=1 GeV through the term Log(m1). As well known that the amplitude must be independent of the choice of mu.
What are the exact expressions of C00 and B0 which are used in numerics in SPheno.
Best regards,
Dris.
Loop functions
Re: Loop functions
Hi,
yes, since the mu-dependence drops out at the end, one can put an arbitrary scale as 1 GeV. I'm not sure, if I understand the rest of your question:
C00 is given by yourself. So, what do you want to know? B0 is calculated including momenta and a quite complicated function which you can find in src/LoopCouplings.f90.
Cheers
Florian
yes, since the mu-dependence drops out at the end, one can put an arbitrary scale as 1 GeV. I'm not sure, if I understand the rest of your question:
C00 is given by yourself. So, what do you want to know? B0 is calculated including momenta and a quite complicated function which you can find in src/LoopCouplings.f90.
Cheers
Florian
Re: Loop functions
I see that B0 is more complected,
You know that for p=0, B0(0,x,y)=-1+Log[x]-xLog[x/y]/x-y, in this case, is it correct to treat B0 as C00 (mu=1GeV)?
Best regards,
Dris.
You know that for p=0, B0(0,x,y)=-1+Log[x]-xLog[x/y]/x-y, in this case, is it correct to treat B0 as C00 (mu=1GeV)?
Best regards,
Dris.
Re: Loop functions
Yes, you just need to set the same scale. On the other side, it might be convenient in your calculation to keep an arbitrary scale mu in the functions and check that it drops out at the very end.
Cheers
Florian
Cheers
Florian