Heavy-Initiated Contributions (Intrinsic)

Todo

clarify normalization in a_s

FONLL

Todo

write about the cancellation and then \(K_{ij}\) use for FONLL in intrinsic charm

\[\begin{split}K_{hh}^{(0)}(Q^2/m_h^2) &= 1\\ K_{hh}^{(1)}(Q^2/m_h^2) &= 2\left(\bar P_{qq}^{(0)}(z) \left(\ln\left(\frac{Q^2}{m_h^2 (1-z)^2}\right) - 1\right)\right)_+\\ K_{hg}^{(0)}(Q^2/m_h^2) &= 0\\ K_{hg}^{(1)}(Q^2/m_h^2) &= 2 P_{qg}^{(0)}(z) \ln\left(\frac{Q^2}{m_h^2}\right)\\ K_{hg}^{(0)}(Q^2/m_h^2) &= 0\\ K_{hg}^{(1)}(Q^2/m_h^2) &= 2 P_{gq}^{(0)}(z) \left(\ln\left(\frac{Q^2}{m_h^2 z^2}\right) - 1 \right)\end{split}\]

The \(K_{ij}\) are always relative to the matching threshold, not to the mass itself.

Nevertheless the effect of choosing an arbitrary threshold \(\mu^2\) different from the mass \(m^2\) is only the appearance in \(K_{ij}\) of an extra term proportional to the splitting functions \(P_{ij}\):

\[K_{ij}^{(1)}(\mu^2) = K_{ij}^{(1)}(m^2) + \log(\frac{\mu^2}{m^2}) P_{ij}^{(0)}\]

see 22, [BBB+16].