renormalisation condition Eq. (\[mufix\]) is equivalent to $$\left.S(p,\mu)\right|_{p^2 = \mu^2} = 
  \left. \frac{1 + \gamma_4}{2}\frac{1}{iK} \right|_{K = \kappa}\, ,$$ to zeroth order in $1/m_R$. Typically we choose $\kappa$ to be on the negative imaginary $K$ axis, as the heavy quark propagator asymptotes to the bare propagator as $K \rightarrow -i\infty$, as can be seen from Fig. \[fig1\].

Heavy quark DSE
---------------

To illustrate the derivation of the renormalised heavy quark DSE, we choose Landau gauge ($\xi = 0$ in Eq. (\[gprop\])) and, for the time being, work with the rainbow or bare vertex approximation $$\Gamma_\mu(p,q) = \gamma_\mu.$$ Using Dirac trace identities to project out from Eqs. (\[DSEqn\]) to (\[Sigpr\]) a pair of coupled integral equations gives $$\begin{aligned}
\lefteqn{A'(p^2,\Lambda^2) = }\nonumber \\
& & \!\!\!\!\!\! 1 + \frac{4Z_1}{3p^2}  
  \int^\Lambda \frac{d^4q}{(2\pi)^4}\, \left[p\cdot q + 
  2\frac{p\cdot(p - q) q\cdot(p - q)}{(p - q)^2}\right] 
   \Delta\left[(p - q)^2\right] \frac{A(q^2)}{q^2 A^2 + B^2},  \label{ASDE}\end{aligned}$$ $$B'(p^2,\Lambda^2) = 4Z_1  \int^\Lambda \frac{d^4q}{(2\pi)^4}\, 
   \Delta\left[(p - q)^2\right] \frac{B(q^2)}{q^2 A^2 + B^2}.  \label{BSDE}$$ Substituting $$\frac{A(q^2)}{q^2 A(q^2)^2 + B(q^2)^2} = 
 \frac{1}{2m_R}\,\frac{1}{iK' + \Sigma(K')}  + O\left(\frac{1}{m_R^2}\right) ,$$ $$\frac{B(q^2)}{q^2 A(q^2)^2 + B(q^2)^2} = 
 \frac{1}{2}\,\frac{1}{iK' + \Sigma(K')}  + O\left(\frac{1}{m_R}\right) ,$$ into Eqs. (\[ASDE\]) and (\[BSDE\]) gives $$\begin{aligned}
m_R\left[A'(p^2,\Lambda^2) - 1\right] & = & 
    \frac{2Z_1}{3} \int^\Lambda 
       \frac{d^4k'}{(2\pi)^4}\,\left[1 