It is unclear whether the references are due to uncertainty about the
definition of the Lorentzian or to the fact that \cref{eq:model-psd}
is also peaked and therefore \cref{eq:lorentzian} a potential
-substitute for \cref{eq:model-psd}. \citet{florian95}
-likely \cref{are} using \cref{eq:lorentzian}, as the slope of the
+substitute for \cref{eq:model-psd}. \citet{florin95}
+likely \emph{are} using \cref{eq:lorentzian}, as the slope of the
fitted \PSD\ in their figure 2, has a slope at $f=0$.
Using \cref{eq:model-psd}, the derivative would have been zero, as we
can see by using the chain rule repeatedly,