What does this equation mean?

A second, purely structural check follows from the same shape function without any new arithmetic. Because a derivative-coupled detector’s response carries higher powers of $\Omega$ in its numerator than a monopole-coupled detector’s does — the two instruments filter the same field differently, as Moustos’s coupling-type comparison already established [ 9 ] — their normalized spectra are provably different functions of $\Omega$ pointwise, even on the identical worldline at the identical acceleration. Two distinct, non-coincident probability densities have nonzero total variation distance between them as a matter of definition, with no integral needing to be evaluated to know this. So $D_{\rm det}$ between a monopole-coupled and a derivative-coupled detector on the same trajectory is strictly positive — exactly the “unmatched apparatus” failure mode the treaty’s second clause exists to catch, demonstrated here in closed form rather than merely asserted.