= Solution
Write $S=U_z$ and $N^2=d\bar b/dz$. The <gradient Richardson number> and <Flux Richardson number> are
$$
\boxed{\mathrm{Ri}_g=\frac{N^2}{S^2},\qquad \mathrm{Ri}_f=-\frac{\overline{w'b'}}{-\overline{u'w'}S}=-\frac{B_0}{P}.}
$$
The first compares the restoring buoyancy gradient with mean shear; the second is the fraction of shear production consumed by stable buoyancy work. With this convention, both are positive for stable mixing and negative for unstable stratification. In local steady equilibrium, $\epsilon=P+B_0=P(1-\mathrm{Ri}_f)$, so a positive dissipation requires $\mathrm{Ri}_f<1$. The inviscid <Miles–Howard theorem> concerning $\mathrm{Ri}_g\geq1/4$ is a linear stability criterion, not a universal equality for turbulent fluxes.
Use the gradient-diffusion closures $-\overline{u'w'}=K_mS$ and $B_0=-K_bN^2$. The <turbulent Prandtl number> is $\mathrm{Pr}_t=K_m/K_b$, giving
$$
\boxed{\mathrm{Ri}_f=\frac{K_b}{K_m}\mathrm{Ri}_g=\frac{\mathrm{Ri}_g}{\mathrm{Pr}_t}.}
$$
Under the ideal <Reynolds analogy> $K_m=K_b$, the two Richardson numbers coincide. It is an additional closure, not an automatic consequence of the definitions. If the buoyancy-gradient similarity function is defined by $N^2=-B_0\phi_h/(\kappa zu_*)$, then
$$
\boxed{\mathrm{Ri}_g=\zeta\frac{\phi_h}{\phi_m^2},\qquad \mathrm{Ri}_f=\frac{\zeta}{\phi_m},\qquad \mathrm{Pr}_t=\frac{\phi_h}{\phi_m}.}
$$
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