 Solution

For a hydrostatic Boussinesq approximation, introduce reduced pressure $\pi =p^{\prime }/{\rho }_0$ and buoyancy $b=-g{\rho }^{\prime }/{\rho }_0$. The geostrophic balance and perturbation hydrostatic balance equations areCommuting the mixed derivatives gives the height-coordinate thermal wind relationsEquivalently $u_{g,z}=g{\rho }_y^{\prime }/(f{\rho }_0)$ and $v_{g,z}=-g{\rho }_x^{\prime }/(f{\rho }_0)$. In an atmosphere the corresponding ideal-gas form is conveniently derived from geopotential $\Phi$ on pressure surfaces: $f{u}_g=-{\Phi }_y{|}_p$, $f{v}_g={\Phi }_x{|}_p$, and ${\Phi }_{\ln p}=-R_{a}T$. HenceSince $d\ln p/dz=-g/(R_{a}T)$, this gives the vertical shear at a fixed horizontal location,The horizontal temperature derivatives here are on isobars; replacing them by fixed-height derivatives is a small-slope approximation. This qualification avoids mixing compressible and Boussinesq density gradients.

For an illustrative poleward temperature drop $40\mathrm{K}$ across $3000\mathrm{km}$, mean $T\sim 250\mathrm{K}$, $f\sim 10^{-4}{\mathrm{s}}^{-1}$ and depth $10\mathrm{km}$,Adding a modest surface wind gives an upper-level westerly jet of order $50$–$60\mathrm{m}{\mathrm{s}}^{-1}$. The thermal wind fixes vertical shear, not the absolute wind without a boundary or reference value.