= Solution
Use eastward $x$, northward $y$, <pressure coordinate> $p$ increasing downward, and <geopotential> $\Phi=gz$. Horizontal derivatives are taken on an <isobaric surface>. The horizontal <geostrophic balance> and vertical <hydrostatic balance> are
$$
f u_g=-\Phi_y,\qquad f v_g=\Phi_x,\qquad
\Phi_p=-\frac1\rho=-\frac{RT}{p},
$$
where $R$ is the <specific gas constant> of dry air and the final equality uses the <ideal gas law>. At fixed latitude $f$ is independent of $p$. Cross-differentiating gives
$$
f\,\partial_pu_g=-\partial_y\Phi_p=\frac R p\,T_y,
\qquad
f\,\partial_pv_g=\partial_x\Phi_p=-\frac R p\,T_x.
$$
Consequently the pressure-coordinate <thermal wind> equations are
$$
\boxed{\frac{\partial u_g}{\partial\ln p}=\frac Rf\frac{\partial T}{\partial y},
\qquad
\frac{\partial v_g}{\partial\ln p}=-\frac Rf\frac{\partial T}{\partial x}.}
$$
Equivalently, $f\mathbf e_z\mathbin\times\partial_{\ln p}\mathbf u_g=R\nabla_pT$. Notice the sign: increasing eastward <velocity> upward means decreasing eastward <velocity> with $\ln p$, and hence poleward-decreasing <temperature> in the Northern Hemisphere.
For the <stratospheric polar vortex> estimate, the original diagram must be read on its logarithmic pressure axis. Near $60^\circ$ in the winter hemisphere, adjacent $10\ \mathrm{m\,s^{-1}}$ wind contours suggest an increase of roughly $20\ \mathrm{m\,s^{-1}}$ through a pressure decade surrounding $10\ \mathrm{mb}$: for example, about $10\ \mathrm{m\,s^{-1}}$ near $30\ \mathrm{mb}$ and $30\ \mathrm{m\,s^{-1}}$ near $3\ \mathrm{mb}$. These are coarse graphical readings, not a sounding. Thus
$$
\frac{\partial u_g}{\partial\ln p}\simeq
\frac{30-10}{\ln(3/30)}\ \mathrm{m\,s^{-1}}
\simeq-8.7\ \mathrm{m\,s^{-1}}.
$$
With $|f|=2\Omega\sin60^\circ\simeq1.26\times10^{-4}\ \mathrm{s^{-1}}$ and $R\simeq287\ \mathrm{J\,kg^{-1}\,K^{-1}}$, the poleward <temperature gradient> is approximately
$$
\frac{\partial T}{\partial y_{\rm pole}}
\simeq\frac{|f|}{R}\frac{\partial u_g}{\partial\ln p}
\simeq-3.8\times10^{-6}\ \mathrm{K\,m^{-1}}
\simeq-0.42\ \mathrm{K\ per\ degree\ latitude}.
$$
If the vortex edge is taken to span roughly $50^\circ$ to $70^\circ$, its width is $a(20\pi/180)\simeq2.2\times10^6\ \mathrm m$, giving
$$
\boxed{T(70^\circ)-T(50^\circ)\simeq-8.5\ \mathrm K,
\quad\text{a poleward drop of order }10\ \mathrm K.}
$$
The estimate depends on the width assigned to the edge and the graphical <vertical shear>; a range of several to roughly fifteen kelvin is more justified than a precise value. The sign remains a cold winter pole in either hemisphere. The diagram does not by itself prescribe a unique pair of edge latitudes.
At low latitudes, horizontal free-atmospheric <temperature gradients> are usually much smaller than at the winter vortex edge. For a comparable finite wind <vertical shear>, the rearranged <thermal-wind balance>, $T_y=(f/R)u_{\ln p}$, is consistent with weaker <temperature gradients> as $|f|$ decreases. However, <geostrophic balance> loses validity near the equator because the <Coriolis parameter> vanishes; division by $f$ cannot be used there to derive a prediction. The <weak temperature gradient approximation> is better motivated by rapid horizontal <internal gravity wave> adjustment, together with tropical <convection> and the <Hadley cell>. <Thermal-wind balance> describes the balanced extratropical relation and its tendency approaching the tropics; it is not a complete causal explanation of equatorial thermal uniformity.
Back to article page
