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A precise stratification–rotation analogy uses constant $N$, ideal fluid, and independence of one horizontal coordinate. For the stratified system take ${\partial }_y=0$, $v=0$, and allow prescribed forces $(F_x,0,F_z)$. The possibly nonlinear Boussinesq equations areFor the unstratified system choose right-handed coordinates $(X,Y,Z)=(z,y,-x)$, rotation $2\mathbf{\Omega }=N{\mathbf{e}}_Z$, and ${\partial }_Y=0$. Its rotating Euler equations, with the static gravitational and centrifugal contributions absorbed into the background pressure, arewhere $D_R={\partial }_t+U{\partial }_X+W{\partial }_Z$. The identification isThen $D_R=D$, $P_X=p_z$, and $P_Z=-p_x$, proving the component-by-component equivalence. In particular, buoyancy perturbation maps to transverse rotating velocity, and stratified horizontal velocity maps to rotation-axis velocity. The mapping permits arbitrary prescribed forces in the two-dimensional plane, but no transverse force: a transverse force in the rotating problem would create a source in the stratified buoyancy equation. A transverse stratified force would also invalidate $v=0$. Constant $N$ is essential to this constant-rotation equivalence.

A physical free surface introduces extra structure beyond the interior equations. After subtracting hydrostatic pressure, its zero total pressure condition gives $p=g\eta$ to first order, together with the kinematic condition ${\eta }_t=w$ for the stratified vertical displacement. The mapping takes this normal velocity into $U$, whereas an actual rotating free surface normal to its physical vertical $Z$ has normal velocity $W$ and a displacement along $Z$. The image of the stratified horizontal surface is a vertical surface in the rotating coordinates. Consequently the physical free-surface problems are not analogous, although deliberately mapped rigid boundaries or mathematical boundary data can be.

For the first prescribed force, the steady linear stratified equations are satisfied byReal parts are understood. The horizontal pressure gradient balances the force; the vertical pressure gradient is balanced by buoyancy; and $w=0$ satisfies buoyancy conservation and incompressible flow. Thus neither $u(z)$ nor $b(z)$ is fixed by the steady equations alone.

Initial rest and switching on in the unbounded fluid remove this ambiguity by causal pressure adjustment by internal waves. Work with the $e^{imz}$ coefficient and let $\hat{F}(q)$ be the horizontal Fourier transform of $f^{\prime }(x)$. For an abrupt switch at time zero, elimination in the linear equations givesAt $q=0$ the bracket cancels the apparent singularity at every finite time. The persistent pressure term is the symmetric antiderivative of the force: the inverse transform of $-i\hat{F}/q$, interpreted as a Cauchy principal value, is $f(x)-\epsilon /2$. The oscillatory part propagates away. To see why its zero-frequency part does not leave an arbitrary offset, note that $\Omega (q)\sim cq$ with $c=N/m$; its long-wave contribution consists of oppositely moving fronts. At fixed $x$, the two far-field values of $f-\epsilon /2$ are $-\epsilon /2$ and $+\epsilon /2$, whose average is zero. Nonzero horizontal wavenumbers give dispersive transients that decay locally. Smooth finite-duration switching changes the transient, not this selected steady limit. ThereforeA spatially constant pressure gauge is immaterial. The long-wave group velocity $c=N/m$ gives the adjustment scale: the elapsed time since switching must satisfy $Nt\gg 1$ and $(N/m)t\gg |x|+x_1$, and exceed the switching duration. These are asymptotic conditions, not an assertion of exact equilibration after a finite time; the full dispersion relation allows tails. The same long-wave limit gives $u_{\infty }=\epsilon m{e}^{imz}/(2N)$ if desired. Equal-amplitude outward fronts leave opposite pressure offsets and the same velocity on their two sides.

For the force depending instead on $y$, any steady linear horizontal momentum balance requires $p_x=f^{\prime }(x)e^{i\ell y}$ and $p_y=0$. Their mixed derivatives disagree wherever $f^{\prime }\ne 0$, since ${\partial }_y{p}_x=i\ell f^{\prime }(x)e^{i\ell y}$ but ${\partial }_x{p}_y=0$. Hence no steady linear velocity field can satisfy the equations, irrespective of whether it is $x$-independent. This statement concerns the linearized problem; nonlinear inertial balances were excluded by the assumptions.

For an unsteady solution put $\mathbf{u}=(-t{\varphi }_y,t{\varphi }_x,0)$ and take $\sigma =0$. Its horizontal divergence is zero. The pressure derivatives needed in momentum balance areTheir compatibility is exactlyThus the Poisson equation supplies an actual pressure and the proposed velocity solves both momentum equations; the vertical momentum and buoyancy equations hold because $w=\sigma =0$. This flow grows linearly in time, so it remains a valid small-disturbance prediction only while its velocities are small.

As $x_1\to 0$ with total force $\epsilon$ fixed, $f^{\prime }\to \epsilon \delta (x)$. Set $\varphi =\Phi (x)e^{i\ell y}$. Away from the source ${\Phi }^{\prime \prime }-{\ell }^2\Phi =0$, and decay at both ends requires $\Phi =A{e}^{-\ell |x|}$. Integrating across the source gives the jump ${\Phi }^{\prime }(0^{+})-{\Phi }^{\prime }(0^{-})=-i\ell \epsilon$. Since this jump is $-2\ell A$, $A=i\epsilon /2$, soThis verifies the distributional source as well as the decay condition.

The corresponding rotating problem is not expected to reproduce this evanescent accelerating flow. Its forcing varies in $Y$, precisely the coordinate whose independence was required in the stratification–rotation analogy. Moreover, $x$ maps to the rotation-axis coordinate $-Z$. At small Rossby number, the Taylor–Proudman theorem gives ${\partial }_Z{\mathbf{u}}_R=0$ in the unforced exterior regions. Such nonzero columns cannot decay with $|Z|$, hence cannot reproduce an $e^{-\ell |x|}$ tail. There is also a direct unsteady obstruction to copying the proposed flow: in a homogeneous fluid rotating about the mapped $x$ axis, a mode with nonzero axial wavenumber has inertial wave frequency of magnitude $N|q|/\sqrt{q^2+{\ell }^2}$, rather than the zero-frequency horizontal vortical response of the stratified problem. The steady or slowly varying exterior rotating response must be columnar or accompanied by outgoing inertial waves; the decaying stratified solution is outside the partial analogy.