Taylor–Proudman theorem

In fluid mechanics, the Taylor–Proudman theorem (after G. I. Taylor and Joseph Proudman) states that when a solid body is moved slowly within a fluid that is steadily rotated with a high $Ω$ , the fluid velocity will be uniform along any line parallel to the axis of rotation. $Ω$ must be relatively large compared to the movement of the solid body in order to make the coriolis force large compared to the acceleration terms.

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That this is so may be seen by considering the Navier Stokes equations for steady flow, with zero viscosity and a body force corresponding to the Coriolis force, which are:

$\rho({\mathbf u}\cdot\nabla){\mathbf u}={\mathbf F}-\nabla p$

where ${\mathbf u}$ is the fluid velocity, $ρ$ is the fluid density, and $p$ the pressure. If we now make the assumption that the advective term may be neglected (reasonable if the Rossby number is much less than unity) and that the flow is incompressible (density is constant) then the equations become:

$2\rho\Omega\times{\mathbf u}=-\nabla p$

where $Ω$ the angular velocity vector. If the curl of this equation is taken, the result is the Taylor–Proudman theorem:

$({\mathbf\Omega}\cdot\nabla){\mathbf u}={\mathbf 0}.$

To derive this, one needs the vector identities

$\nabla\times(A\times B)=(A\cdot\nabla)B-(B\cdot\nabla)A+B(\nabla\cdot A)-A(\nabla\cdot B)$

and

$\nabla\times(\nabla p)=0$ .

Note that $\nabla\cdot{\mathbf\Omega}=0$ is also needed.

The vector form of the Taylor–Proudman theorem is perhaps better understood by expanding it into its coordinate components:

$\Omega_x\frac{\partial {\mathbf u}}{\partial x}=0$

$\Omega_y\frac{\partial {\mathbf u}}{\partial y}=0$

$\Omega_z\frac{\partial {\mathbf u}}{\partial z}=0$

Now choose coordinates in which $Ω x = Ω y = 0$ and then the equations reduce to

$\frac{\partial{\mathbf u}}{\partial z}=0,$

if $\Omega_z\neq 0$ . Note that the implication is that all three components of the velocity vector are uniform along any line parallel to the z-axis.