3
$\begingroup$

With $u, v$ denoting the components of wind velocity along zonal (x) and meridional (y) direction, atmospheric (relative) vorticity is usually defined by

$$\zeta_z = \frac{\partial v}{\partial x} -\frac{\partial u}{\partial y} $$

This is the z-Component of the 3D-Vorticity, defined by

$$\vec \zeta = \vec \nabla \times \vec v$$

However, atmospheric flow is not on a flat earth but on a sphere. E.g. a westerly wind follows the curved earth's surface and is better described in spherical coordinates. There, the vertical component of the 3D-vorticity is obtained by carrying out the curl operation in spherical coordinates (see here). This would give

$$\zeta_z = \frac{1}{r \sin \theta } \left[ \frac{\partial }{\partial \theta} (u \sin \theta ) + \frac{\partial v}{\partial \varphi}\right]$$

$u, v$ are the local horizontal velocities along the unity vectors $\vec e_\varphi$ (zonal) and $-\vec e_\theta$ (meridonal). Note that here latitude $\theta $ is measured from pole and not, as usual, from equator.

By expanding this expression, we get

$$\zeta_z = \frac{1}{r \sin \theta } \left[ \sin \theta \frac{\partial u }{\partial \theta} + u \frac{\partial \sin \theta }{\partial \theta} + \frac{\partial v}{\partial \varphi}\right]$$

On the local tangential plane we have $dx = r \sin \theta d \varphi $ and $dy = -r d \theta $

so the derivatves with respect to $\varphi$ and $\theta$ can be replaced by derivates with respect to x and y. This gives finally:

$$\zeta_z = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} + \frac{u}{r} \cdot \frac{\cos \theta}{\sin\theta} $$

which is clearly not in accordance with the first equation: There is an extra term proportional to u. Why do we have this discrepancy and why is this not mentioned in standard textbooks?

$\endgroup$

1 Answer 1

2
$\begingroup$

The best way to see this extra term is to consider a bug moving along a latitude circle at constant speed $|u|$ in a constant direction arounf the poles. If the bug is moving along a geodesic then there is no vorticity, but the latitude circle is not a geodesic except at the Equator ($\cos\theta=0$). So movement along any other latitude circle will entail some vorticity, even with the vector components remaining constant.

The vorticity term with no component gradients has two additional features. First, we see that it is an odd function of displacement from the Equator; the Northern and Southern Hemispheres give opposite signs. This corresponds to the latitude circle deviating in opposite ways from the geodesic in the two hemispheres. If the "bug" in our example were to be taken as following the rotation of Earth itself, it would appear to be turning clockwise if it were seen from the South Pole (so it woukd be in the Southern Hemisphere) but counterclockwise if seen from the North Pole.

Second, the term goes to infinity as we approach either pole ($\sin\theta\to0$). The closer our bug gets to a pole, tge more angular velocity it would need to attain a given nonzero target $u$ value, and the vorticity in the absence of gradients follows suit. Natural motions near the pole a spherical body (including Earth) either have decreased speed in the latitude direction (balancing out the increased $\cot\theta$ factor) or quickly move away from the pole (the vorticity is then balanced out by nonzero derivative terms).

$\endgroup$
6
  • $\begingroup$ Thanks for answering my question. I'm just a little amazed that this term is completely disregarded in atmospheric dynamics, because relative vorticity is typically about $10^{-5} \ldots 10^{-4 } s ^{-1}$ and with a windspeed u=50kn (25m/s) at 45° latitude I would get for my additional term about $4 \cdot 10^{-6} s^{-1}$. Although this is relatively small compared to typical values ​​of relative vorticity, it is not orders of magnitude smaller that one should ignore without even thinking about it. $\endgroup$
    – MichaelW
    Nov 11 at 12:39
  • $\begingroup$ Just calculated: with the same windspeed as above but at a latitude of 85° instead, this extra term would be $4 \cdot 10^{-5} s^{-1}$. In this case it has already the same order of what we normally expect as relative vorticity. What does it mean? More over, you say, that winds along latitude direction are decreased near the poles. Didn't you mean zonal direction W-E. Latitude direction is N-S. $\endgroup$
    – MichaelW
    Nov 11 at 12:49
  • 2
    $\begingroup$ @MicaelW For the 85° case consider also whether the other terms grow. You get rapid coordinate changes near the poles versus temperate/tropical latitudes. $\endgroup$ Nov 11 at 13:03
  • 1
    $\begingroup$ To be honest, I identified right now, that very close to the poles my interpretation about u und v breaks down completely and is more or less senseless in a practical sense: Lets say I'm located just 10 m from the pole. Then a wind in pure zonal direction cannot exist anymore. What is zonal at one point is already nearly meridional a few meters away. So what does it mean for meteorology? u and v are typically introduced with a cartesian coordinate system in mind, that is overlaid onto the spherical grid in a good approximation. At the poles this view is completely useless. $\endgroup$
    – MichaelW
    Nov 11 at 13:18
  • 1
    $\begingroup$ An alternative view: think of the geidesic (great circle) tangent to tge flow. It makes an angle (between the local latitude and 90°) with the equatorial plane. This angle is not singular at the poles. $\endgroup$ Nov 11 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.