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 Does the Rise or Fall of an Air Parcel Generate Vorticity?

When smoke rise there is often a twisting motion. The funnel of a tornado or water spout also often displays a twisting pattern. Could it be that in the process of rising (or falling) that vorticity is created. The following attempts to answer that question.

## Preliminary Analysis

Consider a parcel of air at latitude φ or a spherical Earth of radius a. The distance x from that point to the axis of the Earth is acos(φ). Let Ω is the angular velocity of the Earth. The west=to=east velocity of the parcel is then Ωx and its angular momentum per unit mass is Ωx².

If the parcel of air rises ot a height H above the Earth's surface then its distance to the axis is x'=(a+H)cos(φ). In order to preserve angular momentum the velocity V at the top of its rise must be such that

#### Vx' = Ωx² and thus V = Ωx²/x' which is equivalent to V = Ω(acos(φ))²/[(a+H)cos(φ)]or V = Ωa²cos(φ)/(a+H) = Ωacos(φ)/(1+H/a)

Now consider two parcels of air; one at latitude φ1 and the other at latitude φ2. The angular velocities Ω1 and Ω2 at the top of their rises are then

#### Ω1 = V1/(a+H)cos(φ1) = Ω/(1+H/a)² and Ω2 = V2/(a+H)cos(φ2) = Ω/(1+H/a)²

Therefore Ω1 and Ω2 are equal and no relative vorticity is created by the rise of the parcels of air. Thus the answer to the title question is "No, the rise or fall of parcels of air does not generate relative vorticity." This answer however only applies to a perfectly spherical Earth.

## An Elliptical Cross-Section

Let the equation of the cross-section of the Earth be given by

#### (x/a)² + (y/b)² = 1 and thus y = b(1−(x/a)²)½

The slope of the tangent line to the ellipse (dy/dx) is given by:

#### 2(x/a) + 2(y/b)(dy/dx) = 0 and hence dy/dx = −(x/y)(b/a)

Let ψ be the angle of the line perpendicular to the tangent line to the ellipse. By the well-known theorem of analytical geometry the product of the slope of a line and the slope of the perpendicular to that line is equal to −1. Thus the slope of the perpendicular to the elliptical cross-section is given by:

#### tan(ψ) = (y/x)(a/b) = tan(φ)(a/b) and cos(ψ) = [1/(1+tan²(ψ))]½ = (x/a)/[(x/a)²+(y/b)²]½

Thus the distance x' of the parcel from the top of its rise of H is given by

#### x' = x + Hcos(ψ) and hence x'/x = 1 + (H/x)cos(ψ)

If the equation for the ellipse is divided through by (x/a)² the result is:

#### 1 + (y/x)²(a/b)² = 1/(x/a)² and since (y/x)=tan(φ) this reduces to x = a/[1+(a/b)²tan²(φ)]½

Therefore the angular velocity of the parcel of air at latitude φ, Ωφ, is dependent upon φ through the following change of dependence:

#### Ωφ = Ω/(x'/x)² x'/x = 1 +(H/a)/[1+(a/b)²tan²(φ)]

This means that some vorticity is created in a parcel of air as it rises from the Earth's surface to a height H.

## The Scale of the Vorticity Created by the Rise of a Parcel of Air

For the Earth

• Ω = 2π radians per 24 hour day = 7.272×10-5 radians per second
• a=3963.189 and b=3949.901 and thus (a/b)=1.003364

Let φ=30° and H=5 miles. Then

#### x'/x=1+(1.2616×10-3/[1+(1.006747)(0.333333)]=1.0009446.

Thus (x'/x)²=1.00189 and Ωφ=7.2583×10-5 radians per second.

On the other hand at φ=31° , Ωφ=7.2586×10-5 radian per second while at φ=29°, Ωφ=7.2580×10-5. This means that in a 24 hour period a parcel originally at 29° lags 0.13° of latitude compared to the parcel originally at 31°. Thus a parcel orginally at 30° of latitude will be turning 0.065 degrees per day. The direction of the turn in the Northen Hemisphere is clockwise. Thus some vorticity is generated but not a whole lot.

Although the vorticity generated in a rise or fall is virtually infinitesimal the amount may be enough to break the symmetry in the flow an establish a vortex. A parcel of air moving into a lower pressure zone at a higher altitude could expand horizontally and this would also establish a clockwise turning in the Northern Hemisphere. For more on this topic see Wind direction.

(To be continued.)