Relation of Local Pressure Tendency to

Motion of Isobars For Synoptically-scaled Systems

The total derivative of pressure is

dp/dt = ∂p/∂t + [u ∂p/∂x + v ∂p/∂y + w ∂p/∂z)] (1)

Normally, pressure changes experienced by the moving parcel are dominated by the vertical motion term (example, discussed in class). But synoptic scaling of this equation for quasi-horizontal flow (say, at the ground) eliminates this far right term.

Rewriting this equation including the above approximations and solving for the local tendency to make it useful for local weather forecasting results in

∂p/∂t = -[u ∂p/∂x + v ∂p/∂y] + dp/dt (2)

This states that the barometric changes observed at a fixed location (a weather station) are due to a combination of advection and the pressure changes experienced by the air parcels as they move along.

Switching to the natural coordinate system

∂p/∂t = -V ∂p/∂s + dp/dt (3)

Now we will try to relate equation (3) to the motion of isobars for synoptically-scaled systems. Consider two isobars spaced a distance ∆x, with a weather station A located on the eastern-most isobar.

Assume that the isobars are moving eastward such that they move ∆x in time increment ∆t. Then the speed of the isobar is given as

∆ x / ∆ t = c = ∆ s / ∆ t (4)

where the right hand side has been converted to the natural coordinate system.

The instantaneous change in pressure observed at A, (∆p/∆t)_a , will be proportional to

(∆p/∆t)_a = -c (∆p/∆s) (5)

but ONLY if the values on the isobars stay the same; if the lows and highs do not change get larger or smaller (“change shape”).

To test this, pretend that ∆s = 100 km and that ∆p/∆s = -4 mb /100 km (lower pressure towards west). Say also that the isobars (the pressure pattern) are moving eastward at 100 km/h. Conceptually, one would expect that the pressures would rise 4 mb in that one hour. That is precisely the result mathematically by substitution in (5).

If the pressure pattern changes (the low deepens), each nearly concentric isobar will have a greater circumference etc. and the shape of the pressure field will change). In other words,

(∆p/∆t)_a = -c (∆p/∆s) + Correction factor (deepening or filling) (6)

In equation (6), taking the limit as t and x -> 0 we really have

∂p/∂t = -c ∂p/∂s + Correction factor (7)

Deepening or filling will lower or raise the pressure of parcels “everywhere” by a factor (∆p/∆t)’ which is the same as considering the pressure change observed at the center of the low (or high). After taking the limit as t -> 0 then we have

∂p/∂t = -c ∂p/∂s + (dp/dt )_s (8)

where (dp/dt )_s is the total change in pressure experienced by all the air parcels due to deepening or filling of the pressure pattern. Note that in this case this is the same as dp/dt in equation (3).

For short time frames, you can consider the system to be non-changing and (dp/dt )_s = 0.

Equation (8) then reduces to the equation you could use to answer one of the homework problems for next week, Petterssen’s formula as outlined by Bluestein (p. 51, Eq. 2.1.38). .

∂p/∂t = -c ∂p/∂s (9)