A practical understanding of dynamic adjustment

How large is 2L_R? That depends primarily on the depth of the disturbance and to a lesser degree on the lapse rate and absolute vorticity. The first two of these factors enter the picture because they affect the gravity wave speed. The third enters because it affects the inertial time scale.

2L_R  =  gravity wave speed  X  inertial period

How fast is the gravity wave speed?	Practical impact
External wave: An external wave extends through the depth of the entire atmosphere, like a tsunami does in the ocean. Depending on the model top height, a model may generate an external wave traveling at several hundred meters per second!	This makes 2LR on the order of 10,000 km! Some information about the disturbance is spread very far very rapidly, and it is wind information that is retained.
Internal wave: The gravity waves of greatest practical importance are internal waves. These include things like mountain waves and the waves that you see in some cloud streets and in raw (not time-averaged) profiler data. A hump-shaped perturbation profile leads to a gravity wave of "equivalent depth" 2H, where H is the depth of the hump. Such an internal wave has a group velocity (energy propagation) of speed where N is the Brunt-Vaisala frequency. lapse rate N at 0°C dry adiabatic 0 s-1 8 K/km 0.008 s-1 5 K/km 0.013 s-1 isothermal 0.019 s-1 The gravity wave speed is complicated by a variety of additional factors.	For N=.01 s-1, cg ~ 3.2 H m/s, where H is given in km. The gravity wave speed, and thus the Rossby radius, increases proportionally with the depth of the disturbance. The gravity wave speed, and thus the Rossby radius, increases with stability by around a factor of two from steep lapse rates to isothermal conditions. Waves corresponding to a variety of depths H may be generated by one disturbance. In the real atmosphere, a disturbance of arbitrary shape would generate a complete spectrum of waves resulting in adjustments to the mass and wind field corresponding to different depths. Broad vertical structure of the disturbance is adjusted with a large Rossby radius Fine vertical structure is adjusted with a small Rossby radius In the model, only a discrete set of vertical depths is possible, so the adjustment details are somewhat different.

How long is the inertial period?

Practical impact

The classic expression for the inertial period, 2/f, only applies in the absence of a horizontal pressure gradient. In uniform horizontal shear, substitutes for the Coriolis parameter, f, where is the absolute vorticity following a parcel (such as on an isentropic surface in unsaturated conditions). For a vortex in solid body rotation, the Rossby radius just uses . There is no general formula for more complicated flows, but a generally reasonable approach is to go with

The Rossby radius is smaller for a cyclone and larger for an anticyclone, typically by a factor of around two compared to average conditions The Rossby radius is larger at lower latitudes, with 20° latitude a factor of two larger than at 45° latitude when the relative vorticity is very small

Putting these pieces together, we find the critical length scale for a disturbance is

which comes out to this most usable simple form in kilometers

where

N	is the Brunt-Vaisala frequency (~.008 s-1 for steep lapse rates of 8 K/km, ~.02 s-1 for isothermal conditions)
H	is the disturbance depth in km
fo	is 10 x 10-5 s -1 (10 "units" on your vorticity map)
f	is the Coriolis parameter (6 x 10-5 s-1 at 25°N, 11 x 10-5 s-1 at 50°N)
	is the absolute vorticity

Thus, whether a feature is dynamically "large-scale" or "small-scale" depends on its stability, its depth, and local and planetary contributions of vorticity.

Example	Result
Vortex 200 km in width at latitude 40° N depth of 5 km contains a 40-unit (40 x 10-5 s-1) vorticity maximum moderate lapse rates	The wavelength L is the full crest-trough-crest wavelength, which is around twice the disturbance width. Across this 400 km distance, the average vorticity is much closer to 20 units. Using 20 units for the vorticity, the formula gives 2LR ~ 700 km. (H=5, N~0.01 s-1, f~fo, absolute vorticity = 2fo) 400 km << 700 km, so the vortex is "small": Once the forcing that spawned it ends, it will weaken rapidly over a period of hours The mass field will end up in balance with residual weak vortex winds and its influence may spread out some There will be little effect of any sort more than around 700 km from the vortex center
Suppose the above vortex is filled with a flat cloud deck, the top of which experiences intense radiative cooling. Suppose that the region of intense cooling is 500 m thick base of the cooled region has steep lapse rates top of the cooled region has weak lapse rates or even an inversion	The difference between this example and the one above is that now H=0.5 km, giving 2LR ~ 70 km. Also, the steep lapse rate near the base of the cooled region further reduces the length scale there, while the opposite affect occurs toward the top of the cooled layer. 400 km >> 70 km, so this feature is dynamically large: After the clouds dissipate, the cool layer will be retained by the dynamics, though it may get smeared vertically by radiative processes Winds will come into balance with the pancake-shaped cool layer, meaning the vortex will develop a ring of slightly stronger winds above this layer (consider the thermal wind around the edge of the cool region)

Remember, this geostrophic adjustment process is in addition to other forcing. As long as a feature is being forced, it will continue to exist, with its structure as determined by the forcing and gradually modified by the adjustment process. Also remember that vertical shear and horizontal deformation can shred a feature over time, even if the dynamics permit a thermal or wind perturbation to otherwise be retained. This is just one more tool in your bag, not the answer to all forecast problems.