<p>The Rossby radius of deformation (<span class="math-container">$\lambda_{R}$</span>) is a length scale over which  adjustment will occur while a system approaches geostrophic equilibrium.  In my class noted I have this defined as the distance at which buoyancy becomes as important as rotation.</p>
<p>Some examples are given <a href="http://glossary.ametsoc.org/wiki/Rossby_radius_of_deformation" rel="noreferrer">here</a> but do not show their derivation.</p>
<blockquote>
<p>The distance that cold pools of air can spread under the influence of the Coriolis force.</p>
<p>A cold pool will initially spread out toward and under warmer air because of higher pressure under the cold, denser air. However, as the spreading velocity increases, the Coriolis force will increasingly turn the velocity vector until it is parallel, rather than perpendicular, to the pressure gradient. At this point, no further spreading will occur and the winds will be in geostrophic equilibrium. The final equilibrium distance traveled by the edge of the cold air equals the external Rossby radius of deformation, <span class="math-container">$\lambda_{R}$</span>:</p>
<p><span class="math-container">$$\lambda_{R} = {\sqrt{(gH\Delta\theta/\theta_0)}\over f_c}$$</span></p>
</blockquote>
<p>I'm not familiar with the particular derivation used to come up with the 1000-2000 km <span class="math-container">$\lambda_R$</span> for synoptic weather patterns but it likely starts with Rossby waves and a wave solution for their propagation.</p>
<p>One example derivation for <span class="math-container">$\lambda_{R}$</span> I can think of is in a shallow water model in a rotating frame from Holton (2004) p 211.  In his setup he starts with the height (<span class="math-container">$h'$</span>) of his shallow fluid with a discontinuity along <span class="math-container">$x$</span> defined as <span class="math-container">$h' = h_0 \mathrm{sign}(x)$</span> where sign(<span class="math-container">$x$</span>) is the sign of x.  As the fluid equilibrates, the heights will change at the discontinuity and at some distance from the discontinuity.  How far away?  That is what the Rossby radius of deformation tells us.  For Holton's shallow water model, he finds <span class="math-container">$\lambda_{R} \equiv f_0^-1\sqrt{gH}$</span> where <span class="math-container">$f_0$</span> is the Coriolis parameter, <span class="math-container">$g$</span> is gravity and <span class="math-container">$H$</span> is the mean depth of the fluid.  At distances less than <span class="math-container">$\lambda_{R}$</span>, we will observe the height changes but further than <span class="math-container">$\lambda_{R}$</span> we will not notice any changes in fluid height.  In this case we can interpret the Rossby radius of deformation as the length scale over which the height field changes during geostrophic adjustment.</p>

CC BY-SA 3.0
2014-07-11T16:43:53.270Z
2014-07-11T16:43:53.270Z
2020-06-18T08:25:19.090Z
casey
<p>In one of his presentations, Sandy MacDonald (of NOAA and ESRL) talks about the Rossby Radius of Deformation in connection with the size of synoptic weather systems: he describes how it sets the diameter of "large-scale" systems to be 1000-2000 km across, so that if we want to even out meteorological effects, we need to consider areas with diameter larger than that.</p>

<p>How is the Rossby Radius of Deformation used to determine that scale of 1000 - 2000 km?</p>

<meteorology><synoptic><dynamics>
How is the Rossby Radius used to determine that synoptic weather systems are 1000-2000km across?
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# How is the Rossby Radius used to determine that synoptic weather systems are 1000-2000km across?

(Asked by: username_0 on 2014-07-06)

In one of his presentations, Sandy MacDonald (of NOAA and ESRL) talks about the Rossby Radius of Deformation in connection with the size of synoptic weather systems: he describes how it sets the diameter of "large-scale" systems to be 1000-2000 km across, so that if we want to even out meteorological effects, we need to consider areas with diameter larger than that.

How is the Rossby Radius of Deformation used to determine that scale of 1000 - 2000 km?

**Tags:** meteorology, synoptic, dynamics


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## 1 answers

### Answer by username_1 on 2014-07-11. Score: 6

The Rossby radius of deformation ($\lambda\_{R}$) is a length scale over which adjustment will occur while a system approaches geostrophic equilibrium. In my class noted I have this defined as the distance at which buoyancy becomes as important as rotation.

Some examples are given [here](http://glossary.ametsoc.org/wiki/Rossby_radius_of_deformation) but do not show their derivation.

> The distance that cold pools of air can spread under the influence of the Coriolis force.
>
> A cold pool will initially spread out toward and under warmer air because of higher pressure under the cold, denser air. However, as the spreading velocity increases, the Coriolis force will increasingly turn the velocity vector until it is parallel, rather than perpendicular, to the pressure gradient. At this point, no further spreading will occur and the winds will be in geostrophic equilibrium. The final equilibrium distance traveled by the edge of the cold air equals the external Rossby radius of deformation, $\lambda\_{R}$:
>
> $$\lambda\_{R} = {\sqrt{(gH\Delta\theta/\theta\_0)}\over f\_c}$$

I'm not familiar with the particular derivation used to come up with the 1000-2000 km $\lambda\_R$ for synoptic weather patterns but it likely starts with Rossby waves and a wave solution for their propagation.

One example derivation for $\lambda\_{R}$ I can think of is in a shallow water model in a rotating frame from Holton (2004) p 211. In his setup he starts with the height ($h'$) of his shallow fluid with a discontinuity along $x$ defined as $h' = h\_0 \mathrm{sign}(x)$ where sign($x$) is the sign of x. As the fluid equilibrates, the heights will change at the discontinuity and at some distance from the discontinuity. How far away? That is what the Rossby radius of deformation tells us. For Holton's shallow water model, he finds $\lambda\_{R} \equiv f\_0^-1\sqrt{gH}$ where $f\_0$ is the Coriolis parameter, $g$ is gravity and $H$ is the mean depth of the fluid. At distances less than $\lambda\_{R}$, we will observe the height changes but further than $\lambda\_{R}$ we will not notice any changes in fluid height. In this case we can interpret the Rossby radius of deformation as the length scale over which the height field changes during geostrophic adjustment.


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