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Why don't cold fronts and other steep-gradient weather effects just dissipate? Why do they last so long? Why doesn't the heat dissipate toward the cooler region?

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    $\begingroup$ The short answer is because there are multiple forces at work (pressure gradient force, coriolis force, centrifugal force, and (minorly) friction). Wind (and through it temperature/moisture/vorticity/etc) does not simply flow from high pressure to low pressure the way you'd imagine because of the way these forces work against each other. aos.wisc.edu/~aalopez/aos101/wk11/HLsfc.jpg In essence, the weather we observe is the result of air wanting to "just dissipate" but being forced to do something else. $\endgroup$
    – DrewP84
    Commented Apr 16, 2014 at 2:16
  • $\begingroup$ @DrewP84: Hrm, yeah, I kind of realised that as I was copying the (rather old) question from the area51 site. It's possible that it's un-answerable as-is (too broad?), or maybe it could be improved. Thoughts? Maybe a slightly expanded version of your comment would serve as an answer. $\endgroup$
    – naught101
    Commented Apr 16, 2014 at 2:34
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    $\begingroup$ I think the question is a good one. I am sure there is someone out there that can explain it better than I did. I chose to give my short answer in the mean time. Fluid dynamics equations were never my strong point. $\endgroup$
    – DrewP84
    Commented Apr 16, 2014 at 2:38

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Good question. It would seem without much thought, that air in a high pressure zone should move towards the air in low pressure zone, resulting in a disappearance of the pressure gradient. The reason this does not occur is because there other forces (or pseudo-forces) at work, acting in addition to the force resulting from the pressure gradient. I will outline such forces below:

1. Pressure Gradient:
This is the force resulting from the pressure difference between a high and low pressure zone. We term this a pressure gradient, because the pressure change is continuous rather than a discrete change between a high and low pressure zone. We can model the acceleration in a particular direction resulting from a pressure gradient using the equation:

$$\alpha=\frac{-1}{\rho}\frac{dP}{dz}$$

where $\alpha$ is the acceleration at a given point, $\rho$ is the density of air at that point, and $dP/dz$ represents a small change in pressure over a small change in horizontal distance. More generally, we can model the acceleration vector in 3-dimensions using the equation:

$$\vec{\alpha}=\frac{-1}{\rho}\vec{\nabla}P$$

Now if the pressure gradient was the only force at work, it is obvious from above that the acceleration would be directed from high pressure areas to low pressure areas, ultimately resulting in a dissipation of such gradients.

The next pseudo-force to consider however is the Coriolis effect.

2. Coriolis Effect:
The Coriolis effect influences winds away from the equator that move horizontally. Such winds in the northern hemisphere are deflected to the right, whereas winds in the southern hemisphere are directed to the left. This is a result of the Earth's rotation. (More information on this effect can be in the question and answer here). The acceleration resulting from the Coriolis psuedo-force is given by the following equation:

$$\boldsymbol{a}_C=-2\Omega\times\boldsymbol{v}$$

where $\Omega$ represents the angular velocity of the Earth and $\boldsymbol{v}$ represents the velocity of the wind. The cross product here is of significance and indicates that the deflection of the Coriolis effect will be at right angles to the direction of the wind's velocity. More details of deriving the specific result for the Earth at different angles of latitude can be found here.

So how does this Coriolis effect stop wind from moving from high pressure to low pressure? Well imagine that wind begins to move North (in the northern hemisphere) from a high pressure region to a low pressure region. Due to the Coriolis effect, this wind will be deflected to the right, and will continue to be deflected in such a manner until the pseduo-force resulting from the Coriolis effect exactly balances the force due to the pressure gradient (ignoring friction for the time being). At this time, we say the wind is said to be in geostrophic balance. The wind therefore is no longer moving directly from the high pressure region to the low pressure region, and it is for this reason that the pressure gradients do not dissipate straight away. (See friction force below) This can be represented by the picture below, and I will include the formula when we have access to mathjax (note the Coriolis term is represented a bit differently in this diagram, but don't worry I'll explain how it is the same when mathjax is added - basically it just a certain direction component of my more generalized vector above):

enter image description here

3. Friction

As I mentioned previously, the geostrophic balance presumes the absence of friction. In reality, friction acts to slow the flow of wind, in turn lessening the influence of the Coriolis effect. Thus ultimately the wind does tend to spirally slightly inwards towards the low pressure zone. The effect of friction is more noticeable in the lower atmosphere, and in the upper troposphere the geostrophic motion approximation is more accurate, and therefore the pressure gradients will take longer to dissipate in the upper atmosphere than the lower atmosphere.

The force of friction is given by:

$$F = cV$$

where $c$ is a constant, and $V$ is the velocity of the wind.

4. Gravity

Pressure gradients can also be sustained vertically, due to the influence of gravity. When the force of gravity balances the pressure gradient, this situation is known as hydro-static wind balance, and is represented by the equation:

$$dP/dz = -{\rho}g$$

where $\rho$ is density of air, and $g$ is the acceleration due to gravity, which is approximately $9.8 m s^{-2}$.

Net Effect:

While I have already answered the question, I have decided include the equation combining all 4 forces for completeness. Combining all these forces acting on the wind, the net acceleration of wind can be determined by the equation:

$$\frac{D\boldsymbol{U}}{Dt}=-2\Omega\times\boldsymbol{U}-\frac{1}{\rho}{\nabla}p+\boldsymbol{g}+\boldsymbol{F}_r$$

where $\boldsymbol{U}$ represents the velocity of the wind, and $t$ represents time. The quantities in bold are vectors, and act in a specified direction. (e.g. $\boldsymbol{g}$ acts vertically, where as $\boldsymbol{F}$ acts in the opposite direction to $\boldsymbol{U}$). Below is in image of this:

enter image description here

This image shows how the forces act (excluding gravity) around high and low pressure zones. PGF is the pressure gradient force, CF is the psuedo-Coriolis force and F is the frictional force opposing the velocity of the wind. Note that the wind moves slightly in wards towards the low pressure region, rather than perpendicular to the gradient predicted by geostrophic motion. This is due to the frictional force.

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    $\begingroup$ This only explains wind, not why gradients persist. You really should include an analysis of the frontogenetical function and the role of the ageostrophic circulation about a front (in 3 dimensions) and the role of deformation in the cross front direction. While the given answer is not incorrect, it doesnt really answer the question. $\endgroup$
    – casey
    Commented Apr 25, 2014 at 11:44
  • $\begingroup$ @casey, yes you are right my post is not complete having left out those additional concepts. I am planning on editing the post to include these later when I have more time. Thanks for pointing it out. $\endgroup$
    – Kenshin
    Commented Apr 25, 2014 at 13:20
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The reason that the gradients persist is that atmospheric processes are part of an engine, driven by the energy of the sun. If the sun were to burn out, then the gradients would then in fact dissipate. But until then, the sun will heat the air at the surface near the equator, while the cold air sinks at the poles, creating a circulation around the earth (see Hadley cells). The fronts are the boundaries of the eddies formed in the circulation.

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  • $\begingroup$ Short, simple and sweet. Should be the correct answer. $\endgroup$ Commented Nov 23, 2018 at 19:26

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