# Why do cold-core lows slope towards the cold air with heigth? How to show mathematically that wind intensifies with height in this case?

I understand that from the hypsometric equation, in a cold core low the pressure gradient must increase with height, because in the center the isobars will be closer together:

quente = hot, frio = cold

What I don't understand is why in practice the low slopes towards the cold air with height:

Also, I wonder how I can show mathematically (I guess from the thermal wind relationship + the hypsometric equation?), that in a cold core low, the wind must increase with height.

• Please provide English only images. I do not know what quente or frio means Oct 12 '20 at 2:53
• frio means cold and quente means hot :-) Oct 12 '20 at 11:03
• Perhaps the geostrophic wind equation is another option towards showing why the wind increases? Oct 12 '20 at 11:08
• Off the top of my head at this late hour, I can't reason out the slope towards cold. Well, to be honest, you've really just about entirely explained why it slopes towards the cold (just consider more what you said at the start)... but why it slopes at all (or in other words: why it slopes towards warm air at low levels) is less clear to me at this moment. My guess is it may have something to do with that thermal wind relationship you mentioned. Hopefully people can give good hints (without full answers if this is a homework question?) :-) Oct 12 '20 at 11:19
• @JeopardyTempest "The cold air in association with a developing surface mid-latitude cyclone is generally located to the northwest of the surface low. Therefore, heights (the trough axis) will tilt with height from the surface to the upper level (tilts toward the northwest / tilts toward the cold air)." theweatherprediction.com/habyhints/128 Oct 12 '20 at 13:07

I've managed to show why wind speed increases with height in a cold core low, as follows. Considering low is in the Southern Hemisphere, the wind blows clockwise around the low. In natural coordinates, the normal vector points outwards. Since the center is colder than the surroundings, we have

$$\frac{\partial T}{\partial n}> 0$$

from the thermal wind relationship,

$$\frac{\partial V_g}{\partial ln p}\boldsymbol{\widehat{t}}= \frac{-R}{f}\boldsymbol{\widehat{k}} \times\nabla_p{T}=\frac{-R}{f}\boldsymbol{\widehat{k}} \times\frac{\partial T}{\partial n}\boldsymbol{\widehat{n}}=\frac{R}{f}\frac{\partial T}{\partial n}\boldsymbol{\widehat{t}}$$

Where $$\boldsymbol{\widehat{t}}$$,$$\boldsymbol{\widehat{n}}$$ and $$\boldsymbol{\widehat{k}}$$ are the tangent, normal (in the horizontal plane) and vertical unit vectors.

Since $$f < 0$$ and $$\frac{\partial T}{\partial n} > 0$$, we have $$\frac{\partial V_g}{\partial ln p} < 0$$, which implies $$\frac{\partial V_g}{\partial z} > 0$$, i.e., wind intensifies with height in a cold low.

I still have not understood why the low is displaced towards cold air with height. Since the wind intensifies with height, I would expect the low to be further ahead at higher levels than at the surface.