Background T field is nudged by small perturbations

These deviations from a zonally symmetric background T field on the ground act like PV anomalies of the same sign.

PV anomalies induce geopotential anomalies with the opposite sign so, the structure of anomalies along the ground follows a form like this:

Now how will these change in the atmosphere above the ground?

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'(x, p = ps) = Acoskx

In an atmosphere in which there are no PV anomalies aside from those at the ground and others at the tropopause,

everywhere else. Therefore

And we have an equation that dictates how geopotential anomalies must behave in the atmosphere. So what is the answer? To find out, we must find the vertical structure G(p):

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q'= 0

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q'=1

f0

∂ 2Φ'

∂x 2+∂ 2Φ'

∂y 2

⎛

⎝ ⎜

⎞

⎠ ⎟+f0σ

∂ 2Φ'

∂p2= 0

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'(x, p) =G * coskx

Substituting this expression for geopotential into the governing equation q’=0, we learn that G must obey the constraints imposed by this ordinary differential equation:

The only function that will satisfy this equation is an expontential one, whose derivatives (d2G/dp2) are proportional to the original function (G) itself.

where

and a and b are coefficients determined by boundary conditions.

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d2G

dp2−k 2σ

f02G = 0

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G = acosh rp + bsinhrp

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r =k σ

f0

Because all temperature perturbations must die by the time they reach the top of the atmosphere (p=0), by the hydrostatic equation all first derivatives of geopotential must do the same:

This means that dG/dp must be 0 there too (all pressure dependence is contained in this function G). So

This can only be true if b is 0.

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∂'

∂p= 0 at p = 0

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dG

dp= r asinhrp + bcoshrp( ) = 0 at p = 0

Putting all this together, we now know the complete functional form of our geopotential perturbations at all points in the atmosphere

where the coefficient a is the amplitude of our perturbations.

That’s the math. But what does it mean? How our initial perturbations from the ground change with pressure is entirely controlled by the hyberbolic cosine function. That only helps if we know what this function looks like. And don’t forget:

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'(x, p) = acosh(rp) *cos(kx)

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r =k σ

f0=

2π σ

f0L

Geopotential anomalies: ’ = a cosh (p√/Lf0) * cos (2x/L)

Cut L in half: ’ = a cosh (2p√/Lf0) * cos (2*2x/L)

Divide L by 4: ’ = a cosh (4p√/Lf0) * cos (4*2x/L)

A wide L (double original): ’ = a cosh (p√/2Lf0) * cos (2x/2L)

Geopotential anomalies: ’ = a cosh (p√/Lf0) * cos (2x/L)

’ = a cosh (rp) * cos (kx); vg’ = (-ak/f0) cosh (rp) * sin (kx)

A wave in the T field on the tropopause will also produce geopotential anomalies that decay as you move down through the atmosphere.

But notice!+T’ on the tropopause gives +’ because it is like –PV’.

If the horizontal scale is smaller, the rate at which the perturbations fall off with height is greater, so these are less able to influence what happens on the other boundary.

And this one on the tropopause is traveling west relative to the background flow, which is blowing east.