NETC Abstract

A Toy Model of the Instability in the Equatorially Trapped Convectively Coupled
Waves on the Equatorial Beta Plane

J Andersen and Z Kuang

Partially based upon Andersen, J. A., Z. Kuang, A toy model of the instability in the equatorially trapped convectively coupled waves on the equatorial beta plane, Journal of Atmospheric Sciences, 65,3736-3757, (2008)
http://swell.eps.harvard.edu/~joe/files/AndersenJA08.pdf


The equatorial atmospheric variability shows a spectrum of significant peaks in the wavenumber–frequency domain. These peaks have been identified with the equatorially trapped wave modes of rotating shallow water wave theory. This paper addresses the observation that the various wave types (e.g. Kelvin, Rossby, etc.) and wavenumbers show differing signal strength relative to a red background. It is hypothesized that this may be due to variations in the linear stability of the atmosphere in response to the various wave types depending on both the specific wave type and the wavenumber. A simple model of the convectively coupled waves on the equatorial beta plane is constructed to identify processes that contribute to this dependence. The linear instability spectrum of the resulting coupled system is evaluated by eigenvalue analysis. This analysis shows unstable waves with phase speeds, growth rates, and structures (vertical and horizontal) that are broadly consistent with the results from observations. The linear system, with an idealized single intertropical convergence zone (ITCZ) as a mean state, shows peak unstable Kelvin waves around zonal wavenumber 7 with peak growth rates of ~0.08 /day (e-folding time of ~13 days). The system also shows unstable mixed Rossby–gravity (MRG) and inertio-gravity waves with significant growth in the zonal wavenumber range from -15 (negative indicates westward phase speed) to +10 (positive indicates eastward phase speed). The peak n =0 eastward inertio-gravity wave (EIG) growth rate is around one-third that of the Kelvin wave and s at zonal wavenumber 3. The Rossby waves in this system are slightly unstable, and the Madden–Julian oscillation (MJO) is not observed. Within this model, it is shown that in addition to the effect of the ITCZ configuration, the differing instabilities of the different wave modes are also related to their different efficiency in converting input energy into divergent flow. This energy conversion efficiency difference is suggested as an additional factor that helps to shape the observed wave spectrum.

It is hypothesized that the MJO is better represented as a "moisture mode" where convection is much more directly coupled to variations in the atmospheric moisture content than to variations in atmospheric temperature (as in the "convective modes"). To this end, I will also discuss some preliminary results investigating this hypothesis and how it may be incorporated into convective wave models.

Can you see this trend in this data?

The data comes from ECMWF reanalysis projected/interpolated onto cloudsat orbits (y), and filtered OLR (x). The line in red is the result of a linear regression from matlab.

Because I'm a Nerd

I was queued up at whole foods behind a lady who was taking forever... and she took even longer once the cashier realized that she had transposed two digits between the register and the credit card charge - instead of $152.73, she charged the lady $125.73. When she realized this, she had to charge the lady an additional $27. Which took even more time, between explaining and doing...

All this is beside the point, except that I noticed that 27, while not round per se, was 3^3. I wondered if transpositions always gave interesting differences.

A little experimentation in my head lead me to conjecture that, rather than a cubic, the difference between two numbers that had a pair of adjacent digits swapped was multiple of 9.

A little more thought yielded a proof that is probably inelegant (but I'm a physicist not a mathematician, or a miracle worker!) and is also probably old news to all (but I don't care to look it up).

imagine a number n represented by ....def, where d, e, f are integers and the number is such that n = f+10*e+100*d + etc

now, consider the number m = ...dfe

m-n = e+10*f+100*d - (f+10*e+100*) = (1-10)*e + (10-1)*f = 9*(f-e)!

It is obvious that this still holds if there are digits to the right of our pair of transposees - everything written above is just multiplied by sufficient powers of ten.

What about when the pair of digits is separated by intervening digits?

eg m = ...fed

this can be considered the result of three transpositions:

def --> dfe --> fde --> fed

Each transposition provides a multiple of 9 to the difference, and so the statement still holds.

we can prove the statement for general separations through induction -

consider the difference between

...abc.....def

and

...aec.....dbf

where there are j+1 digits between the b and the e

consider the following chain of swaps:

...abc.....def --> ...acb.....def --> ...ace.....dbf --> ...aec.....dbf
This chain consists of two swaps of adjacent digits and one of digits seperated by j digits. All of these are known (or assumed) to satisfy the "multiple of nine" rule so the sum of the differences also obeys the rule.

QED!

I haven't proven anything in ages... it's actually a lot more fun when it isn't on an exam (and when its something trivial)

Obama on Science

Nature has an "interview" with the candidates on science (McCain's "responses" are taken from older statements - he declined to be involved).

http://www.nature.com/news/2008/080924/full/455446a.html

The final question was relevant to me:

Would it make sense for more overseas students who receive PhDs at American universities to stay in the country and contribute to its research base and its wealth? What immigration reforms would you support?

Obama: I believe that we must enact comprehensive immigration reform to restore our economic strength, relieve local governments of unfair burdens stemming from an inefficient federal immigration system, ensure that our country and borders remain secure and allow a path to citizenship for the 12 million undocumented immigrants who are willing to pay a fine, pay taxes, and learn English. A critical part of comprehensive immigration reform is turning back misguided policies that since 9/11 have turned away the world's best and brightest from America. As president, I will improve our legal permanent resident visa programmes and temporary programmes to attract some of the world's most talented people to America.

McCain, as a senator from Arizona, has long been involved in immigration issues, mainly through strengthening federal security at land border crossings. He supports immigration reforms to allow more highly skilled workers to stay and work in the United States after graduation.

Shame neither of them really answered the question - The question sounded like it was about F and J visa students, but they turned it into an illegal immigration question.

Another reason to be vegetarian...

People should consider eating less meat as a way of combating global warming, says the UN's top climate scientist.

Rajendra Pachauri, who chairs the Intergovernmental Panel on Climate Change (IPCC), will make the call at a speech in London on Monday evening.

UN figures suggest that meat production puts more greenhouse gases into the atmosphere than transport.

I was vegetarian quite some time before I realized this, but it is another good reason to stop eating meat.

Wallace and Hobbs 2.8 - Second Law and Entropy

2.8.1 Carnot Cycle

In a thermodynamic context, a cycle refers a series of changes undergone by a "working substance" which does external work and transfers heat energy while eventually returning to its initial conditions, ready to make another cycle.

As the internal energy at the beginning and end of the cycle is the same, the net heat taken in by the system must equal the net work done.

Efficiency is defined as:

\eta = (Work done)/(Heat taken in)

Ideal heat engine - a working substance in a cylinder with insulating walls and a conducting base, fitted with a frictionless, insulated piston to which varying loads can be applied. Also, an insulating stand that the cylinder can rest upon and become totally thermally isolated, as well as two infinite reservoirs of heat one "hot" @ T_1 and the other "cold" at T_2 (T_1>T_2). By infinite, we mean that heat can flow into or out of wither reservoir into the working substance with out changing the temperature of the reservoir.

Cycle as follows:
1) Start in state "A" at temperature T2. The system is places on the Stand and compressed (adiabatically) via the piston until it reaches T1 - state "B".
2) The system is then transferred to the warm reservoir, where it extracts heat (in quantity Q1), expanding isothermally to state "C".
3) The cylinder is returned to the stand and expands adiabatically until the temperatrue returns to T2, state "D".
4) The cylinder is placed on the cold reservoir and compressed isothermally to its original volume, expelling heat Q2 in the process, returning to state "A".

This is best visualized in PV space - the net work done in the process is the area contained within the cyclic path of the system.

Example 2.14: Show that, for a Carnot engine, the ratio of heat absorbed to heat expelled is equal to the ratio of the reservoir temperatures.

the heat absorbed is given by:

Q1 = \int_B^C pdV
= \int_B^C RT_1dV/V
= RT_1 ln(V_C/V_B)

similarly,

Q2 = RT_2 ln(V_D/V_A)

(note change of sign, as heat flows in other direction here)


For the adiabatic paths, we have:

p_AV_A^gamma = p_BV_B^gamma
p_CV_C^gamma = p_DV_D^gamma

for the isothermal paths, we have

p_BV_B = p_CV_C
p_DV_D = p_AV_A

combining, we get:

V_C/V_B = V_D/V_A

Thus,

Q1/Q2 = T_1/T_2


Following Carnot's cycle in reverse tranfers heat from the cold reservoir to the hot reservior while taking in external energy in the form of work - this is a refrigeration machine.

2.8.2 - Entropy
Passing reversibly from one adiabat to another along an isotherm (temp T), heat goes in or out of a system (Q_rev). However, it can be shown that Q_rev/T is an constant for a given pair of adiabats. This quantity is the "difference" between the two adiabats and is called the entropy.

More generally, when a system passes from one state to another, the entropy changes:

s_2 - s_1 = \int_1^2 dq_rev/T

We can also see that

ds = dq/T = c_p d theta/theta

so,

s = c_p ln(theta) + const

Example 2.15: Calculate the change in entropy when 5g of water at 0C is raised to 100C and converted to steam at this temperature.

S_373 - S_273 = \int_{273}^{373} dQ/T

where dQ = m*c(T)*dT for the liquid water part of the integration, while the system remains at a constant temperature during the vaporization.

assuming c is constant, we have:

\DeltaS_{liquid} = 20.9 * ln(373/273) = 6.5J/deg

the additional entropy flow during vaporization is:

m*L/373 = 30.2 J/deg

so the total entropy gain by the water is

36.7J/deg


Consider the change in entropy during the Carnot cycle. There is no entropy flow during the adiabatic paths. For BC, the entropy of the working fluid increases by Q_1/T_1. Similarly, during DA, the entropy decreases by Q_2/T_2. As we have shown, these tow amounts are equal (but of opposite sign) so the entropy of the working fluid is unchanged, consistent with the definition of a cycle. Entropy is transferred between the reservoirs, but a negligible amount compared to their infinite extent.

2.8.3 Clausius-Clapeyron

CC eqn can be derived from the Carnot cycle. Describes the change in saturation vapor pressure of a liquid with temperature (or melting point of a solid with pressure).

Consider a working substance made up a liquid in equilibrium with saturated vapor.
Let state A be the initial state with vapor pressure e_s-de_s at temperature T-dT. Adiabatic compression to B, with e_s, T, can be achieved with an infinitesimal compression.
Then, place the system in contact with a reservior at T and expand it until a unit mass of liquid has evaporated. In this process, the pressure remains constant at e_s and the system passes to state C. The change in volume can be expressed in terms of the specific volumes of the gas and liquid:

\DeltaV = (\alpha_g - \alpha_L)

and the heat absorbed is equal to L, the latent heat of vaporization.

Now, we return tot he insulating stand and make an infinitesimal expansion back to pressure e_s-de_s, causing the temperature to fall again to T-dT. Finally, place the system on the heat sink at temperature T-dT and isothermally compress back to the initial state, condensing a unit mass of liquid.

From earlier results:

Q1/T1 = Q2/T2 = (Q1-Q2)/(T1-T2)

Q1 - Q2 is the net work done by the system, and thus is given by:

Q1-Q2 = (alpha_g-alpha_l)de_s

and we also have Q1 = L and T1-T2 = dT

so we have

L/T = (alpha_g-alpha_l)de_s/dT

or

de_s/dT = L/[T(alpha_g-alpha_l)]

The calculation can be repeated for a mixture of solid and liquid, for T being the melting point at pressure p and we get an equivalent equation

dT/dp = T(alpha_l - alpha_s)/L

Example 2.16 Calculate the change in the melting point of ice when the pressure chanes from 1atm to 2atm (alpha_Ice = 1.0903x10-3m^3/kg, alpha_Water = 1.0010x10^-3m^3/kg)

dT = T(alpha_W -alpha_I) dp/L = -0.007deg

Ice's melting point reduced for increases in pressure - unusual, related to the fact the ice is less dense than water

Example 2.17 Derive an expression of L(T) in terms of the specific heats of the liquid and the vapor.

When a unit mass of liquid vaporized, the entropy is increased:

delta s = L/T

taking d/dT of both sides:

ds_v/dT - ds_l/dT = 1/T dL/dT - L/T^2

Tds_v/dT - Tds_l/dT = dL/dT - L/T

dq_v/dT - dq_l/dT = dL/dT - L/T

thus:

dL/dT - L/T - c_v - c_l

2.8.5 Generalized Second Law

Tds >= du + p d\alpha

equality applies to reversible, equilibrium transformations.
inequality for irreversible.

Wallace and Hobbs 2.7 Static Stability

2.7.1 Unsaturated air.

If the lapse rate of the atmosphere in a region is less than the dry adiabatic lapse rate and a parcel is lifted adiabatically from the bottom of the region to the top (assuming that the parcel is never saturated in this path), whent he parcel reaches the top, it will have cooled according to the dry adiabatic lapse rate and thus it will be cooler than its surrounds.

Cooler air at the same pressure will be more dense and thus the parcel will experience a downwards restoring force due to its negative buoyancy. The reverse also holds for a parcel moved from the top downwards.

Generally, if \Gamma < \Gamma_d, the atmosphere is "positively stable" (for unsaturated air) and vertical mixing is inhibited.

Example 2.12 Derive the formula for the restoring force on a parcel displaced a small distance away from rest in a stable atmosphere.

let primes denote the parcel's variables.

for the atmosphere, in hydrostatic equilibrium

dp/dz = - \rho g

the acceleration of the parcel when it is raised upwards is

a = -(\rho'-\rho)*g/\rho

(negative sign indicates downward accel.)

a = -g(p/R) * (1/T' - 1/T)/[(p/R)*(1/T)]
a = -gT(1/T' - 1/T) = -g(T/T' -1)
a = -g/T' * (T-T') = -g/T' * [(T_0 - \Gamma) - (T_0 -\Gamma_d)]*\delta z
a ~ -g/T (\Gamma_d - \Gamma)*\delta z

Harmonic oscillation!

A Layer of air with an increasing temperature (\Gamma <1) is an inversion - a very stable layer of air.

\Gamma = \Gamma_d -- neutral stability - a displaced parcel remains neutrally buoyant, as long as it never becomes saturated!

if \Gamma > \Gamma_d, the air becomes unstable - a parcel displaced upwards experiences an upwards force, because it becomes warmer than its environment. These conditions do not last, as the instability causes a redistribution of heat bringing the atmosphere back to a neutral stability.

Instability can persist int he presence of strong heating from below, eg close to a surface.

Example 2.13 Show that d\theta/dz >0 in the environment is equivalent to positive static stability.

This can be shown mathematically, but logic should suffice - when d\theta/dz >0, an upwards displaced parcel, moving adiabatically, and thus conserving \theta, will have a lower \theta than its surrounds. But, both the parcel and the environment are at the same pressure, so this is equivalent to the parcel being cooler than its surrounds in actual temperature, and thus negatively buoyant.

QED

2.7.2 - Saturated air.

When a parcel becomes saturated as it moves, it no longer follows a dry adiabat - instead it follows a moist adiabat - the stability of a saturated atmosphere is then detemrined by moist adiabats

2.7.2 - Conditional can convective instability

When the actual lapse rate of the atmosphere is somewhere between the dry and moist lapse rates, a parcel of air might be stable for small displacements, but large enought displacements to cause the parcel to become saturated can result in the parcel becoming warmer than its surrounds.

The lifted parcel cools at the dry adiabat until the LCL is reached. At this point, it begins to follow a moist adaibat - cooling slower. If it is lifted far enough, it reaches the LFC (level of free convection) where it is positively buoyant.

In a typical atmosphere, there is a point near the tropopause where the parcel ceases to be buoyant - the LNB, indicating the approximate height reached by the convection.

In practice, mixing with air from the surrounds tends to reduce the buoyancy of the parcel, limiting the height it reaches.

An atmosphere where \Gamma is between the dry and moist lapse rates is referred to as conditionally unstable.