Joseph Andersen's Clouds and Waves and Stuff....

Because I'm a Nerd

noreply@blogger.com (Joe) — Tue, 30 Sep 2008 21:40:00 +0000

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

noreply@blogger.com (Joe) — Thu, 25 Sep 2008 21:37:00 +0000

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...

noreply@blogger.com (Joe) — Sun, 07 Sep 2008 17:30:00 +0000

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

noreply@blogger.com (Joe) — Thu, 07 Aug 2008 20:26:00 +0000

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

noreply@blogger.com (Joe) — Wed, 06 Aug 2008 03:06:00 +0000

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.

Wallace and Hobbs 2.6 - Water Vapor in the Air

noreply@blogger.com (Joe) — Mon, 04 Aug 2008 20:37:00 +0000

2.6 Water Vapor in the Air

2.6.1 Moisture Parameters

Mixing ratio: w = m_v/m_d - typically g_H2O/kg_air

Example 2.7: What is the partial pressure of water vapor, with mixing ratio 5.5g/kg with total pressure 1026.8mb?

e = (m_v/M_w)*p/(m_d/M_d + m_v/M_w) = w/(w+\eps) *p = 9mb

Example 2.8: Calculate the virtual temperature at T=30^oC with w=20g/kg

T_v = T(\eps+w)/(\eps+w*\eps) = T(1+ 0.61w) = 33.69^oC

Saturation vapor pressure - equilibrium state between evaporation and condensation if there is enough moisture. The partial pressure of the moisture in this situation is the saturation vapor pressure - depends only on temperature - increases rapidly with increasing temp

Saturation mixing ratio - the associated mixing ratio:

w_s = \rho_vs/\rho_d = e_s/(R_vT)/[(p-e_s)/(R_dT)] = 0.622 e_s/(p-e_s) ~0.622 e_s/p

Dew point - the temperature that a sample of air must be cooled to at constant pressure to arrive at saturation.

Example 2.9
For air at 1000mb, 18^oC and w = 6g/kg, what is the relative humidity and dew point? Using Pseudo-adiabatic charts - can locate saturation mixing ratio for this point - w_s = 12.9g/kg - so RH = 46.5%. The dew point can be located by finding the temperature at this pressure for which 6g/kg is the saturation value - 6.4^oC

Lifting condensation level - the level to which a parcel must be lifted adiabatically so that the pressure and temperature drop to the point where the vapor in the parcel is at saturation.

2.6.2 Saturated- and Pseudo-adiabatic processes
When lifted, a parcel's temperature decreases at the dry lapse rate until condensation occurs - at the LCL. Further lifting results in condensation, releasing latent heat and reducing the rate of cooling. If all the condensates are carried by the parcel, the process is still adiabatic and reversible: saturated adiabatic. If the condensates fall out immediately, then the process is irreversible and pseudo-adiabatic. The amount fo heat carried by the condensates is usually small and the pseudo-adiabatic and saturated adiabatic lapse rates are essentially the same.

2.6.3 The Saturated Adiabatic Lapse Rate = consider a lifted parcel undergoing condensation in a saturated-adiabatic manner:

dq = cpdT + gdz

the heat released by condensation is dq = -Ldw_s

thus,

dT/dz = -(L/c_p) dw_s/dz - g/c_p = -(L/c_p)(dw_s/dT)(dT/dz) - g/c_p

so,

\Gamma_s == -dT/dz = \Gamma_d/[1+(L/c_p)(dw_s/dT)]

This depends upon temperature and pressure and is always less than \Gamma_d as the second term in the denominator is always positive.

Example 2.10
A parcel is intially @ 15^oC, 1000mb and T_dew = 2^oC. What is it's LCL, and the temperature at that level? If it is lifted a further 200mb, what is its final temperature and how much moisture is condensed out?

Again, this is solved using a pseudo-adiabatic chart. The initial point is located, and the amount of moisture is located from the dew point ~4.4g/kg The intersection of the saturation line from the dew point and the dry adiabat from the actual point determines the LCL at 830mb As we lift further, we follow a saturated adiabat to 630mb, where the temperature of the parcel is -0.5^oC and the saturation mixing ratio is 1.8g/kg - thus 2.6 g/kg of moisture must have condensed during the ascent.

2.6.4 Equivalent potential temperature

\theta_e = \theta exp(Lw_s/c_pT)

the potential temperature of the parcel if all the vapour within is condensed and all the latent heat released go to thermal energy of the dry air. This quantity is conserved in both dry- and saturated- adiabatic motions.

2.6.5 Normand's rule
On a pseudo-adiabatic chart, the LCL of an air parcel is found at the intersection of the potential temperature line which passes through the locus of the parcel and the the pseudo-adiabat that passes through the wet-bulb temperature of the parcel at the same pressure.

2.6.6 Effects of irreversible condensation processes
If any of the products of condensation are allowed to fall out as precipitation, the latent heat gained by the parcel will remain when the now drier parcel returns to its original level. The Net effect is increase in T and \theta, reduction in w but no change in \theta_e

Example 2.11 - a parcel at 950mb has a temperature of 14^oC and a mixing ratio of 8g/kg. What is the wet-bulb temperature? The air parcel is lifted to 700mb and 70% of the condensate is lost before it returns to 950mb. what is the new temperature, potential temp, wet bulb temp, mixing ratio.

Again, this is solved using the chart.

Wallace and Hobbs 2.5 - Adiabatic Processes

noreply@blogger.com (Joe) — Mon, 04 Aug 2008 17:15:00 +0000

2.5 Adiabatic Processes - changes of state (pressure, volume, temperature) without heat added to or lost from the system.

eg. Adiabatic compression of a gas - compression does work so the internal energy of the system increases, raising the temperature. This temperature is higher than that that the system would have if it was compressed isothermally, so the resulting pressure is also higher

2.5.1 Air Parcels

When considering vertical motion and mixing of air, it is often useful to consider the behaviour of a "parcel" - an infinitesimal volume of gas that is: thermally insulated from the environment, stays at the same pressure as its surrounds and is moving infinitesimally slowly, so that KE is negligible.

2.5.2 Adiabatic Lapse rate

Consider a parcel lifted (or depressed) in the atmosphere, adhering to the assumptions above - how does its temperature vary with height?

Assuming a hydrostatic atmosphere and only adiabatic processes only, the parcels will satisfy:

d(c_pT + \Phi) = 0

c_p dT/dz = -d\Phi/dz

\Gamma = -dT/dz = g/c_p = 9.8 deg/km

dry adiabatic lapse rate.

2.5.3 Potential Temperature

\theta is defined as the temperature a parcel would have if it was adiabatically moved from its present environment to a reference pressure level (usually sea level/1000mb)

From the first law:

c_pdT - \alpha dp = 0

using the ideal gas law,

c_p/R dT/T = dp/p

or, integrating from the reference level to the present level,

(c_p/R)ln(T/theta) = ln(p/p0)

theta = T(p_0/p)^(R/c_p)

Under adiabatic processes, the potential temperature of a parcel is conserved.

2.5.4 - The pseudoadiabatic chart

The potential temperature lends itself to graphical solution. See charts in W&H

Example 2.6 - a parcel at 400mb has a temperature of 230K - what will be its temperature if it is moved adiabatically to 600mb?

Reference to charts (or calculation) shows that this parcel has a potential temperature of 300K. It will still have this theta at its final position. Reversing the calculation, or again referring to the chart shows that the parcel now has a temperature of 260K

Wallace and Hobbs 2.4 - Latent Heat

noreply@blogger.com (Joe) — Mon, 04 Aug 2008 17:11:00 +0000

2.4 Latent Heat

Heat energy is absorbed/released without change of temperature when a substance changes phase - this is referred to as latent heat, and is due to the change in internal energy associated with the alteration of the configuration of the internal parts of the system.

The most important latent heats from our perspective are those of water going from gas to liquid and liquid to solid phases

L_vap = 2.5x10^6 J/kg

L_fus = 3.34x10^5 J/kg

Wallace and Hobbs 2.3 - The First Law of Thermodynamics

noreply@blogger.com (Joe) — Sun, 03 Aug 2008 22:24:00 +0000

2.3 The First Law of Thermodynamics

Internal energy of a body (or parcel of gas) - potential and kinetic energy of the constituent atoms: relative configurations and temperature respectively.

consider a body that takes in a certain amount of heat energy q. The body may then do an amount of work, w, and be left with excess energy q-w. Conservation of energy implies that the internal energy, u, must increase by this amount:

u2-u1 = q-w

or, for infinitesimal energy increments

du = dq-dw

2.3.1 Joule's Law

When a gas expands without either doing work or taking in (or loosing) heat, the temperature of the gas is constant (strictly only for an ideal gas, as inter-particle energies can change in real gasses, leading to a change in temperature to balance).

2.3.2 Specific Heats

c_v = dq/dT |_\alpha
c_p = dq/dT |_p

c_p = c_v + R
dq = c_pdT -\alpha dp
dq = c_vdT - p d\alpha

2.3.3 Enthalpy

consider heat added to a material at constant pressure:

dq = (u_2 - u_1) + p(\alpha_2 - \alpha_1) = (u_2+p\alpha_2) - (u_1 + p\alpha_1) = h_2 - h_1

h == u + p\alpha

h is the enthalpy per unit mass of the material.

If the parcel is moving in a hydrostatic atmosphere:

dq = d(h+\Phi)

so, h+\Phi remains constant if the parcel moves adiabatically

Wallace and Hobbs 2.2 - The Hydrostatic Equation

noreply@blogger.com (Joe) — Sun, 03 Aug 2008 20:02:00 +0000

Thre is an upwards force on a thin horizontal slab of air in the atmosphere due to the pressure gradient across the slab - the pressure below is slightly larger than the pressure above, leading to a net force.

This force is usually almost exactly canceled by the downwards gravitational force acting on the slab:

-A dp/dz \delta z = g \rho A \delta z

dp/dz = -g\rho

since p(\infty) = 0;

p(z) = \int_z^infty g\rho dz

2.2.1 Geopotential

geopotential, \Phi, at a point is defined as the work against gravity required to raise a 1kg mass from sea level (or some other reference altitude) to that point.

We can then define a geopotential height:

Z = \Phi(z)/g_0 = (1/g_0)*int_0^z g dz

where g_0 is the standard gravitational acceleration (~9.8 m/s/s). Geopotential height can be used as a vertical coordinate, most usefully when energy plays an important role in the system.

It is generally inconvenient to deal with density, as it is difficult to measure in the field - so, eliminating \rho from the hydrostatic equation:

dp/dz = -pg/R_dT_v

integrating between two levels (1,2) leads to:

Z_2-Z_1 = (R_d/g_0)*int_p2^p1 T_v (dp/p)

2.2.2 Scale height and hypsometric equation

For an isothermal atmosphere, the above equation equation becomes:

Z_2 - Z_1 = H ln(p2/p1)

where H = R_dT_v/g_0 = 29.3T_v is the scale height of the atmosphere.

Below the turbopause, the atmosphere is (chemically) well mixed: the pressures and densities of the various species fall off uniformly, with a scale height of ~8.5 km

Above the turbopause, the veritcal distribution is more strongly controlled by diffusion and each species has its own scale height which is inversely proportional to the molecular weight of that species.

Example 2.2: If the ratio of O to H at Z=200km is 10^5, what is the ratio at Z=1400km, assuming that the atmosphere is isothermal in this region with a temperature of 2000K?
At these heights, because the atmosphere is very thin, the radiative cooling is very inefficient and the atmosphere becomes very "hot" to maintain radiative balance.

Call the ratio RR

RR(1400km) = p_O(200km)*exp(-1200km/H_O) / p_H(200km)*exp(-1200km/H_H)

RR(1400km) = RR(200km) * exp(-1200km*(1/H_O - 1/H_H))

RR(1400km) = 2.47

As the temperature usually varies with height, we can define a mean virtual temperature \bar{T_v} w.r.t. ln p

2.2.3 Thickness and constant pressure surfaces.

Z_2-Z_1 is the thickness of the intervening layer, and is proportional to the mean T_v of the layer. Roughly speaking, if you imagine the air between two levels as being unable to cross the level boundaries, upon heating, the air expands, pushing the levels apart.

Example 2.3: Calculate the thickness of the 1000mb - 500mb layer in a) the tropics where \bar{T_v} is 9^oC and b) the polar region, where \bar{T_v} = -40^oC

\DeltaZ = (R_d\bar{T_v}/g_0)ln(1000/500) = 20.3\bar{T_v}

a) \bar{T_v} = 282K, \DeltaZ = 5725m
b) \bar{T_v} = 233K, \DeltaZ = 4730m

Some more examples:

From the surface up to the tropopause, a hurricane's core is warmer than the surroundings - consequently, the pressure surfaces are pushed further and further down as one descends from the tropopause. The intensity, as measured by the drop of pressure from average pressure for each altitude decreases with height - warm core lows have the greatest intensity near the ground and diminish with altitude.
Some upper air lows do not extend all the way down to the ground. The downwards deflection aloft must be countered by an upwards deflection of isobaric surfaces at lower altitudes, which requires a cold core. A strong enough cold core, combined with a weak enough (or absent) upper warm anomaly can lead to surface high pressures.
Most mid-latitude disturbances have a cold core disturbance (low pressure aloft, high at sea level) to the west of a warm core disturbance. The position of the greatest pressure depression slopes westward with height.

Example 2.4
Calculate the geopotential height of the 1000mb surface when the pressure at sea level is 1014mb. Use H_0 = 8km

Z_2 - 0 = H_0ln(p_2/p_1) = H_0 ln (1.014) ~H_0*(0.014) = ~112m

Example 2.5
Derive the relationship for Z(p), in terms of p_0 and T_0, given a constant lapse rate \Gamma

T = T_0 - \Gamma z

Hydrostatic equation:

dp/p = -g/R dz/T
dp/p = -g/R dz/(T_0 - \Gamma z)

integrating from sea level up:

ln(p/p0) = -g/R \int_0^z dz/(T_0 - \Gamma z)

ln(p/p0) = -(g/R\Gamma) ln[T_0/(T_0 - \Gamma z)]

p = p0*-(T_0 -\Gamma z)/T_0]^(g/R\Gamma)

z = (T_0/\Gamma)[1-(p/p_0)^(R\Gamma/g)]

Wallace and Hobbs 2.1 - The Gas Laws

noreply@blogger.com (Joe) — Sun, 03 Aug 2008 19:40:00 +0000

2.1: The ideal gas law holds approximately for most gasses in most atmospheric conditions:

pV = mRT

where p is the pressure exerted by the gas (or partial pressure for a component of a mixture of gasses), V is the volume occupied, m is the mass, R the gas constant (in this formulation R depends upon the type of gas or gasses present) and T is the temperature in Kelvin.

Personally, I prefer the version I know from high school chemistry: pV = nRT - with n the number of moles of gas, then R is a true constant: R* - but stating things in terms of mass and mass density might make more sense in some situations.

The apparent molecular weight of dry air is 28.97 g/mol, so the gas constant for dry air is:

R_{dry air} = R* /M_{dry air} = 287 J/deg/kg

Example: calculate the density of water vapour at 20^oC which exerts a partial pressure of 9mbar.

Or, in more useful units 900Pa @ 293K

e \alpha = R_v T
900 \alpha = R*/(M_v)*293
\alpha = 150m^3/kg
\rho = \alpha^{-1} = 6.67x10^{-3}kg m^{-3}

2.1.1 Virtual Temperature

In some cases, when dealing with a mixture of air and water vapour, it is more convenient to retain the gas constant for dry air and use a ficticious temperature that corrects for the effect of the moisture in the gas equations.

Consider a mixture thusly:

\rho = (m_d + m_v)/V = \rho_d + \rho_m

Applying the ideal gas law to the two components to find the partial pressures:

e = R_v\rho_vT
p_d = R_d\rho_dT

p = p_d+e

combining leads to:

\rho = (p-e)/R_dT + e/R_vT
\rho = (p/R_dT) * [1-(e/p)*(1-\eps)]

(where \eps is the mass fraction of moisture to dry air)
so, we can rewrite this as:

p = R_d\rhoT_v

where

T_v = T/[1-(e/p)(1-\eps)]

is the virtual temperature - the temperature a sample of dry air would have if it had the same density and pressure as the given sample of moist air. As can be seen, T_v > T, so moist air is "warmer" than other wise identical dry air - that is, a sample of moist air is less dense than an otherwise identical sample of dry air.

Freeman Dyson Takes a swipe at climate change science

noreply@blogger.com (Joe) — Mon, 26 May 2008 19:34:00 +0000

From Realclimate.org

In the New York Review of Books, Freeman Dyson reviews two recent ones about global warming, but his review is mostly shaped by his own rather selective vision. 1. Carbon emissions are not a problem because in a few years genetic engineers
will develop “carbon-eating trees” that will sequester carbon in soils.

There is a lot of criticism of Dyson's article here - a lot of it probably valid. But some people seem to be writing off everything Dyson says based upon his involvement with the pie-in-the-sky Orion Project.

There was nothing intrinsically wrong with the Orion concept. It might not be the best engine design we have (and would probably never be usable in our atmosphere) but it is a good idea.

We should take our own advice, and stick to talking about our areas of expertise - leave the nuclear physics to the nuclear physicists ;).

Generally, I imagine that the trees Dyson talks about are probably somewhere in our future - the question is whether the trees would come quick enough to stop problems if we just go on with business as usual.

There is no reason not to pursue ideas like this - all ideas should be pursued to some degree. The danger lies in reliance upon one solution alone.

Tropical tropospheric warming

noreply@blogger.com (Joe) — Mon, 26 May 2008 18:54:00 +0000

Warming maximum in the tropical upper troposphere deduced from thermal winds - Robert J. Allen & Steven C. Sherwood - Nature Geoscience (subscription required)

The apparent lack of a tropical tropospheric warming signal in observations has been a problem with climate change predictions for some time. However, the direct temperature record has many problems - there have been many changes in the observation mechanisms:

non-climatic artifacts due to station relocations, observation time changes and radiosonde type or design changes...

leading to difficulties in interpretation. Earlier studies have attempted to correct for these artifacts with uncertain amounts of success. Instead, Allen and Sherwood use the thermal wind balance relationship to back calculate spatial temperature gradients from the significantly more trustworthy wind fields.

The "thermal wind" is a vertical shearing of the geostrophic wind arising from the pressure variations caused by horizontal temperature variations. The geostrophic wind is the wind that balances the Coriolis force with the pressure gradients. (See Wikipedia for thermal wind
and geostrophic wind). These relationships are weak near the equator, where the Coriolis force vanishes, but they appear to hold for winds averaged over long timescales.

This data gives a time series of spatial gradients in temperature, which can be integrated from a (collection of) "trustworthy" point(s) to give the temperature structure of the atmosphere. Allen and Sherwood cautiously report a warming trend in the upper tropical troposphere that is consistent with models showing climate change.

Did You Publish Today?

noreply@blogger.com (Joe) — Sun, 25 May 2008 16:46:00 +0000

This column, I'm sure you realize, dear fellow academics, is not for you. You don't need me to tell you that when you're working it can sometimes look to the rest of the world like you're curled up in front of the fire petting the cat. This column is for your husbands, wives, partners, parents, siblings, friends, and strangers who ask questions like "When are you going to graduate? It's been five years already." Or "Why hasn't that book come out yet? You've been working on it forever!"

Chronicle of Higher Education

Global warming reduces hurricanes?

noreply@blogger.com (Joe) — Sat, 24 May 2008 17:19:00 +0000

Simulated reduction in Atlantic hurricane frequency under twenty-first-century warming conditions - Knutson et. al; Nature Geoscience - subscription required :(

Changes in large-scale climate projected by IPCC used to force regional Atlantic Basin Model to assess hurricane frequency changes that might be driven by climate change.

A high-resolution simulation of the present-day Atlantic basin is nudged on large spatial scales to match the reanalysis data (numerical realizations of the global atmosphere that assimilate the many observations that exist into the models).
Hurricanes simulated by the model are not as intense as observed - probably a result of insufficient resolution. Storm tracks in the model are realistic.

The warm-climate runs show a general reduction in storm frequency. Will Global Warming reduce the hurricanes we have to deal with?

I am not so sure - I see a problem with their methodology: the "warm climate" simulations are a simple rerun of the present day simulations with the day to day and year to year variations unchanged, with just the mean temperature (and probably moisture) profiles increased as predicted by the climate projection models.

They thus make the implicit assumption that the climate variability will be unchanged under global warming, an assumption that seems unlikely and is generally not found to be true in climate projections.

Unfortunately, future variability is a hard thing to account for - the current models are not sufficiently accurate that I would trust any specific details of variability.

Another observation is that near storm rainfall is increased - rainfall comes from condensation, which releases latent heat - a source of energy for the hurricanes - so the hurricanes should perhaps be stronger on average. Even with a reduced frequency, if the number of extreme hurricanes increases, then we are probably worse off, both financially and environmentally - the global ecosystems will have evolved "assuming" a certain distribution of hurricane intensities in an average year - changes to this distribution could change many things: mixing of nutrients and/or silt throughout the oceans, rainfall amounts in rivers and lakes in regions near the eastern coasts of the continents that tend to be fed by the precipitation from the storm systems that form out the decaying Hurricanes.

These are complicated questions that I don't know enough to answer.

I suspect that if this experiment was rerun with an increased warm climate variability, the results might be quite different.

Second PhD Comic post in about 12 hours

noreply@blogger.com (Joe) — Thu, 15 May 2008 14:47:00 +0000

I get exactly where Tajel is coming from in this one - I have the same issue with my Family name - the 'e' near the end throws most people.

And it doesn't help that my Australian accent apparently makes 'e' sound like every other vowel (and some consonants) when I spell it out... The accent also makes 'Joe' sound like John or Jason or James to a lot of Americans (apparently)

SMS data rate is 4x more expensive than data from the Hubble

noreply@blogger.com (Joe) — Thu, 15 May 2008 05:16:00 +0000

You know how the mobile carriers charge you a couple cents to SMS a few characters' worth of text over their network? When you add it up, you're paying about a zillion bucks a meg for that traffic -- seriously! A space scientist from Leicester has calculated that SMS data is four times more expensive than receiving data from the Hubble space telescope.

boing boing

Climate makes profound mark on ecosystems

noreply@blogger.com (Joe) — Thu, 15 May 2008 02:48:00 +0000

SYDNEY: The largest study of its kind reveals that global warming is already having a massive effect on life across the planet – greater even than habitat loss and deforestation.

The study, which analysed nearly 30,000 data sets stretching back to 1970, suggests that warnings spelt out last year by the U.N. underestimated the impact of the problem.

Significant changes

The data set covered phenomena as varied as the earlier leafing of trees and plants; the movement of species to higher latitudes and altitudes in the northern hemisphere in response to warmer weather; the shrinkage of glaciers and melting of permafrost; and changes of bird migrations in Europe, North America and Australia.

The study concludes that "significant changes" are already occurring among natural systems on all continents, with the exception of Antarctica, and in most oceans.

Published today in the U.K. journal Nature, it goes beyond the scope of the landmark report issued by the U.N.'s Intergovernmental Panel on Climate Change (IPCC) in February 2007.

From Cosmos

Feynman playing Bongos

noreply@blogger.com (Joe) — Thu, 15 May 2008 01:02:00 +0000

via Symmetry Breaking"

Recognise any familiar buildings?

noreply@blogger.com (Joe) — Thu, 15 May 2008 00:56:00 +0000

Harvard does the redbrick wonderland, with a gothic-envy hall (only from 19thC) and a tower (but from the ugly 1960) or two.

There is the archetype of leaky modernist just down the road...

(from PhD Comics - its a little old, I bookmarked it awhile ago and forgot to post it)

Microsoft's World Wide Telescope

noreply@blogger.com (Joe) — Wed, 14 May 2008 21:18:00 +0000

The WorldWide Telescope (WWT) is a Web 2.0 visualization software environment that enables your computer to function as a virtual telescope—bringing together imagery from the best ground and space-based telescopes in the world for a seamless exploration of the universe.

Choose from a growing number of guided tours of the sky by astronomers and educators from some of the most famous observatories and planetariums in the country. Feel free at any time to pause the tour, explore on your own (with multiple information sources for objects at your fingertips), and rejoin the tour where you left off. Join Harvard Astronomer Alyssa Goodman on a journey showing how dust in the Milky Way Galaxy condenses into stars and planets. Take a tour with University of Chicago Cosmologist Mike Gladders two billion years into the past to see a gravitational lens bending the light from galaxies allowing you to see billions more years into the past.

This looks pretty cool - I haven't had a chance to boot to the windows side of my laptop recently, so I haven't actually tried it out, but the press seems to be pretty good.

Does this make up for Vista? Only time will tell...

Refuting refutations of the IPCC models of global warming

noreply@blogger.com (Joe) — Mon, 12 May 2008 17:27:00 +0000

There are those who try to point out flaws in the global climate model results (or the models themselves) that predict global warming. These people have a whole spectrum of motivations for this (I imagine) - some wishing to keep up business as usual, some interested in keeping the science honest etc.

Some apparent flaws in the IPCC's projections have been pointed out, but Gavin @ realclimate hs put together a cogent counter argument to various complaints about the projections in comparison with observations of the recent past - that basically comes down to really understanding the statistics of the ensembles of models that make up the projections.

Claims that GCMs project monotonic rises in temperature with increasing greenhouse gases are not valid. Natural variability does not disappear because there is a long term trend. The ensemble mean is monotonically increasing in the absence of large volcanoes, but this is the forced component of climate change, not a single realisation or anything that could happen in the real world.
Claims that a negative observed trend over the last 8 years would be inconsistent with the models cannot be supported. Similar claims that the IPCC projection of about 0.2ºC/dec over the next few decades would be falsified with such an observation are equally bogus.
Over a twenty year period, you would be on stronger ground in arguing that a negative trend would be outside the 95% confidence limits of the expected trend (the one model run in the above ensemble suggests that would only happen ~2% of the time)

Fun demonstration of doing science with complex systems, like climate models

noreply@blogger.com (Joe) — Mon, 12 May 2008 16:41:00 +0000

On the weekend I helped run the workshop described below at an event at MIT and it was a lot of fun. The event was a youth "summit" on the environment, and we used the exercise to talk about the difficulty of working out what parts of the climate (or climate models) were influenced by what other parts - although I imagine you could use this to explain almost any complex system - like biological systems or psychology - and the difficulties of observational science.

THE SYSTEMS ACTIVITY TO BE FACILITATED:

Around six of the students [we had a small group and only used two] are selected to be scientists and are taken to some place where they will not hear your instructions to the rest of the group. Those six are going to try and determine what rules govern the motion of the system [created by the larger group of students]. The larger group of students is given the instructions that everyone in the group is to choose two other people in the group and position themselves equidistant between those two people (what equidistant means will have to be thoroughly explained to the students...typically some of them just don't get it). Then the entire system group will have to practice this a bit before introducing the group of six scientist to the system, so keep the scientists away for as long as it takes. The trick for the system students is that each student has to choose and maintain their respective positions covertly. They cannot talk to the persons they've chosen, look at them or in anyway indicate who they've chosen, to the best of their ability, the entire time the activity is being run.

Once you've introduced the scientists to the system, make sure the system is moving/functioning, and instruct the scientists that they are to try and determine what rule(s) govern the motions of this complex system.The objective is to have them explicitly state that each individual element of the system must maintain itself equidistant between two other elements. They don't necessarily have to guess that the system elements can't indicate to their two object elements that they have been chosen, but that's up to the facilitators of the activity.
After a couple of minutes indicate to the scientists that they can get into the system and perform experiments if will help them determine the governing rules of motion of the system. At this point you may have to remind the system that they must continue to follow the rules...this may cause some physical contact issues, so remind all the students to be civil, but stay as true to the experiment as possible. To facilitate the scientist seeing other examples of how the rules apply. Verbally stop the system and tell every element to switch (you've explained ahead of time that switching means that everyone in the group picks a new pair to position themselves equidistant between). Then start the system over again.

Okay, so the students probably won't guess the rules. If they do great!

If they don't get the answer, without telling the scientists the rule(s), to tie this to climate change by having the facilitators modify the system behavior to demonstrate complex system behavior through modifying the motion of one of the elements, e.g. make one of the elements move in a more confined space.

The point of this demonstration is for the system students to feel what a complex system is and the scientists students to experience what it is like trying to decipher all the rules governing a complex system, using crude tools (eyes and some touch), referring to understanding climate change as very similar (tools = coupled GCMs, observational analysis, proxy data etc.).

It's okay to stop the activity in the beginning to let the scientists ask the system questions, but explain to the system elements that they are supposed to answer the exact question asked...and not to help the scientists...because the questions are the key to science!!

(This came from Tim Barnes at the NCAR/UCAR outreach office)

The three of us (myself, Larissa Back of MIT and Amos Tai of Harvard) running the group joined in as elements and I think we had a lot of fun - our system developed some wild oscillations at one point because of positive feedbacks where people where following each other! One thing we added after a while was slowing down the response by having everyone count to five before they moved after someone they were responding to had finished moving - this helped expose the chain of causality and also seemed to make stable configurations easier to achieve.

The two scientists did get quite close to the rule (with a little guidance) in the ~30 mins the exercise ran for.

Newstopia - Inspektor Herring Kidnapped

noreply@blogger.com (Joe) — Tue, 06 May 2008 20:15:00 +0000

From Newstopia last year. This is probably the best Herring so far (and it looked like they finished it up last week).

Herring is a great running joke, with the odd political reference - look for the waterboarding reference in this one.

AMS Hurricanes 08

noreply@blogger.com (Joe) — Thu, 01 May 2008 19:25:00 +0000

I just gave my talk at the AMS hurricanes talk here in Orlando. I think the reception was pretty good. I got one slightly negative comment/question from the leader of a competing group, which I think I dealt with well. The questions allowed me to bring up the question of whether the waves are unstable at all - maybe they're stochastically forced, forced externally and/or unstable only to finite disturbances.

Also, I've posted a draft of my paper to my website in case anyone was interested enough to have a look.