## Thursday, March 15, 2007

What is local thermodynamic equilibrium

Eli, being a bunny, is somewhat fond of non-technical answers to technical questions. On the other hand, Eli has been known to pettifog with the best of em. Eli clearly needs to add some snark to this, but he asks, thermo snark????

The bean that Chris Essex is pushing is that for a system to be in equilibrium it has to be at the same temperature not only of the surroundings, but also with any radiation passing through it, like the sun's during the day, or cold space at night. Since the sun's radiation is best described by a 6000 K black body emission, that is rather not the case. If you want to really pettifog, you can point out that the temperature of the earth is not exactly that of the atmosphere right above it, so the IR emission from the earth passing through the air is at a slightly different temperature. Worse, the air does not absorb as a black body (In fact air and solar radiation are almost 100% decoupled). If you don't have equilibrium, describing a temperature for the system is somewhere between not trivial and not possible.

But dispair not gentle reader, the anonymice have got it pretty much right, you can describe temperatures for volumes of air and water, and even earth. For homogeneous systems like the atmosphere there is even a simple linear relationship between temperature and energy.

The simplest point of view is that thermodynamics describes the properties of a system at equilibrium, pressure, volume, temperature, etc. if the properties of your non-equilibrium system are the same, then you have a local thermodynamic equilibrium (LTE) in your parcel of atmosphere.

Statistical mechanics is another way to show this by measuring the velocity distribution of the molecules in the parcel of air you are looking at. If this matches a Maxwell-Boltzmann distribution (here is a cute little applet to play with. The more particles, the faster you get to equilibrium and the better the match. There are ~ 2 x 10^19 molecules in one cc of air at the surface) the temperature of the distribution is the temperature of the air. You can measure this to frightening levels of precision if the collision rate is much faster than the rate at which light is absorbed/emitted. Such a situation is described as being in local thermodynamic equilibrium (LTE) and temperatures are meaningful.

Places where this becomes difficult are very high up in an atmosphere where the collision rate is going to zero. Those really are non-equilibrium systems in any significant meaning of the words. For the interested and those who want to know more than they need, a slightly more technical discussion framed in terms of stellar atmospheres can be found here.

Essex' challenge is to show that the distribution of velocities (or among quantum states) of the molecules in the air do not follow a thermal distribution which defines a temperature. Since this has been measured directly, he can't do it. Anonymous said...

If all scientists had taken the view of Essex, there would be no science (and hence no global warming -- by definition, since there would be no thermometer to measure it).

Even the ideal gas laws make assumptions that are not (quite) real.

That is not to say that there is nothing worth studying in the field of non-equilibrium thermodynamics, of course (living things, for example) Anonymous said...

My goodness, this has certainly become one of the most amusing pinata parties in some time. Odd that there's not a Pielke or two for whacking.

Mus musculus anonymous

John A said...

The simplest point of view is that thermodynamics describes the properties of a system at equilibrium, pressure, volume, temperature, etc. if the properties of your non-equilibrium system are the same, then you have a local thermodynamic equilibrium (LTE) in your parcel of atmosphere.

Except of course that this is rubbish. In order for this to approximate to LTE in the hypothetical box of atmosphere would be to compress the time period under which these properties apply and assume no loss or gain of molecules during that time period.

Under such circumstances as in the Earth's atmosphere where pressure varies continuously, as does temperature, such conditions of LTE will never happen.

That's the difference between thermodynamic and non-thermodynamic equilibrium which you've failed to grasp. Anonymous said...

"In order for this to approximate to LTE in the hypothetical box of atmosphere would be to compress the time period under which these properties apply and assume no loss or gain of molecules during that time period."

This discussion obviously does not have much (if anything) to do with reality, does it?

It reminds me a little of the guy who sits at his desk worrying about whether he and his papers are going to fall through the gaps between the atoms in his chair and desk.

In the real world, one makes "good enough" approximations all the time. Without them, we'd all still be living in caves.

Within a parcel of air near the earth's surface, where the pressure at the top of the parcel is very nearly the same as that at the bottom, temperature still means something physical -- at least to most of us.

EliRabett said...

Eli wonders if Essex has ever calculated the deviation from a thermal distribution in a volume of air. Rabett Labs has measured it (for example the CO2 rotational distribution in an IR spectrum when you leave the sample chamber open.

Oh yes, John A will you please ask where Souda, India is? or more likely what actual data set was used. Anonymous said...

The condition for local thermodynamic equilibrium (LTE) is that

"the mean free path of the atom is very small compared to the distance over which the temperature changes,"

From "Thermodynamic Equilibrium, Local and otherwise"

Under such conditions, "the atom will collide many times with other atoms, all of the same temperature, before it can possibly reach some region with a different temperature. In this case, the speed of atoms within some small region may very well be described by a Maxwell-Boltzmann distribution with a definite temperature."

http://spiff.rit.edu/classes/phys440/lectures/lte/lte.html

So, to determine whether the air at the surface of the earth is in LTE, we must determine whether the mean free path of an "air" molecule (Oxygen or nitrogen, for the earth's atmosphere) is small compared to the distance over which temperature changes significantly.

or, what is equivalent: determine how much the temperature changes for an elevation gain that is equal to the mean free path of an oxygen (or nitrogen) molecule at temperature and pressure (1 ATM) typical near the earth's surface.

If this temperature change is very small over the mean free path in question, then we can be confident that we have met LTE.

So, is this temperature change very small over the mean free path in question?

For O2 (oxygen)at one atmosphere and 25 °C, the Mean free path is only 9.7 × 10^(–6)cm

where the ^ means "raised to the power"

http://www.iupac.org/goldbook/M03778.pdf.
For nitrogen (N2) it’s of the same order

So, how much does the temperature of the atmosphere change if we move upward over a distance equal to the mean free path of Oxygen gas at 1 ATM and 25 deg C? --ie, if we move up by 10^(-5)cm

For that we need the adiabatic lapse rate, defined as the negative of the rate of change in temperature with height observed while moving upwards through an atmosphere.

The DALR is a constant + 9.78 °C/km
MALR 4.9 °C/km

http://en.wikipedia.org/wiki/Lapse_rate

Using the larger of the two values above (dry lapse rate), gives 9.78x10^(-5) deg C per cm rise in elevation

or roughly 10^(-9) deg C per every 10^(-5)cm rise in elevation

so, in other words, for dry air at the earth's surface, the temperature changes only by 10^(-9) deg C (1 billionth of a deg C!) for a rise in elevation that is equal to the mean free path of Oxygen at the surface, when the temperature at the surface is about 25C.

Or we can look at it the other way:

Suppose we assume that we gain just enough elevation that the temperature changes by 1/10,000th of one percent (0.000001) of an assumed starting value of 25 deg C at the surface or by 0.000025 deg C.

How much elevation elevation gain will that take?

0.000025 deg/ (9.78 deg / km) = 2.6 x 10^(-6)km = 0.26cm

How many times the mean free path is this?

0.26cm/ (mean free path of O2 at 25C and 1 ATM) = 0.26/(9.7 x 10^(-6)cm) = 26000 times!

or in other words, the change in elevation that produces a temperature change of 1/10,000th of one percent is equal in length to 26,000 times the mean free path!

So, in the above case, the mean free path is a tiny fraction of the distance over which the molecule would have to move to experience a temperature change of 1/10,000th of one percent (a tiny fraction of the initial temperature!)

Recall the criterion for satisfying Local Thermodynamic equilibrium:
"the mean free path of the atom is very small compared to the distance over which the temperature changes,"

To a very high degree of accuracy, the criterion for local thermodynamic equilibrium as described above is met for air at the earth's surface.

Anyone who claims otherwise is either blowing smoke in your face or inhaling it himself (and it sure as hell ain't cigarette or cigar smoke!) Anonymous said...