Thermal reservoirs
The energy in the atmosphere is not conserved (constant).
Every time a molecule in
the atmosphere hits the ground, it has an opportunity to pick up energy
from the molecules in the ground, or lose energy to the molecules in
the ground. Now the ground has, in general, a stupendously larger amount
of energy in it than does the atmosphere. Why? Simply because there
are so many more molecules in the ground than
there are in the atmosphere. The
gases in the atmosphere are, after all,
very much less dense than the rock and soil that make up the ground.
What this means is that the percentage change
in the amount of energy stored in the ground due to
transfer to and from the atmosphere is almost zero.
Note that the absolute amount of energy transferred is
quite large -- it's only the
percent or relative change which is small, and
that only because the total amount of energy stored in the
ground is so much larger.
We then call the ground in this situation a thermal reservoir,
because, just as an ordinary reservoir stores lots of water and
is used to "even out" fluctuations in water supply,
so the ground stores a great deal of thermal energy in this case
and serves to "even out" fluctuations in the energy of the
atmosphere. As for why the energy the ground stores is
called "thermal" energy, check out The Second Law
for an illustration of the difference between
macroscopically-unobservable
energy ("heat") and macroscopically-observable energy ("work").
Statistical Mechanics
How much energy does an molecule in the atmosphere pick up or
lose every time it bounces? That's not easy to say.
The collision with the ground is quite a complicated process on the
microscopic level, as you might imagine. To answer the
question exactly we would need to follow the exact trajectory
of the molecule and track each and every collision with each and
every molecule of the ground on each little bit of dirt and grass
and eyelash of uplooking observer.
However without doing all this work we can
make what turn out to be quite accurate statements
about the probability of an atmosphere molecule picking up
a given amount of energy from a collision with the ground.
Using this statistical information we can actually predict a
number of useful things.
For example, we could argue that the amount of energy each
molecule picks up after a collision with the ground
determines the altitude to which it rebounds --
molecules with more energy reach higher
altitudes before beginning to fall again under the influence of gravity.
Therefore the probability of a molecule
reaching a given altitude after each bounce is
directly proportional to the probability
of that molecule picking up the required amount of energy during its
collision with the ground.
Furthermore, we can argue that
the average density of molecules at a given altitude
is roughly proportional to the probability that any given
individual molecule will
reach that altitude, times the total number of individual
molecules in the atmosphere. (This is like noting that the total number
of people that arrive for a Giant Carpet Sale
is given by the total number of
people that go to Giant Carpet Sales when they hear about them times the
probability that any particular person heard about this particular sale.)
So if we can predict the probability that a given molecule
picks up a certain amount of energy in a collision with the
ground, we should also be able to predict
the average density of the atmosphere as a function of altitude,
that is, exactly what the bar graph on the right side of the simulation shows.
Atmospheric pressure is also directly related to the average density of
the air. Thus we will arrive at a very useful description of the
atmosphere -- its average density and pressure as a function of height
above the ground. That will tell us, for example, how high our airplanes
can fly.
By giving up a completely
accurate view of the motion of each and every molecule in the atmosphere,
and instead
seeking a rough, statistical description of the mechanics
of the atmosphere we are plunging into
the science of statistical mechanics. The point of statistical
mechanics is that it takes utterly impractical problems (predict the
locations of the billion billion billion molecules in the
atmosphere) and offers solutions which are correct "on average"
(predict the average number of molecules at a certain height above
the ground.)
Furthermore these statistical solutions
tend to become more and more reliable as the
numbers of "degrees of freedom" (in this case, number of molecules) becomes
larger and larger, much in the way that it is very difficult to predict
whether any individual person will be sick in a given
year, but quite easy to predict -- if you are, say, an insurance
company -- the average number of people who get sick each year
in some large population. In dealing with the atmosphere
we have an enormous number of molecules,
so many in fact that the solutions offered
by statistical mechanics are far more precise than one could ever
possibly need.
The canonical ensemble
The particular situation where a system of interest (the atmosphere
in this case) is in "contact" (can exchange energy with) a
thermal reservoir has a special name. Because this is the situation
most frequently encountered in reality, it was named the
canonical ("natural") ensemble by Josiah W. Gibbs, the
astonishingly gifted American scientist who completed the mighty edifice
of modern equilibrium statistical mechanics in the last quarter of the
nineteenth century. The term "ensemble" arises because we want to
consider the properties of our simulation of the atmosphere
averaged over time, or, equivalently, the average properties of a
set ("ensemble") of trial runs of the simulation all measured at the
same instant of time.
Why so? Recall we are only going to be able to predict the behaviour of
the atmosphere on average. Thus our results will only be
meaningful when we consider the averaged outcome of many trial runs
of our simulation. Any one, specific simulation is perfectly
capable of not following our predictions of what the "average"
behaviour should be. (As an analogy: the observation that your
next-door neighbor is both a vegetarian and a member of
the Democratic party says essentially nothing about
the average tendency of Democrats toward vegetarianism,
and, conversely, the fact that adult women are on average
shorter than adult men by no means guarantees that
the next man you see -- if you are a woman -- will be taller than you.)
Let's imagine roughly what happens when a molecule hits the ground.
We've noted that the ground is a lot more dense than the air.
Thus each atmosphere molecule crashing into the ground will
typically interact with a very large number of ground molecules
before it escapes back up into the atmosphere.
Let's assume for the sake of argument that it interacts with
N ground molecules, and that in doing so it can end up
with a "quantum" of energy e from each. Let's say
the probability that it gets a quantum of energy from any given
ground molecule is some fixed (unknown) number q, e.g. if
q = 0.5 there's a 50% chance the air
molecule picks up a quantum of energy from each ground molecule.
Now what is the result of a large number of interactions? The
probability P that the air molecule picks up N quanta of
energy is given by q × q × q × . . . × q,
that is, just N factors of q multiplied together. (For
example, supposing q = 0.5, the
probability that the air molecule picks up
one quantum of energy is 0.5, the probability that it picks
up two quanta of energy is 0.5 × 0.5 = 0.25, and so on.)
We may write our general result as:
P = q N = (e log q) N
= e N log q
where we've taken advantage of these two facts about powers
and logarithms:
e log q = q
and:
( ab)c = ab c
Now the probability q must be less than 100% or 1.0, so that the logarithm of
q must be negative (as you'll note if you take
the log of any number less than 1.0 on your calculator).
Using this fact, let's rewrite
our probability so that the
final total energy E = N e shows up on the right-hand side:
P = e - N e ( | log q | / e )
The vertical bars around log q just mean "absolute value", i.e.
| 5 | = | -5 | = 5. Now let's just define
T = e / | log q |
and we can then write our probability simply as:
P = e - E / T
It turns out we've left out some
factors in front that should go in because we have "changed dependent variables"
in this expression from N to E.
But these turn out not to be particularly important --
it's easy to find them just by insisting that the sum of all
P be equal to 1 (i.e. if we add up the
probabilities of all possible final energies
we have to get 100% -- our molecule has to have some energy!).
The important result is that
the probability of finding an air molecule with energy E
after it rebounds from the ground is proportional to
e - E / T. This
(famous) probability distribution is called
the Boltzmann distribution after Ludwig Boltzmann, another giant
nineteenth-century scientist.
(For more information about Boltzmann's contributions, consider visiting
The Happy Molecules.)
Incidentally our derivation of the Boltzmann distribution does not
rely on the original assumption we made that the air molecule could only
pick up a fixed "quantum" of energy e during each collision with
each ground molecule, with a fixed and constant probability q.
We could fancy the argument up considerably by allowing for a range
of possible energies to be picked up with a range of different
probabilities. This would force us to employ
integral calculus in our derivation but not, in the end, change anything
at all about our conclusion.
The important limiting assumption actually
turns out to be our assumption that the
probability of picking of a certain amount of energy on each
collision is independent of the probability of picking up a
given amount of energy on any of the previous collisions.
This is what underlies the Boltzmann distribution.
That this assumption should hold is in turn dictated by our characterization
of the ground as a thermal reservoir -- the amount of energy available
for transfer to the air molecule is so large that it
is essentially unchanged by any amount
already transferred. If the ground were not a thermal reservoir, then
our conclusions would be different.
Temperature
What is this mysterious quantity T that appears in the
Boltzmann distribution? No more or less than
the temperature of the ground! Thus the temperature is in our
way of thinking
intimately related to the probability q of getting a quantum of energy
from each interaction with a ground molecule, as
expressed by this equation:
T = e / | log q |
If the probability of acquiring energy during the collision
is high (q close to 1.0, hence log q close to zero,
hence the inverse of log q big)
then the temperature is high, and if the probability of
acquiring energy is low
(q close to 0.0, hence log q very large, hence
inverse of log q small) then the temperature is low.
Thus the temperature can be considered to be the
availability, per ground molecule, of energy to be transferred to an
air molecule. That is in complete
agreement with our intuitive understanding
of temperature as the "intensity" of energy stored in a body.
Density and pressure profile of the atmosphere
The Boltzmann distribution now tells us the probability of any
given molecule having an energy E. Now if it has energy
E, then the greatest altitude h to which it can
rise is determined by when all of this energy is in the
form of gravitational potential energy. The formula for gravitational
potential energy is simply:
E = m g h
where m is the mass of the molecule and g is a
constant called the "acceleration due to gravity" which
describes the strength of gravity.
Near the Earth's surface g equals 9.8 meters per second per
second, or 32 feet per second per second.
Taking the above expression we can rewrite our Boltzmann distribution
in terms of the maximum height a molecule can reach:
P = e- m g h / T
Now let's assume that the molecule spends most of its time at or
near its maximum possible height. We could argue that this is
the case because the molecule
will be moving the slowest then, as it slows down and just starts to fall,
while it will be found at much lower
altitudes only when it is nearing the end of its fall and is moving
very rapidly. Then we can conclude that the probability of a given
molecule being found at a height h is simply given by the above
expression. Recalling our earlier comments that the total density is
just this probability multiplied by the total number of molecules in
the atmosphere (a large but constant number), we conclude that the
density of the atmosphere must
fall off exponentially. That is, the density
d as a function of height h must be approximately:
d = e - m g h / T
The air pressure will have the same form, since the pressure is
proportional to the density, within our approximations.
What does this function look like? If you plug in a few numbers on
your calculator, or plot the function e-x if you have
a graphing calculator,
you will see that the function looks like this:
Which is to say, a function that rapidly decreases and
then trails off into a long tail.
Thus we expect to see atmospheric density and pressure initially
dropping rapidly with altitude, and then trailing off thinly
up to considerable heights.
And indeed if
you watch the simulation long enough you will see just this.
That is, the bar graph on the right-hand side that measures the
average density of the bouncing molecules on the left-hand side
will eventually look just
like the graph above (turned onto its side). You will also note
the molecules on the left staying pretty close to the ground,
except occasionally taking long leaps up into the top of the picture.
Your experience may
also suggest the exponential form of the atmosphere's density to you.
You will of course know that
atmospheric pressure decreases rapidly with height. Even
at a few thousand feet above sea level the air is noticeably
thinner, and moving to Denver, about a mile above sea level,
requires significant acclimatization. At the top of
Mt. Everest, still only six miles above sea level, the atmospheric
pressure is so low (about two-thirds normal) that it is impossible
for all but the most highly-trained mountain climbers
to survive for long without oxygen masks (actually until the
Austrian mountaineer Reinhold Messner accomplished this very feat
it was not believed possible that anyone could climb
Everest without an oxygen mask.)
Yet you are probably also aware that
the atmosphere trails off to a considerable height above
ground, persisting for hundreds of miles above the Earth's surface.
Military airplanes, after all, such
as the U-2 and SR-71 fly at altitudes above 15 miles.
You probably recall from the news
that Skylab and various Soviet satellites
have fallen to the ground because of friction exerted on them
by the atmosphere, even at their orbiting
altitudes of a 100 to 200 miles. The Space Shuttle
typically orbits at around 150 nautical miles and can easily
detect traces of the atmosphere.
The standard U.S. Air Force
definition of the "end" of the atmosphere and the "beginning" of
outer space is 50 miles altitude, which is quite far up.
Note, by the way, that as you can see from the
simulation and the resulting bar graph there isn't
really a definitive, sharp "end" to the atmosphere, just a gradual
trailing off.
The effect of temperature & mass
Notice that the Boltzmann distribution falls off more slowly as the
temperature increases. (You may have to plot it with a few values
of T to see this quite clearly,
but you can also guess it roughly in your head by asking yourself
what happens to the distribution curve
as T becomes very small and very large.)
Thus if you increase the ground temperature in the simulation
you will see the rate at which the atmosphere becomes less dense
with altitude, as shown by the bar graph, decreases. That is,
the atmosphere expands upward as the ground gets hotter, and
correspondingly contracts downward as the ground cools. (If
you bring the temperature all the way
down to zero, all the molecules will
fall out of the atmosphere onto the ground and form a liquid, and
the poor inhabitants of this simulation Earth will suffocate.
Be responsible.)
By looking at the simulation, you can see once again
the intimate correspondence between temperature and the probability
of energy transfer to air molecules during collisions with the ground.
For you will undoubtably notice that when you drop the temperature
the reason the air starts getting thinner at high altitudes
is that the bouncing molecules are
much less likely during a collision with the ground
to receive a lot of energy, and much more likely to receive
only a little and therefore rebound weakly.
Notice also that the Boltzmann distribution falls off more rapidly
as the mass of the molecules increases. That's simply because the
same amount of energy will heft a heavier molecule to lower
altitudes, when working against gravity.
You can confirm this
with the simulation by manipulating the slider to make half the
molecules heavier than the other half. The bar graph will now include
separate graphs of the heavy and light molecules, and you will indeed
see that the heavy molecules stick closer to the ground.
You might then conclude that the heavier
molecules in the Earth's atmosphere should also
be more likely to be found closer to the ground. Thus one might
think that
the air closer to the ground is richer in oxygen
molecules (weight 5.3 × 10-24 grams) than in
nitrogen molecules (weight 4.7 × 10-24 grams).
But in fact the influence of winds (which we've ignored here)
pretty much wipes out the influence of the small difference
in mass between nitrogen and oxygen, and
no such difference is actually observed.
Nevertheless an interesting and important point emerges
when we consider
the lightest gases such as helium (0.6 × 10-24 grams apiece)
and hydrogen (0.3 × 10-24 grams each).
Given that these two gases are actually
the most common elements in the Universe,
why is it that they are extraordinarily rare in the Earth's
atmosphere? (Helium was in fact first detected [by its light
emission under intense heating] on the surface of the Sun,
whence its name, "helios" being Greek for "sun". The principle
source of helium today is the underground
radioactive decay of uranium.)
The reason for the rarity of the light gases in the
Earth's atmosphere arises out of the discussion above:
these gases are in fact so light that they can
reach enormous altitudes at the typical temperatures of the Earth.
So high can they go, actually, that they reach altitudes
where the force of gravity becomes
too weak to pull them back down at all and they escape
into outer space. All the (large) amounts of hydrogen and helium
on the Earth when it was formed
escaped in this manner, and
only the heavier gases such as oxygen, nitrogen, carbon dioxide,
argon, etc. are now left in the atmosphere.
This turns out to be the
case for all the small planets, moons and asteroids in the solar system,
Venus, Mars, Mercury, etc. The giant planets (Jupiter, Saturn,
Uranus and Neptune), however, do have sufficient gravity to hold
even the light gases, and hydrogen and helium therefore make up a
significant fraction of their mass, which gives them the common
appellation of "gas giants".
So what makes it "ideal"?
Why do we call this a simulation of the "ideal" atmosphere? Because among
other things we
have neglected any and all interactions between the molecules making up
the atmosphere. You'll see in the simulation that they pass right
through each other like ghosts. We are treating the atmosphere as if
it was what is called an "ideal gas", which is a gas made up of genuine
point particles which can only bounce off walls (and in this case, the
ground). Real molecules have a finite size, of course, and so they
do bump into each other, and attract and stick to one another, and
many of these "nonidealities" contribute to the more
complex and interesting behaviour of the real atmosphere. We have
also ignored the chemical reactions that go on in the atmosphere, which
are of enormous importance.