Atoms are very small building-blocks of normal matter. They are the bricks molecules are made of. Because they are so small, we need a new kind of physics, which was invented by many famous scientists of the early twentieth century, like Planck, Bohr, Fermi, Dirac Ehrenfest and so forth. Here I will just discuss some of those theories, which are mentioned in the main-texts. It goes deeper than the "atomic-transitions page". However, since the diffucilty in explaining quantummechanics from scratch, I will also discuss gas-kinetic theories, electro-theories. Historically, this is also the best approach.

Back to home.

19th Century problems; dealing with the very small.

Imagine, you're a 19th century scientist and you open a bottle of parfum in one corner of a room and you wait a moment in another corner. Obviously, when the parfum is strong enough, you'll smell the parfum after a little while. Of course you start wondering about the mechanisms which make the smell of the parfum fill the room. You might device some tests, for instance you take measurements at different distances and you might find that under simular circumstances you always find the same time-intervals, of course within a rough estimate of the error of your experiment. (Even though I'm not completely sure whether or not error estimates were done in the 19th century.)
Now, the only way in which one can try to understand these phenomena, without the exact knowledge of the very small, is to use averaged quantities. One will come easily to the concept of distribution functions. With these functions one can obtain the average velocity of the cloud of parfume, the mean position of the cloud, with the spread around this mean, etc. etc. The distribution in this example is the Maxwell Distribution which I'll elaborate from here.

The Maxwell Distribution Function

Let's take a cubic space filled with molecules which collide with each other and collide with the sides of the space. Since these collisions are random and occure "very" often, it may be taken on forehand that this system is purely statistical. Let's try to develope a statistical distribution which describes the velocities of the molecules. We make the follow assumptions:
* the distance between two following collisions of a molecule is much larger than it's own size. This means that the volume taken in by the molecules is much smaller than the volume of the cubic space.
* the different velocities are statistically independent of each other. So the chance that the velocity in the x direction, v_x, has a value between v_x and dv_x is independent of velocities v_y and v_z.
* there's is no preferred direction, this is the principle of isotropy.
Now, let's assume that we have N molecules in a cubic space with volume V, so the density is N/V = n. The fraction of these molecules with a velocity in the x direction between v_x and v_x + dv_x is given by f(v_x) dv_x and the same for the other two directions. Then the fraction of molecules with a velocity between v and v + dv (bold because it's the vector v with components (v_x, v_y, v_z)!) is: f(v_x)dv_x . f(v_y)dv_y . f(v_z)dv_z = phi(v)dv_x dv_y dv_z. So this function should in the end tell us what fraction of the molecules has a velocity between v and v + dv.

Because of the isotropy the distribution can not be dependent on the direction of v, but only on the length. (Think about this!) Thus phi is a function of v^2 = (v_x)^2 + (v_y)^2 + (v_z)^2. Thus with the above given definition:

phi(v) dv_x dv_y dv_z = f(v_x) dv_x . f(v_y) dv_y . f(v_z) dv_z = phi((v_x)^2 + (v_y)^2 + (v_z)^2) dv_x dv_y dv_z

This can only be mathematically solved when f(v_x) = C . exp(-alpha (v_x)^2) and likewise for f(v_y) and f(v_z), which delivers us:

phi(v^2) = C^3 . exp(-alpha v^2)

How do we find the constant C? This always needs the same trick; by integrating one of the distributions over all space, meaning from -infinity to +infinity we are sure to get a fraction of 1, if we say that "all" is equal to 1. So

[Integral]|+infinity , -infinity|: C exp(-alpha (v_x)^2) = 1

This is a so-called "standard integral" which gives the answer: C = sqrt(alpha/pi).(sqrt is a standard abbreviation for square root.) So now we only need to know alpha. It is wise to assume that the velocity depends on the specific temperature of the system, so it probably contains the temperature, T.

Let's try to evaluate the pressure which colliding molecules induce on one of the sides of the cubic space. Let's take a side which is perpendiculer to the x-axis, so in the y,z plane. (By now, anyone should have enough mathematical feeling to do this one inside ones head, without me giving a picture ; ) ) then we only need the molecules with a positive velocity in the x-direction. Molecules with velocities -v_x and +v_x collide v_x/(2l) times per time interval, with l the length of one side of the cube. With orthogonal collisions the change in impulse is 2mv_x, with m the mass of the molecules. So their total contribution to the force on this side of the cube is then v_x/(2l) . 2mv_x.

Now, the fraction of molecules with velocities between v_x and v_x + dv_x is given by the above distribution so that the total force is given by:

This gives :

F= Nsqrt(alpha/pi (m/l) [integra]|+inf , -inf|:exp(-alpha (v_x)^2) (v_x)^2 dv_x

Again a standard integral. Dividing by the area of the side of the cube gives:

P = F/A = N sqrt(alpha/pi)(m/V) sqrt(pi)/2 alpha^-3/2 = N m alpha^-1/2 V.

Using the well known law P V = NkT, with k the constant of boltzmann, we get:

Alpha = m/2kT.

Now we have f(v_x)dv_x = sqrt(m/(2pi kT)) exp(-m(v_x)^2/(2kT)) dv_x. The exponential is a well known form: exp(-E,kin/E,thermal) which is called a Boltzmann factor.

The distribution funtion for the velocity in three components will then be:

phi(v) dv = (m/(2pi kT))^3/2 exp(-mv^2/2kT) dv_x dv_y dv_z

Realising thatdv_x dv_y dv_z = 4 pi dv we get the final version of the Maxwell distribution function:

phi(v) dv = (m/kT)^3/2 sqrt(2/pi) v^2 exp(-mv^2/2kT) dv

Which looks as follows.



As with any probability distribution one can find average values by calculating the moments of the distribution. The average velocity for instance can be found be computing:

v_av = [integral]|+infinity , 0 (!)|: v phi(v) dv

or the average of the square of the velocity as:

v^2_av = [integral]|+infinity , 0 (!)|: v^2 phi(v) dv

The dispersion of a statistical distribution is given by (v^2)_av - (v_av)^2 The answers are: v_av = sqrt(8kT/(pi m)) and the dispersion: kT/m (3-8/pi) So the higher the temperature, the wider the spread in velocities. Check this in the graph. Also compute the average velocity, after looking up the mass of an oxygen molecule and check whether or not it fits.

The Barometric Function