Types of distributions

Under this name, very generic, I am going to talk about a very interesting topic and, above all, statistics that is the distribution in physics.

And it is that, when we have many equal elements that are “apparently” in the same state we see, if we can catch one by one, that these are not in the same state, but in a distribution of several and whose “average” is the state we see.

There are three types of distributions in physics.

Maxwell-Boltzmann distribution. This distribution is based on distinguishable elements, on molecules. That is, if for example we take a box full of air at a certain temperature and, if we were taking molecules 1 to 1 we would see that not all molecules have the same properties. As I said before, in the explanation, what we macroscopically see is the mean but microscopically there is a distribution of particles to different properties.

Fermi-Dirac distribution. Here we are dealing with indistinguishable elements, for example an electron. And is that in a cloud of electrons or if we have an atom with 2 electrons and if we could put a different flag to each of the electrons (for example, a number) we would see that we are unable to distinguish one electron from another. They are indistinguishable. It is something complicated to understand but it is like that, we can not distinguish them by much that we see and we will not know which is which by much “flag that we put them”. The Fermi-Dirac distribution deals with fermions, one of the pillar particles. The good thing is that the fermions fulfill Pauli’s exclusion principle (two fermions can not have the same quantum numbers). With which, this distribution is about how, for example, electrons are electronically distributed in a metal.

The “technical” explanation is because the fermions are not particles per se but are wavelengths (field theory). Two “equal” fermions (of the same type) have the same wavelength but their separation is so small that it makes it impossible to distinguish between one wave or the other. Then in the previous example, the mere fact of putting a “little flag” would be very complicated since we would be unable to separate one from the other.

Remember that everything vibrates (because everything is energy), with which the atoms also vibrate. The story is that we can distinguish them because the wavelength of the vibration is very separate compared to its frequency. That is, the frequency at which they vibrate is so “small” compared to the distances that allows us to distinguish one from another, something that with fermions (an electron) does not since its wave function, its frequency is similar to size or separation if there are several. It is simple and easy to understand.

Bose-Eintein distribution. If the fundamental particles (according to the field theory) are the fermions and the bosons and, the previous distribution deals with the fermions it is quite obvious to understand that the Bose-Einstein distribution deals with the bosons. The bosons are also indistinguishable particles (for us all bosons are equal) but that, moreover, they do not have to comply with Pauli’s principle of exclusion (there can be more than one with the same quantum numbers, than things). For example, as much as we look at the light that is composed of photons and we are not going to distinguish one from the other (you can see the example of the slit to understand it or understand why a photon can be in two places at the same time and be the same) . A mental mesh but very easy to understand if we get into the theory of fields.

Let’s go back to the MB distribution (Maxwell-Boltzmann). Imagine a room with a gas, for example, air. The air is composed of Oxygen and Nitrogen ($latex O_{2}$ and $latex N_{2}$) so that the molecules of the gas move freely (although with the restriction of the size of the room) by it. The molecules can hit each other and, being confined in a room, will also hit the walls. So far everything is very simple to understand.

Thanks to this (to the blows) the molecules exchange energy and moment between them. That is quite obvious.

The energy of the molecule will be:

$latex E=\frac{1}{2}mv^{2}$

The molecules, when they collide, exert a force that will be:

$latex F=\frac{dp}{dt}$

That is to say, the force will be the variation of its moment in time (obviously).

Now, if we look at the walls of the room they suffer a force for the molecules as we have calculated before and, therefore, a pressure for that will be:

$latex P=\frac{F}{A}$

Where F is the force, as we have calculated before and A is the area of the room, nothing complicated since it is classic mechanics of all life.

Now, there is a direct relationship between the pressure and the temperature in such a way that the pressure varies linearly according to the temperature that it has. At lower temperature, lower pressure and, taking the previous equations, there is no force and therefore there is no moment.

This happens at a temperature of 0K (zero kelvin) and is the intringulis of that scale, that at absolute value 0 there is no movement of molecules, there is no moment and therefore there are no forces. Nice!.

And if we make a graph that shows us the number of molecules and the energy they have, we will see that the distribution offers us a Gaussian bell form. Where most of the molecules are in the area of the bell that is most likely.

Now the beautiful. And it is that the energy is related to the temperature and the temperature that we observe in the room is the average energy of the molecules.

$latex E = kT$

That is to say that what we measure as temperature is energy and the average according to the distribution of MB is the average energy of its Gaussian distribution. Tachan!

We must understand that what we see in the distribution are possibilities and, as such, we must take into account what happens if we vary the energy and therefore the temperature since the number of molecules will always be the same (obviously) and that there must be molecules at all temperatures. This means that the graph, the Gaussian bell will be narrower and higher at low temperatures and flatter and longer at high temperatures.

Finally, the formula that tells us the probability of finding a particle with a certain energy with MB is:

$latex f(E)=\frac{1}{e^\frac{E}{kT}}$

For the distribution of FM (Fermi-Dirac) imagine a hydrogen atom. We know that this atom has an electron that in its fundamental state (the state of minimum energy) is located at level 1s.

As we know, in quantum mechanics atoms have their energy levels quantified and an electron can only be at a level jumping from one to another if it absorbs or emits energy. The levels have maximum amount of electrons according to their quantum numbers (n, main quantum value, l, orbital quantum value, $latex m_{l}$, magnetic quantum value, and s or spin). The quantum values thus referred to are a solution of the Schrödinger equation in polar coordinates and are related such that:

n = 1, 2, 3..
l = 0, 1, 2, 3… (n-1)
$latex m_{l}$ = -l, (-l+1),… ,0, …., (l-1), l
spin =
(1/2) o -(1/2)

So, in level 1 we have the allowed values n=1, l=0 and $latex m_{l}=0 $ and spin (1/2) or (-1/2) with which we have two possible values and therefore, by Pauli’s exclusion, only two possible electrons in that level. That is why it is said that the level is complete if it has the value $latex 1s^{2}$.

Energy level 2, will have $latex 2s^{2}$ as the previous form and $latex 2p^{6}$ as n=2, l=0 (I will do not put the rest), l=1, $latex m_{l}$=-1, 0, 1 and spin (1/2) and -(1/2) which gives a total of 6 electrons so if you have both levels (1 and 2) complete, you will have $latex 1s^{2}, 2s^{2}, 2p^{6}$ electrons totaling 8 electrons in total. That, although it sounds very messy, in the end it is quite simple except to know why they have put those names to the levels (s, p, d, f, g …).

As we said, the “fundamental” hydrogen has $latex 1s^{1}$ 1 electrons. Imagine another hydrogen atom also in its ground state, both with that atom, for example, in spin (1/2). If we put both atoms together to try to form (since the valence layer (the lowest non-complete layer of the atom) a $latex H_{2}$ (which exists and is stable) by Pauli’s exclusion principle when in Equal quantum numbers will not be possible unless one of the electrons is “put” in the conduction layer (which is the next energy level -higher- to the valence layer, which already allows us to understand how electricity works , the diodes and, in short, the driving between elements, by the way).

But really what happens is something a little more fun and has a trick. An electron is not always with positive or negative spin (1/2) but as we measure it, it changes its spin. Sometimes it is positive and sometimes it is negative, but in the end, the probability in a finite time of finding it positive or negative is 50%. It’s the fun of quantum mechanics! With which, in the previous example, there will be a moment when the molecule of $latex H_{2}$ can be formed. Although, we lack a little thing and is that the resulting molecule, the energy levels have only one, the value n = 1 but the energy level is unfolded. This effect is called splitting of the energy level and can be observed when a spectrum of light is made in which the energy band can be seen unfolded and it is very beautiful. Normally the band splitting is observed in very large crystals.

Where, a crystal is not the transparent that we have in the windows but a solid structure where there is a unit called unit cell that repeats itself in space and, as long as the unit cell repeats itself (with certain impurities) through space it is called Crystal shape and crystal. That is why crystallography is not about crystals but about the properties of these structures and is a very interesting branch of physics, geometrically speaking.

Returning to the previous topic, if we take a crystal, the valence bands will be unfolded so that if there are N molecules there will be N bands for an energetic level but, if for example it is an atom like hydrogen that only has 1 electron for being the layer s (remember that there are only 2 electrons in the s) at the end of the whole thing and because there are half (N/2) with spin (1/2) and half (N/2) with spin -(1/2) in reality in that unfolded layer N times it will be totally occupied N/2 times, half of it.

Well, this unfolded layer and the energy fills up to half of that layer is called Fermi energy. And it is the maximum energy level occupied by the electrons of a crystal (or a molecule) at a temperature of 0 Kelvin (it is the definition).

The energy of Fermi tells us how far the valence layer goes and what is the energy of it. This is something very important since it is based on drivers, semiconductors, diodes, electronics and especially the physics of the solid state (a branch of physics) and what is, within the same energy level but in the unfolded free zone it’s the valence layer.

The splitting (the entire layer) usually has very low energetic levels of difference, of the order of 10eV (which is an authentic crap) and therefore, some electrons in the valence layer (remember, what I have said) are easily ” promoted “to the conduction layer simply by small increases in temperature or by collisions with other molecules or, because it is a low energy (the lowest of the material) when we apply current allows the electrons to move with little energy (depends of the material and its valence layer) emitting heat. The quantum explanation of the electric current.

Going back to the distributions and after this roll for you to understand, the probability that an energy level is occupied now is (after a lot of mathematics):

$latex f(E)=\frac{1}{e^\frac{(E-E_{fermi})}{kT}+1}$

Where, $latex E_{fermi}$ is the energy of Fermi (or Fermi energy), what I have told you before and E is the level that we want to know what is the probability of being busy (or not). And that as we see is very similar to the probability of MB.

The nice thing is that if $latex EE_{fermi}$ we have that f(E)=0 or what is the same, they are not occupied then they are free if the energy is above the Fermi energy. Something that is obvious.

Then you will say, what is it for? for the energy levels of splitting, friends. This gives us a level of conductance of the atom and therefore of the crystal. And not only that, since we can calculate the so-called Fermi temperature:

$latex T_{F}=\frac{E_{F}}{k}$

Where k is the Boltzmann constant and that is the temperature at which half are free and half are occupied, something very necessary for electronic devices to be the “perfect” conductance of such material.

With this, if the previous image was a Gaussian bell, now it is “a sawtooth” that varies according to the temperature, something easy to understand.

Let’s finish this billet with the probability of BE (Bose-Einstein). As we have said above and you will have forgotten if the distribution of FD is based on the Pauli principle for fermions, the principle of BE is based on bosons that are not subject to Pauli’s principle of exclusivity as being particles that carry forces (I recommend you to see my explanation of the field theory).

Therefore, the formula is similar to that of FD (Fermi-Dirac) which in turn is similar to that of MB (Maxwell-Boltzmann):

$latex f(E)=\frac{1}{e^\frac{E}{kT}-1}$

Where, the most remarkable thing is that if before it was +1, now it is -1 just because of what I have said that the bosons are not subject to the Pauli exclusion principle.

So, you will ask, if they do not comply with Pauli’s principle of exclusion, bosons will always be at the lowest energy level ?. No, but they will be at the level that corresponds to them energetically and, as you know, there are many elements that can give energy, for example, a change in energy level gives a photon of a certain energy (seen with a certain wavelength) and another change to another energy level gives another photon to another different energy and, therefore, to another different wavelength. Another story is that the change to the same level always gives, in the same space and at the same time, the same type of photon (which is the boson easier to understand).

But from the equation we can show that at 0 Kelvin all the photons (remember that it is a boson) will be at the minimum energy level, at the lowest possible energy level, all. This level is called “condensate” and is the result of the graph that you see at temperature 0 and that has created the Bose-Einstein condensate.

This tells us that at 0 absolute all bosons are concentrated and can be “seen” (read the meaning by not being distributed at different levels) at a single level and that the bosons is a particle that carries energy (of the 4 fundamentals, strong, weak, electronic and gravitational). That is why we are trying to make this condensate as an experiment to observe the graviton or boson that carries gravity (which is one of those that we have left) and, although, if I remember correctly in 2001, the Nobel Prize was given to those who got “almost” the condensate of BE with Rubidio (I miss millions of Kelvin) to get the absolute 0 is something that is currently considered impossible to require infinite energy.

It should be noted that the BE distribution is used for the blackbody radiation and where the electromagnetic boson is derived or for the temperature it serves for the phonons which are the particles that carry the vibration in the crystal lattices. In the end everything fits!.

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