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James Clerk Maxwell (1831-1879)

Maybe the most known example of this property is concerned with wealth distribution in a western society. In this note, the authors discuss a possible source of this behavior by resorting to the powerful approach of statistical mechanics, which enlightens the analogies with the classical kinetic theory of rarefied gases. Among other examples, the distribution of populations in towns and cities is illustrated and discussed.

This paper presents a mathematical investigation of the phenomenon of Pareto tails, which is ubiquitous in socio-economic systems. It consists in the fact that often the equilibrium statistical distribution of a certain physical quantity shows a power law-type decay at infinity.

Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature

This implies that the distribution is fat-tailed, as opposed As a consequence, very high values of that physical quantity may be rare but their probability is in general not so negligible. By appealing to conceptual tools of collisional kinetic theory, the authors show that it is possible to explain the emergence of Pareto tails starting from simple binary, i. The case studies considered in the paper refer to: traders that produce a wealth redistribution by exchanging goods, which is a prototype of Pareto's investigations; educated people who shape the distribution of knowledge depending on the school degree that they hold; people who migrate to or from a city, thereby determining the size of the urban population.

The paper provides a limpid evidence of the fact that the kinetic approach applied to socio-economic systems is a valuable methodological tool to shed light on how complex collective trends are generated by simple individual interactions empirically experienceable in personal lives. This is indeed the same spirit as the one which, in the late , motivated the studies of the Austrian physicist Ludwig Boltzmann, who aimed at explaining the complex concepts of thermodynamics starting from the simple mechanics of colliding gas molecules.

Learn Physics: Learn about Kinetic Theory of Gases

The legacy of his theory is nowadays successfully permeating fields very distant from the original one, as this paper clearly demonstrates. I suggest to accept this paper after minor revisions.

Boltzmann Distributions

Here is a list of suggested amendments. Page 3 probability density. In: Akura, Yuji and Kirman, Alan eds. Economic Foundations of Social Complexity Science. What of it? If the piston were left alone, and nobody held onto it, each time it got banged it would pick up a little momentum and it would gradually get pushed out of the box.

The problem is, how much force? The minus sign is there because, as we compress it, we decrease the volume; if we think about it we can see that if a gas is compressed, work is done on it. How much force do we have to apply to balance the banging of the molecules? The piston receives from each collision a certain amount of momentum. A certain amount of momentum per second will pour into the piston, and it will start to move. To keep it from moving, we must pour back into it the same amount of momentum per second from our force.

Of course, the force is the amount of momentum per second that we must pour in. There is another way to put it: if we let go of the piston it will pick up speed because of the bombardments; with each collision we get a little more speed, and the speed thus accelerates. The rate at which the piston picks up speed, or accelerates, is proportional to the force on it.

Approximate kinetic theory of hard-sphere fluids near equilibrium. I. Formal theory | SpringerLink

So we see that the force, which we already have said is the pressure times the area, is equal to the momentum per second delivered to the piston by the colliding molecules. To calculate the momentum per second is easy—we can do it in two parts: first, we find the momentum delivered to the piston by one particular atom in a collision with the piston, then we have to multiply by the number of collisions per second that the atoms have with the wall. The force will be the product of these two factors. If it is not, the whole theory is wrong, and the piston will start to heat up and things will change, but eventually, when equilibrium has set in, the net result is that the collisions are effectively perfectly elastic.

On the average, every particle that comes in leaves with the same energy. So we shall imagine that the gas is in a steady condition, and we lose no energy to the piston because the piston is standing still. In those circumstances, if a particle comes in with a certain speed, it comes out with the same speed and, we will say, with the same mass.

1. Introduction

How many atoms are hitting? Now we notice a little trouble with this analysis: First, all the molecules do not have the same velocity, and they do not move in the same direction.


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No; of all the atoms, only half are headed toward the piston. As a very simple example let us take helium gas, or any other gas, like mercury vapor, or potassium vapor of high enough temperature, or argon, in which all the molecules are single atoms, for which we may suppose that there is no internal motion in the atom. If we had a complex molecule, there might be some internal motion, mutual vibrations, or something.

We suppose that we may disregard that; this is actually a serious matter that we will have to come back to, but it turns out to be all right. We suppose that the internal motion of the atoms can be disregarded, and therefore, for this purpose, that the kinetic energy of the center-of-mass motion is all the energy there is.

A Molecular Description

So for a monatomic gas, the kinetic energy is the total energy. Incidentally, we can stop here and find the answer to the following question: Suppose that we take a can of gas and compress the gas slowly, how much pressure do we need to squeeze the volume down? So we are going to have some kind of a differential equation: If we start out in a given circumstance with a certain energy and a certain volume, we then know the pressure.

So, we have to solve a differential equation, and we will solve it in a moment. We must first emphasize, however, that as we are compressing this gas, we are supposing that all the work goes into increasing the energy of the atoms inside. Where else could it go? We shall suppose for the present that this is not the case. The word adiabatic is used in physics in several ways, and it is sometimes hard to see what is common about them.

That is, for an adiabatic compression all the work done goes into changing the internal energy. Although we derived it theoretically, this is , in fact, the way monatomic gases behave experimentally. We may give one other example of the kinetic theory of a gas, one which is not used in chemistry so much, but is used in astronomy. We have a large number of photons in a box in which the temperature is very high. The box is, of course, the gas in a very hot star. The sun is not hot enough; there are still too many atoms, but at still higher temperatures in certain very hot stars, we may neglect the atoms and suppose that the only objects that we have in the box are photons.

The momentum and the velocity are in the same direction, and the velocity is the speed of light, so this is the momentum of each of the objects, times the speed of light. That is what is used in an analysis of the contribution of radiation pressure in a star, that is how we calculate it, and how it changes when we compress it.

What wonderful things are already within our power! So far we have not dealt with temperature ; we have purposely been avoiding the temperature.


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As we compress a gas, we know that the energy of the molecules increases, and we are used to saying that the gas gets hotter; we would like to understand what this has to do with the temperature. If we try to do the experiment, not adiabatically but at what we call constant temperature , what are we doing? We know that if we take two boxes of gas and let them sit next to each other long enough, even if at the start they were at what we call different temperatures, they will in the end come to the same temperature.

Now what does that mean? That means that they get to a condition that they would get to if we left them alone long enough!


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What we mean by equal temperature is just that—the final condition when things have been sitting around interacting with each other long enough. Let us consider, now, what happens if we have two gases in containers separated by a movable piston as in Fig.

Comments and Questions

What are the conditions for equilibrium? Obviously, the bombardment from the left side must be such that it moves the piston to the right and compresses the other gas until its pressure builds up, and the thing will thus slosh back and forth, and will gradually come to rest at a place where the pressures are equal on both sides.

What we have to try to prove, eventually, is that the numbers themselves are equal. We must realize that this is not the only condition over the long run, but something else must happen more slowly as the true complete equilibrium corresponding to equal temperatures sets in. To see the idea, suppose that the pressure on the left side were developed by having a very high density but a low velocity.