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There is a book from Tom Carter on entropy. In the Economics I application (page 111), he ingeniously computes that the distribution of fixed amount of M money over N individual tends to

$$p_i = {1 \over T}\, e^{-i/T}$$

The temperature, $T = M/N$, is the average amount of money per individual.

The author calls this a Boltzmann-Gibbs distribution. I wanted to find a plot and compare it with Pareto distribution and Planck black body radiation. The problem is however, and this is my question, Wikipedia does not have it! Actually, there is an article on Boltzmann distribution, which is also called Gibbs distribution, but I see nothing similar to $p_i = 1/T\, e^{-i/T}$ there. At least I cannot see how ${N_i \over N} = {g_i e^{-E_i/(k_BT)} \over Z(T)}$ with $N=\sum_i N_i,$ and partition function, $Z(T)=\sum_i g_i e^{-E_i/(k_BT)}$ can be reduced to it. I do not see a plot $p_i(T)$ either. Are they related?

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    $\begingroup$ Why do you say $p_i \propto e^{-\frac{i}{T}}$ doesn't look like $p(E_i) \propto e^{-\frac{E_i}{T}}$ they look the same to me. $\endgroup$ Dec 15 '13 at 22:59
  • $\begingroup$ Are you confused as to which are the parameters and which are the variables? $T$ is a parameter in both. $\endgroup$ Dec 15 '13 at 23:13
  • $\begingroup$ @ComptonScattering Yes, I do not understand why there is $E_i$ instead of simply $E$ or $i$. But there is also partition function in place of T in the nominator. $\endgroup$
    – Val
    Dec 15 '13 at 23:32
  • $\begingroup$ All probability distributions must sum to 1, this is the purpose of the normalisation is not physically interesting. As for how you label your variables, this is purely purely syntactic preference. If you are considering a set of discrete energy values, you may want to label them with an index as with $E_i$. Saying $p(E_i) \propto e^{-E_i/T}$ is physically no different from saying $p(E) \propto e^{-E/T}$. It is just notation, the former would be read to imply the domain of $p$ is over a set of discrete values. $\endgroup$ Dec 16 '13 at 0:50
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Check the derivation of the Boltzmann distribution from the microcanonical ensemble on the Wikipedia page "Maxwell-Boltzmann Statistics".

We suppose that the "wealth classes" of individuals are discretised, so that, for example, we find the number of individuals with $m_1 = \$500$, the number with $m_2 = \$10000$ and so forth: as an approximation we restrict the wealth classes to discrete values. We don't even have to have the wealth classes equispaced.

Now, look at the derivation on the Wiki page, and there will be a precise analogy between that derivation of the Boltzmann distribution from the microcanonical ensemble, and the derivation of Tom Carter's formula. "Amount of Money Held" $m_i$ forms a precise analogy with "Energy" and "Number of Particles in each state" with "Number of Individuals in each Wealth Class", call it $n_i$. What you'll get is that the number of arrangements, helped by Stirling's formula, is:

$$\log\Omega \approx N\log(N)-N - \sum_j \left(n_j\,\log n_j-n_j\right) = N\log(N) - \sum_j n_j\,\log n_j$$

and it is constrained by the same two constraints: the total number of individuals is constant:

$$\sum_j n_j = N = const$$

and the total amount of money is constant:

$$\sum_j n_j m_j = M = const$$

so that you'll get, in analogy with the BD:

$$p_i = \mathcal{Z}^{-1} e^{-\beta\,m_j}$$

where the partition function $\mathcal{Z}$ and $\beta$ come from the Lagrange multipliers for the two constraints.

The underlying assumption is that all the arrangements of allocating shares of money to indivuals are equally likely, something I highly doubt as wealthier people tend to rig the rules and the conditions to make themselves wealthier! However, let's assume it's true. This formula is general for general, unevenly spread wealth classes. We now need to assume what these wealth classes are. Let's now assume they are evenly spaced: $m_j = j\,\delta$. Then we must calculate $\mathcal{Z}$ to make all the $p_j$ sum up to unity. The result is:

$$p_j = (1-e^{-\delta\,\beta}) e^{-j\,\beta\,\delta}\qquad(1)$$

The mean amount of money $M/N$ held by each individual is

$$\frac{M}{N}=\sum_{j=0}^\infty j \,\delta\,p_j = \frac{1}{e^{\beta\,\delta}-1}\qquad(2)$$

and if we make $\delta$ very small to give us many different money classes, then the (2) approximates to $N/M = \beta \delta$, and (1) approximates to $p_j = \delta\,\beta\,e^{-j\,\beta\,\delta}$, thus we get:

$$p_j \approx \frac{N}{M} e^{-j\frac{N}{M}}$$

which is your formula.

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  • $\begingroup$ I wonder why Tom named it Gibbs rather than Maxwell and why do Maxwell-boltzmann distribution looks like Plank back body radiation rather than exponent? $\endgroup$
    – Val
    Dec 19 '13 at 23:00
  • $\begingroup$ @Val I'm not a thermodynamicist, but to my nonspecialist eyes there seems to be quite a bit of overlap as to whose name is ascribed to what. Boltzmann's and Gibb's names get mixed up quite a bit and really that's how it should be: they both made fundamental contributions to this field. From my readings, Boltzmann was more the crazy and dazzling genius who had those seldom and keen insights that let him see in the dark but, as is so with so many such people, he got a few things wrong: Gibbs was more the thoughtful philosopher who saw how to make the genius's ideas work properly ... $\endgroup$ Dec 19 '13 at 23:15
  • $\begingroup$ ... this kind of collaboration can be found in the history of all the really great ideas in science. I'm fairly sure that the following is a fairly common usage amongst scientists like me who ken thermodynamics but are not experts: "Boltzmann distribution" $p_i=\mathcal{Z}^{-1}\,g_i\,e^{-\beta \,E_i}$ is the one we have been talking about: the name Gibbs is sometimes applied to it when it is thought of as distibution of an ensemble of systems rather than the large particle number limit of distribution of particles in a single system (Gibbs thought about things in this way); ... $\endgroup$ Dec 19 '13 at 23:21
  • $\begingroup$ ..."Maxwell-Boltzmann" is the name of the above distribution when it is applied to the distribution of particle speeds in an ideal gas, where $g_i\propto v^2$ and also $E_i \propto v^2$ so we get $p(v) \propto v^2\,\exp(-m\,v^2/(2\,k\,T))$. But in my experience you need to read from the context carefully, as there doesn't seem to be a universally accepted usage. $\endgroup$ Dec 19 '13 at 23:25

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