# Boltzmann–Gibbs-distribution as resulting from a limiting density of states?

I'm interested in the relation between the probability distribution $p_i$ over states of a system on the one side and the density of states $\rho(\eta)$ of its environment. (Meaning, $\int_{\eta_a}^{\eta_b} \rho(\eta) ~ \mathrm{d} \eta$ is the number of environment states with energies in the interval $[\eta_a, \eta_b]$.)

If the whole (system + environment) is energetically closed ("isolated") with a total energy $E = e + \eta$, but system and environment are in thermal equilibrium (i.e. the whole is described by the microcanonical ensemble), then it holds $$p_i = \frac{ \rho(E - e_i) }{ \sum_i \rho(E - e_i) }.$$

This means, the probability distribution over states of the system is determined by
a) something that only characterizes the energetic structure of the system, the $e_i$s
b) something that only characterizes the energetic structure of the environment, $\rho(\eta)$, and
c) the total energy $E$.

This relation holds generally, for arbitrarily small or large systems and/or environments. Please note that we have not yet taken any limits!

If we now consider the thermodynamic limit, i.e. an environment composed of an infinite number of subsystems, the probability distribution $p_i$ over states of the system becomes the Boltzmann–Gibbs-distribution (aka canonical ensemble) $$p_i = \frac{ \exp(- \beta ~ e_i) }{ \sum_i \exp(- \beta ~ e_i) },$$ where the sum in the denominator is called the partition function. Using the first relation above, this distribution could now be interpreted as corresponding to a limiting density of states of the environment of the form $$\rho(\eta) \propto \exp( \beta \eta )$$ which characterizes the "infinite environment". However, the expression refers to the parameter $\beta$ of the Boltzmann–Gibbs distribution, which represents the temperature and depends on the total energy $E$ (per subsystem). Whereas in the finite case $E$ only serves to connect $\rho(\eta)$ and $p_i$, it here defines $\rho(\eta)$ itself.

To me this suggests that it does not make sense to characterize an infinite environment by a density of states — but maybe there's some way around this? Or is there a mistake in the derivation somewhere else?

When one proves that a small part of a greater system is described canonical ensemble, even though the greater system is described by a microcanonical ensemble, the key point is that the density of states of the greater system has the exponential form you mention, over a certain interval of energy.

Specifically what is important is that $\log \rho(\eta) = {\rm const} + \beta \eta$ over the range $\eta = \langle\eta \rangle \pm \Delta e$, where $\Delta e$ is the energy fluctuation of the small system, and $\langle \eta \rangle = E - \langle e \rangle$ is the expected energy. The value of $\rho(\eta)$ is not really important for energies that are very far outside this range, since those energies never occur.

In practice the form of $\rho(\eta)$ is usually something like $\log\rho(\eta) \approx {\rm const} + N \log \eta$ for some very large $N$ (such a form is found for gases, in particular). This log can be taylor expanded to yield the needed form for the canonical ensemble. For a large environment (one that is essentially in the thermodynamic limit and much larger than the attached system), the first order expansion is very accurate over the relevant energy range.

• Thank you very much for your answer. From what you write I take it that the "limiting density" I was thinking about doesn't really exist generally, but one has to look at a limit of the local density around $E$, and the choice of this $E$ determines the resulting $\beta$. – A. Donda Jan 27 '14 at 17:31
• I did some numerical computations where I arbitrarily fix a $\rho_1(\eta)$ characterizing one environmental element, and then compute the energy density $\rho_N(\eta)$ of the combination of a large number of such elements (without special tricks I get up to about $N = 8 \cdot 10^{6}$ before overflowing double precision). The results suggest that quite independent of the form of $\rho_1(\eta)$, $\rho_N(\eta)$ practically always appears to have the form you mention ($\propto \eta^N$). Do you know of an exact result that states under which conditions this happens? – A. Donda Jan 27 '14 at 17:36
• Hmm, that's an interesting point. I was actually thinking about ideal gases in particular where $\eta^N$ is easy to show. In the general case, what you have when combining elements is that the overall density of states is a repeated convolution of the elements' density of states. If the element density of states has a power law behaviour for energies near the minimum energy, then the convolution will behave as $\eta^N$ for energies near the minimum energy. – Nanite Jan 27 '14 at 21:48
• There is probably some sort of general theorem there, but I'm not sure what it would be called. It is related to the central limit theorem which also concerns the result of taking many convolutions. – Nanite Jan 27 '14 at 21:53
• That's exactly what I've been thinking of, and I did some research, but couldn't find anything. The crucial difference to the standard central limit theorem(s) is that the single variables involved have to be of finite spread (variance), which is not necessarily the case for $\rho_1(\eta)$. It's of course easy to show that $\rho_N(\eta) \propto \eta^N$ if $\rho_1(\eta) = const.$, but I think it's much more general than that. – A. Donda Jan 28 '14 at 0:57

The right relationship between the two (the two distributions are pretty much the microcanonical and canonical ensemble) is that the first distribution – all states with some fixed energy – is almost exactly the same (the difference goes to zero if the environment becomes infinite) as the Boltzmann-Gibbs distribution for an appropriate inverse temperature $\beta$.

However, the right value of $\beta$ has to be carefully adjusted for the two distributions to match in the limit. This appropriate temperature may be essentially extracted from the energy of the environment (or energy of everything, it isn't any different in the infinite-environment limit) divided by its number of degrees of freedom, $\beta=\beta(E/N)$.

• Thanks for the answer, but I think either I misunderstand you or you misunderstood me. – The question is exclusively about a system in thermal equilibrium with an environment, so the microcanonical ensemble does not apply to the system (but only to system + environment). – A. Donda Jan 25 '14 at 14:22
• I'm interested in the relation between the probability distribution over states of the system $p_i$ and the density of states of the environment $\rho(\eta)$, a) in the general case and b) in the case of the thermodynamic limit, where $p_i$ becomes the BGD / canonical ensemble. $\rho$ characterizes only the energetic structure of the environment, and the total energy $E$ is an additional piece of information needed to formulate the relation between $p_i$ and $\rho$. – A. Donda Jan 25 '14 at 14:23
• As far as I can see, the first displayed equation holds generally for a system in thermal equilibrium with an environment, at least a finite one. The question is, does it still make sense in the thermodynamic limit (an infinite environment)? If yes, the third displayed equation should hold. However, $\rho(\eta)$ is supposed to characterize the structure of energy states of the environment generally, but via $\beta$ it depends, as you correctly write, on the temperature / energy per degree of freedom. – A. Donda Jan 25 '14 at 14:24
• I suspect that it is just not possible to characterize an environment with infinitely many degrees of freedom by such a limiting density of states – am I correct? – I'll revise the question along the lines of these comments. – A. Donda Jan 25 '14 at 14:25
• Dear @A.Donda, you wrote that the total energy is given so be sure that the microcanonical ensemble (for the whole system including the environment) is what is relevant. Considering all microstates with a fixed energy (or one in an interval) is how the microcanonical ensemble is defined. This implies some probabilities for the small subsystems of interest and they're almost identical to those from a canonical ensemble at a fixed temperature and the temperature is closely related to the energy of the whole system. – Luboš Motl Jan 25 '14 at 14:38

I'm revisiting my own question after almost a year to add an answer that to my mind clears the matter up. This answer does not add much beyond @Nanite's, but spells out the details. Comments are welcome!

If the environment has an $N$-dimensional state space, for large $N$ its density of states over energy $\eta$ will typically* have the form $$\rho(\eta) = a_N ~ \eta ^ {N-1}$$ ($-1$ because the $\rho$ is the derivative of the state space volume).

For the numerator of the probability distribution $p_i$ it follows $$\rho(E - e_i) = a_N ~ (E - e_i)^{N-1} = a_N E ~ \left (1 - \frac{e_i}E \right )^{N-1}.$$ If for increasing $N$ we keep the total energy per dimension constant, $\frac EN = \epsilon$, $$\rho(E - e_i) = a_N N \epsilon ~ \left (1 - \frac{e_i/\epsilon}{N} \right )^{N-1},$$ the power term becomes $\exp(-\frac{e_i}\epsilon)$ in the limit $N \to \infty$.

However, the limit of the full state density $\rho(E - e_i)$ depends on the behavior of $a_N N$. A typical* behavior appears to be $a_N \propto \frac1{N!}$, and therefore the density converges to 0. An infinite environment can therefore not be meaningfully characterized by a density of states, even locally.

Proper limiting behavior for the probability distribution does arise since $$p_i = \frac{ \rho(E - e_i) }{ \sum_i \rho(E - e_i) } = \frac{ \left (1 - \frac{e_i/\epsilon}{N} \right )^{N-1} }{ \sum_i \left (1 - \frac{e_i/\epsilon}{N} \right )^{N-1} }$$ which becomes the Boltzmann–Gibbs-distribution $$p_i = \frac{ \exp(-\frac{e_i}\epsilon) }{ \sum_i \exp(-\frac{e_i}\epsilon) }$$ for $N \to \infty$ — and we see that the temperature of the small system is determined by the energy per dimension of the environment, $1/\beta = \epsilon$.

What we can do is use this limiting behavior to locally approximate $\rho(\eta)$ around $\eta = N \epsilon$ for large but finite $N$: $$\rho(\eta) \approx a_N \exp(1 - N) \frac N\beta ~ \exp ( \beta ~ \eta ) \quad \propto \exp( \beta ~ \eta )$$ The dependence on $\beta$ now appears as a natural consequence of the circumstance that we approximate locally around $N \epsilon = \frac N\beta$.

*) The weak point of this answer is of course that it assumes a "typical" state density of a high-dimensional environment of the form $$\rho(\eta) = a_N ~ \eta ^ {N-1}$$ or even $$\rho(\eta) = a_0 ~ \frac1{N!} ~ \eta ^ {N-1}$$ with $a_0$ not or only weakly dependent on $N$.