I'm looking for an intuitive understanding of the factor $$e^{-E/kT}$$ so often discussed. If we interpret this as a kind of probability distribution of phase space, so that $$\rho(E) = \frac{e^{-E/kT}}{\int_{0}^{\infty}e^{-E'/kT}dE'}$$ then what, precisely, does this probability correspond to? Is its physical significance obvious? Specifically, why is it largest for energies close to zero?

Edit: I'd like to know how it is possible that for increasing $T$ the lower energy states grow more populated. The naive thing to think is "If I stick my hand in an oven and turn up the temperature, I'm assuredly more likely to be burned." This raises the question, exactly what does it mean for the lower energy states to grow more populated relative to the higher energy ones?

  • $\begingroup$ Think of the atmosphere. Where is most of the air? Near the ground, which is lower in energy. $\endgroup$ – Mark Eichenlaub Apr 23 '13 at 4:03
  • $\begingroup$ The hand in the oven experiment shows decisively that as T increases, it is higher levels that get more populated. See my answer below. I wonder if you have copied a formula wrong, or taken it out of context, or something like that. $\endgroup$ – joseph f. johnson Jun 3 '13 at 22:28

The Maxwell-Boltzmann distributes $N$ particles in energy levels $E_i$ such that the entropy is maximized for a fixed total energy $E=\sum E_i N_i$.

The probability that a particle is in the energy level $E_i$ is proportional to the number of particles in the energy level $E_i$ in this particular arrangement of particles in which entropy is maximized (the Maxwell-Boltzmann distribution), which is $N_i$. It so happens that when we distribute the particles such that entropy is maximized, more particles populate the lower energy levels.

I don't follow the above argument @Ben Crowell - we want to show that more particles are distributed in the lowest energy level. In the above, we write $\sum E_i = N_0 E_0 + E_R$ conclude that probability is maximized if the energies of the particles in the lowest energy state, $N_0 E_0$, is minimized, which occurs for $N_0 = 0$ - the opposite of what was desired.

I'm not sure how to intuitively explain the solution to distributing the particles such that entropy is maximized. If we agree that by distributing the particles as evenly as possible in the energy levels, we will maximize the entropy, we can try:

Suppose we require $\sum N_i E_i \equiv \hat{E}$ and that we have plenty of particles $N$, and that $E_i$ increases with $i$:

  1. Starting from $E_0$, put one particle in each energy level whilst $\sum E_i < \hat{E}$.
  2. We need to distribute the remaining particles. Starting from $E_0$, again put one particle in each energy level whilst $\sum N_i E_i < \hat{E}$. The lowest energy level, $E_0$, now has 2 particles
  3. Repeat 2., until all particles are distributed.

We can see that the lowest energy level will be most populated, and that $N_i$, and hence the probability that a particle is in state $E_i$, decreases with $i$. This algorithm won't exactly reproduce the Maxwell-Boltzmann distribution of particles in the energy levels, but it might help with an intuitive feel of why the lower energy levels are more probable.

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    $\begingroup$ So to restate this answer: 1. Thermal equilibrium corresponds to maximum entropy. 2. Maximum entropy is achieved when the number of microstates is greatest. 3. The number of microstates is greatest when the lowest energy states are the highest populated. Does this sound right? $\endgroup$ – Pricklebush Tickletush Apr 23 '13 at 2:24
  • $\begingroup$ Yes, that's exactly what I was trying to say. But I found it tricky to explain 3. intuitively. $\endgroup$ – innisfree Apr 23 '13 at 9:39

Nice question. Part of this is basic stuff, but the question about why it's at a maximum for low energies is a nice conceptual question that wasn't immediately obvious to me.

The Boltzmann factor is the (unnormalized) probability that a specific degree of freedom will be in a specific state. For instance, say we have some helium gas, and we choose one particular atom. One of this atom's degrees of freedom is its momentum $p_y$ along the $y$ axis. This momentum carries with it a certain energy $E=p_y^2/2m$. The Boltzmann factor tells us the probability of that specific value of $E$ compared to other possible values of $E$. The whole thing is actually simplest in the quantum-mechanical case, where the states are discrete. WP's article on the Boltzmann factor explains the classical case as well.

When we talk about phase space, we usually mean the phase space of the whole system, which includes all its degrees of freedom. In that sense, no, the Boltzmann factor is not a probability distribution over phase space.

BTW, note that the energy scale does not have to start at zero. The lowest possible energy could be negative or positive.

The reason that the lowest energy is most probable is that by taking as much energy as possible out of that degree of freedom, we can give it to the rest of the system. Let's say the rest of the system has some energy $E_R=E_{tot}-E$. The rest of the system then has some number of states $\Omega(E_R)$, which grows as a function of $E_R$. The probability that our chosen degree of freedom has a given energy $E$ is proportional to the number $\Omega$ of ways that the rest of the system can accomodate that $E$. (This is assuming that energy level $E$ is non-degenerate, so that specifying $E$ completely specifies the state of this degree of freedom.) Since this probability increases with $E_R$, it decreases with $E$.

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  • $\begingroup$ I don't think that restricting the Boltzmann factor to a specific degree of freedom is standard terminology (I just checked several books to see whether I had been misusing the term for years). The Boltzmann factor is usually seen as (say, in the classical case) the function $e^{-H/kT}$ on the phase space (where $H$ is the Hamiltonian of the system). It does provide the (unnormalized) probability density associated to the whole system. $\endgroup$ – Yvan Velenik Apr 23 '13 at 7:28
  • $\begingroup$ @YvanVelenik: I don't see how you could apply the Boltzmann factor to the entire system. The energy of the entire system is conserved, so its value isn't probabilistic. $\endgroup$ – user4552 Apr 23 '13 at 18:58
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    $\begingroup$ No, the Boltzmann distribution is used for systems in contact with a reservoir, so the energy is not conserved. (Or, equivalently, for a finite subsystem of an isolated system; but in that case too, the Boltzmann factor takes into account the energy of the full subsystem and provides the probability density on the phase space of the subsystem.) $\endgroup$ – Yvan Velenik Apr 23 '13 at 19:13
  • $\begingroup$ first you say your formula is for the probability of a molecule's having momentum within a certain range. But immeediately you switch to "The Boltzmann factor tells us the probability of that specific value of E compared to other possible values of E" This is the same mistake I explain in my answer: you have to first sum over all possible momenta that lead to the same value of E. There is a huge degeneracy here. $\endgroup$ – joseph f. johnson Jun 3 '13 at 22:21
  • First of all the probability density for a system in a canonical ensemble to be at a given energy (i.e. with respect to a measure $dE$) is not the formula you gave but rather:

$$\rho(E) = \frac{\Omega(E)e^{-\beta E}}{\int_0^{+\infty}dE\:\Omega(E)e^{-\beta E}} = \frac{e^{S(E)/k_B}e^{-\beta E}}{Q}=\frac{e^{-\beta F(E)}}{Q}$$

we thus see that weiht associated to a given energy state is more related to a free energy $F(E)=E-TS(E)$ than the simply the actual energy.

Now, the probability density to be in a given microstate $\mu$ (with respect to some phase space measure $d\mu$) for a system in a contact with a thermostat is:

$$f(\mu) = \frac{e^{-\beta E(\mu)}}{\int d\mu\:e^{-\beta E(\mu)}}$$

  • Secondly, I will try to summarize a bit what has been said by the others but in a general situation working with probabilities nstead of probability densities (there is no big difference).

It is important to realize that when a system is in contact with a thermostat at temperature $T$, you in fact have a small system, say $1$, that interacts with a big one, say $2$. The whole thing can be considered as an isolated system of energy $E$. In thermodynamic equilibrium all microstates of the isolated system are equi-probable and have a probability:

$$p(\mu) \equiv \frac{1}{\sum_{E_1}\Omega_1(E_1)\Omega_2(E-E_1)} \tag{1}$$

Here it is assumed that $E=E_1+E_2$. This is only true if there is no volume interaction between the system $1$ and the system $2$.

Now, the probability for the system $1$ to be in some microstate $\mu_1$ with energy $E_1(\mu_1)$ is nothing but the sum of $(1)$ over all the possible microstates of the system $2$ that ensure that $E_1+E_2=E$ i.e.:

$$p_1(\mu_1) \equiv \frac{\Omega_2(E-E_1(\mu_1))}{\sum_{E_1}\Omega_1(E_1)\Omega_2(E-E_1)} \tag{2}$$

What we see already is that the degeneracy of a microstate of system $1$ is the total number of microstates of system $2$ compatible with the constraint $E=E_1+E_2$. We thus already see that the bigger $E_1(\mu_1)$, the smaller the weight associated to the microstate $\mu_1$ (if we assume that the number of microstate is an increasing function of the energy).

This general result can be made quantitative in the case of small $E_1$ and it gives the Boltzmann weight:

$$p_1(\mu_1) \equiv \frac{e^{-\beta_2 E_1(\mu_1)}}{\sum_{E_1}\Omega_1(E_1)e^{-\beta_2 E_1}} \tag{3}$$

where $\beta_2 = 1/k_B T_2$ tells us how the degeneracy $\Omega_2(E-E_1(\mu_1))$ decreases with $E_1$ for small $E_1$.

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  • $\begingroup$ This answer is hard to understand, you hvae not explained your notation. If $\Omega$ is the volume of the region in phase space which has the same energy, your first formula is okay but then the rest does not follow, since this is, in the simple case I have in mind, the surface area of a sphere of radius E, and it grows with E like the square of E. $\endgroup$ – joseph f. johnson Jun 3 '13 at 22:26
  • $\begingroup$ It is true that I forgot to specify what was exactly $\Omega$ at the beginning. That is because I thought it was a pretty common notation (as in the famous $S=k\ln\Omega$) and I did say it was the degeneracy of a given energy (both mathematically in the first equation and in the text at the end). I don't get your second point, I am not refering to an ideal gas but to a general system in thermal equilibrium. Please specify what is not clear (apart from notation which I shall introduce properly later) in my answer. $\endgroup$ – gatsu Jun 4 '13 at 12:50

The formula is wrong. To be more precise, it does not apply to the distribution of molecules with energy $E$ in an ideal gas in equilibrium at temperature $T$. You have to integrate it over all micro-states with the same energy, which bring in a factor of $E^2$ in front of the exponential.

The canonical distribution is the probability distribution of points in phase space, not of energy levels. So to get the probability of an energy level, you have to integrate over all momenta which have the same energy.

The Maxwell--Boltzmann distribution for the probability that a molecule (picked at random from that ideal gas) will have a velocity with coordinates between $u+du$, $v+dv$, and $w+dw$ is $$({hm\over \pi}) ^{\frac 32} e ^{-hm(u^2+v^2+w^2) }du dv dw.$$

But many different velocities (u,v,w) will yield the same energy so we have to integrate over a sphere, which leads to

$$\rho (E) = 4\pi ({hm\over \pi}) ^{\frac 32} c^2 e^{-hmc^2} dc $$ where $m$ is the mass of the molecule, $h$ is proportional to the inverse temperature, and $c$ is the speed of the molecule which is associated to the kinetic energy $E$.

This probability distribution obviously has a hump, it is not monotonic. Its average value is $$2\over \sqrt{\pi hm}$$

which increases with increasing temperature.

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First, the Boltzmann distribution applies only to canonical systems, that is, systems having constant $N$, $V$, and $T$ where $N$ is the number of particles, $V$ is the system volume, and $T$ is the system temperature.

Let us assume that the energy of the system is the sum of the energies of the individual particles. And let us also assume that the energy structure for a single particle is $O, a, b, c...$, where $a$, $b$, and $c$ etc. are positive numbers.

For this case the energy of the system $E$ is given by $$ E = \frac{3}{2} NkT $$ where $k$ is Boltzmann's constant.

Now if all the particles are in the ground state with 0 energy, the average temperature must be 0. But that's impossible since the temperature of the system is fixed at $T$. So the question is: is there an arrangement of particles in the energy levels that will give the proper average energy for the given $T$. Boltzmann's equation does that.

Now what happens if we heat the system? Clearly the temperature will be higher so we must have fewer particles in low energy levels and more in high energy levels. Since there is no upper limit to energy levels, this can always be done.

If we cool the system a similar thing happens, but we can't go lower in energy than the ground state. So the ground state must have more and more particles in it.

By the way, the Boltzmann equation applies only at equilibrium. It does not describe what goes on while the temperature is changing.

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This is a common misunderstanding and the answer is simple: the Boltzmann factor does not give you the probability of finding a particle within $dE$ of $E$. This probability is actually given by:


So you see there is an extra factor of $E^{1/2}$ and there are very few particles at low energy.

The reason is subtle and is essentially because the Boltzmann factor is the probability density per unit volume of phase space, not per unit energy. I will try to explain it in simple terms.

Let's work in terms of velocity rather than energy because it is much easier to picture the reason this way. In terms of velocity, the Boltzmann factor is:

$exp(-m {\bf{v}^2} / {2kT})$

The Boltzmann factor is actually defined to give you the number of particles $dN$ (or equivalently the probability) within $d^3v$ of $\bf{v}$ via:

$dN=exp(-m {\bf{v}^2} / {2kT})d^3v$

Now if we imagine velocity space (which for our simple example is the same thing as "phase space") in spherical coordinates, we can picture thin spherical "shells" at each value of $v$, each shell being of thickness $dv$. In this case the volume occupied by particles with velocity $v$ increases as $v$ increases because the thin shell's volume increases due to it having a large radius in velocity space (the radius is just $v$). Since in spherical coordinates $d^3v=v^2dv$ we can write the above equation as:

$dN=exp(-m {\bf{v}^2} / {2kT})v^2dv$

which shows you that there are very few particles at low velocity. The most likely place to find particles is around $\sqrt(2kT/m)$ and that's because there is more volume in velocity space there - hence more particles can fit in!

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  • $\begingroup$ I am not sure the formula you opened with is true in the general case. A more precise formula would be I think $exp(-E/kT)\rho(E)dE$ where $\rho(E)$ is the density of states at energy level $E$, so that $\rho(E) dE$ is the number of states between energy level $E$ and $E+dE$. It translates into $v^2 dv$ in the kinetic gas example, but is more general than what you have written. $\endgroup$ – itamarhason Feb 1 '16 at 12:34
  • $\begingroup$ Other than that, I think your answer adds important information to the conversation, although it doesn't close it in my opinion, since you could still ask why the low-energy states are more populated with respect to their relative density (which is lower). $\endgroup$ – itamarhason Feb 1 '16 at 12:39

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