How can I intuitively understand the Boltzmann factor?

It is known that for a system at thermal equilibrium described by the canonical ensemble, the probability of being in a state of energy $E$ at temperature $T$ is given by the Boltzmann distribution: $$\tag{1} P(\text{state with energy } E) \propto e^{-E/kT}.$$

I have no problems with the mathematical derivation of this result, but I do wonder if there is a way to understand what (1) means intuitively.

In other words: Is there a way to "guess" that the correct distribution for such a system is given by (1), using solely physical arguments?

• This factor could be founded by physical arguments. See this – Nogueira Jan 24 '15 at 20:18
• For clarification: is the question you are asking about Why should the distribution be exponential? Why should $\beta$ be $(k_BT)^-1$ or are you asking about both (the title seems to ask about the second, however in question itself you only ask about the first)? – kristjan Jan 24 '15 at 20:23
• A very simplistic physical example (from which one might generalize) is that of the barometric formula, where one derives the exponential form for the density of air (and thus, the probability of being in a particular energy state). – alarge Jan 27 '15 at 21:53
• @Nogueira That's certainly a beautiful argument (one of the grandest pieces of technical writing ever IMO - if you mean section 2), but I think the OP is after something like Feynman's argument. – WetSavannaAnimal Jan 27 '15 at 22:27
• Dear glance: The Feynman Lectures on Physics derive the law from the ideal gas law and the exponential distribution of pressure with height of a gas at uniform temperature in a constant gravitational field, and then argue that the law is general. See whether this is what you are after feynmanlectures.caltech.edu/I_40.html – WetSavannaAnimal Jan 27 '15 at 22:30

What is most intuitive to me is to look at what we're asking. What is the probability that we find the system in a state with total energy $\hat{E}$? It is just the fraction of all possible states that have total energy $\hat{E}$, i.e.

$$p(E) \equiv \frac{\int \delta\left(E(\Omega)- \hat{E}\right)d\Omega}{\int d\Omega}\sim N_\hat{E}$$

But $N_\hat{E}$ is just $$e^{\ln N_\hat{E}}\equiv e^{S_S(\hat{E})/k_B}$$ where $e^{S_S(\hat{E})}$ is the number of states in which our system has total energy $\hat{E}$, which is the same as the number of states in which the rest of the universe has dropped to the energy $E_{tot}-\hat{E}$ by giving $\hat{E}$ to our system, i.e. $S_S(\hat{E}) = S_U(E-\hat{E}) \approx S_U(E) - \hat{E}\frac{\partial S_U(E)}{\partial E}$.

To bring it together, I would do the usual demonstration that at equilibrium, it must be the case that in a simple system that exchanges heat $\frac{\partial S}{\partial E}$ must be the same value in every zone of the system.

Furthermore, if one zone of the system has a smaller value of this than the others, the system will tend to bring more energy to that zone.

Therefore $1/\frac{\partial S}{\partial E}$ is naturally identified with our notion of temperature, i.e. $1/\frac{\partial S}{\partial E} \equiv T$.

Then we clearly have $$p(\hat{E})\sim N_\hat{E} \equiv e^{S_S(\hat{E})/k_B} \propto e^{-\hat{E}/k_BT}$$ This always makes it clear to me that we're just taking sub-volumes of the total set of microstates and asking how big it is next to the collection of all microstates.

Also, picking the sub-volumes out with the $\delta$-function naturally generalizes to all other scenarios, that are usually classified as different ensembles. Here it is just a different constraint relation inside the $\delta$-picker.

If there's another thing we want to constrain, maybe the number of particles, then the $\delta$ installs a factor of $\frac{\partial{S}}{\partial N}$ upstairs, along with the $\frac{\partial{S}}{\partial E}$.

This view has been the most useful to me as it unifies all the descriptions, and makes explicit what is actually being done.

• This explains how we get to the result, but it doesn't seem to give an intuitive justification of the result. The answer I'm looking for is more something like: it must be an exponential for this and this reason, the argument of the exponential must be this for this reason and that for that reason and so on – glS Feb 10 '15 at 13:56
• I think the bit about pruning phase space with $\delta$-functions, i.e. demanding that we perform an extreme weighting of some extensive property, explains why "argument of the exponential must be this for this reason". But, what counts as "intuition" depends on what we are willing to take as proven. Can you be more specific about what kinds of things you're comfortable taking for granted? I.e. you want to get from A to B in a way that's less abstract, so what can A be? – joshsilverman Feb 15 '15 at 16:12

Ok. Let's put in this way:

1. We know the average energy: $\langle E\rangle=\sum P_k E_k$
2. $P_K$ is a probability distribution: $\sum P_k=1$
3. We don't know nothing more about the system.

The question is, what probability distribution $P_k$ capture the previous assertions in an unbiased way? The answer is, the distribution that maximize the Shannon's entropy $$H(P_1...P_k)=-k_b\sum P_kln(P_k)$$ and respect the constraints (1) and (2).

To do this we use the Lagrange multipliers. The multiplier correspondent to the constraint (1) is the temperature. The constraint (2) gave us the partition function.

The boltzmann constant have the task of coupling the entropy to temperature. We use a euler basis for the logarithms and the exponentials. If we change that we need to change the boltzmann constant.

Here, temperature appear as average energy tax associated to an variation of entropy. If the maximum of $H(P_1...P_k)$ for some $\langle E \rangle$ is $S$ , then $$d \langle E\rangle=TdS$$

Note that we can make entropy an adimensional quantity, and uses the Boltzmann constant as conversor of kelvin to joules.

How we can interpret the result?

We have the probability $P(E)=e^{-E/kT}$. This mean that

$$\frac{k_bT}{P(E)}\frac{dP(E)}{dE}= -1$$.

This equation tells us that for each temperature we have a fixed scale of energy. We can see this by doing $k_bT=1$. In the scale defined by the temperature we can see that the probability of energy $E$ have a maximum in $E=0$ (This zero energy is defined by the normalization,i.e. partition function of the system). We can think that this exponential represent the thermal fluctuation of the energy in some zero energy defined by the system.

• My problem with this is that using Lagrange multipliers is really not intuitive. You get the correct result, as you do with other methods, but still I do not get any intuition into why the result should be what it is. What I'm looking for is more like an argument explaining why the distribution must be exponential and why the exponential must have that particular form. – glS Feb 5 '15 at 18:40
• The intuition of Lagrange multipliers is actually very simple. We want to maximizes the Shannon's entropy maintain the average energy intact. We know the average energy of the system because this is accessible to us (by measurement procedure) but we only know this! So we need to increase the Shannon's entropy with some restrictions, and this is the role of Lagrange multipliers. – Nogueira Feb 6 '15 at 0:16
• We may note here that the Lagrange multipliers tells us how the entropy (maximum of Shannon's entropy restricted by average energy) changes with average energy. Then, we note that this multipliers tells us how the information(entropy) and energy (average energy) are related. – Nogueira Feb 6 '15 at 0:17
• We can quickly state that this distribution is the most unbiased probability distribution that have a well defined average energy. – Nogueira Feb 6 '15 at 0:19
• It is simple to understand what Lagrange multipliers do, not so much to understand the result they give. – glS Feb 10 '15 at 13:59

When deriving a result (whether physically or mathematically), it can be helpful to first find out the limits of the result. Then if your alleged explanation "explains" the result in regions where it doesn't hold, you know your explanation is faulty.

So, the Boltzmann factor for the canonical ensemble. We have to assume that the particle number isn't changing, and we have to assume the volume isn't changing. Having both of these is actually quite restrictive. For instance you could have a gas of hydrogen, but to preserve particle number they need to be cool enough to avoid any possibility of fusion (even through tunnelling, which in the sun is how it happens since even in the sun it is cold relative to fusion being classically achievable just from KE overcoming PE). But to preserve volume you have to have room for all your hydrogen, which means they all have to be near the ground state, because Rydberg atoms (highly excited but not quite ionized hydrogen) can get very very very large if the principle quantum number is insanely large. If you want this to never ever ever happen you'd have to have so little energy available that even if every atom was low in kinetic energy, that there isn't enough energy left over to use as internal energy within a hydrogen to ionize even one single hydrogen. So technically we'd only expect Boltzmann to hold for very very cold hydrogen.

So looking at $TdS=dU+PdV-\mu dN$, we can see that in those situations, $dV=0$ because no change in volume and $dN=0$ because no change in particle number, so $dS=dU/T$, now we are most of the way there.

So since we don't expect it to hold exactly, the real question is why does it work well at all. It must be that the effects of the violations are small, transitory, or their net effects cancel. So for instance if the hydrogen atoms only overlap a little bit (in space and time) we might be able to ignore it. If there is an equilibrium amount of free electrons, ionized hydrogen, and neutral hydrogen then we might be able to ignore some of them if the effects are small. If we have a population of hydrogen at different states, and even though some are large on average there is plenty of room, we can get that.

One way to see this physically is to imagine a whole bunch of roughly identical regions each with some gas, enough that there are a very large number of hydrogen atoms in the many regions all put together. Then we can try to find an equilibrium for how many electrons, ions, and hydrogen, how many are large on average. Then we can try to take those typical values and see how realistic it is to ignore certain abilities.

In the end, taking no change in particle number, no change in volume, and a relationship between entropy and probability leads to a Boltzmann factor. So if you want to see it physically, focus on the relationship between entropy and internal energy in the absence of change in volume or particle number.

• this explains the limits of the validity of the Boltzmann distribution. But it does not seem to help understanding why the Boltzmann factor is the way it is. – glS Feb 10 '15 at 14:03

I find this the simplest argument. It does not explicitly involve entropy.

• the multiplicity of two systems is the product of their multiplicities
• in thermodynamic equilibrium the multiplicity is at its maximum
• the fractional change of multiplicity with internal energies ($\beta = \frac{1}{\Omega} \frac{d\Omega}{dE}$) is equal for both systems.
• identify $\beta = 1/kT$

So far this is standard. I suspect what follows is a bit too dubious, because I have not seen it anywhere else.

• regard a small system with non-degenerate energy levels exchanging energy with a large reservoir
• write the expression above as a differential equation for $\Omega(E)$ and adjust the sign for the energy of the small system : $\frac{d\Omega}{dE} = -\beta \Omega$
• solution for the multiplicity of the combined system in terms of large-system $\beta$ and the small-system energy $E$: $\Omega \propto e^{-\beta E}$

I can see problems with this derivation, but I do not see why it would be so much worse than many other arguments in statistical physics.

A numerical example: systems not too far from equilibrium with our environment have a $\beta$ of about 4 % per meV. This value can be computed from the barometric formula. Now contemplate a harmonic oscillator with $\hbar \omega = 1$ meV with a reservoir at $\beta = 0.04$/meV. So for the $n$th excited state, the reservoir will have its multiplicity reduced by a factor of $0.96^n$, the negative exponential which gives the probability of the Boltzmann factor.