I have difficulty understanding a certain concept with the derivation of the Maxwell-Boltzmann distribution $f(v)$ function from Boltzmann statistics. The derivation starts with the Boltzmann statistics formula that gives the probability for an energy $E_x$:

$$p(E_x) = \frac{\exp(-E_x/kT)}{\sum_{i=0}^\infty \exp(-E_i/kT)}$$

From this formula the Maxwell-Boltzmann distribution function is derived:

$$f(v) = \left(\frac{m}{2\pi kT}\right)^{3/2} \cdot \exp\left({\frac{mv^2}{2kT}}\right)$$

I understand that this formula gives the probability density of a particle with velocity $v + \mathrm{d}v$. The reason that the formula gives the probability density for velocities between $v$ and $v + \mathrm{d}v$ is because velocity and thus energy are considered continuous, and therefore just 1 specific value of energy or velocity would have a probability of 0. Therefore, a range of speeds must be taken to give a non-zero probability.

  • If one specific value of energy or velocity should give a probability of 0, how come the formula for the Boltzmann statistics gives a non zero probability for a specific value of energy $E_x$?
  • And how can one reason a formula that considers a specific value of energy or velocity to be zero from a formula that doesn't? Is there a way to explain this?

I don't find this to be a duplicate of my other question. My other question states a misconception about translational energy being continuous in general. This question states how the derivation is done between two formulas, one considering energy to be continuous and the other doesn't.

  • 1
    $\begingroup$ Read here: en.wikipedia.org/wiki/Borel%E2%80%93Kolmogorov_paradox $\endgroup$
    – obackhouse
    Jun 12, 2018 at 22:07
  • 1
    $\begingroup$ Possible duplicate of Derivation of Maxwell–Boltzmann statistics $\endgroup$
    – Mithoron
    Jun 12, 2018 at 22:11
  • 2
    $\begingroup$ @Mithoron his question 2 seems to boil down to "how can there be a finite number of permutations of a continuous continuous value". This question boils down to "how can we say a given energy has zero probability even though the Boltzmann equation gives a nonzero probability". They both in part stem from the same misconception about how to treat probability densities, but I don't think they are the same question. $\endgroup$
    – Tyberius
    Jun 13, 2018 at 20:37
  • 1
    $\begingroup$ What is the probability of picking a real number in the interval $\ce{[0,1]}$? $\endgroup$
    – getafix
    Jun 23, 2018 at 14:00

2 Answers 2


The probability of any one specific velocity is not zero, it's just very, very small. Infinitismal, or tending towards zero, is how we usually put it. The reason is because the spacing between allowed translational states is inversely proportional to the size of the container, and when you are talking about a macroscopic container (a bottle of gas, say, or beaker of liquid) then from the atomic point of view this is gargantuan and the spacing between allowed translational energies very, very small. (If you want to get a feel for this, look up the allowed energies for a particle in a box. You'll see that the size of the box goes in the denominator and as the box gets bigger the spacing between energy levels gets smaller and smaller. You can amuse yourself by calculating the exact difference between the 1st and 2nd energy levels for, say, an H atom in a box 1cm x 1 cm x 1cm. It's really small!)

So small that to do the sum over states in your top formula is a gigantic pain, quite challenging. But if you convert it to an integral, the integral can be done pretty easily. So that's what we do. We replace a huge sum, over 10^20 terms or more, each of size 10^-20 or less, with an integral. Think of it as the reverse of when you learned integral calculus, and in order to figure out an integral you replaced it with a large sum -- and let the size of each term go to zero while the number of terms went to infinity.

Here, you are taking a sum which has a very large number of terms, each of which is very small, and replacing it with an integral. It's not exact, but it's very good, and it becomes perfect in the limit of an infinite-sized container, and anyway the errors for ordinary-sized containers are ridiculously small, unmeasureable.

  • $\begingroup$ Thanks for your answer. I have two questions if you don't mind. 1. With "the sum over states", do you mean the sum over each state's probability? 2. Regarding my top formula in my OP, if you fill in a value of say 5 for the energy Ex, would the formula then give the probability for particles having an energy of exactly 5.00000000...... or would it give the probability of particles with energy that is within a certain range of 5? If it's the latter, how large is that range regarding my top formula then? $\endgroup$
    – Phy
    Jun 17, 2018 at 10:54
  • $\begingroup$ @JohnnyGui 2. The numerator in the probability is for a single $E_x$ only. If you wanted to consider some energy "width", you would need to add up all probabilities, one for each possible energy, within that width. 1. "Sum over states" is referring to the denominator, which is the partition function. The article says it well: $\endgroup$ Jun 23, 2018 at 1:32
  • 2
    $\begingroup$ @JohnnyGui "In classical mechanics, the position and momentum variables of a particle can vary continuously, so the set of microstates is actually uncountable. In classical statistical mechanics, it is rather inaccurate to express the partition function as a sum of discrete terms. In this case we must describe the partition function using an integral rather than a sum." $\endgroup$ Jun 23, 2018 at 1:33
  • $\begingroup$ Got it. Does this mean that, clascially, we can NOT use the first formula in my OP to calculate probabilities (without being too inaccurate)? $\endgroup$
    – Phy
    Jun 23, 2018 at 8:52
  • $\begingroup$ @JohnnyGui like pentavalentcarbon said, in "classical statistics" the concept of number of microstates cannot be defined at all; the statistical weight is simply the volume in phase space $\Delta p \Delta q$. In a "quasi-classical" sense, given $s$ degrees of freedom, we can write down the number of states $\Delta \Gamma = \frac{\Delta p \Delta q}{(2 \pi \hbar)^s}$ $\endgroup$
    – getafix
    Jun 23, 2018 at 14:12

As has been mentioned already, your concern is valid: from a mathematical perspective, the probability that a continuous random variable $E$ takes a specific value $E = \epsilon$ is indeed zero. How do we rectify this issue?

(Note that the probability distribution $P$ and the probability density $f$ mentioned here and in the problem statement refer to the velocity vector $\mathbf{V}$ and not the energy $E$, so they should be parametrized with respect to $\mathbf{V}$ and not to $E$. With regards to notation, we will take the upper-case variables $\mathbf{V}$ and $E$ to be random variables and the lower-case variables $\mathbf{v}$ and $\epsilon$ to be specific values that those variables can take.)

  1. Correspondence principle. Assume independence in $x$, $y$, and $z$, so that $$P(\mathbf{V}=\mathbf{v}) = P_1(V_x = v_x)P_1(V_y = v_y)P_1(V_z = v_z),$$ where $P_1$ is the one-dimensional probability distribution of velocities. Let our system exist in a finite cubic box of volume $L^3$, and consider the limit as $L \to \infty$. We can partition all of phase space so that each point of the probability distribution $P(\mathbf{V}=\mathbf{v})$ is associated with cubes of identical volume in phase space. As $L \to \infty$, the number of points grows (the quantum of translational energy becomes smaller), and the volume associated with each point shrinks. In the continuum limit we can replace each point by its associated infinitesimal cube; i.e., we substitute $$P(\mathbf{V}=\mathbf{v}) \to P(\mathbf{v}-\boldsymbol{\delta} < \mathbf{V} < \mathbf{v}+\boldsymbol{\delta}) \approx f(\mathbf{v})\times(2\delta)^3$$ for $\boldsymbol{\delta}$ a constant vector with every element $\delta$, where $2\delta$ is the spacing between neighboring points. The purpose of this entire construction is to show that $P(\mathbf{V}=\mathbf{v}) \propto f(\mathbf{v}) \propto e^{-\beta mv^2/2}$, for then the proportionality constant can be recovered by a simple normalization.

  2. Classical statistical mechanics. An important result from this subject is simply that $$f(\mathbf{r},\mathbf{p}) \propto \exp(-\beta\mathcal{H}(\mathbf{r},\mathbf{p})),$$ which yields the desired probability density directly. It's hard for me to elaborate, because I haven't found a satisfactory classical proof of this result. An interesting approach is here, but it has its own problems.

In either case, it's formally inappropriate to go from a discrete to continuous model without likewise going from a probability distribution to a probability density. Nevertheless, the physical motivation and ideas behind such a switch should be rather clear.

  • $\begingroup$ I have a hard time understanding how it is concluded that P(V = v) is proportional to exp(-Bmv^2/2) and I can't find a satisfactory source that explains this. For example, where does the pi come from in the distribution function f(v) when treating the distribution classically? $\endgroup$
    – Phy
    Aug 11, 2018 at 16:58
  • $\begingroup$ @JohnnyGui, reading over my original response again, part 1. seems possibly incorrect (and at least quite confusing) and probably needs to be modified, so I wouldn't put too much stock into it. The $\pi$ comes from normalization of the probability density. $\endgroup$ Aug 11, 2018 at 18:26
  • $\begingroup$ Thanks for your response. Is there a way to explain what the normalization entails? $\endgroup$
    – Phy
    Aug 11, 2018 at 18:30
  • $\begingroup$ @JohnnyGui, the normalization condition is that $\int\text{d}v\,f(v) = 1$. Since we know $f(v) = Ne^{\beta\epsilon}$ for a constant $N$, the normalization condition is an equation by which $N$ can be identified. $\endgroup$ Aug 11, 2018 at 18:33
  • 1
    $\begingroup$ @JohnnyGui, you can, kind of. Your first equation is an equation over microstates. Classical statistical mechanics works on phase space. The semiclassical association between microstates and phase space is to partition phase space up into little hypercubes of side length $h$, and treat each little hypercube as a microstate. Then we have $P(\omega) \approx f(v) h^3$, where $\omega$ denotes a microstate, and also $P(\omega) \propto e^{-\beta\epsilon(x,v)}$ from your first equation, so then $f(v) \propto e^{-\beta\epsilon(x,v)}$. And this comment should really have been my answer. $\endgroup$ Aug 13, 2018 at 21:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.