Finite probability from discrete distribution and zero probability from continuous distribution in Maxwell-Boltzmann statistics

Question

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?

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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.

Answer 1

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.

Answer 2

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.