# Flaw in alternative approach to determine ideal gas speed distribution

If we assume the energy of particles in an ideal gas follows a Boltzmann distribution, then the energy distribution function can be defined as below: , where k_B is the Boltzmann constant

Since the energy of particles in an ideal gas are assumed to only consist of translational kinetic energy (as they don't interact), , where v is the speed of the particle

However, for a certain level of kinetic energy, there is only one speed that can be associated with it, as speeds take positive value and hence there is a one-to-one relationship between the speed of a particle and its kinetic energy.

By this reasoning, the distribution function of energy should be proportional to the distribution function of velocity. In other words, the velocity distribution function is also proportional to a Boltzmann factor, with a constant factor of proportionality.

However, the Maxwell-Boltzmann speed distribution shows the velocity distribution having the following form: Although I can follow the derivation of the Maxwell-Boltzmann distribution, I fail to see why the line of reasoning I described in previous paragraphs is wrong. Any help would be much appreciated. Thank you!

Indeed, there is just one speed for any energy, but the quantity of interest is given by the number of available states at a certain energy, which is not given by the speed, but by the velocity - a particle moving along the x-axis is certainly in a different state than a particle moving along the y-axis with the same speed.

Velocity is a vector and if you do a quick drawing of a 3D plot, where the three axes are given by the x-, y- and z-components of the velocity, you will find that all vectors that add up to the same speed form the surface of a sphere. If you take a moment to consider how the number of possible vector component combinations that sum up to the same speed scales with the total speed, you should be able to identify the additional factor in the Maxwell-Boltzmann distribution.

• hutchphd
Thank you!

The flow is that you don't take the word "distribution" seriously. The Maxwell distribution can be given in various forms. You have to start with the fundamental definition as a phase-space distribution function
$$f(\vec{x},\vec{p})=\frac{1}{(2 \pi \hbar)^3} \exp \left (-\frac{\vec{p}^2}{2m k_{\text{B}} T} \right).$$
Which denotes the number of particles at place ##(\vec{x},\vec{p})## per unit of phase space. If you want to transform this to a velocity distribution you have to integrate over space (which gives simply a factor ##V##) and to transform from momenta to ##\vec{v}=\vec{p}/m##. Now ##\mathrm{d}^3 v=\mathrm{d}^3 p/m^3## and thus
$$f_2(\vec{v})=\frac{m^3 V}{(2 \pi \hbar)^3} \exp \left (-\frac{m \vec{v}^2}{2 k_{\text{B}} T} \right).$$
Now you want the energy distribution. Using
$$E=\frac{m}{2} \vec{v}^2\; \Rightarrow \; \mathrm{d}E=m v \mathrm{d} v = \sqrt{2mE} \mathrm{d} v$$
you get
$$\mathrm{d}^3 v = v^2 \mathrm{d} v \mathrm{d} \Omega=\sqrt{\frac{2 E}{m}}\frac{1}{m} \mathrm{d} E.$$
Integrating over the solid angle leads finally to
$$f_3(E)=\frac{V\sqrt{2Em^3}}{2 \pi^2 \hbar^3} \exp \left (-\frac{E}{k_{\text{B}}} \right).$$

• Cthugha