Kinetic Energy in Maxwell-Boltzmann distribution

So I was trying to arrange the expression of a Maxwell-Boltzmann velocity vector distribution into a kinetic energy distribution. Since we know that $E= \frac{1}{2} mv^2$ we can rearrange this expression to give $v^2=\frac{2E}{m}$. If we insert this expression in a Maxwell-Boltzmann distribution: $$f(v)=\left(\frac{m}{2\pi K_BT}\right)^\frac{3}{2} \exp\left(\frac{-m(v_x^2 + v_y^2 + v_z^2)}{2K_BT}\right)$$

we get: $$f(E)=\left(\frac{m}{2\pi K_BT}\right)^\frac{3}{2} \exp\left(\frac{-E}{K_BT}\right)$$

But Wikipedia says that the kinetic energy distribution is: $$f(E)=2\left(\frac{E}{\pi}\right)^\frac{1}{2}\left(\frac{1}{K_BT}\right)^\frac{3}{2} \exp\left(\frac{-E}{K_BT}\right)$$ What am I missing here?

• Have you followed the derivation from the Wiki article?
– Nemo
Nov 3, 2017 at 18:27
• Yes, but they derive that expression by some concepts that I never heard of like phases-space volume of momenta. I wanted to derive that expression from the normal M-B velocity distribution.
– BSD
Nov 3, 2017 at 19:29
• You wrote the same $f(v)$. In order to get the energy distribution $g(E)$, you must proceed from $f(v)dv=g(E)dE$. Jul 2, 2021 at 9:26

Kinetic energy distribution $f_E(E)$ is a function that gives probability per unit energy interval.

Since kinetic energy $E$ is a function of speed, this distribution is related to speed distribution $f_v(v)$, but its value at any energy $E$ is not merely proportional to value of $f_v(v(E))$. The derivation of $f_E(E)$ from $f_v(v)$ relies on the substitution theorem from the integral calculus.

For positive speeds $v_1,v_2$ and corresponding energies $E_1=\frac{1}{2}mv_1^2, E_2=\frac{1}{2}mv_2^2$, the probability that particle has kinetic energy in the interval $(E_1,E_2)$ is the same as the probability that it has speed in the interval $(v_1,v_2)$. We transform the expression of the second probability in the following way:

$$P = \int_{v_1}^{v_2}f_v(v)dv => (subst. theorem) => \int_{E_1}^{E_2}f_v(v(E))v'(E) dE$$ where

$$v(E) = (2E/m)^{1/2}$$ and $v'(E)$ is derivative of this function at value $E$.

The expression of the first probability is

$$P = \int_{E_1}^{E_2}f_E(E)dE.$$

Since we can choose $v_1,v_2$ arbitrarily close to each other, it must be

$$f_E(E) = f_v(v(E))\frac{dv(E)}{dE}.$$

Substituting

$$f_v(v) = \left(\frac{m}{2\pi k_BT}\right)^{3/2} 4\pi v^2 e^{-\frac{mv^2}{2k_B T}}$$

we obtain $$f_E(E) = \left(\frac{m}{2\pi k_BT}\right)^{3/2} 4\pi 2E/m \cdot e^{-\frac{E}{k_B T}} \cdot (2/m)^{1/2}\frac{1}{2}E^{-1/2}$$

and after simplification $$f_E(E) = \left(\frac{1}{\pi k_BT}\right)^{3/2} 2\pi \cdot E^{1/2} e^{-\frac{E}{k_B T}}.$$

• Your answer makes total sense to me, except for the fact that I dont get how $v_1$ and $v_2$ being possibly really close implies that identity. That identity comes from a theorem in integral calculus, irregardless of the limits considered (so long as continuity is ensued in such procedure, and the substitution considered satisfies some minor requirements). Nov 4, 2017 at 4:12
• @JohannLiebert, the substitution theorem only states that the integral over $v$ has the same value as the integral over $E$, it does not state that $f_E(E)$ defined in probability theory equals $f_v(v(E))v'(E)$. Only by taking a special case $v_1-v_2 \rightarrow 0$ for which the substitution formula has to be valid too, that formula becomes obvious. This is because then the integrals can be approximately expressed as $f_E(E_1)(E_2-E_1)$ and $f_v(v(E_1))v'(E_1)(E_2-E_1)$. Nov 4, 2017 at 22:29
• I will rephrase what I meant, that the integral is equal is a well known result reviewed in Analysis courses, a good source is Rudins Principles of Analysis page 131: notendur.hi.is/vae11/%C3%9Eekking/… on the other hand, my point lies here: since $f_E$ and $f_v$ are distributions they are never related directly to physical quantities and thus it makes no sense to separate them from their role inside the respective integral representation. Nov 8, 2017 at 4:51
• I agree, $f_E,f_v$ are distributions and their use is under the integration sign. In case $f_v$ is the Maxwell-Boltzmann distribution, both distributions are regular and can be expressed as functions of $E,v$. The substitution theorem states that both integrals above are the same, a corollary is that the function $f_E(E)$ can be expressed as $f_v(v(E))v'(E)$. Nov 8, 2017 at 23:12

You should think in terms of changing your integration variable. I think you should procede this way: first of all write the integral in polar coordinates and integrate the angular variables. Then change your variable of integration from $v$ to $E=\frac12mv^2$. I actually haven't done it but it should work. That's basically because you want to preserve the probability and to set $p(v)dv=p(E)dE$.

• And as a hint: $dE=mvdv=\sqrt{2mE}dv$
– Nemo
Nov 3, 2017 at 21:53