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?

  • $\begingroup$ Have you followed the derivation from the Wiki article? $\endgroup$
    – Nemo
    Nov 3, 2017 at 18:27
  • $\begingroup$ 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. $\endgroup$
    – BSD
    Nov 3, 2017 at 19:29
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    $\begingroup$ You wrote the same $f(v)$. In order to get the energy distribution $g(E)$, you must proceed from $f(v)dv=g(E)dE$. $\endgroup$ Jul 2, 2021 at 9:26

2 Answers 2


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

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


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

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

  • $\begingroup$ 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). $\endgroup$ Nov 4, 2017 at 4:12
  • $\begingroup$ @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 $f_v(v(E_1))v'(E_1)(E_2-E_1)$. $\endgroup$ Nov 4, 2017 at 22:29
  • $\begingroup$ 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. $\endgroup$ Nov 8, 2017 at 4:51
  • $\begingroup$ 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 $f_v(v(E))v'(E)$. $\endgroup$ 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 $p(v)dv=p(E)dE$.

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

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