= Maxwell speed distribution
{c}
{title2=$g(v)$}
= Maxwell-Boltzmann speed distribution
{c}
{synonym}
The Maxwell speed distribution is the <probability density function> for the magnitude $v=|\mathbf v|$. Unlike a signed <velocity> component, <speed> cannot be negative:
$$
g(v)=4\pi v^2
\left(\frac{m}{2\pi k_BT}\right)^{3/2}
e^{-mv^2/(2k_BT)}
=\sqrt{\frac2\pi}\frac{v^2}{s^3}e^{-v^2/(2s^2)},
\qquad v\geq0.
$$
For $0\leq a<b$, the <probability> of a <speed> in $[a,b]$ is $\int_a^b g(v)\,dv$, and $\int_0^\infty g(v)\,dv=1$. The density has units of inverse <speed>, so its height alone is not a <probability>. https://openstax.org/books/university-physics-volume-2/pages/2-4-distribution-of-molecular-speeds[OpenStax, University Physics Volume 2, §2.4]{ref}
All <velocities> with magnitudes from $v$ to $v+dv$ occupy a thin <spherical shell> of volume $4\pi v^2\,dv$ in <velocity> space. The <isotropy> of the <Maxwell-Boltzmann velocity distribution> makes its density constant over each shell, giving $g(v)=4\pi v^2 f(v\mathbf e_x)$ for any fixed unit direction $\mathbf e_x$. A larger radius offers more directions and more volume. The increasing shell factor initially wins over the decreasing <Boltzmann factor>, so $g(0)=0$ and the <most probable speed> is positive. This is the geometric reason that the peaks of the vector and <speed> densities disagree.
With the <dimensionless quantity> $u=v/s$, every such gas has the same rescaled <speed> density:
$$
g(v)\,dv=h(u)\,du,
\qquad h(u)=\sqrt{\frac2\pi}u^2e^{-u^2/2},\quad u\geq0.
$$
For comparison, $w=v_x/s$ has the standard <normal distribution> density $\phi(w)=e^{-w^2/2}/\sqrt{2\pi}$ on the whole real line. The <Maxwell speed distribution> is neither the positive half of this curve nor the distribution of $|v_x|$: it combines three components.
\Image[/maxwell-velocity-versus-speed.png]
{title=One signed velocity component peaks at zero, while speed peaks above zero because of spherical shells}
The first two <moments> of the <Maxwell speed distribution> give the <mean> and <root mean square>:
$$
\langle v\rangle=\sqrt{\frac8\pi}\,s,\qquad
\langle v^2\rangle=3s^2,\qquad
v_{\rm rms}=\sqrt{\langle v^2\rangle}=\sqrt3\,s.
$$
Consequently the <mean> translational <kinetic energy> is $m\langle v^2\rangle/2=3k_BT/2$, as required by the <equipartition theorem>. https://openstax.org/books/university-physics-volume-2/pages/2-2-pressure-temperature-and-rms-speed[OpenStax, University Physics Volume 2, §2.2]{ref} The <mean> <speed> weights fast particles more than the density peak does, and the <root mean square> weights them more strongly still.
Changing <temperature> or particle <mass> only changes $s\propto\sqrt{T/m}$. At fixed <mass>, four times the <temperature> doubles every characteristic <speed>; at fixed <temperature>, four times the <mass> halves them. A stretched density becomes lower so that its area stays one. For two species at the same <temperature>, equal <mean> translational <kinetic energy> therefore means different typical <speeds>.
\Image[/maxwell-speed-temperature-mass.png]
{title=Fourfold temperature doubles the speed scale, while fourfold particle mass halves it}
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