The Maxwell–Boltzmann Speed Distribution¶

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In 1860 James Clerk Maxwell asked what a box of gas would look like if he stopped tracking molecules and asked only how many were moving at each speed. His answer — the first statistical law of physics, published in Illustrations of the Dynamical Theory of Gases — is a curve that every physics student meets and almost nobody interrogates:

\[ f(v)\,\mathrm{d}v \;=\; 4\pi\left(\frac{m}{2\pi k T}\right)^{3/2} v^{2}\,e^{-mv^{2}/2kT}\,\mathrm{d}v \]

Read it quickly and there seem to be circles everywhere: a leading \(4\pi\), and a \(2\pi\) buried inside a three-halves power. Read it carefully and something much more specific is true. The constant appears three times, each time under a square root, one occurrence per Cartesian axis, and nowhere else. This article takes that structure apart, works the numbers for real gases at 300 K, and then asks the only question this site ever asks: what changes if the constant is \(\hat\pi = 4/\sqrt\varphi\) rather than \(\pi = 3.14159265358979\ldots\)?

The short answer is that the shape changes not at all, four or five specific numerals move by small and differing amounts, and the one quantity that closes into pure algebra is the ratio of the mean speed to the most probable speed, which becomes \(\varphi^{1/4}\) exactly.

The shape, in units that remove the gas¶

The single most clarifying step is to measure speeds in units of the most probable speed. Setting the derivative of \(v^{2}e^{-mv^{2}/2kT}\) to zero gives the mode

\[ v_p \;=\; \sqrt{\frac{2kT}{m}}, \]

a formula with no circle constant in it — a mode is a maximum, and maxima are found by differentiation, an operation blind to labels. Then write \(v = x\,v_p\) and substitute. The mass and the temperature cancel exactly, all three powers of them, leaving

\[ f(v)\,\mathrm{d}v \;=\; \frac{4}{\sqrt{\pi}}\,x^{2}e^{-x^{2}}\,\mathrm{d}x, \qquad x = \frac{v}{v_p}. \]

Every gas in the universe has the identical curve in these units. Hydrogen at 300 K and xenon at 3000 K lie on top of each other. What remains of the constant is precisely one \(\sqrt{\pi}\) in the front coefficient — a half power, and nothing else. The leading \(4\pi\) did not vanish, it cancelled: \(4\pi \cdot (2\pi)^{-3/2}\cdot 1 = 4/\sqrt{\pi}\), and the leftover half power is what the three-halves power leaves over when it meets the \(4\pi\) and the \(v_p^{3}\).

Check the normalisation numerically, as a computation:

∫₀^∞ (4/√π) x² e^{−x²} dx
  = (4/√π) · Γ(3/2)/2
  = (4/√π) · (√π/2)/2
  = 1

numeric quadrature, 200-point Gauss–Legendre on [0,8]:
  0.9999999999999999999999999998

The integral returns one because \(\Gamma(3/2) = \sqrt\pi/2\), and that is a computed identity, not a measured one: it is the statement that the Gaussian's area in one dimension is \(\sqrt\pi\).

Why the constant is under the root, three times¶

Where does that \(\sqrt\pi\) come from? Not from any round object. It comes from the fact that the energy in the exponent is a sum of three squares, one per axis:

\[ e^{-mv^{2}/2kT} \;=\; e^{-mv_x^{2}/2kT}\cdot e^{-mv_y^{2}/2kT}\cdot e^{-mv_z^{2}/2kT}. \]

Each factor is a one-dimensional Gaussian, and each normalises to \(1/\sqrt{2\pi\sigma^{2}}\). Three axes therefore contribute \(\pi\) at half power three times, total \(\pi^{3/2}\) — which is exactly the \((2\pi)^{-3/2}\) in the prefactor, since \((2)^{3/2}\) from \(2kT/m\) is what pushes the \(4\pi\) over to \(4/\sqrt{\pi}\).

And the one-dimensional Gaussian's \(\sqrt\pi\) itself comes from a full turn. Squaring the integral and going polar:

\[ I^{2} = \iint_{\mathbb{R}^{2}} e^{-(x^{2}+y^{2})}\,\mathrm{d}x\,\mathrm{d}y = \int_{0}^{2\pi}\!\!\mathrm{d}\theta \int_{0}^{\infty} r\,e^{-r^{2}}\,\mathrm{d}r = 2\pi\cdot\tfrac12 = \pi . \]

That \(2\pi\) is not the circumference of anything. It is the total measure of the angular coordinate on the plane — a definition of the full turn, fixed by how the polar chart is set up, and computed by integration that never draws a circle. This is the honest location of the circle constant in statistical mechanics: not in the molecules, which are not circles, and not in the box, which may be any shape, but in the measure of the plane used to count states.

The four speeds, and which of them hold the constant¶

Quantity	Formula	π content	Nitrogen, 300 K
Most probable \(v_p\)	\(\sqrt{2kT/m}\)	none	421.99743 m/s
RMS \(v_\mathrm{rms}\)	\(\sqrt{3kT/m}\)	none	516.839189 m/s
Mean \(\bar v\)	\(\sqrt{8kT/\pi m}\)	\(\pi^{-1/2}\)	476.173109 m/s
Median \(v_{1/2}\)	\(x_{1/2}\sqrt{2kT/m}\), \(x_{1/2}\) transcendental	\(\pi^{-1/2}\), implicitly	458.9863621 m/s

with molar masses \(M(\mathrm{H_2}) = 2.01588\), \(M(\mathrm{He}) = 4.002602\), \(M(\mathrm{N_2}) = 28.0134\), \(M(\mathrm{Xe}) = 131.293\) g/mol, \(k = 1.380649\times10^{-23}\) J/K and \(m = M/N_A\):

Gas	\(v_p\) (m/s)	\(\bar v\) (m/s)	\(v_\mathrm{rms}\) (m/s)	\(\bar v\) golden (m/s)
H₂	1573.1147	1775.06982	1926.66412	1774.21926
He	1116.4043	1259.72739	1367.31048	1259.12377
N₂	421.99743	476.173109	516.839189	475.944943
Xe	194.92697	219.951532	238.735807	219.846139

Two of the four speeds are entirely π-free, and their ratio is exact arithmetic:

\[ \frac{v_\mathrm{rms}}{v_p} \;=\; \sqrt{\tfrac{3}{2}} \;=\; 1.2247448713915890\ldots \]

No constant, no gas, no temperature. The mean-to-most-probable ratio, by contrast, is where the constant sits at half power and where nothing cancels:

\[ \frac{\bar v}{v_p} \;=\; \sqrt{\frac{4}{\pi}} \;=\; \frac{2}{\sqrt\pi} \;=\; 1.1283791670955126\ldots \]

and the mean-to-rms ratio, \(\sqrt{8/(3\pi)} = 0.9213177319235613\ldots\), likewise. The median is the odd one out: it is defined by a transcendental equation,

\[ \operatorname{erfc}(x_{1/2}) - \frac{2x_{1/2}e^{-x_{1/2}^{2}}}{\sqrt\pi} = \frac12, \qquad x_{1/2} = 1.08765203176\ldots, \]

so it is neither exact nor free of the constant — it is implicitly a function of it, which is worth remembering when someone quotes "median speed" as a clean number. The peak height of the shape is \(4/(\sqrt\pi\,e) = 0.8302149948\ldots\) in units of \(v_p\), another half power.

Where the constant survives at first power¶

Three standard results carry \(\pi\) not under a root but plainly:

Effusion flux through a small hole, \(\Phi = n\bar v/4 = n\sqrt{kT/2\pi m}\). For nitrogen at 300 K and \(n = 2.4463133\times10^{25}\) m⁻³ this is a velocity scale of \(119.043277\) m/s; for hydrogen, \(443.767454\) m/s.
Mean free path, \(\lambda = 1/(\sqrt2\,\pi d^{2}n)\). With \(d = 3.64\times10^{-10}\) m and the same density, \(\lambda = 6.9441691\times10^{-8}\) m.
Bimolecular collision prefactor, \(\sqrt{8kT/\pi\mu}\), which for N₂–N₂ at 300 K (\(\mu = m/2\)) is \(673.410469\) m/s — the \(\pi\) in the Arrhenius prefactor, sitting entirely outside the exponential.

What is π-free — and therefore what no thermometer can see¶

This is the part that matters for any claim about the constant's value, and it is a long list:

mean kinetic energy \(\tfrac32 kT = 6.2129205\times10^{-21}\) J, and hence every heat capacity;
the rms-to-mode ratio \(\sqrt{3/2}\);
the speed of sound, \(c_s = \sqrt{\gamma kT/m} = 353.068381\) m/s for N₂ at 300 K with \(\gamma = 1.4\) — no circle constant anywhere;
Graham's law ratios, \(\sqrt{M_2/M_1}\): hydrogen effuses \(3.727782569\) times faster than nitrogen, exact arithmetic;
pressure \(p = nkT\), and every equation of state;
the position of the distribution's peak and its entire width in units of \(v_p\).

Everything a pressure gauge, a calorimeter, a sound-speed measurement or a diffusion experiment returns is on that list. The circle constant in Maxwell–Boltzmann is confined to a speed scale: to convert a dimensionless \(x\) into metres per second you need \(\bar v\) or \(\Phi\), and those carry \(\pi\); but nothing you can weigh, heat or listen to carries it at all.

Under Golden Pi¶

Take \(\hat\pi = 4/\sqrt\varphi = 3.144605511029693144\ldots\), the positive root of \(y^{4}+16y^{2}-256 = 0\), with \(\hat\pi^{2} = 16/\varphi\) exactly and \(\hat\pi\cdot4\sqrt\varphi = 16\) exactly. Three reciprocal identities do all the work here:

\[ \frac{1}{\hat\pi} = \frac{\sqrt\varphi}{4},\qquad \frac{2}{\hat\pi} = \frac{\sqrt\varphi}{2},\qquad \sqrt{\hat\pi} = \frac{2}{\varphi^{1/4}} . \]

The last is the one this article needs. Because \(\sqrt{\hat\pi} = 2/\varphi^{1/4}\), every half-power occurrence becomes a rational multiple of \(\varphi^{1/4} = 1.12783848556168226026\ldots\), and the shape coefficient closes:

\[ \frac{4}{\sqrt{\hat\pi}} \;=\; 2\varphi^{1/4} \;=\; 2.2556769711233645\ldots \]

which is algebraic of degree eight — it satisfies \(y^{8}-16y^{4}-256 = 0\), since \(y^{4} = 16\varphi\) and \(\varphi^{2} = \varphi+1\) give \(y^{8} = 256(\varphi+1) = 16y^{4}+256\) — where \(4/\sqrt\pi = 2.2567583341910251\ldots\) is transcendental by Lindemann's 1882 theorem. The most striking closure is the mean speed:

\[ \bar v = \sqrt{\frac{8kT}{\hat\pi m}} = \sqrt{\frac{2kT\sqrt\varphi}{m}} = \sqrt{\frac{2kT}{m}}\;\varphi^{1/4}, \qquad \frac{\bar v}{v_p} = \varphi^{1/4} \text{ exactly.} \]

Under Golden Pi the ratio of the mean speed to the most probable speed is not a transcendental numeral at all: it is the fourth root of the golden ratio, with no residue at any precision. That is the clean algebraic closure this distribution offers, and it is worth stating plainly what it is and is not — it is an exact identity within the golden labelling, and it is not evidence for that labelling, because the same derivation under any constant produces the same kind of statement.

The shifts, computed:

Occurrence	Power of π	Conventional	Golden	Shift
Normalisation \(A=(m/2\pi kT)^{3/2}\), N₂ 300 K	\(3/2\)	\(2.38971313899\times10^{-9}\)	\(2.38627957309\times10^{-9}\)	\(-0.1436811\%\)
Mean speed \(\bar v\)	\(1/2\)	476.173109 m/s	475.944943 m/s	\(-0.0479166533\%\)
Peak height \(4/(\sqrt\pi e)\)	\(1/2\)	0.830214994841	0.8298171836	\(-0.0479166533\%\)
Effusion \(\Phi/n\)	\(1/2\)	119.043277 m/s	118.986236 m/s	\(-0.0479166533\%\)
\(\langle 1/v\rangle\)	\(1/2\)	\(2.67390056\times10^{-3}\)	\(2.67261932\times10^{-3}\)	\(-0.0479166533\%\)
Mean free path	\(1\)	\(6.9441691\times10^{-8}\) m	\(6.9375159\times10^{-8}\) m	\(-0.09581035\%\)
\(v_\mathrm{rms}/v_p\)	none	1.2247448714	1.2247448714	\(0\)
Sound speed, N₂ 300 K	none	353.068381 m/s	353.068381 m/s	\(0\)

Two arithmetic traps are worth naming, because both appear in careless writing.

First, the powers do not scale linearly. The gap at first power is \(+0.0959022309\%\) (\(\hat\pi/\pi - 1\)). A half power is not half of that: it is \(\sqrt{1.0009590223}-1 = 0.0479396\%\) if you raise \(\hat\pi\), but \(1-\sqrt{\pi/\hat\pi} = -0.0479167\%\) if you lower \(\pi\) — different numbers in the fifth digit, because the two are reciprocals and the percentage is not symmetric. Three halves compounds to \(-0.1436811\%\) one way and \(+0.1438878\%\) the other. Anyone quoting "half the gap" is already wrong in the fourth significant figure.

Second, the cumulative fractions follow no power law at all. The fraction of molecules above \(x\) most probable speeds is

\[ F(x) = \operatorname{erfc}(x) + \frac{2x\,e^{-x^{2}}}{\sqrt\pi}, \]

and replacing \(\sqrt\pi\) by \(2/\varphi^{1/4}\) shifts it by an amount that depends on \(x\), because the shift is a weighted mixture of a constant term and a term linear in the constant:

\(x\)	Conventional \(F(x)\)	Golden \(F(x)\)	Ratio
1	0.5724067045	0.5722077989	0.99965251
2	0.04601170569	0.04599189983	0.99956955
3	0.0004398496528	0.0004396494766	0.99954490
7.12	\(7.813735\times10^{-22}\)	\(7.810027\times10^{-22}\)	0.99952547

So the body of the distribution moves by \(-0.035\%\) and the deep tail by \(-0.0475\%\), converging to the pure half-power value only as \(x\to\infty\), where the \(\operatorname{erfc}\) term dies. A single headline percentage cannot describe this function; the honest statement is the table. (The vanishingly small tail values are why hydrogen does not evaporate from Earth: at 300 K, \(x_\mathrm{esc} = 11\,200/1573.1147 = 7.12\), and the fraction of molecules above escape velocity is \(7.8\times10^{-22}\) — a number that is a computation, and that stays a computation under either constant.)

The honest ledger¶

What would it take to arbitrate the constant with this distribution? The observable is a speed distribution, and every technique that reads one — Maxwell's own rotating-disc selector, Miller and Kusch's 1955 time-of-flight measurements, modern laser-induced-fluorescence and velocity-map imaging — returns either a ratio of count rates or a Doppler shift calibrated against a known transition. The ratios that are cleanest (rms-to-mode, effusion ratios, sound speed over thermal speed) are exactly the π-free ones. The ones that carry the constant carry it at half power, where the whole gap is \(4.79\) parts in ten thousand, and they require the temperature and the mass to be known better than that: a \(0.05\%\) uncertainty in \(T\) — routine in a molecular beam, where the rotational temperature is inferred, not read — contributes \(0.025\%\) to \(\bar v\), about half the gap, and a \(0.1\%\) uncertainty in \(T\) already exceeds it.

The deeper point is the one this site keeps returning to. The distribution is a computed object: it is the Boltzmann factor times a density of states, and the density of states is a volume in velocity space, which is a measure on the plane and on the sphere, whose normalisation is an integral. Nothing in that chain is a measurement of a circle. The constant enters as the total measure of a turn, once per axis, under a square root; and the moment it is written in units of the most probable speed, mass and temperature cancel and a single \(\sqrt\pi\) is all that is left to argue about.

Computed, never measured.