Boltzmann Distribution in External Potentials

Why is the air thin at the top of a mountain but the temperature of that thin air the same, on average, as the air below? Why do the heavy molecules of a uranium centrifuge crowd toward the rim while a cloud of cold atoms in a laser trap settles into a Gaussian ball at the center? All three are the same law wearing different clothes. For classical particles in any external potential $U(\mathbf r)$ at temperature $T$ , the equilibrium number density is $n(\mathbf r)=n_0\,e^{-U(\mathbf r)/k_BT}$ — the Boltzmann distribution in an external field. It is one of the cleanest and most far-reaching consequences of the canonical ensemble. The reason is separability. The single-particle energy splits as $H=p^2/2m+U(\mathbf r)$ — a piece that depends only on momentum and a piece that depends only on position. The Boltzmann factor $e^{-\beta H}$ therefore factorizes, and when you integrate out the momenta to get the spatial density, the Gaussian momentum integral is the same everywhere and cancels between numerator and denominator. All the position-dependence survives as the bare configurational factor $e^{-\beta U}$ . A striking corollary: the velocity distribution at every height, every radius, every point is the identical full Maxwellian at temperature $T$ . The field does not cool the gas where the potential is high — it only thins the crowd; it does not slow the survivors. This one profile is a workhorse across physics. With $U=mgz$ it is the barometric law and the isothermal atmosphere; with $U=-\tfrac12 m\Omega^2 r^2$ it is the ultracentrifuge and isotope enrichment; with a buoyancy-corrected gravity it is Perrin's colloidal sedimentation, the 1908 experiment that measured Avogadro's number and settled the reality of atoms; with an electrostatic potential it seeds Debye-Hückel screening. The equilibrium condition behind all of them is not constant pressure but constant chemical potential: the rise in $U$ is exactly offset by the fall in $k_BT\ln n$ .

Key takeaways

For classical particles in an external potential $U(\mathbf r)$ at temperature $T$ , the equilibrium density is $n(\mathbf r)=n_0\,e^{-U(\mathbf r)/k_BT}$ — the Boltzmann distribution in a field.
It follows because $H=p^2/2m+U(\mathbf r)$ separates: the momentum Gaussian is position-independent and cancels, leaving the bare configurational Boltzmann factor.
The velocity distribution is the identical Maxwellian at every point; the field changes only the local density prefactor, never the temperature.
Special cases: $U=mgz$ gives the barometric law and scale height $H=k_BT/mg$ ; $U=-\tfrac12 m\Omega^2 r^2$ gives the centrifuge, where density increases outward; buoyancy-corrected gravity gives Perrin's sedimentation.
Equilibrium means constant total chemical potential $\mu=k_BT\ln(n\lambda_T^3)+U$ , not constant pressure.
The law is exact only for a dilute, classical, single-species gas; interactions replace it with the density-functional statement $\mu_{\text{loc}}[n]+U=\text{const}$ .
Key idea: because the Hamiltonian separates into kinetic and potential parts, the momentum integral factors out and cancels, so the spatial density inherits exactly the configurational Boltzmann factor $n(\mathbf r)\propto e^{-U(\mathbf r)/k_BT}$ — one law that becomes the atmosphere, the centrifuge, and the sedimenting colloid by choosing $U$ .

The core insight

Because the single-particle Hamiltonian separates into kinetic and potential parts, the canonical distribution factorizes; the momentum integral contributes a position-independent constant that cancels, so the spatial density inherits the bare configurational Boltzmann factor $n(\mathbf r)\propto e^{-U(\mathbf r)/k_BT}$ — while the velocity distribution stays a uniform Maxwellian.

Where this lives today

The Boltzmann profile in a field is a daily instrument across science. In atomic physics, a thermal cloud of atoms in a harmonic optical or magnetic trap has the Gaussian density $n(r)\propto e^{-r^2/2\sigma^2}$ with $\sigma=\sqrt{k_BT/m\omega^2}$ ; measuring the cloud radius after time-of-flight expansion is the standard thermometry of ultracold gases, and the departure of the central density from this Gaussian is how experimenters detect the onset of Bose-Einstein condensation, where quantum statistics break the classical law. In separations science, the centrifugal version drives both uranium enrichment (gas centrifuges of $\mathrm{UF_6}$ ) and the analytical ultracentrifuge that weighs proteins and DNA by their sedimentation equilibrium. In soft-matter and colloid science, the buoyancy-corrected profile still calibrates particle sizes and interactions. And in electrochemistry and biophysics, ions in an electrostatic potential follow $n_\pm\propto e^{\mp q\phi/k_BT}$ ; linearizing this and feeding it into Poisson's equation is the Poisson-Boltzmann (Debye-Hückel) theory of screening, the backbone of electrolyte, membrane, and colloid modeling.

The physical picture

Weight each microstate by $e^{-E/k_BT}$ . Two configurations that differ only by lifting one particle from height $0$ to height $z$ differ in energy by $mgz$ , so the raised one is less likely by exactly $e^{-mgz/k_BT}$ . The density is nothing more than the relative population of those configurations — hence the exponential fall-off. The profile is a tug-of-war: energy wants every particle at the potential minimum, entropy (the sheer volume of accessible space) wants them spread everywhere, and the Boltzmann factor is the precise truce, set by the single scale $k_BT$ . When $U$ climbs by one $k_BT$ , the density drops by a factor $e$ . The subtle, beautiful part is what does not change. Because the momentum integral is identical at every position, the local velocity distribution is the same Maxwellian everywhere — the sparse gas at altitude is not colder, just rarer. Equilibrium is enforced not by equal pressure (the pressure clearly falls with height) but by equal chemical potential: $\mu=k_BT\ln(n\lambda_T^3)+U$ is constant, the increase in $U$ paid for exactly by the decrease in $k_BT\ln n$ . This is the statement that survives interactions and quantum statistics, where the simple exponential does not.

Figure

The canonical density rho ∝ exp(−βH) factorizes: the momentum Gaussian (same at every height) times the spatial factor exp(−βU). At high U there are fewer particles, but each still has the identical Maxwell velocity spread.

The second figure is the punchline: it is one law, but the choice of $U$ flips its behaviour completely. Uniform gravity ( $U=mgz$ ) makes the density fall with height; the centrifugal potential ( $U=-\tfrac12 m\Omega^2 r^2$ ) has the opposite sign and makes the density grow outward, throwing the heavy species to the rim; a harmonic trap ( $U=\tfrac12 m\omega^2 r^2$ ) peaks the density at the center as a Gaussian ball. Same exponential $e^{-U/k_BT}$ , three opposite profiles — the shape of the density is a direct fingerprint of the shape of the potential.

Figure

Punchline: one law n ∝ exp(−U/kT), three potentials, three opposite profiles. Gravity (U=mgz) decays with height; centrifuge (U=−½mΩ²r²) grows outward; harmonic trap (U=½mω²r²) is a Gaussian peaked at center.

The core equations

The starting point is the single-particle canonical distribution in phase space. The key structural fact is that the energy is a sum of a momentum part and a position part, so the exponential is a product.

\rho(\mathbf r,\mathbf p)=\frac{1}{Z_1}\exp\!\left[-\frac{1}{k_BT}\left(\frac{p^2}{2m}+U(\mathbf r)\right)\right]=\frac{1}{Z_1}\,e^{-\beta p^2/2m}\,e^{-\beta U(\mathbf r)} \qquad (1)

where: $\rho$ — normalized probability density in phase space; $p^2/2m$ — kinetic energy; $U(\mathbf r)$ — external potential energy per particle; $Z_1$ — single-particle partition function; $\beta=1/k_BT$ . Reading it: the two factors are independent, so momentum and position statistics are decoupled — a particle's speed tells you nothing about where it sits, and vice versa. Integrating equation (1) over all momenta gives the spatial number density — the central result of the article.

n(\mathbf r)=n_0\,\exp\!\left[-\frac{U(\mathbf r)}{k_BT}\right] \qquad (2)

where: $n(\mathbf r)$ — number density at position $\mathbf r$ ; $n_0$ — the density at the reference point where $U=0$ ; $k_BT$ — the thermal energy scale. Reading it: the density falls by a factor $e$ every time $U$ rises by $k_BT$ . Where $U$ is negative (as in the centrifuge), the exponent is positive and the density rises — the same equation, opposite behaviour. Specialized to uniform gravity, equation (2) becomes the barometric law with an $e$ -folding scale height.

n(z)=n_0\,e^{-mgz/k_BT},\qquad H=\frac{k_BT}{mg} \qquad (3)

where: $m$ — mass of one molecule; $g$ — gravitational acceleration; $z$ — height; $H$ — the scale height, the distance over which the density drops by a factor $e$ . For air near the ground $H\approx8.4\,\mathrm{km}$ , which is why atmospheric pressure roughly halves every $\sim5.5\,\mathrm{km}$ . Because $H\propto1/m$ , heavier molecules have a shorter scale height and settle lower. Equilibrium is expressed most deeply not by the pressure but by the constancy of the total chemical potential, the statement that survives interactions and quantum statistics.

\mu(\mathbf r)=k_BT\,\ln\!\big[n(\mathbf r)\,\lambda_T^3\big]+U(\mathbf r)=\text{const} \qquad (4)

where: $\mu$ — total chemical potential; $\lambda_T=h/\sqrt{2\pi m k_BT}$ — the thermal de Broglie wavelength; the constant is the same everywhere in equilibrium. Reading it: differentiating $\mu=\text{const}$ reproduces the profile $n\propto e^{-\beta U}$ for an ideal gas, and it also reproduces the hydrostatic balance $\nabla p=-n\nabla U$ .

Derivation AThe density profile as a momentum marginal of the canonical distribution

The goal is to prove that the spatial density is exactly $n(\mathbf r)\propto e^{-U(\mathbf r)/k_BT}$ by starting from the full canonical phase-space distribution and integrating out the momenta, showing rigorously that the momentum integral is position-independent and therefore cancels. The strategy is to write the density as the momentum marginal of $\rho(\mathbf r,\mathbf p)$ , factor the Boltzmann exponential into a momentum piece and a position piece, and observe that the momentum Gaussian in the numerator and denominator are identical Gaussians that cancel to the last decimal.

Step 1. Write the number density as the momentum marginal of the canonical distribution, normalized to $N$ particles: $n(\mathbf r)=N\dfrac{\displaystyle\int d^3p\;e^{-\beta(p^2/2m+U(\mathbf r))}}{\displaystyle\int d^3r'\,d^3p\;e^{-\beta(p^2/2m+U(\mathbf r'))}}$ .

Step 2. Factor each exponential, $e^{-\beta(p^2/2m+U)}=e^{-\beta p^2/2m}\,e^{-\beta U}$ , and pull the position factor out of the momentum integral. The numerator becomes $e^{-\beta U(\mathbf r)}\displaystyle\int d^3p\;e^{-\beta p^2/2m}$ ; the denominator becomes $\big(\displaystyle\int d^3r'\,e^{-\beta U(\mathbf r')}\big)\big(\displaystyle\int d^3p\;e^{-\beta p^2/2m}\big)$ .

Step 3. Evaluate the momentum Gaussian: $\displaystyle\int d^3p\;e^{-\beta p^2/2m}=\left(\displaystyle\int_{-\infty}^{\infty}dp_x\,e^{-\beta p_x^2/2m}\right)^3=(2\pi m/\beta)^{3/2}=h^3/\lambda_T^3$ . Crucially this is a number — it has no dependence on $\mathbf r$ whatsoever. Why the momentum integral cancels. Because the kinetic energy depends only on $\mathbf p$ and not on $\mathbf r$ , the same Gaussian factor appears in both the numerator and the denominator. It is a position-independent constant, so it divides out exactly — this is the entire content of the result, and it is why the field never touches the velocity distribution.

Step 4. Cancel the momentum integrals to obtain $n(\mathbf r)=N\dfrac{e^{-\beta U(\mathbf r)}}{\displaystyle\int e^{-\beta U(\mathbf r')}\,d^3r'}=n_0\,e^{-\beta U(\mathbf r)}$ , equation (2). For a column of height $L$ in uniform gravity the denominator is $A\,H(1-e^{-L/H})$ , and $\int_0^L n(z)\,A\,dz=N$ holds identically — normalization confirmed. Verified: that the momentum integral $\int e^{-\beta p^2/2m}d^3p=(2\pi m/\beta)^{3/2}$ is position-independent and cancels, leaving $n(r_2)/n(r_1)=e^{-\beta(U_2-U_1)}$ , is confirmed symbolically in verify_aq-sm-boltzmann-distribution-external-potential.py.

Go deeper · Derivation B · 1 minThe three faces of the law — gravity, centrifuge, and Perrin's sedimentation

Derivation BThe three faces of the law — gravity, centrifuge, and Perrin's sedimentation

The goal is to show that a single substitution of $U(\mathbf r)$ into equation (2) generates the barometric atmosphere, the ultracentrifuge (with density growing outward), and Perrin's colloidal sedimentation, and to extract Avogadro's number from the last. The strategy is to plug each physical potential into $n\propto e^{-\beta U}$ and read off the profile, paying close attention to the sign of the centrifugal term and to the buoyancy correction that replaces the bare mass in a fluid.

Step 1. Gravity, $U=mgz$ : equation (2) gives $n(z)=n_0\,e^{-mgz/k_BT}$ , the barometric law of the isothermal atmosphere with scale height $H=k_BT/mg$ . This is the uniform-gravity instance owned by the barometric-formula article; here it is one substitution into the general law.

Step 2. Centrifuge, in the rotating frame the effective potential is the centrifugal energy $U_{\text{eff}}(r)=-\tfrac12 m\Omega^2 r^2$ — note the minus sign. Then $n(r)\propto e^{-\beta U_{\text{eff}}}=e^{+m\Omega^2 r^2/2k_BT}$ , which increases with radius. Heavier species pile up faster at large $r$ : the separation factor between masses $m_1<m_2$ at radius $R$ is $\exp[(m_2-m_1)\Omega^2 R^2/2k_BT]$ . Why the sign flips the profile. The centrifugal potential is a potential well that deepens outward (energy decreases with $r$ ), the exact opposite of gravity's potential hill. Equation (2) faithfully turns a deepening well into a rising density — the same law, inverted by the sign of $U$ .

Step 3. Perrin's colloid: a micron-scale sphere in a fluid feels not its bare weight but its excess weight over the displaced fluid (Archimedes). Replace $m$ by the buoyancy-corrected mass $m^*=\tfrac43\pi a^3(\rho_p-\rho_f)$ , so $n(z)=n_0\,e^{-m^*gz/k_BT}$ — a microscopic atmosphere of visible spheres with a scale height of tens of micrometres.

Step 4. Solve for Avogadro's number. Since $k_B=R/N_A$ , the measured ratio gives $N_A=\dfrac{R\,T\,\ln[n(z_1)/n(z_2)]}{m^*g\,(z_2-z_1)}$ . Every quantity — temperature, particle size (hence $m^*$ ), height difference, and the counted density ratio — is measurable under a microscope, which is exactly how Perrin obtained $N_A\approx6\times10^{23}$ . Verified: the centrifuge profile $e^{+m\Omega^2 r^2/2k_BT}$ with $dn/dr>0$ , the separation factor, the buoyancy-corrected mass $m^*=\tfrac43\pi a^3(\rho_p-\rho_f)$ with a tens-of-µm scale height, and $N_A=R/k_B=6.022\times10^{23}$ , are all confirmed in the verify script.

Limiting cases

The single profile $n(\mathbf r)=n_0\,e^{-U(\mathbf r)/k_BT}$ specializes cleanly, and its high- and low-temperature limits bracket every application.

Case	Result	Meaning
High $T$ / weak field ( $U\ll k_BT$ )	$n(\mathbf r)\approx n_0\big(1-U/k_BT\big)$	Nearly uniform; the gas barely feels the potential and fills the container.
Low $T$ / strong field ( $T\to0$ )	$n(\mathbf r)\to N\,\delta(\mathbf r-\mathbf r_{\min})$	All particles collapse into the potential minimum (until quantum zero-point motion intervenes).
Linear potential $U=mgz$	$n(z)=n_0\,e^{-z/H}$ , $H=k_BT/mg$	Exponential atmosphere; $H\approx8.4\,\mathrm{km}$ for air.
Centrifugal $U=-\tfrac12 m\Omega^2 r^2$	$n(r)\propto e^{+m\Omega^2 r^2/2k_BT}$	Density grows outward; basis of isotope enrichment.
Harmonic trap $U=\tfrac12 m\omega^2 r^2$	$n(r)\propto e^{-r^2/2\sigma^2}$ , $\sigma^2=k_BT/m\omega^2$	Gaussian cloud; the standard profile of trapped cold atoms.

Common trapThe field cools the gas where the potential is high

It is tempting to think that a high potential region is "harder to reach" and therefore colder — but the field does not change the temperature anywhere. Because the Hamiltonian separates, the velocity distribution at every point is the identical Maxwellian at the single temperature $T$ ; the potential only depletes the number of particles, never their speed distribution. Two further traps: first, $U$ must be a genuine single-particle external potential — never smuggle the total $N$ -body interaction energy into the exponent, or you double-count and get nonsense. Second, the plain exponential holds only for a dilute, classical, single-species gas; when interactions matter the correct equilibrium statement is that the total chemical potential $\mu_{\text{loc}}[n]+U(\mathbf r)$ is constant, which reduces to $n\propto e^{-\beta U}$ only in the ideal limit.

Worked examples

Example 1 — Perrin's sedimentation and the measurement of Avogadro's number (enumerate, apply, boundary number). Perrin (1908) suspended gamboge spheres of radius $a=0.2\,\mu\mathrm m$ and density $\rho_p=1.2\times10^3\,\mathrm{kg/m^3}$ in water ( $\rho_f=1.0\times10^3\,\mathrm{kg/m^3}$ ) at $T=293\,\mathrm K$ and counted their number density versus height under a microscope.

Step 1. Enumerate the effective mass. Buoyancy leaves only the excess weight: $m^*=\tfrac43\pi a^3(\rho_p-\rho_f)=\tfrac43\pi(2\times10^{-7})^3(200)=6.70\times10^{-18}\,\mathrm{kg}$ .

Step 2. Apply the Boltzmann profile $n(z)=n_0\,e^{-m^*gz/k_BT}$ . The sedimentation scale height is $H_{\text{sed}}=k_BT/(m^*g)=(1.381\times10^{-23})(293)/[(6.70\times10^{-18})(9.81)]=6.2\times10^{-5}\,\mathrm m$ .

Step 3. Boundary number. The colloidal "atmosphere" has a scale height of $H_{\text{sed}}\approx62\,\mu\mathrm m$ — a distance easily resolved under a microscope. Inverting $N_A=R\,T\,\ln[n(z_1)/n(z_2)]/[m^*g\,(z_2-z_1)]$ from the counted density ratio gives $N_A\approx6\times10^{23}\,\mathrm{mol^{-1}}$ , in agreement with the modern $6.022\times10^{23}$ . This is the experiment that made atoms undeniable and won Perrin the 1926 Nobel Prize.

Example 2 — Isotope enrichment in a gas centrifuge. A gas centrifuge separates $^{235}\mathrm{UF_6}$ (mass $\approx349\,\mathrm{u}$ ) from $^{238}\mathrm{UF_6}$ (mass $\approx352\,\mathrm{u}$ ) by spinning at rim speed $v=\Omega R=500\,\mathrm{m/s}$ at $T=320\,\mathrm K$ .

Step 1. In the rotating frame $U=-\tfrac12 m\Omega^2 r^2$ , so each species has $n(r)\propto e^{+m\Omega^2 r^2/2k_BT}$ and the heavier one accumulates faster at the rim.

Step 2. The single-stage separation factor at the rim is $\alpha=\exp\!\big[(m_2-m_1)\Omega^2 R^2/2k_BT\big]=\exp\!\big[(m_2-m_1)v^2/2k_BT\big]$ . With $\Delta m=3\,\mathrm u=3(1.66\times10^{-27})=4.98\times10^{-27}\,\mathrm{kg}$ : $\alpha=\exp\!\big[(4.98\times10^{-27})(500)^2/(2\cdot1.381\times10^{-23}\cdot320)\big]$ .

Step 3. Boundary number. The exponent is $(4.98\times10^{-27})(2.5\times10^5)/(8.84\times10^{-21})=0.141$ , so $\alpha\approx e^{0.141}\approx1.15$ per stage — a mere $15\%$ enrichment per pass. This tiny single-stage factor is exactly why enrichment cascades require thousands of centrifuges in series, a direct, quantitative consequence of the Boltzmann profile with a centrifugal potential.

Origins · 1 minsedimentation-diffusion equilibrium and how this was discovered

sedimentation-diffusion equilibrium

statistical-mechanics concept · Boltzmann 1868, Perrin 1908–1909

Sedimentation equilibrium is the balance between an external potential (gravity, buoyancy-corrected, or centrifugal) driving particles down and thermal diffusion spreading them out, producing the density profile $n(\mathbf r)\propto e^{-U(\mathbf r)/k_BT}$ . Boltzmann established the distribution of particles in a conservative field in the 1860s–70s. Jean Perrin turned it into a decisive experiment in 1908–1909: by counting micron-scale gamboge and mastic spheres at different heights under a microscope, he measured the sedimentation profile of a visible "atmosphere" and extracted Avogadro's number, confirming the molecular-kinetic theory against its skeptics and earning the 1926 Nobel Prize in Physics. The same equilibrium, spun up, is the analytical ultracentrifuge of Svedberg and the gas centrifuge of isotope enrichment.

History

1868

— Boltzmann derives the distribution of particles in a conservative external field.

1901

— Laplace's isothermal-atmosphere barometric formula is a long-standing special case.

1908

— Perrin measures the sedimentation profile of colloidal spheres and extracts Avogadro's number.

1909

— Perrin publishes "Mouvement brownien et réalité moléculaire," settling the reality of atoms.

1926

— Svedberg's ultracentrifuge and Perrin's Nobel Prize cement sedimentation equilibrium as a quantitative tool.

Going further

References. Pathria & Beale, Statistical Mechanics, 3rd ed., §3.5 and §3.7 · Kardar, Statistical Physics of Particles, Ch. 4 (canonical ensemble, ideal gas) · Reif, Fundamentals of Statistical and Thermal Physics, §7.3, §7.10 (barometric formula) · Landau & Lifshitz, Statistical Physics Part 1, §38 (the Boltzmann distribution in an external field) · Sethna, Statistical Mechanics, Ch. 3.

Provenance. All equations are standard textbook results. A verifier should confirm: (i) the factorization of $\rho$ and the momentum-integral cancellation leaving $n(r_2)/n(r_1)=e^{-\beta(U_2-U_1)}$ ; (ii) the barometric formula and scale height $H=k_BT/mg\approx8.4\,\mathrm{km}$ for air at $288\,\mathrm K$ ; (iii) the centrifuge profile $e^{+m\Omega^2 r^2/2k_BT}$ with density increasing outward and the isotope separation factor; (iv) the buoyancy-corrected mass $m^*=\tfrac43\pi a^3(\rho_p-\rho_f)$ giving a sedimentation scale height of tens of µm; (v) $N_A=R/k_B=6.022\times10^{23}$ . All checks are in verify_aq-sm-boltzmann-distribution-external-potential.py. Constants: $k_B=1.380649\times10^{-23}\,\mathrm{J/K}$ , $R=8.314\,\mathrm{J\,mol^{-1}K^{-1}}$ , $N_A=6.022\times10^{23}\,\mathrm{mol^{-1}}$ .

Builds on: Classical Ideal Gas · Maxwell-Boltzmann Velocity Distribution · Thermal de Broglie Wavelength Leads to: Barometric Formula · Equipartition Theorem · Virial Theorem in Statistical Mechanics · Sackur-Tetrode Equation

How is this different from the barometric formula?

The barometric formula is the special case of this law for uniform gravity, $U=mgz$ : it gives $n(z)=n_0\,e^{-mgz/k_BT}$ and the scale height $H=k_BT/mg$ for an isothermal atmosphere. The article you are reading owns the general law $n(\mathbf r)\propto e^{-U(\mathbf r)/k_BT}$ for an arbitrary external potential $U$ . Choosing $U=mgz$ recovers the atmosphere; $U=-\tfrac12 m\Omega^2 r^2$ gives the centrifuge; $U=\tfrac12 m\omega^2 r^2$ gives a trapped cloud; a buoyancy-corrected gravity gives Perrin's sedimentation. The barometric formula is one substitution into this general result.

Why does the momentum integral cancel so cleanly?

Because the kinetic energy $p^2/2m$ depends only on momentum and not on position. When you compute the density as the fraction of particles at a given $\mathbf r$ , the momentum integral $\int e^{-\beta p^2/2m}d^3p$ appears identically in the numerator (particles at $\mathbf r$ , any momentum) and in the denominator (particles anywhere, any momentum). Being a position-independent constant, it divides out exactly, leaving only the configurational factor $e^{-\beta U(\mathbf r)}$ . This same cancellation is why the velocity distribution is the identical Maxwellian everywhere — the field never touches the momenta.

Does a gas in a field have a position-dependent temperature?

No. In thermal equilibrium the temperature is uniform by definition — that is what "isothermal" means and what equilibrium enforces. What varies with position is the density, through the Boltzmann factor $e^{-U/k_BT}$ , not the temperature. Because the Hamiltonian separates, the velocity distribution at every point is the same full Maxwellian at the one temperature $T$ ; the sparse gas at high altitude is rarer, not colder. A real atmosphere does have a temperature gradient, but that is because it is not in true thermal equilibrium (it is convective and driven by sunlight), which is exactly why the isothermal barometric formula is only an idealization.

Why does the density increase outward in a centrifuge but decrease with height under gravity?

Because the two potentials have opposite shapes. Gravity's potential $U=mgz$ is a rising hill — energy increases with height, so the Boltzmann factor $e^{-U/k_BT}$ falls and the density decreases upward. The centrifugal potential in the rotating frame is $U=-\tfrac12 m\Omega^2 r^2$ , a well that deepens outward — energy decreases with radius, so the exponent is positive and the density rises toward the rim. The same law $n\propto e^{-U/k_BT}$ produces opposite profiles because the sign of $U$ is opposite. This is precisely why centrifuges concentrate heavy species (which feel the deeper well) at the outer wall.

Can I apply n ∝ e^{−U/kT} to a strongly interacting fluid?

Not directly. The single-particle exponent contains only the external potential $U$ ; it ignores forces between particles. For a strongly interacting fluid the correct equilibrium statement is that the total chemical potential $\mu_{\text{loc}}[n]+U(\mathbf r)$ is constant, where $\mu_{\text{loc}}[n]$ is the local chemical potential of the interacting fluid at density $n(\mathbf r)$ . For an ideal gas $\mu_{\text{loc}}=k_BT\ln(n\lambda_T^3)$ , which inverts to the simple exponential; but for a non-ideal fluid $\mu_{\text{loc}}$ depends nonlinearly on the local density, so the profile is no longer a pure exponential. This constant-chemical-potential condition is the starting point of classical density-functional theory.

How does this connect to the Maxwell-Boltzmann velocity distribution?

They are the two halves of the same factorized canonical distribution. Writing $\rho(\mathbf r,\mathbf p)\propto e^{-\beta p^2/2m}\,e^{-\beta U(\mathbf r)}$ , the momentum factor $e^{-\beta p^2/2m}$ is the Maxwell-Boltzmann velocity distribution, and the position factor $e^{-\beta U(\mathbf r)}$ is the spatial density profile of this article. Because they multiply, they are statistically independent: knowing where a particle is tells you nothing about its speed. The velocity distribution is the same Maxwellian at every point in the field, while the density varies — the field acts entirely through the position factor.

Why is Perrin's experiment considered such strong evidence for atoms?

Because it turned an invisible molecular law into a visible, countable one. Before 1908 the atomic hypothesis was still contested by influential skeptics. Perrin realized that micron-scale colloidal spheres large enough to see under a microscope obey the same Boltzmann distribution $n\propto e^{-m^*gz/k_BT}$ as molecules — just with a buoyancy-corrected mass and a scale height of tens of micrometres instead of kilometres. By counting the spheres at different heights he measured $k_B$ (through $k_B=R/N_A$ ) and obtained Avogadro's number, $N_A\approx6\times10^{23}$ , agreeing with values from utterly different methods. That agreement across independent routes was decisive evidence that molecules are real, discrete objects.