--- title: "The Microcanonical Ensemble" book: "University Physics — Year 3" subject: physics language: en chapter: 16 exercises: 12 source: https://one-course.com/books/physics/5/en/chapter/16-the-microcanonical-ensemble --- # Chapter 16 — The Microcanonical Ensemble A cubic centimetre of air holds $2.5\times10^{19}$ molecules. No computer will ever integrate their equations of motion, and no experiment could supply the initial conditions — yet the gas obeys laws of splendid simplicity, and the thermodynamics of the Year 1 volume described them without ever mentioning a molecule. This chapter builds the bridge: *statistical physics*, which derives the laws of heat from mechanics plus honest counting. The whole edifice stands on one postulate — an isolated system is equally likely to be in any microscopic state compatible with its energy — made legitimate by Liouville’s theorem ([Chapter 2](https://one-course.com/books/physics/5/en/chapter/2-hamiltonian-mechanics#ch-b3-hamiltonian-mechanics)) and made powerful by the sheer size of the numbers: when configurations are counted in units of $10^{10^{20}}$, “overwhelmingly probable” and “certain” become indistinguishable, and probability hardens into law. Out of the counting come entropy, temperature, pressure and the second law — and, at the chapter’s end, the reason a rubber band pulls back. ## 16.1 Microstates, macrostates, and the postulate **Definition 16.1 (Microstates and macrostates).** A *microstate* is a complete microscopic specification of a system — every quantum number of every particle (or, classically, every position and momentum, counted in phase-space cells of $h^3$ per particle, [Proposition 7.5](https://one-course.com/books/physics/5/en/chapter/7-the-schrodinger-equation-in-three-dimensions#prop-b3-schrodinger-three-dimensions-counting)). A *macrostate* is what thermodynamics can see: energy $E$, volume $V$, particle number $N$, magnetisation… The *multiplicity* $\Omega(E, V, N)$ is the number of microstates wearing the same macrostate. **Theorem 16.2 (The fundamental postulate).** An isolated system in equilibrium is found with equal probability in each of its $\Omega(E, V, N)$ accessible [microstates](#def-b3-microcanonical-ensemble-states). All of equilibrium statistical physics follows from this single sentence. Its credentials: Liouville’s theorem shows the uniform distribution over the energy shell is the one that dynamics preserves — no flow ever bunches phase-space probability — and a century and a half of consequences have never disagreed with an experiment. **Proof.** *Admitted at this level.* ∎ **Example 16.3 (A magnet of NNN coins).** $N$ independent spins, each up or down: $2^N$ [microstates](#def-b3-microcanonical-ensemble-states). The [macrostate](#def-b3-microcanonical-ensemble-states) “$n$ up” has [multiplicity](#def-b3-microcanonical-ensemble-states) $\Omega(n) = \binom Nn$, overwhelmingly peaked at $n = N/2$ with relative width $\sim1/\sqrt N$. For $N = 100$: the all-up state is one; the balanced [macrostate](#def-b3-microcanonical-ensemble-states) holds $1 \times 10^{29}$. For $N = 10^{20}$, deviations of even $10^{-8}$ in the up-fraction are suppressed by factors like $\eu^{-10^4}$: the system *is* at its peak, and what we call equilibrium is the [macrostate](#def-b3-microcanonical-ensemble-states) with the most [microstates](#def-b3-microcanonical-ensemble-states). ![The multiplicity of the spin system against its up-fraction, for growing N: the peak sharpens as 1/√ N. At N 1020 the “distribution” is, for every practical purpose, a single value — macroscopic definiteness out of microscopic democracy.](https://one-course.com/images/onecourse/chapters/physics-5/b3-microcanonical-ensemble/fig-8d09d83e4dda.svg) *The [multiplicity](#def-b3-microcanonical-ensemble-states) of the spin system against its up-fraction, for growing $N$: the peak sharpens as $1/\sqrt N$. At $N \sim 10^{20}$ the “distribution” is, for every practical purpose, a single value — macroscopic definiteness out of microscopic democracy.* ## 16.2 Entropy is a count **Definition 16.4 (Boltzmann entropy).** The *statistical entropy* of a [macrostate](#def-b3-microcanonical-ensemble-states) is $$ S = k_{\text{B}}\ln\Omega , \qquad k_{\text{B}} = 1.38 \times 10^{-23}\,\mathrm{J}/\mathrm{K} . $$ The logarithm makes entropy *additive* where multiplicities *multiply*: for independent systems $\Omega = \Omega_1\Omega_2$ and $S = S_1 + S_2$; Boltzmann’s constant calibrates the count to the kelvin-and-joule units thermodynamics had already chosen. This $S$, we will verify, is the entropy of the Year 1 volume’s second law — now with a microscopic meaning: entropy measures, in logarithmic currency, *how many ways* the [macrostate](#def-b3-microcanonical-ensemble-states) can be realised. **Proposition 16.5 (Entropy of the ideal gas).** Counting the quantum states of $N$ identical atoms in a box ([Proposition 7.5](https://one-course.com/books/physics/5/en/chapter/7-the-schrodinger-equation-in-three-dimensions#prop-b3-schrodinger-three-dimensions-counting), one cell of $h^3$ per atom in [phase space](https://one-course.com/books/physics/5/en/chapter/2-hamiltonian-mechanics#def-b3-hamiltonian-mechanics-hamiltonian), divided by $N!$ for identity — [Exercise 14.10](https://one-course.com/books/physics/5/en/chapter/14-identical-particles#exo-b3-identical-particles-10)) gives $$ S = Nk_{\text{B}}\Big[\ln\Big(\frac VN\,\frac1{\lambda_T^3}\Big) + \frac52\Big] , \qquad \lambda_T = \frac{h}{\sqrt{2\pi mk_{\text{B}}T}} $$ (the *Sackur–Tetrode formula*; $\lambda_T$, the thermal de Broglie wavelength, is the size of an atom’s wave packet at temperature $T$). Two remarkable audits: the formula is extensive *only* thanks to the $N!$; and its absolute value — including the $h$ — matches calorimetric entropies of real gases ([Exercise 16.11](#exo-b3-microcanonical-ensemble-11)): classical thermodynamics, measured in the nineteenth century, already knew Planck’s constant without knowing it. **Proof.** *Admitted at this level.* ∎ ## 16.3 Temperature is a derivative of counting **Theorem 16.6 (Equilibrium and temperature).** Let two systems exchange energy, the total $E = E_1 + E_2$ fixed. The joint [multiplicity](#def-b3-microcanonical-ensemble-states) $\Omega_1(E_1)\,\Omega_2(E - E_1)$ is maximal — overwhelmingly so — at the partition where $$ \frac{\partial S_1}{\partial E_1} = \frac{\partial S_2}{\partial E_2} . $$ Defining $$ \frac1T = \frac{\partial S}{\partial E}\bigg|_{V,N} , $$ equilibrium is *equality of temperature*, and energy flows spontaneously from high $T$ to low $T$ because that direction increases the total count. The same logic applied to volume and particle exchange yields $P/T = \partial S/\partial V$ and $-\mu/T = \partial S/\partial N$: all of thermodynamics’ intensive quantities are derivatives of the count. **Proof.** Maximise $\ln\Omega_1 + \ln\Omega_2$ over $E_1$: the stationarity condition is the stated equality. Sharpness: the peak’s width is $\sim E/\sqrt N$ ([Exercise 16.7](#exo-b3-microcanonical-ensemble-7)), so the maximum is the observed state. If $\partial S_1/\partial E_1 > \partial S_2/\partial E_2$, moving energy *into* 1 raises $S_{\text{tot}}$: the colder body (larger $\partial S/\partial E$) absorbs — heat flows hot to cold as a counting statement. Check on the ideal gas: $S \propto \tfrac32 Nk_{\text{B}}\ln E + \cdots$ gives $1/T = \tfrac32 Nk_{\text{B}}/E$, i.e. $E = \tfrac32 Nk_{\text{B}}T$ — the kinetic-theory result of the Year 1 volume, recovered from pure counting. ∎ ![Two systems sharing energy: the joint count against the split. The equilibrium partition — equal temperatures — is not merely the likeliest; for macroscopic N it is the only one ever observed.](https://one-course.com/images/onecourse/chapters/physics-5/b3-microcanonical-ensemble/fig-e51dc6c55d1d.svg) *Two systems sharing energy: the joint count against the split. The equilibrium partition — equal temperatures — is not merely the likeliest; for macroscopic $N$ it is the only one ever observed.* ## 16.4 The second law, and stranger things **Proposition 16.7 (Irreversibility is arithmetic).** A [constraint](https://one-course.com/books/physics/5/en/chapter/1-lagrangian-mechanics#def-b3-lagrangian-mechanics-coordinates) released (a wall removed, a valve opened) can only enlarge the accessible count: $\Omega$ grows, $S$ grows — the second law of the Year 1 volume, now as a statement about probability. Reversals are not forbidden but *discounted*: a gas of $N$ molecules re-gathering into the left half has probability $2^{-N}$, and for one cubic centimetre of air the waiting time exceeds the age of the universe by a factor with $10^{19}$ digits. The arrow of time, at this level, is the direction in which counts increase. **Partial proof.** Removing a [constraint](https://one-course.com/books/physics/5/en/chapter/1-lagrangian-mechanics#def-b3-lagrangian-mechanics-coordinates) makes every formerly accessible [microstate](#def-b3-microcanonical-ensemble-states) still accessible and adds new ones: $\Omega$ cannot decrease. The free-expansion count: each molecule doubles its accessible volume, $\Omega \to 2^N\Omega$, so $\Delta S = Nk_{\text{B}}\ln2$ — exactly the thermodynamic result for isothermal free doubling. ∎ **Example 16.8 (Negative temperature).** A system whose energy is *bounded above* — $N$ spins in a field, energy from $-N\epsilon$ to $+N\epsilon$ — has $\Omega(E)$ rising then *falling*: beyond the balanced point, adding energy reduces the count. There $\partial S/\partial E < 0$: the temperature is *negative*. Such states exist (population inversions — the working state of every laser medium — and spin systems prepared by field reversal) and they are not cold but *hotter than every positive temperature*: energy flows from any $T < 0$ body into any $T > 0$ body. The honest ordering of temperatures runs $+0, \dots, +\infty \equiv -\infty, \dots, -0$: what $1/T$ orders naturally, $T$ garbles ([Exercise 16.9](#exo-b3-microcanonical-ensemble-9)). ![Entropy of the two-level spin system against energy. On the rising flank temperature is positive; at the crest it passes through infinity; on the falling flank — population inverted — it is negative, and hotter than anything on the left.](https://one-course.com/images/onecourse/chapters/physics-5/b3-microcanonical-ensemble/fig-f7da2bd2ce38.svg) *Entropy of the two-level spin system against energy. On the rising flank temperature is positive; at the crest it passes through infinity; on the falling flank — population inverted — it is negative, and hotter than anything on the left.* **Method 16.9 (Microcanonical craft).** (1) Enumerate: what is a [microstate](#def-b3-microcanonical-ensemble-states) here, and what does the [macrostate](#def-b3-microcanonical-ensemble-states) fix? (2) Count $\Omega$ — combinatorics for discrete units, phase-space volume over $h^{3N}N!$ for gases. (3) Take the logarithm early: Stirling ($\ln N! \approx N\ln N - N$) turns products into tractable sums. (4) Differentiate $S$: energy derivative for $T$, volume for $P$, number for $\mu$; length or magnetisation derivatives for tensions and fields ([Problem 16.1](#pb-b3-microcanonical-ensemble-1)). (5) Trust the peak: fluctuations are down by $1/\sqrt N$, so replace “most probable” by “equals” and apologise to no one. ![Smoke stirred in a box, cut by a laser sheet: micro-motion beyond any bookkeeping, statistics taking over — the moment a mechanics problem becomes an entropy problem.](https://one-course.com/images/onecourse/chapters/physics-5/b3-microcanonical-ensemble/img-1f6f21fcd18d.jpg) *Smoke stirred in a box, cut by a laser sheet: micro-motion beyond any bookkeeping, statistics taking over — the moment a mechanics problem becomes an entropy problem.* ## 16.5 Exercises **Exercise 16.1 ★.** Four spins. (a) List the multiplicities $\Omega(n)$ of the [macrostates](#def-b3-microcanonical-ensemble-states) “$n$ up”. (b) The probability of each [macrostate](#def-b3-microcanonical-ensemble-states) under the postulate. (c) The entropy (in units of $k_{\text{B}}$) of each. (d) Which [macrostate](#def-b3-microcanonical-ensemble-states) is “equilibrium”, and how bad an approximation is “the system is surely there” at $N = 4$? **Solution of Exercise 16.1.** (a) $1, 4, 6, 4, 1$. (b) Over $16$: $6.25\%$, $25\%$, $37.5\%$, $25\%$, $6.25\%$. (c) $S/k_{\text{B}} = 0$, $\ln4$, $\ln6$, $\ln4$, $0$. (d) $n = 2$, holding only $37.5\%$ of the probability: at $N = 4$, “equilibrium” is merely a plurality — the sharpness that licenses thermodynamics is bought with large $N$. **Exercise 16.2 ★.** Stirling in practice. (a) Compare $\ln N!$ with $N\ln N - N$ for $N = 10$ and $N = 100$ (use $\ln10! = 15.10$, $\ln100! = 363.7$). (b) Show the relative error falls like $\ln N/N$. (c) Use Stirling to show $\ln\binom N{N/2} \approx N\ln2 - \tfrac12\ln N +$ const. (d) Why is dropping the $\tfrac12\ln N$ utterly safe in a mole? **Solution of Exercise 16.2.** (a) $N = 10$: $13.0$ against $15.1$ ($14\%$ off); $N = 100$: $360.5$ against $363.7$ ($0.9\%$). (b) The neglected term is $\tfrac12\ln(2\pi N)$: relative error $\sim\ln N/N$. (c) Apply Stirling to the three factorials. (d) At $N = 10^{23}$ the dropped term is $\sim26$ against $N\ln2 \sim 10^{23}$: twenty-two orders below the leading one. **Exercise 16.3 ★.** (a) Show that for two independent systems $S = S_1 + S_2$. (b) A system’s $\Omega$ doubles: by how much does $S$ rise — and what physical act ([Proposition 16.7](#prop-b3-microcanonical-ensemble-second-law)) does that correspond to for one particle? (c) Compute the entropy, in $k_{\text{B}}$ and in J/K, of a shuffled deck of 52 cards ($\ln52! \approx 156.4$). (d) Why is card entropy thermodynamically negligible while molecular entropy is not? **Solution of Exercise 16.3.** (a) $\ln(\Omega_1\Omega_2) = \ln\Omega_1 + \ln\Omega_2$. (b) $\Delta S = k_{\text{B}}\ln2$ — one particle offered a doubled volume. (c) $S = 156\,k_{\text{B}} = 2.2 \times 10^{-21}\,\mathrm{J}/\mathrm{K}$. (d) Thermodynamic entropies carry factors of $10^{23}$: all the shuffling in every casino on Earth is invisible next to a breath of warm air. **Exercise 16.4 ★.** Free expansion, audited. One mole doubles its volume into vacuum. (a) $\Delta S$ from the counting argument. (b) The probability of observing the gas back in the original half. (c) Estimate the time scale for such a fluctuation given molecular rearrangement times of $10^{-10}\,\mathrm{s}$ — and compare with the age of the universe ($4 \times 10^{17}\,\mathrm{s}$). (d) In what precise sense is the second law “only” probabilistic? **Solution of Exercise 16.4.** (a) $\Delta S = N_{\text{A}}k_{\text{B}}\ln2 = R\ln2 = 5.76\,\mathrm{J}/\mathrm{K}$. (b) $2^{-N_{\text{A}}} \sim 10^{-1.8\times 10^{23}}$. (c) Even sampling configurations every $10^{-10}\,\mathrm{s}$, the expected wait dwarfs $4 \times 10^{17}\,\mathrm{s}$ by a factor whose *exponent* has twenty-three digits. (d) Not impossible — unwitnessable: the law is probabilistic in principle and absolute in any world that contains observers with finite patience. **Exercise 16.5 ★★.** Ideal-gas counting. Starting from $\Omega \propto \dfrac{V^N}{N!\,h^{3N}}\times$ (momentum-shell volume $\propto E^{3N/2}$): (a) show $S = Nk_{\text{B}}[\ln V + \tfrac32\ln E] - k_{\text{B}}\ln N! +$ const. (b) Show that without the $N!$, doubling the system (both $V$ and $N$) would *not* double $S$ — the extensivity failure of [Exercise 14.10](https://one-course.com/books/physics/5/en/chapter/14-identical-particles#exo-b3-identical-particles-10). (c) With Stirling, put the result in the form $S = Nk_{\text{B}}[\ln(V/N) + \tfrac32\ln(E/N)] +$ const$\cdot N$. (d) Which physical inputs fixed the two “consts” history left undetermined (quantum cell size; particle identity)? **Solution of Exercise 16.5.** (a) Take the logarithm of the product. (b) Doubling $V, N$ at fixed $E/N$: without $N!$, $S$ gains an extra $N\ln2$-type excess — the Gibbs failure. (c) Stirling on $\ln N!$ delivers the $V/N$ and $E/N$ combinations. (d) Planck’s cell $h^3$ fixes the entropy’s zero; identity of particles fixes the $N!$ — two quantum inputs hiding in classical thermodynamics. **Exercise 16.6 ★★.** Thermodynamics from derivatives. Using the form of [Exercise 16.5](#exo-b3-microcanonical-ensemble-5)(c): (a) compute $1/T = \partial S/\partial E$ and recover $E = \tfrac32 Nk_{\text{B}}T$; (b) compute $P/T = \partial S/\partial V$ and recover $PV = Nk_{\text{B}}T$; (c) derive the [adiabatic invariant](https://one-course.com/books/physics/5/en/chapter/2-hamiltonian-mechanics#prop-b3-hamiltonian-mechanics-adiabatic): at constant $S$ and $N$, show $VE^{3/2}$ is fixed, i.e. $TV^{2/3}$ — and match to the $PV^{5/3}$ of the Year 1 volume; (d) state what has been achieved: which empirical laws just became theorems of counting. **Solution of Exercise 16.6.** (a) $1/T = \tfrac32 Nk_{\text{B}}/E$. (b) $P/T = Nk_{\text{B}}/V$: the ideal-gas law, from counting. (c) Constant $S$: $\ln(V/N) + \tfrac32\ln(E/N)$ fixed, so $VE^{3/2}$ constant; with $E \propto T$: $TV^{2/3}$ constant, equivalent to $PV^{5/3}$. (d) The gas law, the energy–temperature relation and the adiabatic exponent — three empirical pillars of the Year 1 volume — are now theorems. **Exercise 16.7 ★★.** Sharpness of equilibrium. Two equal ideal-gas blocks share total energy $E$. (a) Show $\ln[\Omega_1\Omega_2]$ is maximal at the even split. (b) Expand to second order in the imbalance $x = (E_1 - E_2)/E$ and show the probability is $\propto\eu^{-3Nx^2/4}$ (each block $\tfrac32 N$ [degrees of freedom](https://one-course.com/books/physics/5/en/chapter/1-lagrangian-mechanics#def-b3-lagrangian-mechanics-coordinates) — keep factors loose). (c) For $N = 10^{22}$: the r.m.s. imbalance. (d) The temperature difference that imbalance represents, for gas at $300\,\mathrm{K}$: could any thermometer see it? **Solution of Exercise 16.7.** (a) Symmetry (equal blocks). (b) Each $\ln\Omega \propto \tfrac32 N\ln E_i$: expanding about the even split gives a Gaussian in $x$ of variance $\sim1/N$. (c) $x_{\text{rms}} \sim 10^{-11}$. (d) $\Delta T/T \sim x_{\text{rms}}$: nano-nanokelvins — fluctuations exist, and are unmeasurably tame at laboratory scale. **Exercise 16.8 ★★.** The two-level (Schottky) solid: $N$ sites, each of energy $0$ or $\epsilon$; $n$ excited. (a) $\Omega(n)$ and $S(n)$. (b) With $E = n\epsilon$ and Stirling, derive $$ \frac1T = \frac{k_{\text{B}}}{\epsilon}\, \ln\frac{N - n}{n} , \qquad\text{hence}\qquad \frac nN = \frac{1}{\eu^{\epsilon/k_{\text{B}}T} + 1} . $$ (c) Sketch $E(T)$ and the heat capacity: a bump (the *Schottky anomaly*) near $k_{\text{B}}T \sim \epsilon/2$ — why does $C$ vanish at both ends? (d) Where has the occupation formula’s shape appeared before, and where will it return ([Chapter 19](https://one-course.com/books/physics/5/en/chapter/19-quantum-statistics#ch-b3-quantum-statistics))? **Solution of Exercise 16.8.** (a) $\Omega = \binom Nn$, $S = k_{\text{B}}\ln\binom Nn$. (b) $1/T = \partial S/\partial E = (k_{\text{B}}/\epsilon)[\ln(N-n) - \ln n]$; solve for $n/N$. (c) $E(T)$ rises from $0$ to the saturation $N\epsilon/2$; $C = \dd E/\dd T$ vanishes at low $T$ (no quantum can be afforded) and at high $T$ (both levels equally full, nothing left to absorb): a bump between — the calorimetric fingerprint of any two-level population, used to find magnetic impurities in solids. (d) It is the thermal two-level occupation — and, with $\epsilon \to E - \mu$, it will reappear verbatim as the Fermi–Dirac distribution. **Exercise 16.9 ★★.** Below zero. For the spin system of [Example 16.8](#ex-b3-microcanonical-ensemble-negative) with level splitting $\epsilon$: (a) show $T < 0$ exactly when more than half the spins are up (population inverted). (b) Compute $T$ for a 60:40 inversion with $\epsilon = 10^{-4}\,\mathrm{eV}$. (c) Show energy always flows from negative-$T$ to positive-$T$ bodies (compare $\partial S/\partial E$). (d) Why can a system with unbounded energy (a gas) never reach $T < 0$? **Solution of Exercise 16.9.** (a) $1/T \propto \ln[(N-n)/n] < 0$ exactly when $n > N/2$. (b) $T = \epsilon/[k_{\text{B}}\ln(2/3)] = -2.9\,\mathrm{K}$. (c) A negative-$T$ body *gains* entropy by losing energy, a positive-$T$ body gains by receiving: both counts grow when energy flows negative $\to$ positive — the inverted system is the universal donor, i.e. hotter than everything. (d) With unbounded energy, $\Omega(E)$ never turns over: $\partial S/\partial E$ stays positive at any finite energy. **Exercise 16.10 ★★★.** Mixing, quantitatively. Two different gases ($N$ each, volumes $V$) are joined. (a) Compute $\Delta S$ by counting. (b) Same gas: show, $N!$’s included, $\Delta S = 0$. (c) Helium-3 and helium-4 are different: joining them *does* create $2Nk_{\text{B}}\ln2$ — yet no heat flows and no work is done: where is the entropy increase physically (what would separating them again cost)? (d) The minimum work to re-separate at temperature $T$: express it, and name the modern industry built on paying it (isotope separation). **Solution of Exercise 16.10.** (a) Each gas doubles its accessible volume: $\Delta S = 2Nk_{\text{B}}\ln2$. (b) The $N!$’s of the merged populations exactly absorb the apparent gain: $\Delta S = 0$. (c) The entropy increase is the *loss of sortedness*: no macroscopic variable moved, but re-separating now requires work — the increase is stored as a future bill. (d) $W_{\min} = T\Delta S = 2Nk_{\text{B}}T\ln2$; isotope-separation plants (uranium centrifuge cascades, helium-3 recovery) pay this thermodynamic minimum many times over in practice. **Exercise 16.11 ★★★.** The tables knew about $h$. Sackur–Tetrode for argon ($m = 6.6 \times 10^{-26}\,\mathrm{kg}$) at $300\,\mathrm{K}$, $1\,\mathrm{bar}$ ($V/N = k_{\text{B}}T/P$): (a) compute $\lambda_T$; (b) compute $S/N k_{\text{B}}$; (c) compare with the calorimetric value $S_{\text{molar}} = 154.8\,\mathrm{J}/(\mathrm{mol}\,\mathrm{K})$, i.e. $S/Nk_{\text{B}} = 18.6$; (d) explain what is being tested: which two quantum inputs sit inside a number measured with ice calorimeters before 1912? **Solution of Exercise 16.11.** (a) $\lambda_T = 1.6 \times 10^{-11}\,\mathrm{m}$. (b) $V/N = k_{\text{B}}T/P = 4.1 \times 10^{-26}\,\mathrm{m}^{3}$: $V/N\lambda_T^3 = 1.0 \times 10^{7}$, so $S/Nk_{\text{B}} = \ln(1.0 \times 10^{7}) + 2.5 = 18.6$. (c) Dead on the calorimetric $18.6$. (d) The absolute entropy contains $h$ (through $\lambda_T$) and the $N!$: heat measurements from the age of steam confirm, to three digits, the quantum of [action](https://one-course.com/books/physics/5/en/chapter/1-lagrangian-mechanics#def-b3-lagrangian-mechanics-action) and the identity of atoms. **Exercise 16.12 ★★★.** Fluctuations you can see. The number of molecules in a small volume $v$ of gas fluctuates; the probability of a relative deviation $\delta$ is $\propto\eu^{-\bar n\delta^2/2}$ ($\bar n = Nv/V$ — accept this Gaussian, or derive from the binomial). (a) For $v$ a cube of side $550\,\mathrm{nm}$ at atmospheric density: $\bar n$ and the r.m.s. $\delta$. (b) Why are these permanent few-per-mille density ripples at optical scales exactly what [Problem 13.1](https://one-course.com/books/physics/5/en/chapter/13-perturbation-theory#pb-b3-perturbation-theory-1) needed to let air scatter at all? (c) At what $\bar n$ (hence what scale) would fluctuations reach $10\%$? (d) Close the loop in one sentence: the blue of the sky is a statement about $\sqrt N$ statistics. **Solution of Exercise 16.12.** (a) $v = (550\,\mathrm{nm})^3 = 1.7 \times 10^{-19}\,\mathrm{m}^{3}$: $\bar n = 4.2 \times 10^{6}$, $\delta_{\text{rms}} = 1/\sqrt{\bar n} \approx 5 \times 10^{-4}$. (b) A perfectly uniform medium would scatter nothing sideways ([Problem 13.1](https://one-course.com/books/physics/5/en/chapter/13-perturbation-theory#pb-b3-perturbation-theory-1), Part III): these irreducible $\sqrt N$ ripples in every optical cell are precisely the disorder that lets the sky exist. (c) $\bar n = 100$: $v \approx (16\,\mathrm{nm})^3$. (d) The sky is blue because air is made of countable molecules whose numbers fluctuate as $\sqrt N$ — Einstein used exactly this to argue molecules were real. ## 16.6 Problem: The entropic spring **Problem 16.1.** Weekend problem — why rubber pulls back Stretch a rubber band and it snaps back — yet its molecules’ bonds are barely deformed and its internal energy barely changes. Rubber’s restoring force is not energy seeking a minimum but *entropy seeking a maximum*: the first great victory of pure counting over mechanism, with the same mathematics running today’s single-molecule DNA experiments. Model: a chain of $N$ rigid links, each of length $b$, each pointing left or right along the pull axis; end-to-end extension $x = (n_+ - n_-)\,b$. **Part I — Counting configurations.** 1. Express $n_\pm$ in terms of $N$ and $x$ , and write the [multiplicity](#def-b3-microcanonical-ensemble-states) $\Omega(x)$ . 2. Where is $\Omega$ maximal, and what does that say about a free chain’s preferred extension? 3. With Stirling, show for $x \ll Nb$: $$S(x) \approx S(0) - \frac{k_{\text{B}}x^2}{2Nb^2} .$$ 4. The chain’s energy is (in this model) independent of $x$ : justify calling any restoring force “entropic”. 5. The r.m.s. extension of the free chain is $b\sqrt N$ (random walk): for $N = 10^4$ and $b = 0.5\,\mathrm{nm}$ , compare the coil size with the stretched length $Nb$ . 6. Real rubber is a network of such chains between cross-links: which single parameter of the model does vulcanisation (more cross-links) change? **Part II — The force of counting.** 7. For a system held at temperature $T$, the tension required to hold extension $x$ is $f = -T\,\partial S/\partial x$ (accept this from $\dd E = T\dd S + f\dd x$ at constant energy): derive $$f = \frac{k_{\text{B}}T\,x}{Nb^2} .$$ 8. [Hooke’s law](https://one-course.com/books/physics/5/en/chapter/3-continuum-mechanics-and-elasticity#thm-b3-continuum-elasticity-hooke) has emerged with spring constant $k = k_{\text{B}}T/Nb^2$ : evaluate it for one chain with $N = 10^{4}$ , $b = 0.5\,\mathrm{nm}$ at $300\,\mathrm{K}$ . 9. The startling factor is $T$ : what should a rubber band under fixed load do when *heated* ? Contrast with a steel spring. 10. Estimate the force to stretch one chain to half its full length, and the force scale $k_{\text{B}}T/b$ at which the linear model fails. 11. A rubber band of [cross-section](https://one-course.com/books/physics/5/en/chapter/15-scattering-theory#def-b3-scattering-theory-cross-section) $1\,\mathrm{mm}^{2}$ contains $\sim10^{14}$ effective chains in parallel: estimate its spring constant and compare with experience. 12. Why does rubber stiffen (not soften) with temperature, gram for gram, while metals soften? **Part III — The Gough–Joule kitchen.** 13. Stretch a rubber band quickly (adiabatically) against your lip: it warms. Explain with the entropy budget: stretching *reduces* configurational entropy, so where must entropy (heat) go at constant total? 14. Release it quickly: it cools. Complete the symmetric argument. 15. A weighted rubber band is gently heated (hair dryer): which way does the weight move, and why is this the clean signature of entropic elasticity? 16. Design the counter-experiment with a steel spring: what happens instead, and why (which kind of elasticity)? 17. An engine: a wheel with rubber spokes, heated on one side, turns steadily. Trace one cycle’s logic (heated spokes contract, unbalancing the wheel). 18. The same $f = k_{\text{B}}Tx/Nb^2$ law, measured by optical tweezers on single DNA molecules (with $b \approx 100\,\mathrm{nm}$ , $N \approx 500$ for a bacterial genome fragment), gives forces in which range? (Compute $k_{\text{B}}T/b$ .) Why did this make the model *directly* testable on one molecule? **Part IV — The moral.** 19. The chain model has no interactions and no energy scale, yet produces a force law: state precisely where the force “comes from” in the microcanonical language of this chapter. 20. Why does the force vanish at $T = 0$ — and what does that say about the nature of elasticity in a world without thermal agitation? 21. Compare the entropic spring constant’s $T$ -linearity with the ideal-gas pressure’s $T$ -linearity: show both are the same phenomenon (counting configurations against a [constraint](https://one-course.com/books/physics/5/en/chapter/1-lagrangian-mechanics#def-b3-lagrangian-mechanics-coordinates) ) in different clothes. 22. The heated band lifts a weight, doing real work: trace the energy’s source and route (heat in from the dryer, entropy bookkeeping, work out) and confirm the first law is untouched. 23. The ideal chain ignores self-avoidance and link-bending energy: name one measured feature of real rubber each omission hides (sharp stiffening near full extension; strain-induced crystallisation and hysteresis). 24. Proteins fold, membranes flicker, polymers coil: why is $k_{\text{B}}T$ at body temperature the natural force currency ( $k_{\text{B}}T/\mathrm{nm}$ in piconewtons — compute it) of all soft and living matter? 25. Summarise the named result: a chain of $10^4$ blind links, counted honestly, yields [Hooke’s law](https://one-course.com/books/physics/5/en/chapter/3-continuum-mechanics-and-elasticity#thm-b3-continuum-elasticity-hooke) with $k = k_{\text{B}}T/Nb^2$ , predicts that rubber warms when stretched and lifts weights when heated — entropy acting as a force, verified from bicycle inner tubes to single DNA molecules at $\sim0.04\,\mathrm{pN}$ scales. **Solution of Problem 16.1.** **1.** $n_\pm = \tfrac12(N \pm x/b)$; $\Omega = N!/n_+!\,n_-!$. **2.** At $x = 0$: the free chain coils; full extension has $\Omega = 1$. **3.** Stirling on the binomial, expanded to second order in $x/Nb$: the stated Gaussian entropy. **4.** With $E$ independent of $x$, any pull toward small $x$ can only come from the count: entropic by construction. **5.** Coil $\sim b\sqrt N = 50\,\mathrm{nm}$ against $Nb = 5\,\text{µ}\mathrm{m}$: the free chain is a hundred times shorter than its contour — crumpled almost entirely. **6.** $N$, the number of links between cross-links: vulcanisation shortens the effective chains, stiffening the network. **7.** $f = -T\,\partial S/\partial x = k_{\text{B}}Tx/Nb^2$. **8.** $k = k_{\text{B}}T/Nb^2 = 4.14 \times 10^{-21}/(10^{4} \times 2.5 \times 10^{-19}) = 1.7 \times 10^{-6}\,\mathrm{N}/\mathrm{m}$ per chain. **9.** $k \propto T$: heated under fixed load, the band stiffens and *contracts*, lifting the load; a steel spring merely softens and sags a little. **10.** $f(Nb/2) = k_{\text{B}}T/2b \approx 4\,\mathrm{pN}$; the linear law fails as $f$ approaches $k_{\text{B}}T/b \approx 8\,\mathrm{pN}$, where the chain nears full extension. **11.** $\sim10^{14}$ chains in parallel and $\sim10^{5}$ coil-lengths in series along a centimetre: $k_{\text{band}} \sim k_{\text{chain}} \times 10^{14}/10^{5} \sim 10^{2}$–$10^{3}\,\mathrm{N}/\mathrm{m}$ — the familiar feel of a rubber band. **12.** Rubber’s stiffness *is* $k_{\text{B}}T$ per configuration: more agitation, more recoil; metallic stiffness is bond energy, which anharmonic agitation loosens. **13.** Stretching cuts the configurational count; done fast (no entropy exchanged with outside), the lost configurational entropy must reappear as thermal entropy: the band warms — $1\,\mathrm{K}$-scale, lip-detectable. **14.** Releasing restores configurations; the thermal account refunds the difference: it cools. **15.** The weight *rises*: contraction on heating is the entropic signature (energy-elastic materials expand). **16.** The steel spring lengthens slightly (thermal expansion) and its modulus drops: opposite sign — elasticity of energy, not of counting. **17.** Heated spokes contract, pulling the rim’s mass off-centre toward the cool side; gravity torques the unbalanced wheel; each spoke re-relaxes as it rotates away: a heat engine whose working substance is entropy itself. **18.** $k_{\text{B}}T/b = 4.14 \times 10^{-21}/10^{-7} = 4 \times 10^{-14}\,\mathrm{N} = 0.04\,\mathrm{pN}$: piconewtons and below — exactly the optical-tweezer range, so the force–extension law of one DNA molecule could be traced point by point (and matched, with the refinements of item 23). **19.** From nowhere but the count: at extension $x$ there are fewer [microstates](#def-b3-microcanonical-ensemble-states) than at $x - \dd x$, and thermal agitation drifts the chain toward the bigger count; the “force” is $T$ times the entropy gradient. **20.** At $T = 0$ nothing explores configurations: the entropic force vanishes with the agitation that powers it — elasticity would be purely energetic, and rubber would behave like a limp thread. **21.** Gas: $P = T\,\partial S/\partial V$ with $S \ni Nk_{\text{B}}\ln V$; chain: $f = -T\,\partial S/\partial x$ with the Gaussian count — both are counting pushed against a [constraint](https://one-course.com/books/physics/5/en/chapter/1-lagrangian-mechanics#def-b3-lagrangian-mechanics-coordinates), priced at $T$. **22.** The dryer’s heat enters the band; part is converted to work in the isothermal contraction, the books balanced by the entropy carried in and out — a miniature heat engine, first law intact. **23.** Finite extensibility (the real force diverges near full stretch — the worm-like-chain refinement) and strain-induced crystallisation (alignment orders the chains, giving hysteresis and heat release beyond the ideal model). **24.** $k_{\text{B}}T/\mathrm{nm} = 4.1\,\mathrm{pN}$: the force at which thermal energy and nanometre displacements trade evenly — molecular motors, folding proteins and stretched DNA all operate within an order of magnitude of it. **25.** Ten thousand blind links, counted: [Hooke’s law](https://one-course.com/books/physics/5/en/chapter/3-continuum-mechanics-and-elasticity#thm-b3-continuum-elasticity-hooke) with $k = k_{\text{B}}T/Nb^2$, warmth on stretching, weights lifted by hot air, and single molecules obeying at $0.04\,\mathrm{pN}$ — entropy, acting as a force.