advanced32 min studyLesson 322 in Physics

Find temperature by sharing a fixed energy

Derive thermal equilibrium from state multiplicities and distinguish bounded-spectrum negative temperatures from ordinary cooling.

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What you’ll understand

  • Normalize a microcanonical shell and define its entropy convention.
  • Find the most probable energy division from an entropy derivative.
  • Explain when negative absolute temperature is possible and how energy flows.

An isolated composite system can have exactly fixed total energy while each of its parts has a fluctuating share. The equilibrium division is selected by how many joint microstates realize each share. Temperature emerges as the slope of entropy with energy, rather than an average particle speed imposed in advance. This construction also reveals why a bounded set of energy levels can support negative temperatures, while a free particle's unbounded kinetic energy cannot sustain the same canonical distribution.

Count a narrow energy shell

The microcanonical ensemble assigns equal probability to accessible microstates in a specified narrow energy interval at fixed volume and particle number. For discrete levels this means counting states in that interval; for classical phase space it means integrating a shell with the dimensionless measure. State the interval width before taking a logarithm. A density of states has inverse-energy units and cannot be used as a dimensionless multiplicity without a reference interval or an equivalent convention.

Here entropy means Boltzmann shell entropy. Differentiating its smooth large-system form at fixed volume and number defines inverse temperature. For a finite discrete system derivatives require smoothing or a finite-difference convention, so do not present a macroscopic derivative as an exact small-system measurement. Entropy defined from the cumulative volume below energy is a different finite-system convention. Its differences are particularly important for bounded spectra and must not be silently mixed with the shell convention used here.

W(E,ΔE)=∫EE+ΔEg(E′) dE′,SB=kBln⁡W,1T=(∂SB∂E)V,NW(E,\Delta E)=\int_E^{E+\Delta E}g(E^{\prime})\,dE^{\prime},\qquad S_B=k_B\ln W,\qquad \frac1T=\left(\frac{\partial S_B}{\partial E}\right)_{V,N}
W is a dimensionless shell multiplicity, g is a state density in J⁻¹, Sᴮ is in J/K, and ΔE is a specified narrow interval. The derivative assumes a smooth thermodynamic approximation to the shell entropy.↔ Scroll wide equations horizontally.

Maximize the product of multiplicities

Allow two weakly coupled systems to exchange energy without exchanging particles or changing volume. If the first has energy EA, the second has the fixed total minus EA. Their joint multiplicity is the product of their separate shell multiplicities. Taking its logarithm converts the product into an entropy sum. A stationary interior division has equal entropy slopes, hence equal temperatures. Weak coupling matters because assigning separate additive subsystem energies neglects interaction energy.

A stationary point must also be a maximum to describe stable thermal equilibrium. For ordinary positive heat capacities the entropy curvature is negative, giving a sharply peaked energy division in large systems. Temperature equality alone does not supply this stability check or the rate of equilibration. The statistical argument predicts an equilibrium distribution after accessibility and equilibration assumptions are met; it does not prove that an isolated integrable system explores every state on the relevant time scale.

P(EA)∝WA(EA)WB(Etot−EA),dStotdEA=1TA−1TB,SA′′+SB′′<0\mathcal P(E_A)\propto W_A(E_A)W_B(E_{\mathrm{tot}}-E_A),\qquad \frac{dS_{\mathrm{tot}}}{dE_A}=\frac1{T_A}-\frac1{T_B},\qquad S_A^{\prime\prime}+S_B^{\prime\prime}<0
EA and Etot are energies in joules. Each subsystem has fixed volume and number, coupling energy is negligible, and the last inequality tests stability of an interior entropy maximum.↔ Scroll wide equations horizontally.

A bounded spectrum can have decreasing shell multiplicity

Consider many fixed two-level sites with energies zero and epsilon. If K sites are excited, the energy is K times epsilon and the multiplicity is a binomial coefficient. The multiplicity rises to half excitation and then falls as the system approaches its fully excited state. In the large-site smooth approximation, the shell entropy therefore acquires a negative slope above half filling. This defines negative temperature within the specified bounded subsystem and shell-entropy convention.

A negative-temperature subsystem gives up energy when coupled to an ordinary positive-temperature reservoir. Its inverse temperature is below every positive inverse temperature, so it is hotter in the direction-of-energy-flow sense, not colder than zero kelvin. A free gas with unbounded kinetic energy cannot normalize a negative-beta Boltzmann weight. The binomial lab shows the bounded sites' degeneracy landscape at equal microstate weights; it is not a microcanonical dynamical evolution or a mechanism that prepares a population inversion.

WK=(NK),β=1kBT≃1ϵln⁡N−KK,0<K<NW_K=\binom NK,\qquad \beta=\frac1{k_BT}\simeq\frac1\epsilon\ln\frac{N-K}{K},\qquad 0<K<N
ε is a positive site energy gap in joules. The derivative formula uses Stirling smoothing for large K and N−K; β is in J⁻¹ and is negative for K>N/2.↔ Scroll wide equations horizontally.

Share energy between two smooth state counts

Two weakly coupled systems have shell multiplicities proportional to EA² and EB³ over their relevant positive-energy ranges. Their total energy is 10 J. Find the most probable split and confirm its entropy curvature is negative.

  1. The combined log multiplicity is 2 lnEA+3 ln(10−EA) plus constants. Its derivative is 2/EA−3/(10−EA).

  2. Set that derivative to zero: 2(10−EA)=3EA, so EA=4 J and EB=6 J. Their entropy slopes are both kB/2 J, giving equal temperatures.

  3. The second derivative of log multiplicity is −2/EA²−3/EB². At the split it is −2/16−3/36=−5/24 J⁻², strictly negative. The maximum is therefore stable within these assumed smooth multiplicities.

The result

The most probable division is 4 J and 6 J, rather than equal energies. Equal temperature fixes entropy slopes, and the negative curvature confirms an interior maximum.

Explore the model

Interactive lab

Interactive lab · probability

View a bounded multiplicity landscape

For fresh runs Reset and set BOTH p=0.5, trials n=10; then Reset and set p=0.5, n=20. Predict that equal-weight state counts peak near half excitation and are symmetric under K↔n−K. The exact binomial bars are proportional to multiplicities, not a fixed-energy microcanonical distribution or thermalization trajectory.

View a bounded multiplicity landscapeThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forMultiplicity decreases above half excitation, permitting a negative shell-entropy slope for a bounded spectrum.

Check your understanding

Within the declared large-system shell-entropy convention, a two-level population above half excitation has:

Try it yourself

A large two-level system has 75% of its sites excited and gap ε. Find βε in the smooth approximation and determine whether its canonical population favors higher or lower energy.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Insert the excited and unexcited fractions in the logarithmic slope formula.

Compare with the solution

βε=ln(0.25/0.75)=−ln3. The Boltzmann factor therefore increases with energy, favoring the excited level by a factor of three per microstate. Normalization is possible because each site has a bounded two-state spectrum. This negative-temperature equilibrium requires the bounded subsystem to remain appropriately isolated from unbounded positive-temperature degrees of freedom.

Take it with you

  • Thermal equilibrium maximizes the product of subsystem multiplicities at fixed total energy.
  • Equal temperature means equal entropy slopes, not equal energies.
  • Negative temperature requires appropriate bounded spectra and a declared entropy convention.

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