advanced32 min studyLesson 324 in Physics

Turn a state sum into an energy integral

Derive density-of-states normalization and use a Laplace transform to distinguish available states from occupied states.

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What you’ll understand

  • Derive the three-dimensional free-particle density of states with internal degeneracy declared.
  • Normalize a continuum energy probability density.
  • Use Gamma integrals to obtain energy moments and their limits.

An energy interval near one joule may contain a different number of states from an equal-width interval near two joules. The Boltzmann factor assigns a weight to each state, while the density of states counts how many such states are available. Their product determines an energy distribution. We will derive this counting factor from momentum space, keeping its units and internal degeneracy visible, and then use it to calculate partition integrals and energy moments.

Count momentum cells and change variables

In a large box with periodic boundaries, allowed wavevectors occupy a cubic grid with spacing two pi divided by side length. The number of states in a thin spherical shell is its momentum-space volume divided by the cell volume, multiplied by any independent internal degeneracy. For a free nonrelativistic particle energy is momentum squared divided by twice the mass. Converting the shell thickness to an energy interval introduces a Jacobian that changes the power of energy.

In three dimensions the one-particle density of states is proportional to square root of energy. Its integral below an energy cutoff grows as energy to the three-halves power. These are different functions. Neither is a probability density until an ensemble weight and normalization are supplied. The continuum approximation neglects the discreteness of low box levels and boundary corrections. It works when many relevant levels lie within thermal energy scales, but a special ground state may still need separate treatment.

g1(E)=gsV4π2(2mℏ2)3/2E,N1(E)=∫0Eg1(E′) dE′g_1(E)=\frac{g_sV}{4\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2}\sqrt E,\qquad \mathcal N_1(E)=\int_0^E g_1(E^{\prime})\,dE^{\prime}
g₁ has units J⁻¹, gs is an integer internal degeneracy, V is in m³, m in kg, and E≥0 is kinetic energy in joules. The formula is for a large three-dimensional free-particle box.↔ Scroll wide equations horizontally.

The partition function is a weighted count

The one-particle canonical partition function is the integral of density of states times the Boltzmann factor. Dividing that same product by the partition function gives a normalized energy probability density. High energy has more available states but a smaller weight per state. The balance can place the most probable energy away from the ground even though every individual high-energy state is less likely than a lower-energy state. Confusing a state probability with an energy-band probability misses this degeneracy effect.

Suppose more generally that the relevant state density is a constant times energy to the power s minus one. The partition integral is a Gamma integral and scales as beta to the power minus s. The mean energy is s times thermal energy, and its variance is s times thermal energy squared. The assumptions are a continuum extending from zero upward and a positive s. Extra thresholds, gaps, finite upper limits, or energy-dependent internal structure modify the integral.

Z1=∫0∞g1(E)e−βE dE,pE(E)=g1(E)e−βEZ1,g1(E)=CEs−1 ⟹ Z1=CΓ(s)β−sZ_1=\int_0^\infty g_1(E)e^{-\beta E}\,dE,\qquad p_E(E)=\frac{g_1(E)e^{-\beta E}}{Z_1},\qquad g_1(E)=CE^{s-1}\ \Longrightarrow\ Z_1=C\Gamma(s)\beta^{-s}
Z₁ is dimensionless and pE has units J⁻¹, so pE dE is a probability. C carries the units needed for g₁; s>0 and β>0 ensure the idealized integral converges.↔ Scroll wide equations horizontally.

Energy moments reveal what the continuum retained

For one free particle in three dimensions s equals three halves. The mean energy is therefore three halves times thermal energy, while the most probable kinetic energy is only one half times thermal energy. The mode depends on the variable used: a momentum distribution and a speed distribution have different Jacobians and therefore different maxima. A most probable value cannot be transferred between variables without changing the probability measure.

For N independent dilute particles, the one-particle partition function builds the fixed-number many-particle function with its Gibbs factor. That factor affects free energy but not the temperature derivative of translational energy. The quadratic lab can illustrate the kinetic-energy mapping from one momentum component to energy, including its changing slope. It cannot count three-dimensional momentum shells or render the resulting Gamma energy distribution; those are supplied by the derivation and the partition integral.

⟨E⟩=sβ=skBT,Var⁡(E)=sβ2,Emode=(s−1)kBT(s>1)\langle E\rangle=\frac{s}{\beta}=sk_BT,\qquad \operatorname{Var}(E)=\frac{s}{\beta^2},\qquad E_{\mathrm{mode}}=(s-1)k_BT\quad(s>1)
These are for the normalized continuum density proportional to E^(s−1)e^(−βE), with a temperature-independent spectrum. For one three-dimensional free particle s=3/2; moments are in J and J².↔ Scroll wide equations horizontally.

Evaluate a simple continuum partition integral

An illustrative system has state density g(E)=E/ε₀² for E≥0, where ε₀ is a fixed energy scale. At kBT=2ε₀, find Z, mean energy, variance, and the mode of its energy distribution.

  1. This density has s=2 and C=1/ε₀². Since Γ(2)=1, Z=(kBT)²/ε₀²=4, a dimensionless weighted state count.

  2. The mean is s kBT=4ε₀. The variance is s(kBT)²=8ε₀², not 8ε₀; a fluctuation variance has squared-energy units.

  3. The mode is (s−1)kBT=2ε₀. It lies below the mean because the distribution has a long positive-energy tail. The model is a declared density of states, not the three-dimensional free-particle square-root law.

The result

Z=4, mean energy=4ε₀, variance=8ε₀², and mode=2ε₀. The distinction between density, probability, mean, and mode remains explicit.

Explore the model

Interactive lab

Interactive lab · function

Inspect the kinetic-energy change of variables

For fresh runs Reset and set BOTH coefficient a=0.5, position x=1; then Reset and set a=0.5, x=2. Treat x as normalized momentum and y as one-component kinetic energy. Predict y=0.5 and 2 and slopes 1 and 2. This parabola is not a density-of-states or partition-integral solver.

Inspect the kinetic-energy change of variablesThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. 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 forThe energy map is quadratic, so changing variables requires a nonconstant Jacobian.

Check your understanding

For one nonrelativistic free particle in a large three-dimensional box, which statement describes g₁(E)?

Try it yourself

A continuum system has constant density g(E)=1/ε₀ for E≥0. At kBT=3ε₀, find Z, mean energy, and variance. Is its energy distribution maximal at a positive 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

Use s=1 and recall that its density is an exponential beginning at zero.

Compare with the solution

Z=kBT/ε₀=3. The mean is 3ε₀ and variance is 9ε₀². The normalized distribution decreases monotonically with energy, so its maximum is at E=0, not at its mean. The s>1 interior-mode formula is replaced by the boundary maximum when s=1.

Take it with you

  • A density of states counts available states per energy interval.
  • Boltzmann weighting and normalization turn state density into an energy distribution.
  • Jacobians determine both density powers and the location of distribution maxima.

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