Boltzmann Entropy

Equivalent forms

S = -k_B \sum_i p_i \ln p_i

W = e^{S / k_B}

A single logarithm turns combinatorics into thermodynamics — the deepest bridge between the microscopic and macroscopic worlds.

Symbol	Name	SI unit	Dimension	Range
$S$	Entropyoutput Thermodynamic entropy of the system	J/K	M·L²·T⁻²·Θ⁻¹	0 – 1e-20
$W$	Number of microstates Count of equally probable microscopic configurations	dimensionless	1	1 – 1e+25

Unit systems

Symbol	SI	CGS	Imperial
$S$	J/K	erg/K	BTU/°R
$W$	dimensionless	dimensionless	dimensionless

Where it holds

Valid for isolated equilibrium systems with equally probable microstates. For non-uniform probability distributions use the Gibbs form

S = -k_B \Sigma p_i \ln p_i

.

Dimensional analysis

$[S] = [k_B][\ln W] = (J/K)$ (dimensionless $) = J/K \checkmark$

Discovery

Ludwig Boltzmann · 1877

Boltzmann linked thermodynamic entropy to the logarithm of the number of microstates, bridging mechanics and thermodynamics. The formula S = k log W is engraved on his tombstone in Vienna.

Try this

Why does shuffled cards never re-order themselves?

A system has 1.0e20 equally probable microstates. Compute its entropy in J/K.

Research status: stable

Real-world applications

Information theory (Shannon entropy is the same formula with $k_B \to 1/\ln 2)$
Black hole entropy via Bekenstein-Hawking formula
Protein folding thermodynamics
Lattice defect counting in semiconductor physics

Common misconceptions

Entropy is NOT 'disorder' — it's the log of phase-space volume; a crystal can have higher entropy than a glass at the same temperature
Boltzmann's W is a count of microstates, not a probability
Entropy is extensive only because W multiplies for independent subsystems

Experimental verification

Confirmed by residual entropy measurements (e.g.,

ice's S_{0} = 3.4\,\mathrm{J}/(mol\cdot K)

matches Pauling's microstate count) and third-law extrapolation in low-temperature calorimetry.

Derivation

Consider an isolated system with W equally probable microstates.

Demanding entropy be additive for independent systems

(S_{12} = S_{1} + S_{2})

but microstates multiplicative

(W_{12} = W_{1}\cdot W_{2})

forces

S = k\cdot \ln (W)

.

The constant k is fixed by matching dQ/T from classical thermodynamics, giving Boltzmann's constant k_B.

Limiting cases

W = 1

⟶

S = 0

A perfectly ordered single-microstate system has zero entropy — the third law of thermodynamics.

W \to \infty

⟶

S \to \infty

Infinite microstates correspond to infinite disorder.

What if…

What if W doubles?

Entropy increases by $k_B\cdot \ln 2 \approx 9.57e-24\,\mathrm{J/K}$ — a tiny but universal increment per bit of information.

What if all microstates collapse to one

(W=1)

?

Entropy hits zero. This is the third law of thermodynamics: a perfect crystal at absolute zero has zero entropy.

What if microstates aren't equally probable?

Use the Gibbs form $S = -k_B \Sigma p_i \ln p_i$ , which reduces to Boltzmann's form when $all p_i = 1/W$ .

1

Entropy of 1e20 microstates

Given ·

W:: 100000000000000000000

Find · S

Steps

Compute ln W: $\ln (1e20) = 20\cdot \ln (10) \approx 46.05$
Multiply by k_B: $S = 1.381e-23 \times 46.05$
$S \approx 6.36 \times 10^{-22} J/K$

Result ·

S = k_B \ln W = 1.381e-23 \times \ln (1e20) = 1.381e-23 \times 46.05 = 6.36e-22\,\mathrm{J/K}

2

Entropy of 1 mole of ideal gas microstates

Given ·

W:: 1e+25

Find · S

Steps

$\ln (1e25) = 25\cdot \ln (10) = 57.56$
$S = 1.381e-23 \times 57.56$
$S \approx 7.95 \times 10^{-22} J/K$

Result ·

S = 1.381e-23 \times \ln (1e25) = 1.381e-23 \times 57.56 \approx 7.95e-22\,\mathrm{J/K}

Entropy Change→First Law of Thermodynamics→Ideal Gas Law→