A coarse description can represent many microscopic arrangements. Counting explains why some macrostates are more likely.
By the end: distinguish a state from a category of states and interpret entropy as a count under explicit assumptions.
Specify what you count
A macrostate describes coarse properties such as energy, volume, or how many particles occupy each part of a box. A microstate supplies a more detailed configuration consistent with those properties. Our deliberately small model labels four particles and records only whether each is in the left or right half. LLRR means particles one and two are left while three and four are right. Real gas microstates also include positions and momenta; these sixteen assignments are not a complete gas phase space.
Assume each labeled particle independently has equal probability of being in either equal-sized half. There are two choices per particle, giving sixteen equally likely assignments for four particles. The macrostate “two left” contains six assignments, while “all four left” contains one. A category can be more likely without any one of its individual assignments being more likely.
Choose the detail before assigning probabilities
A box divided into equal halves can support many descriptions of the same gas. One description gives the total number of particles on each side. Another records which labeled particles are left. A still finer classical description gives every particle's position and momentum. The lesson's four-letter assignments occupy the middle level of detail. They deliberately ignore positions within a half and all momenta. Calling them detailed assignments is useful only relative to the coarser occupancy count, not relative to a complete real gas state.
Take the assignment LLRR and draw particles 1 and 2 in the left compartment, with 3 and 4 in the right compartment. Moving particle 1 within the left half changes its position but leaves the assignment unchanged. Exchanging the sides occupied by particles 2 and 3 changes the assignment to LRLR, yet both descriptions still have two particles left. The drawing thus separates two kinds of information loss. Occupancy ignores identities, while the four-letter record ignores the remaining spatial and motion details.
To assign equal weights to the sixteen records, assume each particle independently occupies either equal-sized half with probability one half. For LLRR, multiply four factors of one half, giving one sixteenth. The same multiplication works for LLLL. Equal probability follows from those symmetry and independence assumptions. It does not follow from how orderly a record looks to a reader. Interactions, an external force, unequal compartment volumes, or restrictions on allowed assignments could change the weights and require a different probability model.
Different records can share one occupancy
Particle numbers identify labels, not spatial coordinates. All three schematic records have two particles left; the first two even have the same letter assignment.
The event 'two left' does not identify a single assignment. Choose the two particle labels that appear as L. The possible pairs are 12, 13, 14, 23, 24, and 34. Each pair gives a different record with the same occupancy. Because these six mutually exclusive records each carry probability one sixteenth, their probabilities add to six sixteenths. No record becomes more probable merely by belonging to this category. The category is more probable because there are more ways to satisfy its description.
Contrast that event with 'all on one side.' This phrase includes both LLLL and RRRR, not just one of them. Its probability is two sixteenths, or one eighth. 'All left' alone has probability one sixteenth. A probability question therefore needs an exact event definition before any arithmetic. Changing 'left' to 'either side' changes the set of accepted records. It doubles the probability in this example without changing the particles, compartment sizes, or individual assignment weights in any way.
The labels are bookkeeping devices that make the small counting problem unambiguous. They are not a claim that identical quantum particles have permanent observable identity tags. Real statistical mechanics must treat identical-particle counting and the measure over accessible states consistently. The small assignment model can still teach how a coarse event collects alternatives. Its success at that task does not make sixteen a count of all positions, momenta, and energy possibilities available to four molecules in an actual gas.
Count the records accepted by an event
Each cell has weight 1/16 under the equal independent assignment assumptions. Filled cells name the accepted event; cell areas do not represent real phase-space volumes.
Relate multiplicity to entropy
Multiplicity Ω is the number of allowed microscopic alternatives for a specified macrostate under the counting model. For N independent left/right assignments with k on the left, choose which k labeled particles occupy that side. This gives the binomial count N choose k. For two left among four, there are four choices for the first selected particle and three for the second, but each pair was counted twice. Dividing twelve ordered choices by two gives six pairs. The general formula removes the ordering in the same way. Divide that count by the total number of assignments to get the macrostate probability. The display keeps every four-particle assignment visible.
Boltzmann’s entropy relation uses the natural logarithm of multiplicity, multiplied by kB. The logarithm matters because independent counts multiply while entropies add: if one subsystem has three alternatives and another has four, the combined count is twelve. The relative counting values here illustrate that structure; they are not calibrated absolute entropies for four real gas molecules. Equal probability of detailed states is an assumption tied to the physical constraints, not a rule for every probability model.
Derive the count and normalize it
For four particles with two left, selecting the first left label gives four possibilities and selecting a different second label gives three. That produces twelve ordered selections. But selecting 1 then 2 describes the same occupied pair as selecting 2 then 1. Every pair occurs twice, so divide twelve by two to obtain six. The division removes an artificial selection order. It does not remove actual assignments or halve their probability. The physical record never said which left particle was selected first.
For eight particles with four left, the ordered selection count is eight times seven times six times five, or 1,680. Each set of four labels occurs in 24 orders, since four times three times two times one equals 24. Dividing gives 70 distinct assignments with four left. The other four labels automatically occupy the right, so choosing a second group independently would overcount. The binomial formula packages this reasoning for general particle number and occupancy, including zero particles on one side.
There are 256 total eight-particle assignments, because eight independent binary choices give two multiplied by itself eight times. Exact balance therefore has probability 70 divided by 256, about 27.34 percent. The four-particle result was six divided by sixteen, or 37.5 percent. A larger numerator alone cannot decide which event is more likely when the total number of alternatives also changes. Here the exact-balance multiplicity grows by more than eleven times while the full assignment space grows by sixteen times.
Remove the artificial order of selection
Each cell is one ordered choice of two different labels. Mirrored cells represent the same unordered pair, leaving six pairs rather than twelve.
The entropy relation applies a logarithm to a count, not directly to this normalized probability. For the four-particle balanced category, the displayed value S divided by kB is ln 6, about 1.792. For an all-left category it is ln 1, which is zero. These are counting values within the chosen description and constraints. Zero here does not say that a real gas has zero thermodynamic entropy. Motion and positions within each compartment were absent from the counted records from the start.
Why use a logarithm? Consider two independent copies of the balanced four-particle category. Each has six assignments, so the combined category has 36 pairs of assignments. The logarithm obeys ln 36 equal to ln 6 plus ln 6, about 3.584. Independent counts multiply, while the associated counting entropies add. If a constraint couples the copies, some pairs may be forbidden and the product rule must be reconsidered. Additivity follows from independent accessible alternatives rather than from a visual impression of disorder.
Count and probability remain separate even when both use the same assignments. A category probability requires its count divided by the total count only when those assignments are equally weighted. If one side is energetically favored, the relevant probabilities need weighted sums instead. Conversely, changing the total accessible assignment space can change a category's probability without changing its internal count. Name the category and surrounding set before interpreting multiplicity, its logarithm, or a probability comparison.
Independent alternatives form a product
The 6 by 6 grid contains 36 pairs of balanced assignments. Row and column choices are independent; each subsystem contains four labeled particles.
Work through the mechanism
Probability of a coarse occupancy
N is the number of independently assigned labeled particles. C(N, k) counts ways to choose which k are left. Each complete assignment has probability 1/2ᴺ.
Count choices without counting their order
Factorial means multiply the positive integers up to that number; zero factorial is one. For N = 4 and k = 2, the denominator is the product of two and two. Dividing 24 by that product gives six.
Entropy from multiplicity
Ω is the count consistent with the specified constraints. kB = 1.380649 × 10⁻²³ J/K. The diagram reports S/kB for its coarse counting model, not an absolute entropy calibration of a real gas.
Keep all sixteen detailed assignments visible
Multiplicity 6 of 16; probability 37.5%. Counting entropy S/kB = ln 6 = 1.79.
Each letter refers to one labeled particle, in particle order 1–4. L and R record only which half contains it. Outlined blue cells match the selected left count. All assignments have equal weight under the independent equal-half assumption; selecting a count does not run a random simulation.
Particles on left
Assignments
Probability
0
1
1/16
1
4
4/16
2
6
6/16
3
4
4/16
4
1
1/16
All sixteen assignments have probability 1/16. Macrostate probabilities sum to one.
Explore a category without changing the random model
In the existing assignment display, select 'All left' and inspect which record receives the outline. There should be one, LLLL. Now select 'Balanced: two left' and identify the six records by their pairs of left labels. The outlined count changes, but every record stays visible. This action changes the event being examined, not the underlying equal weights. It neither moves particles through a physical trajectory nor repeatedly samples random outcomes. You are comparing subsets of a fixed set of sixteen assignments.
Next select 'One left, three right.' Predict four outlined records before checking the display, because any one of the four labels can be the left particle. Selecting 'Three left, one right' also gives four, since choosing the sole right label determines the rest. These equal counts arise from exchanging L and R everywhere. With equal halves and no left-right preference, that exchange preserves each assignment's weight. The symmetry predicts matching probabilities for occupancies equally far from the central count.
Add the category probabilities across all five possible left counts. Their numerators are one, four, six, four, and one, with the common denominator sixteen. The sum is sixteen sixteenths, exactly one. This check works because every assignment belongs to one occupancy category and none belongs to two. By contrast, 'at least one left' and 'at least one right' overlap. Adding their probabilities without subtracting their shared assignments would count the mixed records twice and can produce a result above one.
Use the display to calculate that overlap explicitly. 'At least one left' excludes only RRRR and has probability fifteen sixteenths. 'At least one right' excludes only LLLL and also has probability fifteen sixteenths. Their overlap contains the fourteen mixed assignments. Subtract fourteen sixteenths from their sum and the union has probability one, as expected. This is a useful distinction between adding disjoint occupancy categories and adding verbal events whose boundaries may cross, even though both calculations start with the same sixteen records.
Mixed records belong to both events
The schematic regions contain 1, 14, and 1 mutually exclusive assignments. Region area is not proportional to probability. Counts, not overlap geometry, supply the probabilities.
For another check, compute the average left count rather than the most probable count. Multiply each occupancy by its probability and add. The numerator is zero times one plus one times four plus two times six plus three times four plus four times one, which totals 32. Dividing by sixteen gives two particles left on average. That average does not mean every assignment is balanced. Ten of the sixteen records are unbalanced even though the mean occupancy equals the central category's count.
Translate that average into a physical distinction. The mean left fraction is one half, while a single record can place all particles right or all particles left. A larger collection can concentrate near its mean without ever sitting permanently at that mean. The display supplies no collision times or sequence of visits, so it cannot determine how frequently a moving gas changes categories per second. It supplies the probability structure under the stated assumptions, which is different from a dynamical law for reaching equilibrium.
A mean of two does not require balance
Bar heights count assignments on a zero-based scale. The mean left count is two; the ten assignments outside the central bar still have nonzero probability.
Explain typical behavior without a one-way microscopic rule
A macroscopic gas has an enormous number of ways to be spread fairly evenly and comparatively few ways to occupy a tiny region. Starting from an unusually constrained state, removing a partition opens many more accessible configurations. This makes a spread-out state typical without requiring each microscopic trajectory to increase a coarse entropy at every instant. Fluctuations remain possible, especially in small systems.
Do not replace this argument with visual “messiness.” You must specify the constraints and the states being counted. Nor does increasing particle number necessarily increase the probability of exactly half being left. Four particles give six out of sixteen; eight give seventy out of 256, a smaller exact-half probability. What concentrates with large N is the fraction near one half, over an appropriate band, rather than necessarily the single exactly balanced count.
Compare exact balance with a band of fractions
Define 'near balance' precisely before comparing different particle numbers. Take a left fraction between one quarter and three quarters, including both endpoints. For four particles, that means left counts one, two, or three. Their multiplicities add to fourteen, giving probability fourteen sixteenths, or 87.5 percent. Exact balance accepts only the central six records. The model assigns different probabilities to equality and a band. Neither result contradicts the other because the events have different boundaries.
For eight particles, that same fractional band accepts left counts two through six. The corresponding multiplicities are 28, 56, 70, 56, and 28. They sum to 238, so the band probability is 238 divided by 256, about 92.97 percent. Exact balance meanwhile falls to about 27.34 percent. Band probability can grow while exact-equality probability falls. A fixed interval of fractions includes more integer occupancies as particle number increases, even while the distribution narrows in fractional terms.
Absolute fluctuations and fractional fluctuations also differ. For independent equal left-right assignments, the average left count is N divided by two. The standard deviation measures spread about that average. It equals the square root of N divided by two. Dividing by N gives a fractional standard deviation of one divided by twice the square root of N. Four particles have count spread one and fractional spread one quarter. Sixteen particles have count spread two but fractional spread one eighth, so a larger absolute spread accompanies tighter fractional concentration.
The same fractional band accepts more outcomes
Each panel has its own normalized count distribution. Horizontal position is left fraction, and the shaded band includes fractions from 1/4 through 3/4. Heights are probabilities.
Removing a partition changes accessible states rather than forcing each individual trajectory toward a fixed occupancy at every instant. Initially constraining all four labeled particles left permits only LLLL in this coarse assignment description. Removing that constraint permits sixteen records. Balanced occupancy is one category, not the entire newly accessible space. The count ratio sixteen to one concerns releasing the constraint. The multiplicity six concerns conditioning on a particular final occupancy. Confusing those questions would give different entropy comparisons without noticing why.
For a macroscopic dilute ideal gas expanding into twice its original volume at unchanged energy, the position contribution gives an entropy increase proportional to particle number times kB ln 2. Real phase space also includes momentum, and absolute classical counting needs a consistent measure and identical-particle treatment. The simple assignment example explains the factor of two in newly accessible positional alternatives. It cannot calibrate the gas's absolute entropy or replace the reversible reference calculation used for thermodynamic endpoint differences elsewhere in the course.
Removing a constraint opens alternatives
Schematic accessible-set diagram. Initially only LLLL is allowed in this coarse record. After release all sixteen assignments are allowed, of which six are balanced.
An isolated total can relax toward more probable coarse conditions while its energy remains constant. That statement does not require every subsystem's entropy to increase, nor does it prohibit small fluctuations. Compressing a gas or sorting its occupants can reduce the subsystem's accessible alternatives if the surroundings change too. The second-law account belongs to the complete isolated total, including whatever caused the restriction. Counting establishes typicality under stated weights and constraints. It supplies neither an exception to energy conservation nor a relaxation clock.
Exercises
Compare a configuration with a category
Which is more likely in the four-particle model: the single assignment LLRR or the single assignment LLLL? How does that differ from comparing “two left” with “all left”?
Hint
Count how many assignments each phrase names.
Worked answer
Event
Number of assignments
Probability
LLRR alone
1
1/16
LLLL alone
1
1/16
Any two particles left
6
6/16
An event containing more equiprobable states has more probability.
The two named assignments are equally likely, each 1/16. The macrostate “two left” contains six assignments and has probability 6/16. “All left” contains one and has probability 1/16. Confusing a representative configuration with its whole category produces the wrong comparison.
Test the exact-balance claim
Eight independent particles have 70 assignments with exactly four on each side. Compare their exact-balance probability with the four-particle case. Does the larger multiplicity alone settle the comparison?
Hint
The total number of assignments changes too.
Worked answer
Model
Balanced assignments
All assignments
Exact balance
4 particles
6
16
37.5%
8 particles
70
256
27.34%
Normalize before comparing multiplicities from different outcome spaces.
For four particles the probability is 6/16 = 37.5%. For eight it is 70/256 ≈ 27.34%. The numerator grew, but the denominator grew faster. This does not contradict concentration of the fractional occupancy near one half as N grows. Exact balance and a band of near-balanced outcomes are different events.
Sources and further reading
These optional references support the stated laws and model assumptions. The numerical examples and diagrams are authored teaching models, not experimental measurements or engineering specifications. Selected relevant passages were reviewed, not entire textbooks or a comprehensive research literature.
Selected free-expansion entropy calculation and third-law discussion. This section uses qualitative disorder language; our counting model is instead supported by the separate Chemistry entropy section and explicit enumeration.