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Microstates and macrostates, the multiplicity of a two-state system as a binomial coefficient, the fundamental assumption, and the two-state paramagnet.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to tell a microstate from a macrostate, find the multiplicity of a macrostate of coins or of a two-state paramagnet, turn an energy into a count, and find the probability of a macrostate from the fundamental assumption.
You can count arrangements: the number of ways to choose $n$ things from $N$ is the binomial coefficient $\binom{N}{n} = N!/(n!(N - n)!)$, and the probability of an outcome is the number of favorable cases over the total when all cases are equally likely. This lesson uses exactly that arithmetic to lay the foundation of statistical mechanics.
| Term | What it means |
|---|---|
| Microstate | A complete description of every particle: which coins show heads, which dipoles point up. |
| Macrostate | A description by the measurable totals only: how many heads, how much energy. |
| Multiplicity | $\Omega$, the number of microstates in a macrostate. |
| Fundamental assumption | In an isolated system in equilibrium, every accessible microstate is equally probable. |
| Binomial coefficient | $\binom{N}{n} = \dfrac{N!}{n!\,(N - n)!}$, the number of ways to choose $n$ of $N$. |
| Two-state system | A system of $N$ independent parts, each with exactly two states. |
| Two-state paramagnet | $N$ magnetic dipoles in a field $B$, each with energy $-\mu B$ along the field or $+\mu B$ against it. |
| Magnetic moment | $\mu$, the strength of a dipole: its energy in a field $B$ is $\mp\mu B$. |
Toss three coins. There are eight possible results, from HHH to TTT, and each is a microstate: a full specification of every coin. Usually we care only about the macrostate — how many heads — and the macrostates are not equally populated:
| Heads | Microstates | Multiplicity |
|---|---|---|
| 3 | HHH | 1 |
| 2 | HHT, HTH, THH | 3 |
| 1 | HTT, THT, TTH | 3 |
| 0 | TTT | 1 |
The number of microstates in a macrostate is its multiplicity $\Omega$. For $N$ two-state objects with $n$ in one state,
$$\Omega(N, n) = \binom{N}{n} = \frac{N!}{n!\,(N - n)!},$$
and the total over all macrostates is $2^N$.
Statistical mechanics rests on one assumption, the fundamental assumption: in an isolated system in equilibrium, every accessible microstate is equally probable. It cannot be derived; it is justified by the fact that the collisions and interactions inside a real system shuffle it through its microstates without favoring any. With it, probability is just counting:
$$P(\text{macrostate}) = \frac{\Omega(\text{macrostate})}{\Omega(\text{all})}.$$
The physical example is a two-state paramagnet: $N$ magnetic dipoles in a field $B$, each pointing either along the field (energy $-\mu B$) or against it ($+\mu B$). A macrostate is fixed by its total energy $U = \mu B(N_{\downarrow} - N_{\uparrow}) = \mu B(N - 2N_{\uparrow})$, and its multiplicity is $\binom{N}{N_{\uparrow}}$ — the same arithmetic as the coins, now attached to an energy.
Another way: picture
Picture a class of thirty students each flipping a coin at once, and the teacher writing on the board only the number of heads. Every student's individual result is a microstate; the number on the board is the macrostate. Day after day the number hovers around fifteen, and a board reading thirty never appears — not because all heads is forbidden, but because only one arrangement of thirty coins produces it, while about 155 million produce exactly fifteen.
Another way: steps
To build a microstate with $n$ heads among $N$ coins, pick which coins are heads. Pick them one at a time: $N$ choices for the first, $N - 1$ for the second, down to $N - n + 1$ for the last, a product of $N!/(N - n)!$. But picking coin 3 then coin 7 gives the same microstate as picking 7 then 3; each set of $n$ coins has been counted once for every order it can be picked in, which is $n!$ times. Dividing out gives
$$\Omega = \frac{N!}{n!\,(N - n)!}.$$
Two consequences are worth seeing. First, $\binom{N}{n} = \binom{N}{N - n}$: choosing which coins are heads is the same as choosing which are tails, so the multiplicities are symmetric about $n = N/2$. Second, the peak at $n = N/2$ grows enormously as $N$ grows while its relative width shrinks. For $10$ coins the middle macrostate holds $252$ of the $1024$ microstates, a quarter. For $100$ coins it holds $1.01 \times 10^{29}$ of $1.27 \times 10^{30}$, about $8$ percent — smaller as a share, but all the probability is now crowded into counts between about $40$ and $60$. For $10^{23}$ particles the crowding is so extreme that the most probable macrostate is, for every practical purpose, the only one ever seen.
Checks. A multiplicity is a positive whole number. The multiplicities of all the macrostates must add up to $2^N$; for small $N$ that is a quick, complete check, and it is Pascal's triangle row by row. The symmetry $\binom{N}{n} = \binom{N}{N - n}$ catches a reversed count. And a probability for a single count can never exceed the probability of the most likely count, which for $N$ coins is roughly $\sqrt{2/(\pi N)}$ — about $0.25$ for ten coins, $0.08$ for a hundred.
The paramagnet makes the connection to physics explicit. Suppose $100$ dipoles start all aligned with the field, a macrostate of multiplicity $1$, and are then allowed to exchange energy with their surroundings and with each other so that they wander among the microstates of a given energy. Nothing forces them toward disorder. But if the energy allows, say, anywhere from $40$ to $60$ dipoles along the field, the macrostates in that range hold over $96$ percent of all the microstates, and a random walk through microstates will spend $96$ percent of its time there.
That is the whole idea of the second law in miniature: a system moves toward the macrostates with the most microstates simply because there are more of them to be in. With $10^{23}$ particles the imbalance is so large that the reverse — all the air in a room gathering in one corner, a stirred cup of coffee unmixing its cream — is not forbidden but has a probability so small that it would not be seen in many lifetimes of the universe. The next lessons make this precise by giving the logarithm of the multiplicity a name: entropy.
Every multiplicity in this lesson can be read off Pascal's triangle, in which each entry is the sum of the two above it. The rule is the counting argument in miniature: a macrostate of $N$ coins with $n$ heads either has the last coin heads, leaving $n - 1$ heads among the first $N - 1$, or tails, leaving $n$ heads among them. So $\binom{N}{n} = \binom{N - 1}{n - 1} + \binom{N - 1}{n}$.
Writing out the first few rows — $1$; $1, 1$; $1, 2, 1$; $1, 3, 3, 1$; $1, 4, 6, 4, 1$; $1, 5, 10, 10, 5, 1$ — gives a table to check any small count against, and each row adds up to $2^N$. It also shows the peak growing in the middle and the ends staying at $1$: there is always exactly one way to have all heads, however many coins there are, while the middle entries grow almost as fast as $2^N$ itself.
Place $N$ gas molecules in a box and draw an imaginary line down the middle. Each molecule is on the left or the right: a two-state system. If the molecules move freely, every arrangement is equally likely, so the probability that all $N$ are on the left at a given instant is $1/2^N$.
For $N = 10$ that is $1/1024$: take a snapshot every second and you would see it about once in seventeen minutes. For $N = 100$ it is $7.9 \times 10^{-31}$: a snapshot every nanosecond since the Big Bang, about $4 \times 10^{26}$ of them, would still be millions of times too few to expect one. A thimble of air holds about $2.5 \times 10^{19}$ molecules, and $2^{-2.5 \times 10^{19}}$ is a number with some seven quintillion zeros after the decimal point.
The same counting predicts the size of the fluctuations that do happen. The number on the left is spread about $N/2$ with a standard deviation of $\sqrt{N}/2$, so for $10^{19}$ molecules the imbalance is typically a few billion molecules, one part in a few billion. That is why a pressure gauge on a tank reads steadily: the fluctuations are real, but far too small to measure.
An MRI scanner images the body by working with a real two-state system: the protons in its hydrogen atoms, whose magnetic moments point either along or against the scanner's field. In a $1.5$ T clinical scanner the energy difference between the two states, $2\mu B$, is about $4 \times 10^{-26}$ J, while $kT$ at body temperature is $4.3 \times 10^{-21}$ J, a hundred thousand times larger.
So the thermal jostling almost completely randomizes the protons, and the macrostate is barely tilted toward alignment: the excess of aligned protons is about $\mu B/kT \approx 5$ per million. The next unit will derive that number from the Boltzmann factor. The imaging works anyway because a cubic millimeter of tissue contains about $7 \times 10^{19}$ hydrogen nuclei, and five per million of that is still $3 \times 10^{14}$ excess aligned moments — enough to produce a measurable signal when radio pulses flip them. It is also why stronger magnets, $3$ T and $7$ T research scanners, give sharper images: doubling $B$ doubles the tiny imbalance the signal comes from.
Asked whether HHHHHH or HTTHTH is more likely from six fair coins, most people choose the second: it looks random. But each is a single microstate, and every microstate has probability $1/64$. What the intuition is really tracking is the macrostate: "three heads" contains twenty sequences and "six heads" contains one, so a count near half is twenty times likelier than all heads — even though no particular sequence is.
Keeping the two levels apart is the discipline this whole unit needs. The fundamental assumption is about microstates: all equally likely. Everything interesting — why systems drift toward equilibrium, why heat flows from hot to cold — comes from macrostates containing wildly different numbers of them. A related trap is the gambler's fallacy: after five heads, a sixth toss is still fifty-fifty, because the coin has no memory of the macrostate it is building.
List the macrostates of three coins by the number of heads.
$n = 0, 1, 2, 3$
Four possible counts.
Find the multiplicity of three heads and of none.
$\Omega(3) = \binom{3}{3} = 1, \qquad \Omega(0) = \binom{3}{0} = 1$
Only one sequence gives all heads, and only one all tails.
Find the multiplicity of two heads.
$\Omega(2) = \dfrac{3!}{2!\,1!} = \dfrac{6}{2} = 3$
The one tail can be any of the three coins.
Find the multiplicity of one head by symmetry.
$\Omega(1) = \binom{3}{1} = \binom{3}{2} = 3$
Choosing the heads is the same as choosing the tails.
Check the total and find the probabilities.
$1 + 3 + 3 + 1 = 8 = 2^3, \qquad P(2) = \dfrac{3}{8}$
The multiplicities account for every microstate, so each probability is its share of eight.
Find the total number of microstates of ten coins.
$\Omega_{\text{all}} = 2^{10} = 1024$
Each coin doubles the count.
Write the multiplicity of exactly five heads.
$\Omega(5) = \binom{10}{5} = \dfrac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}$
Cancel $5!$ from $10!$ first, leaving five factors on top.
Evaluate the binomial coefficient.
$\Omega(5) = \dfrac{30\,240}{120} = 252$
The largest multiplicity of any count of ten coins.
Find the probability of exactly five heads.
$P(5) = \dfrac{252}{1024} \approx 0.246$
Even the most likely count happens less than a quarter of the time with only ten coins.
Find the multiplicity of exactly seven heads.
$\Omega(7) = \binom{10}{3} = \dfrac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$
Seven heads is three tails, so use the smaller of the two.
Compare the two macrostates.
$\dfrac{\Omega(5)}{\Omega(7)} = \dfrac{252}{120} = 2.1$
Five heads is about twice as likely as seven: with ten coins the peak is broad.
$100$ dipoles sit in a field; $60$ point along it. Find the energy.
$U = \mu B(N - 2N_{\uparrow}) = \mu B(100 - 120) = -20\mu B$
Sixty contribute $-\mu B$ each and forty contribute $+\mu B$.
Write the multiplicity of this macrostate.
$\Omega(60) = \binom{100}{60} = \dfrac{100!}{60!\,40!}$
Choose which sixty dipoles point along the field.
Evaluate it with a calculator.
$\Omega(60) \approx 1.38 \times 10^{28}$
Far too many to list, but perfectly definite.
Find the multiplicity of the half-aligned macrostate.
$\Omega(50) = \binom{100}{50} \approx 1.01 \times 10^{29}, \qquad U = 0$
Fifty along and fifty against: zero energy and the largest multiplicity.
Compare the two.
$\dfrac{\Omega(60)}{\Omega(50)} \approx \dfrac{1.38 \times 10^{28}}{1.01 \times 10^{29}} \approx 0.14$
Ten extra aligned dipoles cut the multiplicity sevenfold.
Find the probability of the half-aligned macrostate if every microstate is allowed.
$P(50) = \dfrac{1.01 \times 10^{29}}{2^{100}} = \dfrac{1.01 \times 10^{29}}{1.27 \times 10^{30}} \approx 0.080$
The single most likely count still has only an eight percent chance.
Find the probability of the fully aligned macrostate.
$P(100) = \dfrac{1}{2^{100}} \approx 7.9 \times 10^{-31}$
Not forbidden, but a random shuffle through microstates would essentially never find it; this imbalance is what the second law will rest on.
Count all the microstates.
$2^6 = 64$
Two outcomes per coin.
Find the multiplicity of two heads.
$\binom{6}{2} = \dfrac{6 \times 5}{2 \times 1} = 15$
Choose which two coins are heads.
Divide the multiplicity by the total.
$5$ fair coins are tossed one after another. Which is more likely: that every coin comes up heads, or one particular mixed sequence that starts heads, tails, tails, heads?
Complete the worked solution: what is the probability of exactly $3$ heads when $9$ fair coins are tossed?
Evaluate the binomial coefficient.
$\dfrac{9!}{3!\,6!} =$ w
It counts the ways to choose which coins show heads.
Count all the microstates.
$2^{9} =$ t
Each coin doubles the number of possible sequences.
Divide the multiplicity by the total.
$P = \dfrac{\Omega}{\Omega_{\text{all}}} =$ p
All microstates of fair coins are equally likely.
Match each macrostate of $7$ fair coins to its multiplicity.
| $1$ | $7$ | $21$ | $128$ | |
|---|---|---|---|---|
| no heads | ||||
| exactly one head | ||||
| exactly two heads | ||||
| any number of heads: every microstate |
A paramagnet of $6$ dipoles sits in a field. Fill in the multiplicity and the energy, in units of $\mu B$, for each number of dipoles along the field.
| multiplicity | energy ($\mu B$) | |
|---|---|---|
| none along | ||
| one along | ||
| two along | ||
| three along |
$8$ fair coins are tossed. What is the probability of exactly $3$ heads? Give it as a decimal.
Answer:
An ideal two-state paramagnet has $14$ dipoles in a magnetic field. Each dipole pointing along the field has energy $-\mu B$ and each pointing against it $+\mu B$. The total energy is $U = 2\mu B$. How many microstates does this macrostate have?
Answer:
A detector watches exactly $10$ radioactive nuclei for one half-life. Each has probability one half of decaying in that time, independently of the others. What is the probability that exactly $5$ of them decay? Give it as a decimal.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A paramagnet of $7$ dipoles sits in a field. Fill in the multiplicity and the energy, in units of $\mu B$, for each number of dipoles along the field.
| multiplicity | energy ($\mu B$) | |
|---|---|---|
| none along | ||
| one along | ||
| two along | ||
| three along |
You can count the microstates of a two-state system and turn counts into probabilities. Explain to someone why every sequence of six coin tosses is equally likely, yet three heads is far likelier than six.
15. Your turn: six fair coins are tossed. What is the probability of exactly two heads?, step 3
$P = \dfrac{15}{64} \approx 0.234$
Multiplicity over the total.