Macrostates and Microstates and their relations with Thermodynamic Probability

Here’s a concept that confused me for months in statistical physics: the difference between macrostates and microstates.

Textbooks make it sound complicated. It’s not. Once you see a concrete example, everything clicks.

Let me show you exactly how it works.

The Setup: Four Particles, Two Boxes

Imagine you have four distinguishable particles. Call them \( a \), \( b \), \( c \), and \( d \). You’re going to distribute them into two identical compartments.

Each particle has a 50/50 chance of landing in either compartment. Simple enough.

Now here’s the question: How many different ways can you arrange these particles?

The answer depends on what you mean by “different.”

Macrostates vs. Microstates

A macrostate describes the overall distribution. How many particles are in each compartment? That’s it. No names, just numbers.

A microstate describes the specific arrangement. Which exact particles are where?

Think of it this way:

• Macrostate: “2 particles in compartment 1, 2 in compartment 2”
• Microstate: “Particles \( a \) and \( b \) in compartment 1, particles \( c \) and \( d \) in compartment 2”

The macrostate tells you the summary. The microstate tells you the full story.

The kicker: Multiple microstates can correspond to the same macrostate. This is where thermodynamics gets interesting.

The Complete Example

Let’s enumerate everything for our four-particle system.

All Possible Macrostates

With four particles and two compartments, there are exactly five macrostates:

General rule: For \( n \) particles in two compartments, there are exactly \( n + 1 \) macrostates.

All Possible Microstates

Now let’s count the specific arrangements within each macrostate:

Total microstates: \( 1 + 4 + 6 + 4 + 1 = 16 = 2^4 \)

Notice something? That’s \( 2^n \) where \( n = 4 \). Not a coincidence. Each particle independently chooses one of two compartments: \( 2 \times 2 \times 2 \times 2 = 16 \).

General rule: For \( n \) particles in two compartments, the total number of microstates is \( 2^n \).

The Counting Formula

The number of microstates for macrostate \( (r, n-r) \) is given by the binomial coefficient:

$$\text{Number of microstates} = \binom{n}{r} = \frac{n!}{r!(n-r)!}$$

For our four-particle example:

• Macrostate (0, 4): \( \binom{4}{0} = 1 \) ✓
• Macrostate (1, 3): \( \binom{4}{1} = 4 \) ✓
• Macrostate (2, 2): \( \binom{4}{2} = 6 \) ✓
• Macrostate (3, 1): \( \binom{4}{3} = 4 \) ✓
• Macrostate (4, 0): \( \binom{4}{4} = 1 \) ✓

The formula works. And notice: \( \binom{n}{r} = \binom{n}{n-r} \), so macrostates (1, 3) and (3, 1) have the same count. Symmetry.

Thermodynamic Probability

Here’s where physics enters.

The thermodynamic probability of a macrostate is simply the number of microstates corresponding to it. We denote it by \( W \) or \( \Omega \).

$$W_{(r, n-r)} = \binom{n}{r} = \frac{n!}{r!(n-r)!}$$

Warning: “Thermodynamic probability” is a terrible name. It’s not a probability in the mathematical sense (it can be greater than 1). It’s a count. But the name stuck, so we’re stuck with it.

For our four-particle system:

The macrostate (2, 2) has the highest thermodynamic probability. This matters because systems tend toward macrostates with higher \( W \). That’s the statistical basis of the Second Law of Thermodynamics.

Actual Probability of a Macrostate

Now let’s calculate real probabilities. The probability of finding the system in a particular macrostate:

$$P_m = \frac{\text{Microstates in this macrostate}}{\text{Total microstates}} = \frac{W}{2^n}$$

For macrostate \( (r, n-r) \):

$$P_{(r, n-r)} = \frac{W_{(r, n-r)}}{2^n} = \binom{n}{r} \times \frac{1}{2^n}$$

For our four-particle system:

Sanity check: The probabilities sum to 100%. Good.

Most Probable Macrostate

Which macrostate is most likely? The one with particles evenly distributed.

For \( n \) particles:

• If \( n \) is even: Most probable is \( (n/2, n/2) \)
• If \( n \) is odd: Most probable are \( ((n+1)/2, (n-1)/2) \) and \( ((n-1)/2, (n+1)/2) \)

Maximum probability:

$$P_{max} = \frac{n!}{\left(\frac{n}{2}\right)! \cdot \left(\frac{n}{2}\right)!} \times \frac{1}{2^n} \quad \text{(n even)}$$

$$P_{max} = \frac{n!}{\frac{n-1}{2}! \cdot \frac{n+1}{2}!} \times \frac{1}{2^n} \quad \text{(n odd)}$$

Least Probable Macrostate

The least likely macrostates are the extremes: all particles in one compartment.

$$P_{min} = \frac{1}{2^n}$$

For four particles, that’s \( 1/16 = 6.25\% \). Doesn’t sound that rare, right?

Now imagine \( n = 100 \) particles. The probability of all 100 being in one compartment:

$$P_{min} = \frac{1}{2^{100}} \approx 10^{-30}$$

For Avogadro’s number of particles (\( n \approx 10^{23} \))? The probability becomes so small it’s effectively zero. You’d wait longer than the age of the universe to see it happen.

This is why the Second Law works. Systems evolve toward probable macrostates not because of some mysterious force, but because improbable macrostates are astronomically improbable.

Key Relationships

Let me summarize the important formulas:

Critical insight: Probability \( P \) is directly proportional to thermodynamic probability \( W \). Higher \( W \) means higher \( P \). Systems naturally evolve toward high-\( W \) macrostates.

Why This Matters

Macrostates and microstates aren’t just academic concepts. They’re the foundation of:

• Entropy: Boltzmann’s famous formula \( S = k_B \ln W \) connects entropy directly to thermodynamic probability
• The Second Law: Systems evolve toward macrostates with more microstates (higher \( W \))
• Temperature: Defined in terms of how \( W \) changes with energy
• Equilibrium: The macrostate with maximum \( W \) subject to constraints

Get this foundation right, and statistical mechanics becomes much clearer.

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