Why Is Entropy an Extensive Quantity?

Title: Why Is Entropy an Extensive Quantity? Category: physics

Short answer: For ordinary macroscopic systems, entropy is extensive: scaling the amount of material, volume, and energy by the same factor scales entropy by that factor. In statistical mechanics, independent subsystems have a combined number of microstates equal to the product of their separate counts. Boltzmann’s logarithm turns that product into an additive sum of entropies.

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Entropy is called extensive because it grows in proportion to the size of a system when you scale up the system without changing its intensive conditions, such as temperature, pressure, or density. It is also additive for independent parts: the entropy of the combined system is the sum of the entropies of the parts.

The statistical-mechanical reason is that independent parts have a multiplicative number of possible combined microstates. Taking the logarithm in the definition of Boltzmann entropy turns that multiplication into addition.

What “extensive” means

An extensive quantity depends on the amount of material in the system. Examples include mass, volume, internal energy, and entropy. If you combine two identical, noninteracting systems in the same state, an extensive quantity for the combined system is twice its value for either one alone.

More generally, for a homogeneous system at fixed composition, scaling all extensive state variables by a factor $\lambda$ gives

$$S(\lambda U,\lambda V,\lambda N)=\lambda S(U,V,N),$$

where $U$ is internal energy, $V$ is volume, and $N$ is the number of particles. This scaling describes making a larger copy of the system while keeping its energy, volume, and particle number per unit amount unchanged.

Why the logarithm makes entropy additive

For a microcanonical system in which the $\Omega$ accessible microstates are equally probable, the Boltzmann entropy is

$$S=k_{\mathrm B}\ln \Omega,$$

where $k_{\mathrm B}$ is Boltzmann’s constant. If two independent subsystems, A and B, have $\Omega_A$ and $\Omega_B$ accessible microstates, then every state of A can be paired with every state of B. Therefore,

$$\Omega_{AB}=\Omega_A\Omega_B.$$

Taking the logarithm gives

$$\begin{aligned} S_{AB}&=k_{\mathrm B}\ln(\Omega_A\Omega_B)\\ &=k_{\mathrm B}\ln\Omega_A+k_{\mathrm B}\ln\Omega_B\\ &=S_A+S_B. \end{aligned}$$

That is the basic statistical reason entropy adds for independent subsystems. The logarithm is essential: the number of possible joint arrangements multiplies, while entropy, defined as its logarithm, adds.

A worked example

Suppose subsystem A has 10 accessible microstates and subsystem B has 20. If they are independent, the combined system has

$$\Omega_{AB}=10\times20=200$$

accessible microstates. Their entropies are

$$S_A=k_{\mathrm B}\ln 10,\qquad S_B=k_{\mathrm B}\ln 20.$$

The combined entropy is

$$S_{AB}=k_{\mathrm B}\ln 200 =k_{\mathrm B}(\ln 10+\ln 20) =S_A+S_B.$$

If you instead make $\lambda$ independent copies of the same subsystem, the number of combined microstates becomes $\Omega^\lambda$. Its entropy is $k_{\mathrm B}\ln(\Omega^\lambda)=\lambda k_{\mathrm B}\ln\Omega$: it grows in proportion to the number of copies.

Why this works for bulk matter

Real pieces of matter are not perfectly independent: particles in neighboring regions interact, and correlations can connect distant parts. For ordinary materials with short-range interactions, however, most particles are in the bulk, while the interface between two pieces contains far fewer particles than the bulk. When a macroscopic system is divided into large parts, interface effects are usually small relative to the total volume. The bulk entropy is therefore additive to a very good approximation, and scales with system size.

The same idea appears in the thermodynamic entropy of an ideal gas. For a monatomic ideal gas in the classical, nondegenerate regime, the Sackur–Tetrode expression has the form

$$S=Nk_{\mathrm B}\left[\ln\left(\frac{V}{N}\left(\frac{4\pi mU}{3Nh^2}\right)^{3/2}\right)+\frac{5}{2}\right],$$

where $m$ is the particle mass and $h$ is Planck’s constant. If $N$, $V$, and $U$ are all multiplied by $\lambda$, the ratios inside the logarithm do not change. The bracket stays the same, while the factor $N$ becomes $\lambda N$. Thus $S$ becomes $\lambda S$.

When additivity needs qualification

For statistically dependent subsystems, the combined entropy need not equal the sum of the separate entropies. In classical statistical mechanics, the relation can be written

$$S_{AB}=S_A+S_B-k_{\mathrm B}I(A:B),$$

where $I(A:B)$ is the mutual information, a measure of correlations between the parts. If the parts are independent, $I(A:B)=0$ and their entropies add. If they are correlated, the joint system has fewer independent possibilities than the product of the separate counts suggests.

Systems with important long-range interactions can also fail to have the usual extensive scaling. In such cases, interface or collective effects may not be negligible, so the standard bulk argument does not apply without modification.

Common mistakes

• “Entropy is extensive just because it is a logarithm.” The logarithm alone is not enough. The key additional fact is that independent subsystems’ numbers of microstates multiply.
• “Entropy always adds exactly.” Exact additivity applies to independent subsystems. Correlations or significant interactions can change the relationship.
• “Extensive means entropy is proportional to temperature.” It means entropy scales with system size when the relevant intensive conditions and composition are held fixed. Entropy can still change with temperature.