Boltzmann's definition of entropy is $\sigma = \log \Omega$, where $\Omega$ is the number of microstates consistent with a given macrostate. If I understand correctly, this means that it only makes sense to speak of the entropy of a physical system with respect to a definition of the "macrostate" of the system. For example, we can consider a system consisting of $N$ spins that can be either up or down. Suppose that the actual system is in the "alternating" microstate $\uparrow \downarrow \uparrow \downarrow \cdots \uparrow \downarrow$. We could stipulate that the macrostate of the system is the total spin excess $s$, and calculate the entropy of the actual physical system using the fact that $s = 0$. Or, instead, we could stipulate that the macrostate of the system is the spin excess $s_1$ of the first $N/2$ particles together with the spin excess $s_2$ of the second $N/2$ particles. In this case, we would calculate a different entropy for the same actual physical system, because this time we would use the extra information $s_1 = s_2 = 0$.

What I'm confused about is that the way I always hear the 2nd law stated is something like "The entropy of any closed system tends to increase," with no mention of the definition of macrostate with respect to which we are calculating entropy. I am wondering what the more precise statement of the 2nd law is, with explicit reference to how macrostates are defined.

My first thought was that perhaps the 2nd law says something along the lines of "With respect to any fixed definition of the macrostate of a closed system, the entropy of the system tends to increase." But this can't quite be right, because we can rig the macrostates so that entropy decreases just by adding in "impossible" microstates. For example, if we consider the system $\uparrow \downarrow \uparrow \downarrow \cdots \uparrow \downarrow$ described above, we could define macrostates as follows: For each integer $s$, we define a macrostate $S_s$, which consists of all microstates of $N$ spins with spin excess $s$, along with a whole bunch of quantum harmonic oscillator microstates. If we choose the number of QHO microstates that we include in $S_s$ carefully, we can ensure that entropy w.r.t. this definition of "macrostate" decreases whenever entropy w.r.t. the straightforward spin-excess definition increases, and vice versa.

So this makes me think that maybe we need some notion of the "possible" microstates of a system, and then the 2nd law should say something like "With respect to any fixed partition of the possible microstates of a closed system into macrostates, the entropy of the system tends to increase." But I'm struggling to even imagine what "possible" would mean here. Maybe this is all completely wrong?

Any help would be appreciated. Thanks!

  • $\begingroup$ Excellent question. And welcome to Physics Stackexchange! $\endgroup$
    – user10851
    Commented Apr 21, 2014 at 6:02
  • $\begingroup$ Nice question, William. Could you elaborate on your model for the partitioning of spin + QHO states into macrostates? $\endgroup$ Commented Apr 21, 2014 at 9:37
  • $\begingroup$ Macrostate cannot be defined by any choice of microstates, but by statement of few macroscopic quantities, e.g. number of particles, volume (magnetic field or moment in case of spins) and temperature. I do not understand why you are considering including QHO states into microstates. If we discuss entropy of the spin set, only the states of those spins are relevant. $\endgroup$ Commented Apr 21, 2014 at 11:05
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    $\begingroup$ @MarkMitchison: The point of the model was not that it is a reasonable way of calculating entropy or thinking about the world, but just that it is one that I don't know how to rule out. I've never seen "macrostate" very carefully defined; it seems the simplest definition is that a macrostate is just a set of microstates. Let $M_s$ denote the number of N-spin microstates with spin excess $s$. Let $A$ be the maximum over $s$ of $M_s$. If we include $2(A - M_s)$ QHO microstates in $S_s$, then we'll achieve the entropy-reversing effect mentioned in the post. I hope that answered your question. $\endgroup$ Commented Apr 21, 2014 at 17:36
  • $\begingroup$ @JánLalinský: That sounds like a great resolution! But I'm struggling to understand what precisely constitutes a "valid" macrostate. Could you please elaborate on the precise definition? Thanks! $\endgroup$ Commented Apr 21, 2014 at 17:38

1 Answer 1


As Jan Lalinsky has already pointed out (and indeed the OP has implied), the concept of macrostate is defined by the macroscopic quantities you have available experimentally. Suppose you know the expectation values of some set of extensive observables $X_i$. Typically these observables will be energy, particle number, average magnetisation, etc. This set defines your thermodynamic state space. In other words, a thermodynamic state (macrostate), is completely defined by a vector of expectation values $\mathbf{x}$ with elements $\mathbf{x}_i = \langle X_i \rangle$. A microstate (quantum state) $\rho$ corresponds to the macrostate $\mathbf{x}$ if $$ \mathrm{Tr}(\rho X_i) = \mathbf{x}_i $$ for all $i$. You should find that as long as you are consistent in your choice of thermodynamic state space (and kinematical assumptions, see below), you cannot distribute microstates among macrostates in such a way that the second law is violated.

OP has pointed out that if the total number of microstates available to the system is allowed to change from one macrostate to the next, then the second law appears to be violable. This is a good point, and illustrates that a further assumption is needed (which is so obvious that it's often left unsaid). Namely, the assumption is that the kinematical laws governing the system do not change from one measurement or macrostate to the next. In particular, this means that the total number of microstates available to the system is fixed. Indeed, this follows from a stronger assumption: that the Hamiltonian does not change in time (apart from any parametric time dependence due to external driving fields). Since the job of statistical mechanics is to infer the microscopic properties from the macroscopic data (or vice versa), it is clearly nonsensical to allow the microscopic Hamiltonian to vary arbitrarily. Really this just amounts to making a set of choices about how to model the system and then sticking to them consistently, which is always necessary for a working mathematical theory of physics.


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