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Entropy, intuitively
26th September, 2026

Where to begin? With Boltzmann.

We can model the world at two levels: macro and micro. A microscopic model-a microstate-will contain the position, momentum, charge, etc, of every constituent particle. A macroscopic model-a macrostate-will just tell you the pressure and volume of the whole. Simple example: roll NN dice. A microstate could be the number on each die, and you could define the sum as a macrostate.

Notice that macroscopic models are abstractions. Reality is actually a soup of fundamental particles and fields, but we can still model it extremely well using macro aggregates. Macrostates, being abstractions, can have *multiple* microstates corresponding to them. There are many ways to roll NN dice to get a sum of 3N3N. Similarly, there are a lot of ways to arrange gas particles to have some total energy UU, but only one such that each particle has a specified momentum, position, etc.

Some macrostates clearly have more microstates corresponding to them than others. With the dice, there's only one roll each for NN or 6N6N, and-again-multiple for a sum of 3N3N.

What affects how many compatible microstates? The conventional explanation is that more disordered macrostate means more compatible microstates, but I prefer the term constrained. The more you constrain the macroscopic value and the more specific you are about it, the less internal arrangements produce that value. If you're perfectly specific, your "macrostate" just becomes a microstate.

Can we quantify this constrained-ness? Boltzmann did, figuring out that

S=kBln⁡ΩS = k_B \ln \Omega

where Ω\Omega is the number of microstates producing a given macrostate, kBk_B is just a constant, and SS is entropy, which actually measures how unconstrained the underlying macrostate is. High entropy means a macrostate that specifies very little about the underlying microtates, thus having many compatible microstates.

Why does this mean the entropy of the universe must increase (the second law of thermodynamics)? Imagine a universe containing just some gas. No energy flows into or out of the gas besides what it started with. We intuitively understand the gas is decidedly not static internally: its particles are constantly moving and colliding. Clearly it has a different microstate at each moment, and it's moving between microstates without energy input. Therefore, to not violate energy conservation, each microstate it moves between must have the same total energy. Let's also assume that they're all equally likely.

The second law of thermodynamics now falls out. Individual microstates are all equally likely, but high-entropy macrostates such as "the gas is in thermal equilibrium everywhere" have orders upon orders of magnitudes more corresponding microstates than low-entropy macrostates like "there is a 200 K200\,\mathrm{K} temperature difference between the two half-volumes." As the gas wanders evenly through all possible microstates, its trajectory on the macro level is overwhelmingly likely to be towards higher-entropy macrostates, because they have vastly many more microstates.

Notice that this makes the second law a statement of vast probability, not about something fundamentally baked into the structure of reality. It's not impossible for heat to flow from cold to hot; for paint in water to spontaneously concentrate back into the dot it was initially. It is impossible (given what we currently know) for matter to exceed the speed of light. It's just that you'd have to wait much, much longer than the age of the universe to watch entropy spontaneously decrease. Unlikely is an understatement, but unlikely isn't impossible.

Going back to the gas universe, notice that a low-entropy state such as an internal temperature gradient has the same total energy as a high-entropy state, but the former is much more useful if we want to extract work out of the gas. Why, if the energy is all still there? It's because useful work follows this pattern of two things at different "potential"- gravitational, electrical, thermal-which causes a flow between them, and then placing something in the path of the flow. Only the lower-entropy state has that gradient. For useful work we almost always care about a difference rather than the absolute value.

Now notice that the state of having a potential difference is inherently more constrained and thus lower-entropy than a state that doesn't have that requirement, so we arrive at the unfortunate result that these gradients we care about are overwhelmingly likely to be erased even if the energy is all still there, because they're low-entropy. Keeping total energy constant, an increase in entropy means a decrease in free energy, the energy available to do useful work. Therefore, because the entropy of the universe increases, its free energy decreases. Heat death simply means zero free energy, or maximum entropy.

I've been talking about entropy from the statistical mechanics angle so far. Now I'm going to do so from the thermodynamics angle. Entropy change is defined as

ΔS=∫ifδQrevT\Delta S = \int_{i}^{f} \frac{\delta Q_{\mathrm{rev}}}{T}

along a reversible path between ii and ff. There's no derivation for dQT\frac{dQ}{T} equaling entropy; that's just how Rudolf Clausius defined it. He also proved it's a state function, and Boltzmann later showed it equals the statistical-mechanics interpretation above.

Consider a fast transfer of QQ joules of heat from a reservoir fixed at T1T_1 K\mathrm{K} to a reservoir fixed at T2T_2 K\mathrm{K}, T1>T2T_1 > T_2. T1T_1 loses QT1\frac{Q}{T_1} entropy, T2T_2 gains QT2\frac{Q}{T_2} entropy, but if T1>T2T_1 > T_2 then QT1<QT2\frac{Q}{T_1} < \frac{Q}{T_2}. Simply transferring heat across a finite temperature difference generates entropy. Like me, you might initially think that slowing this transfer down infinitely makes it reversible, but it doesn't. That makes the heat transferred at each moment an infinitestimal, dQdQ. But it does nothing for the temperature difference; 1T2>1T1\frac{1}{T_2} > \frac{1}{T_1} still holds. To make temperature an infinitesimal as well, we'd have to change the process itself by adding infinitely many intermediary bodies, each differing by dTdT.

Also, notice that adding the same energy via heat to a hot body increases entropy less than doing so with a cold body because the hot body's microstates are already so unconstrained that more freedom doesn't change much.

To go further we need processes. Thermodynamic systems have state: a complete description of them at some moment in time. For a gas, state is (P,V)(P, V). Now imagine each point in 2D space (x,y)(x, y) encodes a state using the values of its coordinates. We've mapped all possible states to a point on the plane. A process is any curve drawn on the plane, including loops.

We also distinguish between irreversible and reversible processes. There are a lot of ways I've heard them described: reversibles happen infinitely slowly and with infinitesimal temperature changes; everything real is irreversible, etc. I think the simplest definition is that reversible processes never change the entropy of their universe, and irreversible processes always increase it, both regardless of cyclicity.

And we want to separate a few things. Now that we can quantify entropy using thermodynamic quantities, we can talk about the entropy change in the system, just the entropy transferred across the system-surroundings boundary, entropy generated in the system but not transferred, or change in the universe as a whole (system + surroundings). Spend a moment on that.

Now we're ready to intuit the tangle of inequalities thermodynamics classes teach.

For any process:

dSsys=Stransfer+Sgen(1)dS_{\text{sys}} = S_{\text{transfer}} + S_{\text{gen}} \tag{1}
Just entropy accounting. Also, assuming heat transfer across the interface is reversible:
dSsurr=−Stransfer(2)dS_{\text{surr}} = -S_{\text{transfer}} \tag{2}
SgenS_{\text{gen}} doesn't matter to the surroundings because it never leaves the system. Therefore:
Suniverse=Ssys+Ssurr=(Stransfer+Sgen)−Stransfer=Sgen(3)\begin{aligned} S_{\text{universe}} &= S_{\text{sys}} + S_{\text{surr}} \\ &= (S_{\text{transfer}} + S_{\text{gen}}) - S_{\text{transfer}} \\ &= S_{\text{gen}} \end{aligned} \tag{3}

By definition:
Sgen=0for reversibles,Sgen>0for irreversibles(4)S_{\text{gen}} = 0 \quad \text{for reversibles}, \qquad S_{\text{gen}} > 0 \quad \text{for irreversibles} \tag{4}
Therefore:
dSuniverse≥0(5)dS_{\text{universe}} \geq 0 \tag{5}
For all cyclics, reversible or irreversible:
dSsys=0(6)dS_{\text{sys}} = 0 \tag{6}
This is because entropy is a state function; dSdS between two states depends only on the difference between them, not on the path used to travel between. Clearly the difference between identical states is zero. You might think this contradicts what I said about irreversibles earlier, but it doesn't: they do always increase the entropy of their universe, but here we're only concerned about the system. Again for all cyclics:
Stransfer≤0(7)S_{\text{transfer}} \leq 0 \tag{7}
Look back at (6). dSsys=0dS_{\text{sys}} = 0 because of cyclicity, so Stransfer=−SgenS_{\text{transfer}} = -S_{\text{gen}}. Sgen≥0S_{\text{gen}} \geq 0, from (4), therefore Stransfer≤0S_{\text{transfer}} \leq 0. Intuitively: to stay cyclic, you have to transfer out as much entropy as you generate. Reversibles generate no entropy, so =0= 0, and irreversibles do, so they have to transfer entropy out, meaning <0< 0.

And that's entropy. I prefer the statistical mechanics definition. We lifeforms are absurdly low-entropy creatures compared to how our constituent atoms would evolve if left alone. If you take all my oxygen, hydrogen, carbon, and nitrogen atoms at face value, there are vastly more ways to arrange them into not-me than me. Boltzmann's formula captures that in Ω\Omega, but I don't currently understand how the dQT\frac{dQ}{T} definition also does, given that they're the same quantity. If you do get it, please shoot me an email. I'm going to close off with an excellent quote I ran into while studying: civilization doesn't consume energy; energy is conserved. It consumes low entropy.

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