One of the biggest confusions I've had in trying to piece this together over the years is in mixing up whether the process of dissipation involves a transition from a higher dimensional space to a lower, or from a lower to a higher. I think it is both depending on how you look at it, but you have to keep straight what space you're talking about and what you mean.

If you look at it from the point of view of quantum mechanics, dissipation comes from the measurement process which involves projection matrices (or projection "operators" more generally) which take many possibilities and collapse them down to one. It's common to hear people use the word "reduction" in phrases like "the reduction of the state vector" to mean measurement in quantum mechanics. And measurement is the only time something irreversible happens, the rest of the laws of quantum mechanics are entirely reversible. So you would think intuitively, that a reduction or a collapse involves going from a higher dimensional space to a lower dimensional space. That's what a projection is mathematically. For example, when you walk in the sun outside and it's not directly overhead, you are followed around by a shadow. Your shadow is a 2-dimensional projection of your 3-dimensional self on the ground. A shadow is one of the simplest kinds of projections, but mathematically a projection refers to anything that reduces a higher dimensional object to a lower dimensional image. That's what the measurement operators used in quantum mechanics do, but because they are acting within the quantum Hilbert space, they project a space of ridiculously large dimensionality down to one of slightly lower dimensionality (but usually, still infinite).

On the other hand, if you look at it from the point of view of thermodynamics, dissipation happens only when entropy increases. The microscopic laws of physics, even in classical mechanics, are completely reversible and non-dissipative. The only irreversibility that comes into play is when the available phase space of a system increases. Let's walk through a concrete example of a mixing process step by step and see why it is irreversible and why it increases entropy.

First, imagine that you just have 3 classical particles in a box. They just bounce around in the box according to Newton's laws of physics. They move in straight lines unless they bounce off of a wall, in which case their angle of refraction equals their angle of incidence, just as a billiard ball bounces off of the wall of a pool table. It's easy to see that these laws are reversible, and that if you applied them backwards, you'd see basically the same thing happening, it's just that all 3 particles would be moving backward along their original paths instead of forward. Nothing weird or spooky or irreversible about it. But now let's conceptually divide that box into a left side and a right side, and keep track of which side each of the 3 particles are in. If the microstate of this system is the exact positions of all 3 particles plus the exact direction that each of them is moving in, then let's call the "macrostate" a single number between 0 and 3 that equals how many particles are on the right side of the box. To get at this number, we can construct a simplified microstate phase space which is a list of 3 booleans specifying which side of the box they are on. For example, if particles A and B are on the left side of the box, and particle C is on the right side, our list would be (left,left,right). If they were all on the right side, it would be (right,right,right). The macrostate can be deduced from the microstate (by summing up the number of right's in our list), but the reverse is not true as some of the macrostates correspond to more than one microstate. For example, the macrostate "2" could either be (left,right,right) or (right,right,left).

The full microstate phase space is what we talked about earlier--it's an 18-dimensional space, 3 times 3 coordinates for position, and 3 times 3 coordinates for momentum. But in order to understand maxing, we only really have to visualize a simplified microstate phase space based on our list of 3 right/left booleans--in order to do so, you need to picture something that looks like a mini Rubik's cube. A regular Rubik's cube consists of 3x3x3 = 27 cubes (if you include the center cube, which doesn't actually have any colors painted on it, and you can't actually see). But they also sell "mini" Rubik's cubes that are only 2x2x2 = 8 cubes. They are much easier to solve, but not completely trivial if I recall. Each of the 8 cubes in a mini Rubik's cube corresponds to one of the 8 microstates of our system: (left, left, left), (left, left, right), (left,right,left), ... etc., ... (right,right,right). Each of the particles can be in one of 2 possible states, but there are 3 particles so the space is 3-dimenionsal. But because we're only concerned with left-versus-right the space is discrete rather than continuous.

Now imagine that instead of 3 particles, we had an entire mole of particles--that is, Avagadro's number of particles, 6.02x10^23 particles. Any quantity of gas that you could fit in an actual box and hold in your hand would realistically have to have at least this number of particles, and probably a lot more! So what happens to our space of states? Now, instead of being 3 dimensional it is 6.02x10^23 dimensional--quite a bit larger. But still each dimension can only have 2 possible states. Our 3-particle system had only 2^3 = 8 microstates. But this system has an unimaginably large number of microstates, 2^(6.02x10^23) states! Equilibrium for this system means that you have allowed the particle to bounce around long enough that they are nice and randomly mixed (roughly equal numbers on the left and the right). The entropy of any system is simply the natural log of the number of accessible microstates it has. If the system is in equilibrium, then nearly all states are accessible so the entropy is log(2^(6.02x10^23)) = 4x10^23. A very small number of those states correspond to most of the particles being on the left, or most being on the right, but these are such a negligible fraction out of the number above, that it doesn't change the answer.

But what if we started the system in a state where all of the particles just happened to be on one side of the box? In other words, in the state (left,left,left,left,...,left,left) where there are 6.02x10^23 left's? This is a very special initial condition, similar to the special state the universe started in which I mentioned earlier. Because there is only 1 microstate where all of the particles are on the left, this state is extremely unlikely compared to the state where roughly equal numbers are on the left and on the right. The entropy of this state is just log(1) = 0. The true entropy of a gas like this is of course more than 0, but for our purposes here, we only care about the entropy associated with its mixing equally on either side of the box. The rest of the entropy is locked up in the much larger microstate phase space mentioned earlier, before we simplified it down to only the part that cares about which half of the box the particles are in.

The main point of all of this, which I'm getting to is... if all of the particles start out on one side of the box, and then later they are allowed to fill the whole box, you've drastically increased the entropy, because there are a lot more possible places for the particles to be. A slightly more complicated version of what we just went through is if you had two different kinds of particles, let's call them blue and red. Imagine all of the red particles started on the left side, and all of the blue particles started on the right side, perhaps because there is initially a divider in between. Then when you lift the divider, the two of them mix with each other, and after it reaches equilibrium, roughly equal numbers of red and blue will be on both sides. This is what is meant by "mixing" in thermodynamics. There are many many more ways in which they could be mixed than the one way in which they could all be on their own sides, so there is a lot more entropy in the mixed state at the end than there was in the separated state in the beginning. Unlike the version of this where only 3 particles were involved, this version is irreversible in the sense that: it's extremely unlikely, and pretty much inconceivable, that you would ever see a mixed box of these particles naturally and spontaneously sort themselves out to all blue on one side and all red on the other, whereas you would find nothing surprising whatsoever if initially unmixed red and blue gases gradually mixed with each other and wound up in a perfectly homogeneously purple mixture at the end.

In this example, the available state space before the particles are allowed to mix involves only 1 state. Afterwards, it involves something with a size on the order of 2 raised to the power of Avogadro's number of states. So the entropy should increase by something on the order of Avogadro's number. It seems like what has happened is that the state space at the beginning was very small and low dimensional, and then at the end it is very large with high dimensionality. Naively, this appears to be exactly the opposite of what happens during a measurement in quantum mechanics. But somehow, that's not the case--what's going on really is exactly the same. I think what's happening is that we're just confusing two different spaces here. And it's a confusion that I've often made in thinking about this. Where we'll have to go from here to resolve this paradox is to discuss open vs closed systems, and to explore a little bit the many worlds interpretation of quantum mechanics, and what happens to entropy in different parts of the multiverse as different branches of the universal wave function evolve forward in time. I'll leave you with one final piece of the paradox which seems directly related to this high-low dimensionality confusion: if the total entropy remains exactly the same in all branches of the multiverse, then you would think that every time a quantum measurement is performed and different branches split off from each other, the entropy would get divided among them and hence be less and less in each branch over time. (Because surely, the dimensionality of the accessible states in one branch is less than the dimensionality of the accessible states in the combined trunk before the split?) And yet, exactly the opposite happens--while the total entropy remains exactly constant, the entropy in each single branch increases more and more every time there is a split!

To be continued...

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