# Does it make sense to ask how many of the molecules you are inhaling Caesar exhaled in his last breath?

There is a nice example of a Fermi estimation question where you ask: How many of the molecules which Julius Caesar (or Jesus, Muhammad, etc. - anyone who died a long time ago) exhaled in their dying breath do you inhale when breathing, on average. If you estimate the number of molecules in the atmosphere and the number of molecules in a human breath, and assume that the molecules of this last breath are evenly spread across the globe, you get an answer in the order or magnitude of 1 - Which is pretty nice.

I'm not sure whether this question is meaningful physically. Statistical mechanics teaches us that molecules (of the same element) are indistinguishable, so it seems that "molecules which Caesar exhaled in his dying breath" is not something well-defined. On the other hand if you replace "you" by "one of the people standing next to him when he died" it looks like the answer must be yes.

What is the difference between these two situations? Is the "thermodynamic" assumption that the molecules are evenly distributed (i.e. the system has "mixed") what makes it impossible to talk about these specific particles anymore?

• Did you try to do a back-of-an-envelope-calculation? – Qmechanic Jan 1 at 19:13
• The question is not about the Fermi estimate. – 0x539 Jan 1 at 19:30
• 2Ox539 Different atoms and moceluse of the same element or compound may be identical and indistinguishable, but that does not affect the individual histories of those molecules. Most of the atoms and subatomic particles you breath in ere not in Julius Caesar's last brath, but possibly some were. It is a historical fact that some of the atoms you are beathing in now were in Julius Caesar's last breath and some were not. And our lack of ability to find out which are which does not affect the fact that they have had different histories. – M. A. Golding Jan 2 at 5:11
• That was a Paper V question in my Physics final in St.Andrews, in 1981! I seem to remember my answer was somewhere in the vicinity of 1! – Brian Drummond Jan 2 at 21:09
• @M.A.Golding I'd add a disclaimer - this is only true insofar as you can factorize the individual atoms at all times throughout their history. Should you pick one of the "bundle" at random, or "mark" all of them as "potentially interesting"? But in the end, it doesn't really matter - it's not the point of the premise, and you don't really need quantum physics to have confusing thoughts about identity, Theseus' ship and all that. The real answer has always been "identity is a human concept, and we use it in a way that is useful, not necessarily physically meaningful". – Luaan Jan 4 at 13:54

This is a classic classical question that is designed to get us to understand how big Avogadro's number is. We know our lungs are much smaller then the atmosphere, but as $$N_A \rightarrow \infty$$, any finite volume will have some of Caesar's air, so in conclusion $$6 \times 10^{23}$$ is a number so large it may be outside the reach of common intuition.

The problem ignores both geophysical processes and quantum issues. The former because atoms have a residence time in atmosphere, so exchange with the hydro and lithosphere drops an unwanted complication....basically an exponential blanket, or exchange with a large reservoir that provides no intuition.

Regarding the indistinguishability of particle, I don't think thermodynamics teaches us that. If atoms where classic but still identical, thermodynamics would be very different because the underlying statistical physics would by different. Identical particles are classically distinguishable.

That quantum particle are indistinguishable is something that really only makes sense if they're not particles but rather field excitations running around in superpositions of position eigenstates with complex probability amplitudes...and that's a complication that is beyond the scope of the problem.

[Aside: so the indistinguishability issues have real consequences. Back when we were aerobraking in Mars's upper CO$$_2$$ atmosphere, we started with the Apollo re-entry software for hypersonic shock formation, and here hypersonic means: the molecules are dissociated. The fact that a nitrogen molecule breaks up into two identical atoms while carbon dioxide does not was initially not included in the change from Earth re-entry to Mars entry, and that affects entropy which affect shock formation which caused the vehicles to "go-long" in the landing ellipse...it has since been fixed].

• "classical" thermodynamics still needs to account for indistinguishibility, else you encounter Gibb's Paradox. So thermodynamics would be different if atoms were distinguishable. – 0x539 Jan 1 at 17:58
• @0x539 I believe that classically identical particles are distinguishable, so we are in agreement. The problem is thanks to the available adjectives in English when describing quantum concepts. – JEB Jan 1 at 19:11
• Note that both of the indistinguishable-particle distributions, for Bose-Einstein and Fermi-Dirac statistics, reduce to the Maxwell-Boltzmann distribution in the high-temperature limit. I think your statement “thermodynamics would be very different because the underlying statistical physics would be different” is a low-temperature statement. – rob Jan 1 at 21:31
• +1 for Partial Caesar Pressure. – JEB Jan 3 at 17:15
• The question mentions molecules, not atoms. So you need to account for all relevant chemical processes, not just geophysical. E.g. ozone formation plays a role, too. – MSalters Jan 4 at 14:36

Statistical mechanics teaches us that molecules (of the same element) are indistinguishable, so it seems that "molecules which Caesar exhaled in his dying breath" is not something well-defined.

There's two different misconceptions in this sentence. Let's break them both down:

1. "Statistical mechanics teaches us that molecules (of the same element) are indistinguishable" - This gets the relationship between these concepts exactly backwards. Molecular indistinguishability is an axiom, not a theorem, of statistical mechanics--it's something that is assumed in order to do statistical mechanics, not something that is demonstrated with techniques from statistical mechanics. (This sentence also ignores the distinction between different molecular forms of the same element; statistical mechanics does NOT assume that, for instance, diatomic oxygen, $$O_2$$, and ozone, $$O_3$$, are indistinguishable!)

2. "so it seems that "molecules which Caesar exhaled in his dying breath" is not something well-defined" - This is a confusion of two different senses of what it means for molecules to be "the same". There are $$O_2$$ molecules I'm breathing right now, and there are $$O_2$$ molecules OP is breathing right now, and they're genuinely different molecules. But if you took a McForge $$O_2$$ molecule and an OP $$O_2$$ molecule, and put them in a lab, there's no experiment you could perform, even in principle, to tell which is which (assuming we didn't ionize them or otherwise give them distinguishing properties.) Yet saying "we can't tell these $$O_2$$ molecules apart" is different from saying "there's only one $$O_2$$ molecule here". Indeed, it would be absurd to claim that OP and I are breathing the exact same molecules of $$O_2$$ right now. We're breathing indistinguishable molecules of $$O_2$$, but that's different.

Just so with the Caesar's-dying-breath molecules--I may not be able to perform an experiment that tells for sure whether a particular molecule I'm breathing is from Caesar's dying breath, but any particular molecule I'm breathing either was, or was not, in Caesar's dying breath. And probabilistically, it's practically certain that I have breathed one of those Caesar's-dying-breath molecules in my lifetime--at least under assumptions necessary to the OP scenario like:

• the molecules have had time to diffuse evenly across the globe

• none of them ever underwent any chemical reactions or decay that might have made them into some other type of molecule since then

• none of them got locked into some rock formation or something that would have prevented them from so diffusing

...and so on, and so forth.

• What is wrong with the following reasoning? "A McForge O2 molecule and an OP O2 molecule are distinguishable, since they are separated in space. Even if you can't pinpoint their exact locations, you can say that one probably is in Region A, while the other probably is in Region B, where A and B don't overlap and are separated by a great distance (e.g. A = McForge's room while B = OP's room)" – Asker Jan 3 at 0:32
• @Asker Key word: "probably". Let's say McForge is in, like, Dallas and the OP is in, like, Tokyo. Then yes, it's very very likely that, if we let the system evolve for a minute and check the $O_2$ molecules, the one in McForge's room and the one in OP's room are the same ones that McForge and OP were just breathing. But it's not certain, because the wave functions propagate through all of space, even to the far side of Earth. So in principle, they're not distinguishable, even though in practice, we can distinguish between them with a very high probability of being correct. – Rivers McForge Jan 3 at 0:49
• @RiversMcForge And less fringely, there are plenty of situations where you lose even that notion of traceability - two individual oxygen atoms can evolve through interactions that make you lose track of which is which even in an ideal scenario, and as time passes (and atoms undergo more and more interactions), it becomes increasingly ridiculous to talk about any kind of "oxygen atom identity". It's reasonable enough to say "None of the oxygen atoms I've breathed in the past five minutes were recently breathed in by OP", but it makes much less practical sense in five days, or years, or decades. – Luaan Jan 4 at 14:00

For a classical system, where we presume we can know the position and momentum of every molecule to infinite precision, there are only two ways to make a molecule identifiable:

1. The molecule has unique extensive properties, like mass, charge, composition, etc..
2. We have traced the time history of the molecule from some point in the past.

For an ideal gas of a fictitious "air" molecule, we can't do anything about the first point -- they all have the same mass, charge, composition, etc.. Even if we treat air as components, we can identify the component molecules as a class, but within the class they are otherwise indistinguishable (i.e. we can tell a nitrogen from an oxygen, but we can't tell two oxygens apart.)

So then we're left with tracing the time history of the molecules. And that is something that we could do (in theory). When molecules enter Caesar's lungs, we mentally tag them with a $$C$$ and when molecules enter Plato's lungs, we mentally tag them with a $$P$$. And then we can begin to evaluate the propagation of $$C$$ and $$P$$ molecules in air molecules over time. If all we do is tag them, then we've added an extensive property to them -- we'll be able to tell $$C$$ from $$P$$, but we won't be able to identify two different $$C$$ molecules uniquely. On the other hand, if we track their histories specifically, then each molecule will be uniquely identified (by, say, it's initial position inside Caesar's or Plato's lungs).

Of course, the pressure of a box composed of air, $$C$$, and $$P$$ molecules will end up with the same pressure as just air, or just $$C$$, or just $$P$$ at the same temperature and volume because they collide and interact just the same. But if we wanted to calculate another property, like "Partial Caesar Pressure", then those distinguishable properties will become a factor in the property evaluation.

• Can we actually track them, even in principle? Following a quantum mechanical collision between a molecule marked $C$ and an (otherwise identical) one that is not, I don't think one can say which outgoing particle is which. – Rococo Jan 1 at 18:25
• @Rococo that's why I had to preface the answer by saying we're only talking about classical interactions where we can identify both position and momentum exactly. If we bring in quantum effects, then we can't uniquely identify particles because we cannot tell where they are or where they are going with enough detail to distinguish them. But for the approximations being asked about, those effects aren't important. – tpg2114 Jan 1 at 20:22
• ah, sorry, wasn't reading clearly. then I suppose this answer is sensible within its domain of applicability (+1), but probably this question is more interesting when going to the real quantum world... – Rococo Jan 2 at 17:07
• @Rococo: no, you can't track even large things like atoms or molecules, which is why I added the Mars EDL vignette to my answer. Swapping 2 identical atoms is the same state, and that actually changes shockwave development in diatomic gases vs. not diatomic gases. – JEB Jan 3 at 17:14
• @Rococo In fact, if we did have a magic marker for atoms that persisted indefinitely without affecting anything other than our magic marker sensor, you would see things like "two molecules bumped into each other and exchanged (identical) atoms". Of course, such magic markers are utter magic and in practice they would actually affect our observations by making such interaction impossible, among many other complications, so all of these are pure thought/math experiments. – Luaan Jan 4 at 14:03

Your intuition is right. The "set of molecules making up Caesar's last breath" is never completely well defined, but it has some approximate validity shortly after Caesar's death. In the long term, after thorough mixing, it's completely meaningless. This is true even in a toy model where you treat the atmosphere as an ideal gas (which I'll do in the rest of this answer).

In a quantum world it is not the case that "any particular molecule I'm breathing either was, or was not, in Caesar's dying breath", as one previous answer claimed. Particle indistinguishability is not a mere practical limitation on our ability to track particles or tell them apart. It's more fundamental. This seems to be a common misconception; two of the three earlier answers seem to suffer from it, as does the most upvoted comment on the question itself. The third earlier answer, which has a lot more upvotes, isn't incorrect but doesn't seem to actually answer the question.

It's probably easiest to understand this problem in the Lagrangian sum-over-histories picture. This isn't usually taught in introductory QM courses, but it is taught in Feynman's popularization of quantum electrodynamics which is well worth reading.

In the sum-over-histories picture you choose quasiclassical initial and final conditions – which in this case can be a bunch of molecules with precise positions and orientations in space – and you calculate the quantum amplitude of a transition from that initial to that final state, over that span of time, as the sum of contributions from all quasiclassical paths between those states. (In state-vector terms this amplitude is $$\langle ψ_f|e^{iHt}|ψ_i\rangle$$, where I'm taking $$ψ_i$$ and $$ψ_f$$ to be position basis states.)

The set of valid paths/histories includes paths that permute the sets of indistinguishable particles in all possible ways.* If the elapsed time between the initial and final states is small enough, and there's a speed-of-light limitation, some permutations are actually impossible, but that doesn't help us much in this problem given the small size of Earth's atmosphere in light-millennia. If the elapsed time is longer but still fairly short, all permutations are possible but the overall amplitude is dominated by paths in which the molecules don't move very far. It's therefore reasonable to say that, shortly after Caesar's death, the molecules from his breath are still, for the most part, nearby. This is not true in any absolute sense. It's not even true in a "statistically likely" sense, as all paths actually contribute to the transition and so all of them "happen". But it's about as true as any other statement you can make about a quantum world.

If the elapsed time is long enough that the atmosphere is thoroughly mixed, then all permutations of the whole atmosphere contribute essentially equally. The molecules are not permuted in any particular, random way. It's more like every molecule in the final state is a combination of every molecule in the initial state. There is no true path from start to finish, no merely-in-practice-unmeasurable correct permutation. This is true of any particular final state, at these long time scales.

* The set of valid paths does not include paths that exchange particles that merely have the same measurable properties (mass, charge, spin) but are not indistinguishable in the precise technical sense used in quantum mechanics.

For example, in a variant of quantum electrodynamics where there are many copies of the electron field (like the fermion generations of the Standard Model, except the masses are also the same, and there are no W bosons to complicate things), if you are given a bunch of electron-like particles, you can experimentally divide them into groups such that all of the particles within each group have Fermi-Dirac statistics, while particles from different groups have classical statistics. The fact that you can't say "which group is which", because they all have the same mass, charge and spin, is not what is meant by particle indistinguishability. The fact that the particles within each group have nonclassical statistics is what's meant by particle indistinguishability.

I'm mentioning this because I think it may be related to the misconception about Caesar's last breath. If there was just one electron in each group, and you let them interact unsupervised for a while and then tried to figure out which one had come from which group, it would probably be impossible in practice, but it is possible in principle, by the rules of quantum mechanics. If you took a bunch of electrons from the same group and let them interact for a while, it's impossible in practice and in principle to say which one was which. The latter case is the one that's relevant in the real world.

• Concerning the second paragraph in the footnote: Is indistinguishability really about non_classical_ statistics? Doesn't statistical mechanics of a classical system also need to account for indistinguishability (the Boltzmann distribution just being a low-occupation number per state / high-temperature approximation of both the Fermi ad Bose distributions)? – 0x539 Jan 3 at 12:25
• @0x539 Do note that you're already talking about things that are starting the path into a quantum understanding of the universe. The classical prediction was wrong, and had to be corrected by understanding that particles are not distinguishable. This is much the same as the few decades later ultra-violet catastrophe paradox. Where exactly you draw the line between "classical" and "quantum" physics is largely arbitrary. It's kind of silly how humans focus on definite boundaries and categories, even though nature really has few such things (if any). – Luaan Jan 4 at 14:12