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Comment by kgwgk | Hacker News Reader

> unless of course you tautologically define entropy as the amount of work the experimenter/humanity can extract from the system

Do you have a different definition? (By the way the entropy is the energy that _cannot_ be extracted as work.) The entropy of the system is a function of the particular choice of state variables - those that the experimenters can manipulate and use to extract work. It’s not a property of the system on its own. There is no “true” entropy anymore that there is a “true” set of state variables for the “true” macroscopic (incomplete) description of the system. If there was a true limiting value for the entropy it would be zero - corresponding to the case where the system is described using every microscopic variable and every degree of freedom could be manipulated.

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Yes, the more regular statistical mechanics theory of entropy doesn't make any reference to an observer, and definitely not to a human observer. In that definition, the entropy of a thermodynamic system is proportional to the amount of microstates (positions, types, momentum, etc. of individual particles) that would lead to the same macrostate (temperature, volume, pressure, etc.). It doesn't matter if an observer is aware of any of this, it's an objective property of the system.

Now sure, you could choose to describe a gas (or any other system) in other terms and compute a different value for entropy with the same generalized definition. But you will not get different results from this operation - the second law of thermodynamics will still apply, and your system will be just as able or unable to produce work regardless of how you choose to represent it. You won't get better efficiency out of an engine by choosing to measure something other than the temperature/volume/pressure of the gases involved, for example.

Even if you described the system in terms its specific microstate, and thus by the definition above your computed entropy would be the minimum possible, you still wouldn't be able to do anything that a more regular model couldn't do. Maxwell's demon is not a physically possible being/machine.

> the entropy of a thermodynamic system is proportional to the amount of microstates (positions, types, momentum, etc. of individual particles) that would lead to the same macrostate (temperature, volume, pressure, etc.). It doesn't matter if an observer is aware of any of this, it's an objective property of the system.

The meaning of "would lead to the same macrostate" (and therefore the entropy) is not an "objective" property of the system (positions, types, momentum, etc. of individual particles). At least not in the way that the energy is an "objective" property of the system.

The entropy is an "objective" property of the pair formed by the system (which can be described by a microstate) and some particular way of defining macrostates for that system.

That's what people mean when they say that the entropy is not an "objective" property of a physical system: that it depends on how we choose to describe that physical system (and that description is external to the physical system itself).

Of course, if you define "system" as "the underlying microscopical system plus this thermodynamical system description that takes into account some derived state variables only" the situation is not the same as if you define "system" as "the underlying microscopical system alone".

> That's what people mean when they say that the entropy is not an "objective" property of a physical system: that it depends on how we choose to describe that physical system (and that description is external to the physical system itself).

I understand that's what they mean, but this is the part that I think is either trivial or wrong. That is, depending on your choice you'll of course get different values, but it won't change anything about the system. It's basically like choosing to measure speed in meters per second or in furlongs per fortnight, or choosing the coordinate system and reference frame: you get radically different values, but relative results are always the same.

If a system has high entropy in the traditional sense, and another one has lower entropy, and the difference is high enough that you can run an engine by transferring heat from one to the other, then this difference and this fact will remain true whatever valid choice you make for how you describe the system's macrostates. This is the sense in which the entropy is an objective, observer-independent property of the system itself: same as energy, position, momentum, and anything else we care to measure.

> I understand that's what they mean, but this is the part that I think is either trivial or wrong. That is, depending on your choice you'll of course get different values, but it won't change anything about the system. It's basically like choosing to measure speed in meters per second or in furlongs per fortnight, or choosing the coordinate system and reference frame: you get radically different values, but relative results are always the same.

I would agree that it's trivial but then it's equally trivial that it's not just like a change of coordinates.

Say that you choose to represent the macrostate of a volume of gas using either (a) its pressure or (b) the partial pressures of the helium and argon that make it up. If you put together two volumes of the same mixture the entropy won't change. The entropy after they mix is just the sum of the entropies before mixing.

However when you put together one volume of helium and a one volume of argon the entropy calculated under choice (a) doesn't change but the entropy calculated under choice (b) does increase. We're not calculating the same thing in different units: we're calculating different things. There is no change of units that makes a quantity change and also remain constant!

The (a)-entropy and the (b)-entropy are different things. Of course it's the same concept applied to two different situations but that doesn't mean it's the same thing. (Otherwise one could also say that the momentum of a particule doesn't depend on its mass or velocity because it's always the same concept applied in different situations.)

> However when you put together one volume of helium and a one volume of argon the entropy calculated under choice (a) doesn't change but the entropy calculated under choice (b) does increase. We're not calculating the same thing in different units: we're calculating different things. There is no change of units that makes a quantity change and also remain constant!

Agreed, this is not like a coordinate transform at all. But the difference from a coordinate transform is that they are not both equally valid choices for describing the physical phenomenon. Choice (a) is simply wrong: it will not accurately predict how certain experiments with the combined gas will behave.

It will predict how other certain experiments with the combined gas will behave. That's what people mean when they say that the entropy is not an "objective" property of a physical system: that it depends on how we choose to describe that physical system - and what experiments we can perform acting on that description.

What would be an example of such an experiment?

By my understanding, even if we have no idea what gas we have, if we put it into a calorimeter and measure the amount of heat we need to transfer to it to change its temperature to some value, we will get a value that will be different for a gas made up of only argon versus one that contains both neon and argon. Doesn't this show that there is some objective definition of the entropy of the gas that doesn't care about an observer's knowledge of it?

Actually the molar heat capacity for neon, or argon, or a mixture thereof, is the same. These are monotomic ideal gases as far as your calorimeter measurements can see.

If the number of particles is the same you’ll need the same heat to increase the temperature by some amount and the entropy increase will be the same. Of course you could do other things to find out what it is, like weighing the container or reading the label.

No, they are not. The entropy of an ideal monatomic gas depends on the mass of its atoms (see the Sackur–Tetrode equation). And a gas mix is not an ideal monatomic gas; its entropy increases at the same temperature and volume compared to an equal volume divided between the two gases.

Also, entropy is not the same thing as heat capacity. It's true that I didn't describe the entropy measurement process very well, so I may have been ambiguous, but they are not the same quantity.

I'll leave the discussion here but let me remind you that you talked (indirectly) about changes in entropy and not about absolute entropies: "if we put it into a calorimeter and measure the amount of heat we need to transfer to it to change its temperature to some value".

Note as well that the mass dependence in that equation for the entropy is just an additive term. The absolute value of the entropy may be different but the change in entropy is the same when you heat a 1l container of helium or neon or a mixture of them from 300K to 301K. That's 0.0406 moles of gas. The heat flow is 0.506 joules. The change in entropy is approximately 0.0017 J/K.

> And a gas mix is not an ideal monatomic gas; its entropy increases at the same temperature and volume compared to an equal volume divided between the two gases.

A mix of ideal gases is an ideal gas and its heat capacity is the weighted average of the heat capacities (trivially equal to the heat capacity of the components when it's the same). The change of entropy when you heat one, or the other, or the mix, will be the same (because you're calculating exactly the same integral of the same heat flow).

The difference in absolute value is irrelevant when we are discussing changes in entropy and measurements of the amount of heat needed to increase the temperature and whether you "will get a value that will be different for a gas made up of only argon versus one that contains both neon and argon".

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