4 ms·

> the entropy of a thermodynamic system is proportional to the amount of microstates (positions, types, momentum, etc. of individual particles) that would lead

by kgwgk 2y ago

> 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".

tsimionescu 2y ago
> 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.
- kgwgk 2y ago
  > 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.)
  - tsimionescu 2y ago
    > 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.
    - kgwgk 2y ago

by kgwgk 2y ago

tsimionescu 2y ago

> 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.

kgwgk 2y ago
> 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.)
- tsimionescu 2y ago
  > 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.
  - kgwgk 2y ago