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This is my definition: https://www.sciencedirect.com/topics/mathematics/equilibrium-macrostate https://www.sciencedirect.com/topics/mathematics/equilibrium...

This is my definition: https://www.sciencedirect.com/topics/mathematics/equilibrium-macrostate https://www.sciencedirect.com/topics/mathematics/equilibrium... "The equilibrium macrostate is that with the most microstates, and this is the state of greatest entropy. A macroscopic flux is most likely in the direction of increasing entropy." >I’m not making up the idea of an equilibrium state being associated with a macrostate. I know you aren't. TBH not even sure why equilibrium needs to be brought up. The topic "is entropy is independent of knowledge?" You disagree with that statement. Macrostate is defined in terms of microstates and is independent of your knowledge of the exact configuration of the microstates. You have some blunt tool like the thermometer that gives you macroscopic data and that hides the microstates from you. But if I had some precision tool that reads the position of every particle I can still identify how that configuration of microstates will influence the blunt tool. The blunt tool hides knowledge, but your knowledge of the microstate does not change the reading on that blunt tool. Because entropy is defined in termms of macrostate it stays the same regardless of which tool you used to do the measurement.

kgwgk 4y ago
> This is my definition See? We agree that there is an equilibrium macrostate. When I fist asked you to clarify what did you mean by macrostate you told me that “it changes with time; even at equilibrium.” We agree that the entropy depends on the macrostate. Note that the macrostate is our description of the system and depends on how we choose to describe it which normally depends on what are the constraints, how it was prepared., etc. It’s not just a property of the position of those balls. It’s because we agree that Gibbs’ entropy is a function of the macrostate that I asked how did you define it in your example. You told me: “Temperature as measured by a thermometer is the macrostate. Every possible configuration of particles (microstates) that causes mercury to rise to a certain degree represents a different macrostate.” I asked “How does your configuration where the balls were near one corner in the cube cause mercury to rise to a different level than the configuration where they occupy a larger volume near the center?” and the answer “The balls have to touch thermometer” doesn’t cut it. The balls don’t touch the thermometer in either case. You seemed to imply that the higher concentration means a different macrostate with lower entropy. Or maybe the low entropy in you example is because the balls are near a corner? Anyway, it would indeed have been easier to say that definition of macrostate included the density of particles in each octant of the cube - or something like that.
- deltasevennine 4y ago
  >See? We agree that there is an equilibrium macrostate. When I fist asked you to clarify what did you mean by macrostate you told me that “it changes with time; even at equilibrium.” I still stand by my statement. Even at equilibrium it can lower in entropy. The equilibrium is simply the highest entropy state. >I asked “How does your configuration where the balls were near one corner in the cube cause mercury to rise to a different level than the configuration where they occupy a larger volume near the center?” and the answer “The balls have to touch thermometer” doesn’t cut it. The balls don’t touch the thermometer in either case. I stated this is pedantism. The concept and intuition remain true. I changed the definition so that it's a volume around the thermometer if the particle is in that volume and heading for the thermometer is counts as a collision. I stated all of this already. >You seemed to imply that the higher concentration means a different macrostate with lower entropy. Or maybe the low entropy in you example is because the balls are near a corner? Yes. The higher concentration has a lower probability of occurring. And occupies a different temperature reading on the thermometer. Each temperature reading is a different macrostate. >Anyway, it would indeed have been easier to say that definition of macrostate included the density of particles in each octant of the cube - or something like that. Sure, Divide the box into a bunch of cubes. If 1 or more particles are in the cube then that cube represents 1, otherwise 0. Add those numbers up and that represents a macrostate. The inuition remains the same. For all particles to be concentrated in 1 cube is a very low probability. And the macrostate will be quite low too. With enough cubes and boxes such a state has a very low probability of occuring. But all of this is, again, independent of your knowledge of where the particles are in each cube.
  - kgwgk 4y ago
    > And occupies a different temperature reading on the thermometer. You are unable to explain how reducing the space occupied by the initial configuration you proposed could change the temperature - which you say is 0 in all those cases - or the macrostate - which you call absolute zero macrostate in all those cases. The entropy would be the same whenever the particules are more concentrated than in your example - even though they would have lower probability of ocurrence. Don't you agree? > With enough cubes and boxes such a state has a very low probability of occuring. Sure. The thing is that if you calculate an entropy using Gibbs formula from the distribution of microstates for a given macrostate the value that you obtain depends on how many cubes are used to define the macrostate. There is no entropy of the microstate - the entropy depends on how you decide to define the macrostate. If the lattice is fine enough, and the particles indistinguishable, in the limit the entropy is zero - the macrostate becomes the same as the microstate and one single microstate is possible. It all depends on the description we make of the system. The "I have balls in a box and I know their positions" example was incomplete. If for example I have N of those balls in equilibrium in an isolated box of volume V and I know the energy E I know the equilibrium macrostate. The energy won't change because it's isolated. The macrostate will not change. The (equi)probability distribution of the possible microstates corresponding to the macrostate won't change. Gibbs entropy which is calculated using that distribution won't change. If by sheer luck all gas particles moved to the exact left side of the container the energy wouldn't change, the macrostate wouldn't change, the entropy wouldn't change - it doesn't matter how unlikely that is. It doesn't matter if you know the position of each particle. The entropy for the thermodynamical system described in the previous paragraph is independent of your knowledge of where the particles are or how unlikely you think their positions are. I agree that you could use alternative ways of defining the macrostate where that happens! (And calculating Gibbs entropy will give different results. One can even get zero entropy using the microstate as macrostate - and one may actually want to do that if the microstate is known!) > Even at equilibrium it can lower in entropy. The equilibrium is simply the highest entropy state. I would understand that you said either: "the equilibrium is the highest entropy state but there may be fluctuations that shift the system temporarily out equilibrium" or "the equilibrium is the highest entropy state and possibly a distribution of states around it." Both concepts are used. When you say that the equilibrium is simply the highest entropy state but the equilibrium entropy can also be lower I'm not sure if it can be read as one of those or you're saying something else entirely.

kgwgk 4y ago

> This is my definition See? We agree that there is an equilibrium macrostate. When I fist asked you to clarify what did you mean by macrostate you told me that “it changes with time; even at equilibrium.” We agree that the entropy depends on the macrostate. Note that the macrostate is our description of the system and depends on how we choose to describe it which normally depends on what are the constraints, how it was prepared., etc. It’s not just a property of the position of those balls. It’s because we agree that Gibbs’ entropy is a function of the macrostate that I asked how did you define it in your example. You told me: “Temperature as measured by a thermometer is the macrostate. Every possible configuration of particles (microstates) that causes mercury to rise to a certain degree represents a different macrostate.” I asked “How does your configuration where the balls were near one corner in the cube cause mercury to rise to a different level than the configuration where they occupy a larger volume near the center?” and the answer “The balls have to touch thermometer” doesn’t cut it. The balls don’t touch the thermometer in either case. You seemed to imply that the higher concentration means a different macrostate with lower entropy. Or maybe the low entropy in you example is because the balls are near a corner? Anyway, it would indeed have been easier to say that definition of macrostate included the density of particles in each octant of the cube - or something like that.

deltasevennine 4y ago
>See? We agree that there is an equilibrium macrostate. When I fist asked you to clarify what did you mean by macrostate you told me that “it changes with time; even at equilibrium.” I still stand by my statement. Even at equilibrium it can lower in entropy. The equilibrium is simply the highest entropy state. >I asked “How does your configuration where the balls were near one corner in the cube cause mercury to rise to a different level than the configuration where they occupy a larger volume near the center?” and the answer “The balls have to touch thermometer” doesn’t cut it. The balls don’t touch the thermometer in either case. I stated this is pedantism. The concept and intuition remain true. I changed the definition so that it's a volume around the thermometer if the particle is in that volume and heading for the thermometer is counts as a collision. I stated all of this already. >You seemed to imply that the higher concentration means a different macrostate with lower entropy. Or maybe the low entropy in you example is because the balls are near a corner? Yes. The higher concentration has a lower probability of occurring. And occupies a different temperature reading on the thermometer. Each temperature reading is a different macrostate. >Anyway, it would indeed have been easier to say that definition of macrostate included the density of particles in each octant of the cube - or something like that. Sure, Divide the box into a bunch of cubes. If 1 or more particles are in the cube then that cube represents 1, otherwise 0. Add those numbers up and that represents a macrostate. The inuition remains the same. For all particles to be concentrated in 1 cube is a very low probability. And the macrostate will be quite low too. With enough cubes and boxes such a state has a very low probability of occuring. But all of this is, again, independent of your knowledge of where the particles are in each cube.
- kgwgk 4y ago
  > And occupies a different temperature reading on the thermometer. You are unable to explain how reducing the space occupied by the initial configuration you proposed could change the temperature - which you say is 0 in all those cases - or the macrostate - which you call absolute zero macrostate in all those cases. The entropy would be the same whenever the particules are more concentrated than in your example - even though they would have lower probability of ocurrence. Don't you agree? > With enough cubes and boxes such a state has a very low probability of occuring. Sure. The thing is that if you calculate an entropy using Gibbs formula from the distribution of microstates for a given macrostate the value that you obtain depends on how many cubes are used to define the macrostate. There is no entropy of the microstate - the entropy depends on how you decide to define the macrostate. If the lattice is fine enough, and the particles indistinguishable, in the limit the entropy is zero - the macrostate becomes the same as the microstate and one single microstate is possible. It all depends on the description we make of the system. The "I have balls in a box and I know their positions" example was incomplete. If for example I have N of those balls in equilibrium in an isolated box of volume V and I know the energy E I know the equilibrium macrostate. The energy won't change because it's isolated. The macrostate will not change. The (equi)probability distribution of the possible microstates corresponding to the macrostate won't change. Gibbs entropy which is calculated using that distribution won't change. If by sheer luck all gas particles moved to the exact left side of the container the energy wouldn't change, the macrostate wouldn't change, the entropy wouldn't change - it doesn't matter how unlikely that is. It doesn't matter if you know the position of each particle. The entropy for the thermodynamical system described in the previous paragraph is independent of your knowledge of where the particles are or how unlikely you think their positions are. I agree that you could use alternative ways of defining the macrostate where that happens! (And calculating Gibbs entropy will give different results. One can even get zero entropy using the microstate as macrostate - and one may actually want to do that if the microstate is known!) > Even at equilibrium it can lower in entropy. The equilibrium is simply the highest entropy state. I would understand that you said either: "the equilibrium is the highest entropy state but there may be fluctuations that shift the system temporarily out equilibrium" or "the equilibrium is the highest entropy state and possibly a distribution of states around it." Both concepts are used. When you say that the equilibrium is simply the highest entropy state but the equilibrium entropy can also be lower I'm not sure if it can be read as one of those or you're saying something else entirely.