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>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;

>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.
- deltasevennine 4y ago
  >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? You're unable to read my explanation which i've already repeated twice. Go back and look it up. I even went with your cubical definition. I don't agree. Entropy is lower when the current macrostate has a low probability of occuring. >There is no entropy of the microstate - the entropy depends on how you decide to define the macrostate The macrostate is defined in terms of possible microstates. Thus Entropy relies on both microstate and macrostate. >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. Wrong. Energy wouldn't change. Macrostate changes. Entropy changes. >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. Isn't this my point? And weren't you against my point? Now you're in agreement? My point was entropy is independent of knowledge. Your point was that it is dependent. >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!) Except you don't have to do this. My point remains true given MY stated definitions of macrostate. >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. I made a true statement that from what I can gather you agree it's true. You're just trying to extrapolate my reasoning behind the statement. It's pointless. Example: If I give you a number, say -1, do I mean i^2 or 0 - 1? because one of those equations is impossible to express in reality.
  - kgwgk 4y ago
    >>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. > Wrong. Energy wouldn't change. Macrostate changes. Entropy changes. That's my thermodynamical system - that I decided to describe using the state variables N, V, E. S_kgwgk = constant I have agreed that you can chose to define macrostates as you wish! (You cannot measure them though. You don't have access to my system. They are purely hypothetical.) And then you can say things like "If the position of the particles in your system is X, S_deltasevennine = whatever." Hopefully you will agree that someone else could choose to define the macrostate differently and say things like "If the position of the particles in kgwgk's system is X, S_anon = somethingelse." Let's imagine that the position of the particles is actually X. What is the entropy of the system then? S_kgwgk? S_deltasevennine? S_anon? They are all different! If you think that you can define macrostates for my system in some arbitrary way (your own words: "I'll choose something arbitrary.") and others can't - or that yours are somehow more real - I really wonder why. And as I said, someone could also come and say "If the position [and velocities] of the particles in kgwgk's system is X, S_vonneumann = 0." [a complete definition of the microstate requires knowing both position and momentum - there might have been some ambiguity about that before but it shouldn't distract us from the main points]

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.

deltasevennine 4y ago
>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? You're unable to read my explanation which i've already repeated twice. Go back and look it up. I even went with your cubical definition. I don't agree. Entropy is lower when the current macrostate has a low probability of occuring. >There is no entropy of the microstate - the entropy depends on how you decide to define the macrostate The macrostate is defined in terms of possible microstates. Thus Entropy relies on both microstate and macrostate. >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. Wrong. Energy wouldn't change. Macrostate changes. Entropy changes. >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. Isn't this my point? And weren't you against my point? Now you're in agreement? My point was entropy is independent of knowledge. Your point was that it is dependent. >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!) Except you don't have to do this. My point remains true given MY stated definitions of macrostate. >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. I made a true statement that from what I can gather you agree it's true. You're just trying to extrapolate my reasoning behind the statement. It's pointless. Example: If I give you a number, say -1, do I mean i^2 or 0 - 1? because one of those equations is impossible to express in reality.
- kgwgk 4y ago
  >>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. > Wrong. Energy wouldn't change. Macrostate changes. Entropy changes. That's my thermodynamical system - that I decided to describe using the state variables N, V, E. S_kgwgk = constant I have agreed that you can chose to define macrostates as you wish! (You cannot measure them though. You don't have access to my system. They are purely hypothetical.) And then you can say things like "If the position of the particles in your system is X, S_deltasevennine = whatever." Hopefully you will agree that someone else could choose to define the macrostate differently and say things like "If the position of the particles in kgwgk's system is X, S_anon = somethingelse." Let's imagine that the position of the particles is actually X. What is the entropy of the system then? S_kgwgk? S_deltasevennine? S_anon? They are all different! If you think that you can define macrostates for my system in some arbitrary way (your own words: "I'll choose something arbitrary.") and others can't - or that yours are somehow more real - I really wonder why. And as I said, someone could also come and say "If the position [and velocities] of the particles in kgwgk's system is X, S_vonneumann = 0." [a complete definition of the microstate requires knowing both position and momentum - there might have been some ambiguity about that before but it shouldn't distract us from the main points]