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I already mentioned that logic can be defined in terms of probability above. Here's the strange part. Let's say we make those probabilities to be 100% thus we

I already mentioned that logic can be defined in terms of probability above. Here's the strange part. Let's say we make those probabilities to be 100% thus we have ordinary logic without probability. Then let's create a physical closed Newtonian system: The interior of a cube with no gravity and a bunch of identical bouncing balls which obey Newtonian physics and therefore logic. Let's say those balls all have a random initial velocity at a random direction but all those balls are initially positioned near one corner in the cube. Thus the balls from a position stand point start with low entropy. I use the term random, but it's not really random as you have perfect knowledge of these numbers, you know what they all are. As time increases, entropy increases. The balls as a system begin to increase in entropy. Entropy continues to rise towards an equilibrium. That is the question. Entropy is a probabilistic phenomenon. It occurs because ball configurations that are spread out have higher probabilities of occurring then balls that are concentrated in a corner. Thus given enough time the balls occupy the higher entropy state. My question is, WHY does this occur. Probability (aka Entropy) is ARISING out of a perfect Newtonian system following perfect logic without any probabilistic extensions. Can entropy be derived from Newtonian physics? The better question is, can probability be derived from Newtonian physics because entropy is in actuality a phenomenon of probability? Any system that uses pure logic (and completely avoids probability in any part of its definition) exhibits this sort of rule as long as you introduce a sort of randomness into initial parameters above. Probability permeates everything.

kgwgk 4y ago
> As time increases, entropy increases. The balls as a system begin to increase in entropy. Entropy continues to rise towards an equilibrium. If you know the positions and velocities of the balls at all times arguably the entropy is always zero. What defintion of entropy are you using?
- deltasevennine 4y ago
  The statistical version. The equation is independent of knowledge. Your logic is flawed. Think about it. Knowing more or less about a system does not change it's entropy. If I knew the position and velocity of every particle in a gas chamber does that somehow lower it's temperature? No. Macro-states summarize perfect information. But with knowledge of perfect information, you still have the ability to summarize it. Or in other words, probability still exists even when you have perfect knowledge of everything and the future.
  - kgwgk 4y ago
    If you mean Boltzmann's formula it will be valid for systems in thermodynamical equilibrium. Is your system in thermodynamical equilibrium? You talk about macrostates. How do you define a macrostate for your initial configuration? Why this one and not another? Does it change over time? Think of it from another angle. You have a box with some of these balls inside - all you know is that they all have the same energy. Do you agree that the macrostate won't change over time and the entropy will remain constant? However, if you knew the positions they will be concentrated in some regions more than in others. That's what you called lower entropy in your example. So what is it, does knowing more about a system change it's entropy or not?
    - deltasevennine 4y ago
      > If you mean Boltzmann's formula it will be valid for systems in thermodynamical equilibrium. Is your system in thermodynamical equilibrium? Gibbs >You talk about macrostates. How do you define a macrostate for your initial configuration? Why this one and not another? Does it change over time? I'll choose something arbitrary. 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. And yes it changes with time; even at equilibrium. >Think of it from another angle. You have a box with some of these balls inside - all you know is that they all have the same energy. Do you agree that the macrostate won't change over time and the entropy will remain constant? No don't agree. If balls are located on one side of the box and the thermometer is on the other side of the box. The thermometer reads nothing. When the balls increase in entropy they collide with the thermometer producing a reading. >However, if you knew the positions they will be concentrated in some regions more than in others. That's what you called lower entropy in your example. Yeah I already don't agree with you, so the rest of your argument is gone. If you define macrostate as total energy in a system then yeah it never changes. But that's not the definition of macrostate. It's just one arbitrary choice you have chosen.
      - kgwgk 4y ago
        > Gibbs Ok. If you have a system at constant energy they are the same (all the microstates have the same probability). If you have a system a constant temperature a microstate can always correspond to different macrostates (it seems you’re not concerned about that ambiguity though). In any case, the entropy is a property of (the ensemble of microstates that conform) the macrostate and not a property of the microstate. For a given macrostate you cannot use Gibbs entropy to talk about low-entropy and high-entropy microstates. > 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. How does your configuration where the balls where near one corner in the cube causes mercury to rise to a different level than the configuration where they occupy a larger volume near the center? > And yes it changes with time; even at equilibrium. What’s your definition of equilibrium? It’s hard for me to see what are we talking about, really. > If balls are located on one side of the box and the thermometer is on the other side of the box. The thermometer reads nothing. One could also say that if the balls are located anywhere within the box the thermometer reads nothing. The thermometer reading exists only when enough time has passed and the mercury-balls composite system is in equilibrium. > When the balls increase in entropy they collide with the thermometer producing a reading. Apparently you say that the entropy increases because the macrostate changes. But the macrostate is undefined until the balls hit the thermometer. But the entropy has to increase for the balls to move. It’s all a bit confusing, don’t you think? Won’t the balls collide with the thermometer even if the entropy doesn’t change? It’s their movement that will take them there. Consider your state - or any other state, for that matter - at t=0 and the state at t=1ps where the balls have advanced slightly. Is the macrostate different? How? Is the entropy different? How?
kgwgk 4y ago
> My question is, WHY does this occur. Probability (aka Entropy) is ARISING out of a perfect Newtonian system following perfect logic without any probabilistic extensions. Probability can arise out of perfect information only when you throw away that perfect information and chose to describe the system probabilistically using imperfect information instead. If you describe fully the state of a classical system - giving the positions and velocities of the particles in this case - there is no uncertainty at all. The volume in the 6N-dimensional phase space of the possible states collapse to a single point. The entropy is a function of the accessible volume in the phase space - it's minimised when the state is perfectly determined (it's defined to an additive constant and this minimum is conventionally set at 0). If you describe the state only partially there is uncertainty in the description. There is a non-zero volume accessible in the phase space. The larger this volume the higher the entropy. Interestingly the volume - and thus the entropy - doesn't change in classical mechanics (Liouville's theorem). In our perfect Newtonian system following perfect logic even if we don't know the precise state - and have a volume in phase space instead - we know how each of those potential states will evolve. There is no irreversibility, no changes in entropy. (Gibbs entropy is also conserved.) Only if we keep adding uncertainty / forgetting information as we go forward will entropy increase.
- kgwgk 4y ago
  > You have not answered my question. You aren't getting it. Rising Entropy is just an example for illustrative purposes, entropy is simply a probabilistic phenomenon. I thought that your question was how probability "appears" from classical mechanics and the answer is that you're putting it in - either by obligation (because you cannot describe the system perfectly) or by choice (because you decide not to do so even if you could). Entropy arises when the classical state of the system is not identified precisely as a point in phase space and you have a volume instead corresponding to the possible physical states. More uncertainty about the state, more volume, higher entropy. Less uncertainty about the state, less volume, lower entropy. No uncertainty about the state, no volume, zero entropy. >> Only if we keep adding uncertainty / forgetting information as we go forward will entropy increase. > Bro. Forgetting stuff doesn't change entropy. We already agreed that entropy is independent of knowledge. We already agreed that the entropy depends on the description that we make of the system. By forgetting stuff or otherwise adding uncertainty you get a worse description than what you could have if you didn't - that means higher entropy. If the uncertainty doesn't increase somehow the entropy remains constant. Entropy is a function of the volume in phase space and that volume is constant in classical -and quantum, for the matter- mechanics. If the equations describing the evolution are perfectly known the uncertainty doesn't have to increase. It doesn't matter how much time passes, the particles cannot be _anywhere_: their possible positions in the future are determined by their possible positions in the past. But this constant volume of possible positions may get more and more spread around even though they used to be in a compact region. See Figure 1.7 here: https://books.google.ch/books?id=DjUgfVFtY6YC&lpg=PA27&ots=aT3UBjrwTc&pg=PA27#v=onepage&q&f=false https://books.google.ch/books?id=DjUgfVFtY6YC&lpg=PA27&ots=a... By not making use of that additional information - possibly because we just can't afford that level of detail - we let uncertainty grow. That's what I meant by adding uncertainty / forgetting information. The volume in phase space representing the possible states is larger. The entropy of the degraded description of the system is higher.

kgwgk 4y ago

> As time increases, entropy increases. The balls as a system begin to increase in entropy. Entropy continues to rise towards an equilibrium. If you know the positions and velocities of the balls at all times arguably the entropy is always zero. What defintion of entropy are you using?

deltasevennine 4y ago
The statistical version. The equation is independent of knowledge. Your logic is flawed. Think about it. Knowing more or less about a system does not change it's entropy. If I knew the position and velocity of every particle in a gas chamber does that somehow lower it's temperature? No. Macro-states summarize perfect information. But with knowledge of perfect information, you still have the ability to summarize it. Or in other words, probability still exists even when you have perfect knowledge of everything and the future.
- kgwgk 4y ago
  If you mean Boltzmann's formula it will be valid for systems in thermodynamical equilibrium. Is your system in thermodynamical equilibrium? You talk about macrostates. How do you define a macrostate for your initial configuration? Why this one and not another? Does it change over time? Think of it from another angle. You have a box with some of these balls inside - all you know is that they all have the same energy. Do you agree that the macrostate won't change over time and the entropy will remain constant? However, if you knew the positions they will be concentrated in some regions more than in others. That's what you called lower entropy in your example. So what is it, does knowing more about a system change it's entropy or not?
  - deltasevennine 4y ago
    > If you mean Boltzmann's formula it will be valid for systems in thermodynamical equilibrium. Is your system in thermodynamical equilibrium? Gibbs >You talk about macrostates. How do you define a macrostate for your initial configuration? Why this one and not another? Does it change over time? I'll choose something arbitrary. 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. And yes it changes with time; even at equilibrium. >Think of it from another angle. You have a box with some of these balls inside - all you know is that they all have the same energy. Do you agree that the macrostate won't change over time and the entropy will remain constant? No don't agree. If balls are located on one side of the box and the thermometer is on the other side of the box. The thermometer reads nothing. When the balls increase in entropy they collide with the thermometer producing a reading. >However, if you knew the positions they will be concentrated in some regions more than in others. That's what you called lower entropy in your example. Yeah I already don't agree with you, so the rest of your argument is gone. If you define macrostate as total energy in a system then yeah it never changes. But that's not the definition of macrostate. It's just one arbitrary choice you have chosen.
    - kgwgk 4y ago
      > Gibbs Ok. If you have a system at constant energy they are the same (all the microstates have the same probability). If you have a system a constant temperature a microstate can always correspond to different macrostates (it seems you’re not concerned about that ambiguity though). In any case, the entropy is a property of (the ensemble of microstates that conform) the macrostate and not a property of the microstate. For a given macrostate you cannot use Gibbs entropy to talk about low-entropy and high-entropy microstates. > 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. How does your configuration where the balls where near one corner in the cube causes mercury to rise to a different level than the configuration where they occupy a larger volume near the center? > And yes it changes with time; even at equilibrium. What’s your definition of equilibrium? It’s hard for me to see what are we talking about, really. > If balls are located on one side of the box and the thermometer is on the other side of the box. The thermometer reads nothing. One could also say that if the balls are located anywhere within the box the thermometer reads nothing. The thermometer reading exists only when enough time has passed and the mercury-balls composite system is in equilibrium. > When the balls increase in entropy they collide with the thermometer producing a reading. Apparently you say that the entropy increases because the macrostate changes. But the macrostate is undefined until the balls hit the thermometer. But the entropy has to increase for the balls to move. It’s all a bit confusing, don’t you think? Won’t the balls collide with the thermometer even if the entropy doesn’t change? It’s their movement that will take them there. Consider your state - or any other state, for that matter - at t=0 and the state at t=1ps where the balls have advanced slightly. Is the macrostate different? How? Is the entropy different? How?

kgwgk 4y ago

> My question is, WHY does this occur. Probability (aka Entropy) is ARISING out of a perfect Newtonian system following perfect logic without any probabilistic extensions. Probability can arise out of perfect information only when you throw away that perfect information and chose to describe the system probabilistically using imperfect information instead. If you describe fully the state of a classical system - giving the positions and velocities of the particles in this case - there is no uncertainty at all. The volume in the 6N-dimensional phase space of the possible states collapse to a single point. The entropy is a function of the accessible volume in the phase space - it's minimised when the state is perfectly determined (it's defined to an additive constant and this minimum is conventionally set at 0). If you describe the state only partially there is uncertainty in the description. There is a non-zero volume accessible in the phase space. The larger this volume the higher the entropy. Interestingly the volume - and thus the entropy - doesn't change in classical mechanics (Liouville's theorem). In our perfect Newtonian system following perfect logic even if we don't know the precise state - and have a volume in phase space instead - we know how each of those potential states will evolve. There is no irreversibility, no changes in entropy. (Gibbs entropy is also conserved.) Only if we keep adding uncertainty / forgetting information as we go forward will entropy increase.

kgwgk 4y ago
> You have not answered my question. You aren't getting it. Rising Entropy is just an example for illustrative purposes, entropy is simply a probabilistic phenomenon. I thought that your question was how probability "appears" from classical mechanics and the answer is that you're putting it in - either by obligation (because you cannot describe the system perfectly) or by choice (because you decide not to do so even if you could). Entropy arises when the classical state of the system is not identified precisely as a point in phase space and you have a volume instead corresponding to the possible physical states. More uncertainty about the state, more volume, higher entropy. Less uncertainty about the state, less volume, lower entropy. No uncertainty about the state, no volume, zero entropy. >> Only if we keep adding uncertainty / forgetting information as we go forward will entropy increase. > Bro. Forgetting stuff doesn't change entropy. We already agreed that entropy is independent of knowledge. We already agreed that the entropy depends on the description that we make of the system. By forgetting stuff or otherwise adding uncertainty you get a worse description than what you could have if you didn't - that means higher entropy. If the uncertainty doesn't increase somehow the entropy remains constant. Entropy is a function of the volume in phase space and that volume is constant in classical -and quantum, for the matter- mechanics. If the equations describing the evolution are perfectly known the uncertainty doesn't have to increase. It doesn't matter how much time passes, the particles cannot be _anywhere_: their possible positions in the future are determined by their possible positions in the past. But this constant volume of possible positions may get more and more spread around even though they used to be in a compact region. See Figure 1.7 here: https://books.google.ch/books?id=DjUgfVFtY6YC&lpg=PA27&ots=aT3UBjrwTc&pg=PA27#v=onepage&q&f=false https://books.google.ch/books?id=DjUgfVFtY6YC&lpg=PA27&ots=a... By not making use of that additional information - possibly because we just can't afford that level of detail - we let uncertainty grow. That's what I meant by adding uncertainty / forgetting information. The volume in phase space representing the possible states is larger. The entropy of the degraded description of the system is higher.