5 ms·

Don't see how the connection is superficial. Even the classical macroscopic definition of entropy as ΔS=∫TdQ can be derived from the information theory perspect

by enugu 2y ago

Don't see how the connection is superficial. Even the classical macroscopic definition of entropy as ΔS=∫TdQ can be derived from the information theory perspective as Baez shows in article(using entropy maximizing distributions and Lagrange multipliers). If you have a more specific critique, it would be good to discuss.

im3w1l 2y ago
In classical physics there is no real objective randomness. Particles have a defined position and momentum and those evolve deterministically. If you somehow learned these then the shannon entropy is zero. If entropy is zero then all kinds of things break down. So now you are forced to consider e.g. temperature an impossibility without quantum-derived randomness, even though temperature does not really seem to be a quantum thing.
- kgwgk 2y ago
  > Particles have a defined position and momentum Which we don’t know precisely. Entropy is about not knowing. > If you somehow learned these then the shannon entropy is zero. Minus infinity. Entropy in classical statistical mechanics is proportional to the logarithm of the volume in phase space. (You need an appropriate extension of Shannon’s entropy to continuous distributions.) > So now you are forced to consider e.g. temperature an impossibility without quantum-derived randomness Or you may study statistical mechanics :-)
  - kergonath 2y ago
    > Which we don’t know precisely. Entropy is about not knowing. No, it is not about not knowing. This is an instance of the intuition from Shannon’s entropy does not translate to statistical Physics. It is about the number of possible microstates, which is completely different. In Physics, entropy is a property of a bit of matter, it is not related to the observer or their knowledge. We can measure the enthalpy change of a material sample and work out its entropy without knowing a thing about its structure. > Minus infinity. Entropy in classical statistical mechanics is proportional to the logarithm of the volume in phase space. No, 0. In this case, there is a single state with p=1 and and S = - k Σ p ln(p) = 0. This is the same if you consider the phase space because then it is reduced to a single point (you need a bit of distribution theory to prove it rigorously but it is somewhat intuitive). The probability p of an microstate is always between 0 and 1, therefore p ln(p) is always negative and S is always positive. You get the same using Boltzmann’s approach, in which case Ω = 1 and S = k ln(Ω) is also 0. > (You need an appropriate extension of Shannon’s entropy to continuous distributions.) Gibbs’ entropy. > Or you may study statistical mechanics Indeed.
    - kgwgk 2y ago
      > possible microstates Conditional on the known macrostate. Because we don’t know the precise microstate - only which microstates are possible. If your reasoning is that « experimental entropy can be measured so it’s not about that » then it’s not about macrostates and microstates either!
    - nyssos 2y ago
      > In Physics, entropy is a property of a bit of matter, it is not related to the observer or their knowledge. We can measure the enthalpy change of a material sample and work out its entropy without knowing a thing about its structure. Enthalpy is also dependent on your choice of state variables, which is in turn dictated by which observables you want to make predictions about: whether two microstates are distinguishable, and thus whether the part of the same macrostate, depends on the tools you have for distinguishing them.
      - kergonath 2y ago
        A calorimeter does not care about anyone’s choice of state variables. Entropy is not only something that exists in abstract theoretical constructs, it is something we can get experimentally.
    - kgwgk 2y ago
      >>> Particles have a defined position and momentum [...] If you somehow learned these then the shannon entropy is zero. >> Entropy in classical statistical mechanics is proportional to the logarithm of the volume in phase space [and diverges to minus infinity if you define precisely the position and momentum of the particles and the volume in phase sphere goes to zero] > [It's zero also] if you consider the phase space because then it is reduced to a single point (you need a bit of distribution theory to prove it rigorously but it is somewhat intuitive). > The probability p of an microstate is always between 0 and 1, therefore p ln(p) is always negative and S is always positive. The points in the phase space are not "microstates" with probability between 0 and 1. It's a continuous distribution and if it collapses to a point (i.e. you somehow learned the exact positions and momentums) the density at that point is unbounded. The entropy is also unbounded and goes to minus infinity as the volume in phase space collapses to zero. You can avoid the divergence by dividing the continuous phase space into discrete "microstates" but having a well-defined "microstate" corresponding to some finite volume in phase space is not the same as what was written above about "particles having a defined position and momentum" that is "somehow learned". The microstates do not have precisely defined positions and momentums. The phase space is not reduced to a single point in that case. If the phase space is reduced to a single point I'd like to see your proof that S(ρ) = −k ∫ ρ(x) log ρ(x) dx = 0
- enugu 2y ago
  > If entropy is zero then all kinds of things break down. Entropy is a macroscopic variable and if you allow microscopic information, strange things can happen! One can move from a high entropy macrostate to a low entropy macrostate if you choose the initial microstate carefully. But this is not a reliable process which you can reproduce experimentally, ie. it is not a thermodynamic process. A thermodynamics process P is something which takes a macrostate A to a macrostate B, independent of which microstate a0, a1, a2.. in A you started off with it. If the process depends on microstate, then it wouldn't be something we would recognize as we are looking from the macro perspective.

by enugu 2y ago

im3w1l 2y ago

In classical physics there is no real objective randomness. Particles have a defined position and momentum and those evolve deterministically. If you somehow learned these then the shannon entropy is zero. If entropy is zero then all kinds of things break down. So now you are forced to consider e.g. temperature an impossibility without quantum-derived randomness, even though temperature does not really seem to be a quantum thing.

kgwgk 2y ago
> Particles have a defined position and momentum Which we don’t know precisely. Entropy is about not knowing. > If you somehow learned these then the shannon entropy is zero. Minus infinity. Entropy in classical statistical mechanics is proportional to the logarithm of the volume in phase space. (You need an appropriate extension of Shannon’s entropy to continuous distributions.) > So now you are forced to consider e.g. temperature an impossibility without quantum-derived randomness Or you may study statistical mechanics :-)
- kergonath 2y ago
  > Which we don’t know precisely. Entropy is about not knowing. No, it is not about not knowing. This is an instance of the intuition from Shannon’s entropy does not translate to statistical Physics. It is about the number of possible microstates, which is completely different. In Physics, entropy is a property of a bit of matter, it is not related to the observer or their knowledge. We can measure the enthalpy change of a material sample and work out its entropy without knowing a thing about its structure. > Minus infinity. Entropy in classical statistical mechanics is proportional to the logarithm of the volume in phase space. No, 0. In this case, there is a single state with p=1 and and S = - k Σ p ln(p) = 0. This is the same if you consider the phase space because then it is reduced to a single point (you need a bit of distribution theory to prove it rigorously but it is somewhat intuitive). The probability p of an microstate is always between 0 and 1, therefore p ln(p) is always negative and S is always positive. You get the same using Boltzmann’s approach, in which case Ω = 1 and S = k ln(Ω) is also 0. > (You need an appropriate extension of Shannon’s entropy to continuous distributions.) Gibbs’ entropy. > Or you may study statistical mechanics Indeed.
  - kgwgk 2y ago
    > possible microstates Conditional on the known macrostate. Because we don’t know the precise microstate - only which microstates are possible. If your reasoning is that « experimental entropy can be measured so it’s not about that » then it’s not about macrostates and microstates either!
  - nyssos 2y ago
    > In Physics, entropy is a property of a bit of matter, it is not related to the observer or their knowledge. We can measure the enthalpy change of a material sample and work out its entropy without knowing a thing about its structure. Enthalpy is also dependent on your choice of state variables, which is in turn dictated by which observables you want to make predictions about: whether two microstates are distinguishable, and thus whether the part of the same macrostate, depends on the tools you have for distinguishing them.
    - kergonath 2y ago
      A calorimeter does not care about anyone’s choice of state variables. Entropy is not only something that exists in abstract theoretical constructs, it is something we can get experimentally.
  - kgwgk 2y ago
    >>> Particles have a defined position and momentum [...] If you somehow learned these then the shannon entropy is zero. >> Entropy in classical statistical mechanics is proportional to the logarithm of the volume in phase space [and diverges to minus infinity if you define precisely the position and momentum of the particles and the volume in phase sphere goes to zero] > [It's zero also] if you consider the phase space because then it is reduced to a single point (you need a bit of distribution theory to prove it rigorously but it is somewhat intuitive). > The probability p of an microstate is always between 0 and 1, therefore p ln(p) is always negative and S is always positive. The points in the phase space are not "microstates" with probability between 0 and 1. It's a continuous distribution and if it collapses to a point (i.e. you somehow learned the exact positions and momentums) the density at that point is unbounded. The entropy is also unbounded and goes to minus infinity as the volume in phase space collapses to zero. You can avoid the divergence by dividing the continuous phase space into discrete "microstates" but having a well-defined "microstate" corresponding to some finite volume in phase space is not the same as what was written above about "particles having a defined position and momentum" that is "somehow learned". The microstates do not have precisely defined positions and momentums. The phase space is not reduced to a single point in that case. If the phase space is reduced to a single point I'd like to see your proof that S(ρ) = −k ∫ ρ(x) log ρ(x) dx = 0
enugu 2y ago
> If entropy is zero then all kinds of things break down. Entropy is a macroscopic variable and if you allow microscopic information, strange things can happen! One can move from a high entropy macrostate to a low entropy macrostate if you choose the initial microstate carefully. But this is not a reliable process which you can reproduce experimentally, ie. it is not a thermodynamic process. A thermodynamics process P is something which takes a macrostate A to a macrostate B, independent of which microstate a0, a1, a2.. in A you started off with it. If the process depends on microstate, then it wouldn't be something we would recognize as we are looking from the macro perspective.