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Themis
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First on the additive constant: it doesn't matter. Since, as you point out, classical thermodynamics defines $\Delta S$, not $S$, the classical entropy has an arbitrary constant attached to it. But it is always $\Delta S$ that we measure or care to calculate, the additive constant is irrelevant. The statistical entropy only needs to match the classical entropy when we take the difference between the same two states.

On the calculation of the statistical entropy – The short answer: By taking the microstate to be a volume element in momentum $\times$ space (phase space) and by assigning equal weight to all microstates with the same energy, volume and number of particles, we end up with a statistical entropy that behaves exactly the same as classical entropy. That's the beauty of statistical mechanics. No additional assumptions, discretizations or other such operations required.

On the calculation of the statistical entropy – The longer answer: We do not discretize the continuous phase space, we integrate over it. The easiest calculationdemonstration is in the canonical partition function: $$ Q = \frac{1}{h^{3N}N!} \int e^{-\mathbf{p}^2/2mkT} d^N \mathbf{p} \int e^{-E_P/mkT} d^{N} \mathbf{r} $$ where $\mathbf{p}$ is the vector of all $(x,y,z)$ components of momentum of all particles and $\mathbf{r}$ is the vector of all positions. Here a volume element of phase space is represented by the product $$ d^N\mathbf{p}~ d^N\mathbf{x} = dr_{1x} dp_{1x} dr_{1y} dp_{1y} dr_{1z} dp_{1z} \cdots $$$$ d^N\mathbf{p}~ d^N\mathbf{r} = dr_{1x} dp_{1x} dr_{1y} dp_{1y} dr_{1z} dp_{1z} \cdots $$ This is a true differential, which is to say that the size of the grid goes to zero. The momentum integral can be solved directly: $$ \frac{1}{h^{3N}} \int e^{-\mathbf{p}^2/2mkT} d^N \mathbf{p} = \left(\frac{2\pi m k T}{h^2}\right)^{3N/2} $$ The integral over space cannot be solved for arbitrary potential energy but in the special case $E_P=0$ (no interactions) we obtain the simple result $$ \int e^{-E_P/mkT} d^{N} \mathbf{r} = V^N $$ COmbiningCombining with the momentum integral we obtain the canonical partition of the ideal gas: $$ Q^\text{ig} = \frac{V^N}{N!}\left(\frac{2\pi m k T}{h^2}\right)^{3N/2} \tag{1} $$ Since the question was about entropy, the entropy in the canonical ensemble iswe calculate it as $$ \frac{S}{k} = \ln Q - \beta \left(\frac{\ln\partial Q}{\partial\beta}\right) $$ Applying this to the partition function of the ideal gas we obtain the an entropy that matches the classical entropy of the ideal gas to within an additive constant. The result can be found in any standard textbook. The point of this lengthier answer was to address the question about how to discretize the phase space: there is no discretization involved.

First on the additive constant: it doesn't matter. Since, as you point out, classical thermodynamics defines $\Delta S$, not $S$, the classical entropy has an arbitrary constant attached to it. But it is always $\Delta S$ that we measure or care to calculate, the additive constant is irrelevant. The statistical entropy only needs to match the classical entropy when we take the difference between the same two states.

On the calculation of the statistical entropy – The short answer: By taking the microstate to be a volume element in momentum $\times$ space (phase space) and by assigning equal weight to all microstates with the same energy, volume and number of particles, we end up with a statistical entropy that behaves exactly the same as classical entropy. That's the beauty of statistical mechanics. No additional assumptions, discretizations or other such operations required.

On the calculation of the statistical entropy – The longer answer: We do not discretize the continuous phase space, we integrate over it. The easiest calculation is in the canonical partition function: $$ Q = \frac{1}{h^{3N}N!} \int e^{-\mathbf{p}^2/2mkT} d^N \mathbf{p} \int e^{-E_P/mkT} d^{N} \mathbf{r} $$ where $\mathbf{p}$ is the vector of all $(x,y,z)$ components of momentum of all particles and $\mathbf{r}$ is the vector of all positions. Here a volume element of phase space is represented by the product $$ d^N\mathbf{p}~ d^N\mathbf{x} = dr_{1x} dp_{1x} dr_{1y} dp_{1y} dr_{1z} dp_{1z} \cdots $$ This is a true differential, which is to say that the size of the grid goes to zero. The momentum integral can be solved directly: $$ \frac{1}{h^{3N}} \int e^{-\mathbf{p}^2/2mkT} d^N \mathbf{p} = \left(\frac{2\pi m k T}{h^2}\right)^{3N/2} $$ The integral over space cannot be solved for arbitrary potential energy but in the special case $E_P=0$ (no interactions) we obtain the simple result $$ \int e^{-E_P/mkT} d^{N} \mathbf{r} = V^N $$ COmbining with the momentum integral we obtain the canonical partition of the ideal gas: $$ Q^\text{ig} = \frac{V^N}{N!}\left(\frac{2\pi m k T}{h^2}\right)^{3N/2} \tag{1} $$ Since the question was about entropy, the entropy in the canonical ensemble is $$ \frac{S}{k} = \ln Q - \beta \left(\frac{\ln\partial Q}{\partial\beta}\right) $$ Applying this to the partition function of the ideal gas we obtain the an entropy that matches the classical entropy of the ideal gas to within an additive constant. The result can be found in any standard textbook. The point of this lengthier answer was to address the question about how to discretize the phase space: there is no discretization involved.

First on the additive constant: it doesn't matter. Since, as you point out, classical thermodynamics defines $\Delta S$, not $S$, the classical entropy has an arbitrary constant attached to it. But it is always $\Delta S$ that we measure or care to calculate, the additive constant is irrelevant. The statistical entropy only needs to match the classical entropy when we take the difference between the same two states.

On the calculation of the statistical entropy – The short answer: By taking the microstate to be a volume element in momentum $\times$ space (phase space) and by assigning equal weight to all microstates with the same energy, volume and number of particles, we end up with a statistical entropy that behaves exactly the same as classical entropy. That's the beauty of statistical mechanics. No additional assumptions, discretizations or other such operations required.

On the calculation of the statistical entropy – The longer answer: We do not discretize the continuous phase space, we integrate over it. The easiest demonstration is in the canonical partition function: $$ Q = \frac{1}{h^{3N}N!} \int e^{-\mathbf{p}^2/2mkT} d^N \mathbf{p} \int e^{-E_P/mkT} d^{N} \mathbf{r} $$ where $\mathbf{p}$ is the vector of all $(x,y,z)$ components of momentum of all particles and $\mathbf{r}$ is the vector of all positions. Here a volume element of phase space is represented by the product $$ d^N\mathbf{p}~ d^N\mathbf{r} = dr_{1x} dp_{1x} dr_{1y} dp_{1y} dr_{1z} dp_{1z} \cdots $$ This is a true differential, which is to say that the size of the grid goes to zero. The momentum integral can be solved directly: $$ \frac{1}{h^{3N}} \int e^{-\mathbf{p}^2/2mkT} d^N \mathbf{p} = \left(\frac{2\pi m k T}{h^2}\right)^{3N/2} $$ The integral over space cannot be solved for arbitrary potential energy but in the special case $E_P=0$ (no interactions) we obtain the simple result $$ \int e^{-E_P/mkT} d^{N} \mathbf{r} = V^N $$ Combining with the momentum integral we obtain the canonical partition of the ideal gas: $$ Q^\text{ig} = \frac{V^N}{N!}\left(\frac{2\pi m k T}{h^2}\right)^{3N/2} \tag{1} $$ Since the question was about entropy, in the canonical ensemble we calculate it as $$ \frac{S}{k} = \ln Q - \beta \left(\frac{\ln\partial Q}{\partial\beta}\right) $$ Applying this to the partition function of the ideal gas we obtain the an entropy that matches the classical entropy of the ideal gas to within an additive constant. The result can be found in any standard textbook. The point of this lengthier answer was to address the question about how to discretize the phase space: there is no discretization involved.

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Themis
Themis
  • 5.9k
  • 8
  • 37

First on the additive constant: it doesn't matter. Since, as you point out, classical thermodynamics defines $\Delta S$, not $S$, the classical entropy has an arbitrary constant attached to it. But it is always $\Delta S$ that we measure or care to calculate, the additive constant is irrelevant. The statistical entropy only needs to match the classical entropy when we take the difference between the same two states.

On the calculation of the statistical entropy – The short answer: By taking the microstate to be a volume element in momentum $\times$ space (phase space) and by assigning equal weight to all microstates with the same energy, volume and number of particles, we end up with a statistical entropy that behaves exactly the same as classical entropy. That's the beauty of statistical mechanics. No additional assumptions, discretizations or other such operations required.

On the calculation of the statistical entropy – The longer answer: We do not discretize the continuous phase space, we integrate over it. The easiest calculation is in the canonical partition function: $$ Q = \frac{1}{h^{3N}N!} \int e^{-\mathbf{p}^2/2mkT} d^N \mathbf{p} \int e^{-E_P/mkT} d^{N} \mathbf{r} $$ where $\mathbf{p}$ is the vector of all $(x,y,z)$ components of momentum of all particles and $\mathbf{r}$ is the vector of all positions. Here a volume element of phase space is represented by the product $$ d^N\mathbf{p}~ d^N\mathbf{x} = dr_{1x} dp_{1x} dr_{1y} dp_{1y} dr_{1z} dp_{1z} \cdots $$ This is a true differential, which is to say that the size of the grid goes to zero. The momentum integral can be solved directly: $$ \frac{1}{h^{3N}} \int e^{-\mathbf{p}^2/2mkT} d^N \mathbf{p} = \left(\frac{2\pi m k T}{h^2}\right)^{3N/2} $$ The integral over space cannot be solved for arbitrary potential energy but in the special case $E_P=0$ (no interactions) we obtain the simple result $$ \int e^{-E_P/mkT} d^{N} \mathbf{r} = V^N $$ COmbining with the momentum integral we obtain the canonical partition of the ideal gas: $$ Q^\text{ig} = \frac{V^N}{N!}\left(\frac{2\pi m k T}{h^2}\right)^{3N/2} \tag{1} $$ Since the question was about entropy, the entropy in the canonical ensemble is $$ \frac{S}{k} = \ln Q - \beta \left(\frac{\ln\partial Q}{\partial\beta}\right) $$ Applying this to the partition function of the ideal gas we obtain the an entropy that matches the classical entropy of the ideal gas to within an additive constant. The result can be found in any standard textbook. The point of this lengthier answer was to address the question about how to discretize the phase space: there is no discretization involved.