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There is quite a big controversy these days about the correct definition of the entropy in the microcanonical ensemble (the debate between the Gibbs and Boltzmann entropy), which is closely related to the question. Everyone agrees that the correct definition of the density matrix is given by $$\rho(E)=\frac{\delta(E-H)}{\omega(E)},$$ where $H$ is the ...


4

Phase transitions are a many-body effects. You can not generate sharp transition with a finite number of degrees of freedom (or particles). However as you add particles the features of the system may become sharper. In the limit of infinitely many particles (thermodynamic limit) you get a truly discontinuous transition. In practice nothing is infinite. The ...


3

The microcanonical ensemble is the (maximum entropy) probability distribution for a given specified total energy. What you've calculated is actually the maximum entropy distribution with no constraints on the energy, which is the same as the canonical distribution at infinite temperature ($\beta = 0$). To correctly calculate the microcanonical entropy, ...


1

You may start from the density matrix $\rho$ of the grand canonical ensemble (assuming $k_BT=1$ as the unit of energy), $$\rho=Z^{-1}e^{-H+\mu n},$$ where $Z=\text{Tr}\,e^{-H+\mu n}$ is the partition function. Then $\langle n\rangle$ and $\langle n^2\rangle$ are defined by tracing with the density matrix $$\langle n\rangle=\text{Tr}\,n\rho\text{, and ...


1

No. Your answer would be correct only for high vacuum - a density so low that collisions between molecules can be ignored. At atmospheric pressure the mean free path of an air molecule is only about a micron. If the heated volume is much larger than this we can ignore diffusion and thermal conduction in the short run. The constant collisions create what ...



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