# Tag Info

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The distribution you use depends on the ''ensemble'' you are working in. There are three kinds of ensembles: Mircocanonical ensemble (or practical: an isolated system): the number of particles $N_s$ and the energy $E_s$ is fixed. This is ONE SPECIFIC REALISATION of the system. You have only one possible value for the energy and number of particles, so the ...

4

Actually, it's exact. The flaw is "regarding the whole system consisting of these two particles, we can also write" $Z = 1 + e^{- \beta E} + e^{-2 \beta E}.$ Assuming the two particles are distinguishable, we have $$Z=\sum_ig_ie^{-\beta E_i}=1+2e^{-\beta E}+e^{-2\beta E}=Z_0^2,$$ with the $2e^{-\beta E}$ since the state of energy $E$ is doubly-degenerate. ...

4

You will almost never encounter a calculation that is intended to account for every detail of a phenomenon with perfect accuracy. That isn't possible, and in fact many times adding more detail to a calculation only takes away from the insight it grants. Why make a complicated calculation when a simple one tells you everything you want to know? Gamow is ...

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In the figure above, consider the different configurations that are possible with 3 particles and 5 energy levels. Dividing by 3! gets the symmetry factor correct only for configurations of type 1 but is wrong for configurations of type 2 and 3. You can see this by explicitly writing out $Z$ and comparing with $z^3/3!$. That is why the OP's statement that ...

3

The factorial factor $1/N!$ is exact but does not apply to all statistics. Consider a two level system and let us call $\xi_i$ the grand-canonical partition function for the energy level $i$ ($i=0$ or $1$) and $z=\mathrm e^{\beta\mu}$ the fugacity. Keep in mind that $\xi_i$ is the partition function for a given energy level. Classical particles are ...

3

It is indeed an approximation, in particular an approximation that is usually good at high enough temperatures. For $N$ distinguishable, non-interacting particles the partition function is $Z(\mathrm{dist.}) = {Z_0}^N$, where $Z_0$ is the single-particle partition function. If the $N$ particles are all in different quantum states then there are $N!$ ...

2

The expression $$k_B \frac{\Omega}{\bar{\Omega}}$$ equals $$k_B\frac{1}{\bar{\Omega}}\frac{d\bar{\Omega}}{dE}$$ which equals $$\frac{dS}{dE}.$$ In thermodynamics, where $S$ is the Clausius entropy, this is equal to $1/T$ where $T$ is the Kelvin temperature. In statistical physics, this expression can be taken as a definition of $1/T$ of a system from ...

2

You should use $U=q\epsilon_1$. With the total number of particles $N$ being constant, we have: $$\frac{\partial S}{\partial q}=\epsilon_1 \frac{\partial S}{\partial E}=\frac{1}{T}\epsilon_1\tag{1}$$ As you said: $$\frac{\partial S}{\partial q}=k_B\ln(N/q - 1)=k_B\ln(N\epsilon_1/q\epsilon_1 - 1)=k_B\ln(N\epsilon_1/U - 1)\tag{2}$$ \to ... 2 Two obvious desirable features of this definition are: When you put two systems next to each other, considering them as one system, the total number of possible microstates \Omega_t is equal to the product of \Omegas of the two systems, \Omega_t=\Omega_1\times \Omega_2. But for this system the entropy is the sum of the entropies, indicating the ... 2 Entropy was first met in classical thermodynamics and was defined as , where Q comes from the first law of thermodynamics and T is the temperature, W work done by the system. Once it was established experimentally that matter at the micro level is discrete, i.e. is composed of molecules the statistical behavior of matter became the underlying ... 2 The answer is no, or at least it is in the classical vacuum sense. I also don't see a rationale for why creating a vacuum would require infinite energy. An explicit construction is to use a solid-phase reactive chemical "getter" to eliminate (nearly) all gas molecules present; in experimental practice, virtually all man-made materials still outgas ... 2 Let me expand on the previous answer. We expand each term separately. We have, \begin{align} \log \left( 2 \cosh x \right) & = \log \left( e ^x + e ^{ - x } \right) \\ & = \log \left[ e ^{ x } ( 1 + e ^{ - 2 x } ) \right] \\ & = x + \log \left( 1 + e ^{ - 2 x } \right) \\ & \approx x + e^{-2x} \end{align} For the second term we have, ... 1 Mustafa's answer gives one important reason for the logarithmic dependence: microstates multiply, whereas we'd like an extrinsic property of a system to be additive. So we simply need an isomorphism that turns multiplication into addition. The only continuous one is the "slide rule isomorphism" aka the logarithm. The base e is arbitrary as you can see from ... 1 I would like to share my thoughts and questions on the issue. The Boltzmann H theorem based on classical mechanics is well discussed in various literatures, the irreversibility comes from his assumption of molecular chaos, which cannot be justified from the underlying dynamical equation. Here I will try to say something on quantum H theorem, the point I want ... 1 I am going to address the question as to why energy and information have time symmetric conservation properties whereas entropy does not. According to the Wikipedia entry on entropy - "The entropy of an isolated system never decreases, because isolated systems spontaneously evolve towards thermodynamic equilibrium, which is the state of maximum entropy." ... 1 I don't know if this answers your question. Have you seen how a cube of ice melts? Focus on one corner and you will see the melting happening on the edges. This is precisely the limit shape that you get from the domino tiling of a hexagon (which can be mapped to the dimer problem). This is called the Wulff shape of a crystal. See also the (theoretical) ... 1 The derivation of the Fermi-Dirac distribution using the density matrix formalism proceeds as follows: The setup. We assume that the single-particle hamiltonian has a discrete spectrum, so the single-particle energy eigenstates are labeled by an index i which runs over some finite or countably infinite index set I. A basis for the Hilbert space of the ... 1 I) If we expect \Omega(E) to depend analytically on the variable \hbar\omega>0 extended to (parts of) the complex plane, then we may regularize by introducing an i\epsilon prescription, and substitute\tag{1} \hbar\omega ~\longrightarrow ~ \hbar\omega (1-i\epsilon). $$The variable$$\tag{2} q~:=~ e^{-i\hbar\omega k}~\longrightarrow ~ ...

1

Here's an semi-formal explanation. Define $$f(N,V,T)=\frac{p(N,V,T)}{kT}.$$ While $f(N,V,T)$ is a function of $N,T$ and $V$, the variables $N$ and $V$ are partially redundant, and only the ratio $\rho=\frac{N}{V}$ is needed, since pressure is an intensive quantity. Thus we can write $$f(\rho,T)=\frac{p(\rho,T)}{kT}.$$ Every smooth multivariate function ...

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A definitive volume, one that I learned from during graduate school, is Kerson Huang's (of MIT, emeritus of the Physics Dept.) Statistical Mechanics. The book covers both classical and quantum computations of the partition function and observables from it, as well as thermodynamics, kinetic theory, transport, superfluids, critical phenomena, and the Ising ...

1

In Landau's theory, the order parameter $M$ should make $G(M)$ minimal. $$\frac{\partial G}{\partial M} = 2 B(T) M + 4 C(T) M^3=0$$ $$\frac{\partial^2 G}{\partial M^2} = 2 B(T) + 12 C(T) M^2 > 0$$ Hence, $M=0$ or $M = \pm M_0 = \pm \sqrt{ - \frac{B(T)}{2 C(T)}}$ When $T$ is higher than the critical point $T_c$, the groud state of system satisfies $M=0$. ...

1

Law of equipartition predicts the heat capacity of gases correctly. It assumes that inter-molecular attraction in gases is negligible (which is true). But for solids, inter-molecular attraction is not negligible, the, how come it still predicts the correct value for molar heat capacity? You have to know what "negligible" means in the context of the ...

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Given an ideal piston/cylinder, starting with the piston completely inserted and zero volume, the work to make a perfect vacuum is simply: (distance the piston moves) X (force) = (distance) X (area of the piston) X (exterior pressure). So the work to make a vacuum of volume V, is V X P, where P is the exterior pressure, such as atmospheric pressure. ...

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It is indeed an approximation. As suresh says, the approximation holds only when each particle is in a different single-particle state, or more precisely, when the number of states in which each particle is in a different single-particle state vastly outweighs the number of states in which more than one particle is in the same SP state. The single particle ...

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First, note that the following statement is untrue: Since the particles are fermions, they will have spin $s=1/2$. Fermions can have any spin in the set $\{\frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \dots\}$, namely in the set of non-negative integers plus a half. Second, my interpretation of the problem is that the author wants you to ignore the spin ...

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