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While a cavity can have modes of arbitrarily high frequency, a phonon can't have a wavelength smaller than the separation between atoms in the solid (well, it could, but it'd be degenerate with phonons of smaller frequency because the discrete Fourier transform is periodic). Hence, there is a maximum possible wavenumber. A physical (or sometimes ...

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The intuitive picture is that phase points move more quickly through regions of phase space where $|\nabla H|$ is higher. As a result if you have a constant-energy ensemble of phase points (a flat "packet"), their phase space area enlarges as they move through high $|\nabla H|$ regions. For the full Liouville's theorem this is not a problem. A ...

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The plot you have for $Z$ looks right. To get some intuition, at low $T$(large $\beta$) the spins are ordered, so the dominant term in the partition function is just $e^{2\beta N J}$, which grows like an exponential with $\beta$. (the Hamiltonian for the Ising model is $H=-\sum_{\langle ij\rangle} \sigma_i\sigma_j$, notice the important minus sign in $H$!) ...

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While we are waiting for clarification: The fact that a completely degenerate gas of fermions has a non-zero kinetic energy means that even at zero temperature the fermions have momentum and can exert a pressure. This is of fundamental importance in astrophysics for explaining the nature of many phenomena including (but not limited to); providing support at ...

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First, note that the equation you use is only valid when all relativistic particles are in thermal equilibrium. The more general equation, which allows for particles with different temperatures, is $$g(T) = \sum_B g_B\left(\frac{T_B}{T}\right)^4 + \frac{7}{8}\sum_F g_F\left(\frac{T_F}{T}\right)^4$$ where $T$ is the photon temperature and $T_B$, $T_F$ are ...

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The grand canonical ensemble allows the number of particles of your system to fluctuate but makes the assumption that it is constant amongst the reservoir and system combined i.e. $$n_{res} + n_{sys} = const$$ For the case of photons this is not true.

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The conclusion is correct: for any system containing a single reaction, there exists an equilibrium (with detailed balance) for any values of the rate constants. I'm not sure what's meant exactly by "equilibrium tools," but it's true to say that there's no real difference in the kinetics of an equilibrium or non-equilibrium system if it only contains one ...

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I learned I can calculate the entropy $$S=-k_B\sum_jp_jln(p_j)$$ where $$p_j$$=probability at j state but I saw that the entropy is also can be calculated by $$S=-k_Bln(Z)$$ I think this equation applicable for both of isolated system and non isolated system these two equations are same ? Given the number of states $\Omega(E,\Delta ... 1 You say "the four quantum numbers", by which I assume you mean$n, l, m, s$of atomic spectroscopy. It's important to realize that this set of quantum numbers only applies in (simple) spectroscopy. It is not the case that any random quantum system is characterized by these four numbers. When a statistical mechanics book says to sum over states$s_1, s_2, ...

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I understand this formula in the following way: $dw$ is a probability measure that aims at being general. It can be thought as being the probability measure for having a quantum system at energy $E_0$. The Dirac mass $\delta(E-E_0)$ specifies in any case the ensemble while $d\Gamma$ specifies the "natural" measure used to count the number of states. From ...

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Following is computation of $E[(\sigma_a - E[\sigma_a])(\sigma_b-E[\sigma_b])]$: $E[(\sigma_a - E[\sigma_a])(\sigma_b-E[\sigma_b])]=\\ E[\sigma_a\sigma_b-\sigma_aE[\sigma_b]-E[\sigma_a]\sigma_b+E[\sigma_a]E[\sigma_b]]=\\ E[\sigma_a\sigma_b]-E[\sigma_a]E[\sigma_b]-E[\sigma_a]E[\sigma_b]+E[\sigma_a]E[\sigma_b]=\\ E[\sigma_a\sigma_b]-E[\sigma_a]E[\sigma_b]$ ...

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Tricritical Ising model belongs to the family of minimal models ($M(5,4)$). There are several different coset constructions that represent them, one of them is the following: $M(m+1,m)=SU(2)_{m-2} \times SU(2)_1/SU(2)_{m-1}$

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That is exactly right. I would say "exactly one spin up" rather than "any spin up", but that is just phrasing. Another way of looking at this is that there is an increase in entropy (of $\log n$ times Boltzman's constant) associated with flipping one spin, which will sometimes overcome the energy cost.

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One way to look at is that the total internal energy is the same as though each particle has $3/5$ of the Fermi energy, even at 0K, or that each fermion has, on average, an energy of 60% of the Fermi energy. Just the fact that the fermions (e.g. electrons) must stack up in their energies, with no "extra" energy from heating the gas etc, gives the gas a ...

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The method I will present here is very general and by training this is the first that comes to my mind. There might be a simpler one though. The idea is to consider another function $G(\Gamma) \equiv e^{-g F(\Gamma)}$ with $g > 0$ and $F(\Gamma) = a \sum_i n_i$. The sum I want to compute is now $S_g \equiv \sum_{\Gamma} G(\Gamma)$. This sum is actually ...

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This is a great question. Actually, the temperature of the gas in the cylinder will increase. This isn't equivalent to a Joule expansion, because the cylinder is specified to be "insulated." I will assume that this means that there is no thermal contact between the surrouding atmosphere and the inside of the cylinder. In the case of Joule expansion, we have ...

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The number $\Omega$ of microstates accessible to the particle system is defined as : $$\Omega=\omega(E)\Delta$$ where $\omega(E)$ is the density of state. To determine this guy, you need first to calculate the number $\chi(E)$ of µ-states corresponding to an energy inferior or equal to $E$. Let us call the associated phase space region ...

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Your requirement that $z\to 1$ (i.e. $\mu\to 0$) already requires the limit $N\to \infty$. If $N$ is finite, the chemical potential tends to a small but finite value at the transition temperature. This ensures that the occupation of the ground state is $\langle n_0\rangle \lesssim N$.

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If we consider temperature to be due to translational motion of the molecules and we assume the system has reached equilibrium, then the velocity distribution of the molecules is given by the Maxwell distribution: $$f(v) = \sqrt{\left(\frac{m}{2\pi k T}\right)^3} 4 \pi v^2 \exp\left(\frac{m v^2}{2 k T}\right)$$ which will give you the velocity ...

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Correlation between two variables (or objects) is, very simply put, how much a change in one variable affects or determines a change in the other. Replacing variables with spins, highly correlated spins would mean that, due to some interactions between them, a change in the direction of one spin will cause a change in the direction of the spin it is ...

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One way to understand it is to write it using the Dirac measure to express the phase space in the microcanonical ensemble (because that's what it is about). In this ensemble the idea is to say that the energy $H(\textbf{r})$ (where $r$ is a point in phase space) is fixed at some value $E$ (actually it belongs to a very small interval $[E,E+\delta E]$). One ...

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What do you mean by temperature? Let's say you have your gas hooked up to a very tiny thermometer. It has a pointer that jiggles and bumps around. What you'd find is that the pointer fluctuations are mostly related to the mechanical evolution of the thermometer itself, not the system under consideration. But that's beside the point. Temperature is not just ...

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Dipole gas. If the dipoles are weak, there are probably many atoms with non-zero dipole moment that can be gas or plasma. If your dipoles are strong, their strength exceeding normal molecular bonds, I would say that the situation would lead to a condensation of the gas to liquid or solid phase. And you could probably break the bonds (and keep it gaseous) ...

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E. T. Jaynes: Probability Theory: The Logic of Science http://omega.albany.edu:8008/JaynesBook.html The book has also printed form. Jaynes also published readable and revealing papers on probability, statistical physics and other physics. Here you can find them: http://bayes.wustl.edu/etj/node1.html

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I think the wrong step was the assumption that entropy increases, where in fact maintaining the temperature would require a an outflow of heat, which means the entropy of the gas is decreasing. To see how this relates to your formula, notice that this decrease in entropy would also increase the enthalpy by the same amount, however enthalpy will also ...

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The correlation function measures, as you would expect, how correlated two random variables are. That is, how often two random variables have similar values. We can construct such a function very simply. Say you are flipping coins, and you want to know if their results are correlated. To quantify things, call "heads" $+1$ and "tails" $-1$. To make ...

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As pointed out in detail in this article, the time variation of dimensional constants has no operational meaning.

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