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4

To determine the upper limit on chemical potential for a gas of $\mathcal N$ bosons, look at the form of the Bose distribution in the grand canonical ensemble with $\langle N \rangle = \mathcal N$. When using the GCE, it's easiest to work at chemical potential $\mu$ and to then choose $\mu(\mathcal N)$ so that $\langle N\rangle(\mu)=\mathcal N$. Each state ...


4

You have to distinguish between "different states" and "number of states" - or, in the words of @Numrok, between "macrostates" and "microstates". The fundamental theorem refers to "accessible micro states". If I have three white balls and two buckets to put them in, I could put two balls in one and one in the other (that is a macro state); there are in fact ...


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The probability for one compartment being empty is actually the probability that at least one compartment is empty. Let's call it $\Pi^{(1)}$, $$ \Pi^{(1)} = \frac{(p-1)^N}{p^N} $$ When you write the probability for any of the compartments to be empty as $$ \Pi = \sum_{j=1}^{p}{\Pi^{(1)}} = p\frac{(p-1)^N}{p^N} $$ you are overcounting. Longer argument: ...


3

First, let me explain what the notation means. The sum over $\{n_\ell\}$ is a sum over all possible values of $n_\ell$, for each possible values of $\ell$ (in other words, $\{n_\ell\}$ specifies the values of $n_1,n_2,n_3,\ldots)$. Then, once these values are fixed, you take the product of all the functions $f_\ell(n_\ell) = \frac{1}{n_\ell!}\bigl( ...


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I would highly recommend the two seminal papers by E. T. Jaynes, http://journals.aps.org/pr/abstract/10.1103/PhysRev.106.620 and, http://journals.aps.org/pr/abstract/10.1103/PhysRev.108.171 Also check out the book by E. T. Jaynes, which has a focus on the foundations in probability but is rather light on applications in physics: ...


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The whole point of statistical physics is that we don't really care about what the microstate could be, we just know that there's a great collection of similar ones that give rise to the same macrostate. I believe what you're questionning about is closely related to the ergodic hypothesis


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The weight for a specific state is $w(\phi)=e^{-E/T+F/T}\equiv e^{-I(\phi)+F/T}$. Due to the definition of the entropy $S$, we have $S=-\int D\phi w\ln w=\frac{1}{T}(\int D\phi wE-F\int D\phi w)=\frac{1}{T}(U-F)$. Then we have $F=U-TS$.


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For both hydrogen and chlorine E = 3/2 kT is only true at very low temperatures since they are diatomic gases. In general you get a contribution of 1/2 kT to the energy for every quadratic degree of freedom. For a monatomic gas that is 3 translational degrees of freedom, hence 3*1/2 kT. For a diatomic gas there are in addition 2 rotational, 1 vibrational ...


2

If operators $a_j$, $a^\dagger_j$ correspond to the system's orthonormal natural orbitals $\phi_j({\bf x})$, such that $$ \int{d{\bf x}\; \phi^*_j({\bf x})\phi_k({\bf x})} = \delta_{jk}, \;\;\;\sum_j{\phi^*_j({\bf x})\phi_j({\bf x'})} = \delta({\bf x} - {\bf x'})\\ \hat\psi({\bf x}) = \sum_j{\phi_j({\bf x})\;a_j},\;\;\; a_j = \int{d{\bf x} \;\phi^*_j({\bf ...


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Let us use the definition $\langle\eta(s)\eta(t)\rangle=\Gamma\gamma^2\delta(t-s)$. First of all, $C(s,t)$ depends on $t$ because $$C(s,t)=\Gamma\min(s,t).$$ It is clear, from causality, that $\frac{\delta x(t)}{\delta \eta(s)}=0$ if $t<s$. If $t>s$, compute the difference $\delta x(t)$ caused by two realisations of noise that differ only at time $s$ ...


2

You can think of the chemical potential as the amount of free energy needed to add one additional particle to the system. Because the ground state of a BEC is degenerate and can hold an infinite number of particles, there's no energy cost to add another particle to that state. So, $\mu = 0$.


2

Very interesting question. As you wrote yourself in your Edit it is hard to describe water via the ideal gas model. You have to introduce at least two important improvements of your ideal gas: Dipole - Dipole - Interaction instead of no interaction. Let's call this pair potential $V_d$ and note that for two given molecules $V_d$ is not only distance but ...


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The general Pearson correlation between two variables is defined as $$ \textrm{cor}(X,Y) = E[XY] - E[X]E[Y] $$ up to a denominator containing the standard deviations of the distributions of the two variables. In some field theories the expectation values of the variable itself (one-point function) vanishes, therefore oftentimes the above definition reduces ...


2

The second version is the first one but with the observables shifted by their respective averages: $$\langle \tilde A\tilde B\rangle = \langle (A-\langle A\rangle)(B-\langle B\rangle)\rangle = \langle AB\rangle - \langle A\rangle\langle B\rangle$$ Oftentimes in e.g. numerical studies, it is easier to just sample the average product of the original ...


2

In statistical mechanics and field theory, the second type is referred to as a "connected" correlation function. You sometimes see the notation $$ g_{\text{c}}(\mathbf{x}-\mathbf{x'},t-t') = \langle s_1(\mathbf{x},t) s_2(\mathbf{x}',t') \rangle_{\text{c}}\,\text{,} $$ where $\langle\ldots\rangle_{\text{c}}$ indicates that the product of the averages should ...


2

It says in the notes [...]the factor of 8 arises because we are only counting the states in the quadrangle with positive $n_x$, $n_y$ and $n_z$. The eigenfunctions of a particle in a box are $$ \Psi(x) \propto\prod_{j\in{x,y,z}} \sin\left(\frac{n_j}{L_j}x_j\right) $$ with $n_j\in\mathbb{N}$. Choosing $n_j$ a negative integer does not yield a different ...


2

Let us start with an example, called Langevin paramagnetism, where the magnetic moment is described classically, as a vector in three dimensions. Calling $\vec\mu$ this moment, $\vec B$ the magnetic induction and $\theta$ the angle between $\vec\mu$ and $\vec B$. The probability density of the angle $\theta$ is $\rho(\theta)=\frac{1}{\mathcal Z}\exp(\beta\mu ...


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While a funny-looking coincidence, this is not a valid alternative expression for entropy in general, since the entropy of a probability distribution (which are what rigorously hides behind the strange word "macrostate") is more generally given by $$ S = - k_B \sum_i p_i\ln(p_i) \tag{1}$$ and becomes only $$S = k_B \ln(\Omega) \tag{2}$$ in the case of a ...


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The uses of this two theories are completely different. Statistical Mechanics is used to see how by modelling the behavior of microscopic constituents you can predict the macroscopic phenomenas that you observe. On the other hand Many Body Theory uses first principle techniques to see what happens microscopically when you have large no of particles in your ...


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I see at least 3 reasons. It is only in this limit that macroscopic observables become deterministic. It is only in this limit that one has equivalence of the different statistical ensembles. It is only in this limit that one has sharp phase transitions (genuine singularities of thermodynamic potentials). (The first two properties may fail to hold even ...


2

At equilibrium, the system would be in that macrostate which would have the maximum multiplicity or the largest number of microstates; that would correspond to gas totally dispersed over the whole volume $V\;.$ This is wrong, and based on a misunderstanding of terminology. You have microstates, and macrostates. A macrostate assigns a probability to each ...


1

A CFT is still a QFT, and the way to put it at finite temperature is standard for any quantum system - you take your Hamiltonian $H$ and compute $Z=\mathrm{tr}\,e^{-\beta H}$, where the trace is over the Hilbert space of states living on $\mathbb{R}^{d-1}$ if your CFT is in $d$ dimensions. The thermal correlators are computed in a similar way, ...


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Assume each atom can be in only two states, n = 1 and n = 2, where its energy is respectively E1 and E2. In a gas, you have many atoms. At any given time, some of them are in state n = 1 and the rest are in state n = 2. The number of atoms in a state n is the "population of the state n ".


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In nuclear and particle physics, the primary data of interest are scattering cross sections. In these data, the structure of the particle(s) of interest shows up through various "resonances" in your data (i.e. big spikes in the data set). These resonances correspond to states of the system of particle(s) (e.g. nucleus, proton, etc.) of interest. The width of ...


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I think you were right. The viscous term in the NS equations cannot be derived from the Boltzmann equations. If you derive the conservation laws from the Boltzmann equations using first order approximation, you will get an force term, which should include the pressure, viscous forces and external forces shown in the NS equations. I think the approximation ...


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You say: We shouldn't care about how we reached that equilibrium In fact the entire point of the example is that we do. I shall try to explain why. Jaynes and Gull both work in the framework of Bayesian inference (I can recommend the introductory text: http://www.amazon.co.uk/Data-Analysis-A-Bayesian-Tutorial/dp/0198568320. The title may seem ...


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Regardless of the system, Cv will be proportional to the variance of energy. If you have peaks at higher energies, that will increase its value. But at high enough energies the occupation of those states will be so low they won't significantly affect the variance. In this case the variance of the distribution isn't just the width² of one of the peaks, you ...


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I think you have the wrong idea when you ask how specific heat is "defined". In computational physics, the starting point is an experimental measurement that one could measure, or at least, a physical quantity that one might care about ... and then the question is, "how do I compute it?" The wrong approach is to have in mind a certain formula. You should ...


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A simple way to see this is by considering the fact that the probability to transition from a particular initial state to a particular final state is the same as for the inverse process where one considers the transition from the final to the initial state. This is because the square of the modulus of the matrix element is the same for both cases. This ...


1

A macrostate is a set of microstates. Some microstates are thermal, others are not. Without the assumption of being in thermal equilibrium you can't assume anything since any possible microstate is possible. And lots of possibilities macrostates could be picked. Usually you want to group your macrostates according to a state variables such as pressure, ...



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