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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

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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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( ...


2

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 ...


2

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 ...


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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

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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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 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

Umm...OK well lets see what happens. Lets let $s = \beta\varepsilon$, where $\varepsilon$ is some fixed energy and $$ Z\left(\frac{s}{\varepsilon}\right) = \zeta(s) $$. To get some kind of idea for what kind of system $Z$ describes we need to find the energy levels of the system and to do that we need to express $Z$ in the form $\sum_{i} e^{-\beta E_i}$. ...


2

The 2nd Law of Thermodynamics is based on an overwhelmingly extensive body of empirical evidence on how thermodynamic systems behave. There are many different statements of the 2nd Law, and all of them are equivalent to one another. Once one of these has been specified, all the other follow. One such statement says that heat cannot flow spontaneously from ...


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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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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 ...


2

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 ...


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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 ...


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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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Most things along those lines are just simply the central limit theorem / law of large numbers. For example, when you compress a gas in an insulated container using a piston, its temperature goes up. Why? Because the moving piston accelerates gas molecules that bounce off it. And why does the temperature always go up by the same amount? Because there are ...


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In ideal gas model, temperature is the measure of average kinetic energy of the gas molecules. In the kinetic theory of gases random motion is assumed before deriving anything. If by some means the gas particles are accelerated to a very high speed in one direction, KE certainly increased, can we say the gas becomes hotter? Do we need to ...


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From a mathematical perspective this means that it is not differentiable. The problem is that you need the discreteness to be able to count states. If you replace the discreteness by something smooth you get something differentiable, but your definition of entropy no longer makes sense. This is just one of the points where mathematicians cringe, but it works ...


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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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The reason why heat cannot flow from cold to warm is that the change in entropy will become negative, and that doesn't happen in a closed system. Negative entropy is by definition not possible. Here's why: Alternative example: gas in a box I think entropy gets a little more intuitive if we think of it in terms of statistical mechanics. If we imagine a box ...


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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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