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Yes, your three-body expression is consistent with the BBGKY hierarchy. Regarding $n$-body correlations, there may be a way to express the collision integrand in terms of $g$ and $f$ functions, but I do not know it and can not remember ever seeing one. That said, I'll heartily upvote any answer that manages to write it out. Most treatments of the problem ...


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I propose this : As long as the detailed balance is respected, the system can be studied with equilibrium thermodynamic. Therefore we can write the free energy of the solution, for exemple $ F=A \log(A) + B \log(B) + S \log(S) + \chi_{A S} A S + \chi_{B S} B S + \chi_{A B} A B $ ($S$ the solvent, $\chi$ the interaction parameters), plug the equilibrium ...


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Bose particles cannot be identified as different in a given state, whereas boltzmann particles can (even though both types can occupy a given energy state with more than one particle). Thus boltzmann statistics need to take into account the permutations ($n!$) of the $n$ particles into a given state, in contrast to the bose particles (which are not ...


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REMARK. Perhaps I wrongly interpreted the question. I interpreted it as if were referred to the total volume of phase space. The answer is negative if the question regards general changes in time of topology of the total space of phases and if you do not impose any generic restriction on the topology of the spaces, like compactness (see the final ...


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Here's something I believe is a simple proof. Unfortunately it uses a little bit of cohomology. Consider the canonical 2-form in extended phase space $T^*M \times \mathbb{R}$ $$\omega = \sum_{i=1}^N dq_i \wedge dp_i - dH(\vec{q},\vec{p},t) \wedge dt ,$$ where $N = dim(M)$. A function $f: M \to M$ is said to be a canonical transformation iff $f^* \omega = ...


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The only temperature that would make sense in your question is the temperature of the thermal distribution of the photons (if the photon gas is thermalized). So let us assume we have a photon gas at equilibrium. Its chemical potential is $\mu=0$ because photons have no ground state. Consider now the atom. One of the characteristics of an atom is that the ...


3

To add to Whelp's Answer. Even though the $$\bar{d}Q=T\,dS$$ does not hold in an irreversible process, it still gives us something general. Consider a thermodynamic system linked to a system of reservoirs and which can only exchange heat and nothing else with the reservoirs. Let's call the thermodynamic system an engine for our convenience. Then the engine ...


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By the main theorem of connectedness in general topology, continuous maps preserve connectedness. Time evolution of Hamiltonian systems preserves connectedness because it is continuous. I think it is independent of from Liouville's theorem, it just requires the proving Hamiltonian time evolution is continuous. This is just a formal way of restating ...


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This relation is not true for general processes. For a closed system, the general relation is $\delta Q \leq TdS$, as is illustrated by the Clausius Theorem (http://en.wikipedia.org/wiki/Clausius_theorem). Another way of writing it is $dS=\delta Q/T + dS_{irr}$ where $ dS_{irr}$ is the entropy change due to irreversibilitiy in the closed system ...


1

Interesting question. I would have thought that if you were aware of the exact number of energy states and the populations thereof, you could apply boltzmann statistics to each of the levels in order to fit an appropriate temperature to the population in each state. This temperature, if comparable amongst the included levels, would therefore require that the ...


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Since you have the expression of $\Omega$ your work is almost done. First remember that the entropy for a micro-canonical (fixed energy) system at thermal equilibrium is given the very famous Boltzmann's formula : $$S=k_B\,ln(\Omega)$$ Then, simply use the Stirling's approximation to evaluate $ln(N!)\approx Nln(N)-N$ (because $N>>1$, i.e. very long ...


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John Baez has freely available book based on a series of internet articles articles. I is very readable and definitely starts from scratch.


31

How about lifting a litre or kilogram of water? Lifting a kilogram of water $235.3\text{ m}$ (772 feet in sensible units) involves a potential energy change:$$PE=mgh=1\times9.8\times235=2.31 \text{ kJ}$$ Warming a kilogram of water, (specific heat $4.179\text{ kJ/kg/K}$), through a temperature change of $0.556\text{ K}$ (1 degree Farenheit) ...


0

First law of thermodynamics states that the change in internal energy, $dU$, of a system is equal to $\delta Q + \delta W$. The only way to separate the contributions from the two is to arrange the experiment so that one of the quantities is zero. If the change is adiabatic so that there is no heat exchange, $\delta Q = 0$ and hence $dU = \delta W$. On the ...


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Consider two connected systems $A$ and $B$ in the microcanonical ensemble. Calling $E$ the total (fixed) energy we have $$\Omega(E)=\Omega_A(E_A)\Omega_B(E_B)=\Omega(E_A)\Omega_B(E-E_A).\tag{1}$$ This only states that number of states of the whole system is the product of the number of states of the subsystems but here is really the key, since everything ...


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I am not sure that it is true in general. In your equation, $x$ might not be a single continuous variable in real number line. It may consist of a set of variables such as interacting spin system. It can also be discrete variable, and in the most general case, it is just a set of states in some model system. The Laplace transform is not well defined in ...


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I guess, one could start by considering the microcanonical and canonical ensembles as entirely different concepts. For they represent different statistical ensembles: in the former the energy of the system is fixed while in the second the energy can span all possible values allowed by the energy spectrum of the system with some penalty attributed to high ...


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Suppose you have some radioactive material with a half life $\tau_{1/2}$. What that term "half life" means is that the amount of material $m(t)$ you have left after a time $t$ is $$m(t) = m(0) \exp[ -t / \tau_{1/2}] . \qquad (*)$$ However, the material is made up of discrete atoms and each one decays in a random way. Therefore, it's not 100% guaranteed ...


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Let's consider just one cycle of the Szilard engine. Aside from discussion of energy free state polling, one of the main points of Bennett paper (if you mean Charles Bennett, "The Thermodynamics of Computation: A Review", Int. J. Theo. Phys., 21, No. 12, 1982) here is that you must build a finite state machine (a very simple three-state machine) as a minimal ...


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In contrast to the other answers, I would like to mention that it is possible to compute rigorously the value of the critical temperature of the two-dimensional Ising (and Potts) model, without computing explicitly the free energy (which is in any case not possible for general Potts models). In the Ising case, this has been known for a long time, and there ...


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I guess you can use the Van der Waals equation and some estimates of the molecule volume and intermolecular attractive forces. The parameters of the critical point depend on these characteristics, so you need to assess them.


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There are four state kets $|\uparrow_1\uparrow_2\rangle,|\uparrow_1\downarrow_2\rangle,|\downarrow_1\uparrow_2\rangle,|\downarrow_1\downarrow_2\rangle$ representing four possible configuration of the two spins, the eigenvectors of the Hamiltonian is the linear combination of them. There will be four eigenvectors and four eigenenergy. You can also obtain the ...


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Actually, all of your references are correct. In a MD simulation, for homogeneous systems, you can calculate the ensemble average of a thermodynamic property in several different ways. You can simply average over all the particles in your system during one time step. You can average over a single particle over many time steps. You can average over all ...


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Ok, I think I figured it out: Take the fermionic expression, first perform an integration by parts, then the substitution that I tried above and only then use the classical approximation: $$ P=T\int\frac{d^3p}{(2\pi)^3}\ln\left[1+e^{-\beta(\omega-\mu)}\right]\\ =\frac{T}{2\pi^2}\int_0^\infty dp\;p^2\ln\left[1+e^{-\beta(\sqrt{p^2+m^2}-\mu)}\right]\\ ...


2

An order parameter distinguishes two different phases (or orders). In one phase the order parameter is zero and in another phase it is non-zero. It does not have to be macroscopic. For example, in the BCS theory of superconductivity the order parameter is called the gap $\Delta$. It can be interpreted as the binding energy of a Cooper pair, namely two ...


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This is a good question. I can only answer half of it: The eigenvalues of $M^{l}$ are some times complex. Take a look at the RG flow of Einstein gravity close to its UV fixed point. The two relevant couplings (newton's constant and cosmological constant) spiral around the fixed point as they move away from it. The real part of the critical exponents ...


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Any increasing sequence $(\Lambda_n)_{n\geq 1}$ of finite subsets of $\mathbb{Z}^d$, $d\geq 2$, such that $\bigcup_{n\geq 1} \Lambda_n =\mathbb{Z}^d$ will do. All sequences $(\mu_{\Lambda_n}^+)_{n\geq 1}$ of finite-volume Gibbs measures in $\Lambda_n$ with $+$-boundary condition converge to the same infinite-volume Gibbs measure $\mu^+$, under which there is ...


1

One could interpret the update steps as possible discrete steps in a fictitious time and in that case the transitions represent dynamics on the state space of a Markov chain. As an example, there is the relaxational non-conservative Glauber dynamics and the magnetization conserving Kawasaki dynamics which are used to simulate Ising and related systems. The ...


0

The difference is that one expansion is quasi-static (the reversible one) while the other is spontaneous because of a dramatic change of the external constraints (the irreversible one). In the quasi-static case, you start off indeed in the state where gas pressure equates external pressure. An external operator then slightly decreases the outside pressure ...


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Yes. The Hamiltonian is your $H$. $p_r=0$ because $\frac{\partial L}{\partial \dot r}=0$ with $L:=\frac{m}{2}(R^2\dot\theta^2+R^2\sin^2\theta\dot\phi^2)$. The integral is over the 4-dimensional phase space: $(\theta,\phi,p_{\theta},p_{\phi})$, because the particles just move over the 2D surface of the sphere. $J=1$ regardless of the coordinate system you ...


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2) Put a gas in a sealed box with an attached vacuum chamber. Remove the partition and allow the gas to fill both chambers. The entropy of the system has increased and yet no heat flowed into/out of the system. But heat has flowed from the box to the vacuum chamber! (Whatever heat is!)


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Consider an isolated system with a large number of degrees of freedom. An example could be a quantum computer that is able to compute the exact time evolution of a gas of 10^23 molecules undergoing an ideal free expansion. In such a system, time evolution is always unitary and thus reversible. Clearly the entropy of the quantum computer is equal to zero and ...


0

It's impossible. According to Blotzmann distribution http://en.wikipedia.org/wiki/Boltzmann_distribution , the number of microstates decreases as energy increase. If the number of microstates for energy you choose is right, then the stable state of the system should have infinite energy, and it's impossible.


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I think that you are really interested in the $q$-state clock model, which is similar to the Potts model, and is defined as follows. Fix an integer $q\geq2$. For each $i\in\mathbb{Z}^d$, let $$ \theta_i \in \bigl\{\frac{2\pi}{q} k\,:\, k\in\{0,1,\ldots,q-1\}\bigr\}, $$ and define the spin at site $i$ by $$ \mathbf{S}_i = (\cos\theta_i,\sin\theta_i) . $$ The ...



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