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Frustrated total internal reflection is an optical phenomenon. It's such a close analogue to quantum tunneling that I sometimes even explain it to people as "quantum tunneling for photons". But you can calculate everything about it using classical Maxwell's equations.

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There is quite a big controversy these days about the correct definition of the entropy in the microcanonical ensemble (the debate between the Gibbs and Boltzmann entropy), which is closely related to the question. Everyone agrees that the correct definition of the density matrix is given by $$\rho(E)=\frac{\delta(E-H)}{\omega(E)},$$ where $H$ is the ...

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Standard Monte Carlo samples the canonical (NVT) ensemble. So it maintains constant temperature but the potential energy is free to fluctuate - both up and down. This will only seem odd if you incorrectly imagine the equilibrium state of a system to correspond to that with the minimum energy. The equilibrium state is actually determined by the minimum free ...

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Phase transitions are a many-body effects. You can not generate sharp transition with a finite number of degrees of freedom (or particles). However as you add particles the features of the system may become sharper. In the limit of infinitely many particles (thermodynamic limit) you get a truly discontinuous transition. In practice nothing is infinite. The ...

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There are two ways to see the analogy between the "quantum diffusion" and classical diffusion. The first one, I think the easier one is comparing the Schrödinger equation with the diffusion equation: $$i \partial_t \psi = -\sum \partial_{xx} \psi$$ (forgetting all the $\hbar,m$ factors) When you transform $t \to -i \tau$ you get the usual diffusion equation ...

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There are several possible answers to this. One is to look at Shannon's definition of the entropy, $$H = -\sum_i p_i \log p_i,$$ and note that it has the form of an expectation: $H$ is the expected value of $-\log p_i$, so it makes sense to give a name to that latter quantity. This is nice if you understand the value of the entropy. In Shannon's paper ('A ...

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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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Just to give an account of some of the most popular approaches that I have met so far about out of equilibrium thermodynamics and corresponding generalized definitions of entropy and thermodynamic potentials. On one end of the spectrum, on can follow a statistical inference approach to statistical mechanics in its very foundation (as it has been proposed ...

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As for "a classical analog to quantum mechanical tunneling", theoretically one can jump over a classical barrier having lesser kinetic energy than the potential energy one's mass would have at the top of the barrier. In fact, in the course of a high jump, one can bend over the barrier in such a way that one's center of gravity will be outside of the body and ...

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Consider the path integral: $$\int Dx \exp\left(i\int\left(\frac{m\dot{x}^2}{2}-V\right)dt\right)$$ You can consider paths in "imaginary time" by performing a Wick rotation $t\to i\tau$ $$\int Dx \exp\left(-i^2\int\left(\frac{m}{2}\left[\frac{1}{i}\frac{dx}{d\tau}\right]^2-V\right)d\tau\right)$$ \int Dx ... 3 The microcanonical ensemble is the (maximum entropy) probability distribution for a given specified total energy. What you've calculated is actually the maximum entropy distribution with no constraints on the energy, which is the same as the canonical distribution at infinite temperature (\beta = 0). To correctly calculate the microcanonical entropy, ... 3 When people say "accessible microstates" it means "microstates consistent with a set of constraints or conditions which you are supposed to keep in the back of your mind". The most common example of a constraint is a fixed amount of energy. A closed physical system with exactly one constraint, fixed total energy, is called the microcanonical ensemble. ... 2 Was initially posted as a comment. (Comment removed now) The micro-states are changing nonetheless, a different point in the phase space of your system, so the system is evolving, even though the two states are part of the same macro-state. Finally remember that the Metropolis probability criterion for accepting moves is (in one of the simplest schemes of ... 2 Statistical mechanics relies on a probabilistic understanding of the world and as such one needs to define a probability space. In classical statistical mechanics the probability space consists of a domain which is the set of all possible microstates (that is the position and velocity vectors of all the particles in the system) and a probability measure ... 2 The definition of a physical concept can be a differential form but can’t be the difference of functions. \Delta S=S_{final}-S_{initial} is an equation but not the definition of entropy. Thermodynamics itself now can hardly explain “what is the entropy really" , the reason please see bellow. 1.Clausius’ definition \begin{align}dS=\left(\frac{\delta ... 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 ... 1 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 ... 1 This is a very difficult question. In one dimensions a two body scattering is the minimum interaction you can have. In multiple dimensions you can have more bodies scattering as the minimum interaction. The Yang Baxter equation is formally,(R\otimes \mathbf{1})(\mathbf{1}\otimes R)(R\otimes \mathbf{1}) =(\mathbf{1}\otimes R)(R\otimes ...

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Yes, a high magnetic field can massively reduce the entropy of the dipoles for a given temperature. When the dipoles enter a field they become hotter (analogous to a gas when its volume is compressed) so they dissipate heat into their environment (into lattice vibrations and other degrees of freedom). Once the dipoles have returned to thermal equilibrium ...

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Yes, the efficiency is then higher than 1 but that is fine because, although it can be described by a negative temperature, it does not per se correspond to a heat reservoir but rather to a very peculiar (statistically) excited state of a macroscopic system that is just waiting to release this energy. Now, this is not conceptually so problematic overall, ...

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Here is a proof following Ojima, "Lorentz Invariance vs. Temperature in QFT", Letters in Mathematical Physics (1986) Vol. 11, Issue 1 (1986) 73-80. The first two pages of the paper are available for free here, but the website wants money for more of the paper. (Click the orange "Look Inside" button if the paper doesn't open automatically.) Fortunately, the ...

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No, this is talking about correlations between s random particles. The s-particle distribution function is a 2*d*s (so 6s in 3 dimensional space) dimensional PDF that statistically describes s particles. For s=1, this is just the normal density in phase space. For s=2, this might show, for example, that more often than not two particles are traveling away ...

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"Accessible microstates" is related to things like activation energy and metastability. For example, think about a diamond sitting on a table at room temperature and atmospheric pressure. There are microstates for the diamond in which the carbon atoms are rearranged into a piece of graphite. But those microstates will essentially never occur no matter how ...

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The underlying reason is purely mathematical (apart from the fact that the relative probability of a state is $\exp(-\beta E)$), so'll give a mathematical answer. The idea is that you want to sample the state space, meaning that you want to randomly generate states so that the generated states follow a probability distribution, in your case the Boltzmann ...

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It's because you don't want to just find an optimum, but to sample the entire distribution. The MH algorithm satisfies a property called "detailed balance", and for me this video is the most intuitive explanation I've seen.

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Since you found the critical point via numerical simulations, you probably have little analytical insight into its properties. This makes it hard to extract the central charge, because it often appears in expressions combined with speed of sound or other quantities (e.g. in the free energy for a 2d CFT). So you need to find a universal quantity, easily ...

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What did your book mean by "low" temperature ? In this case I guess it's room temperature, which is really not close to absolute 0. Anyway, note that what matters is not the absolute entropy but the variations between two states, and even at low temperatures entropy variations can be high between two different spatial configurations. To answer your ...

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I assume that your question is: "How do I compute the helmholtz free energy given the dependence of the entropy on the intensive parameters $S(E,N,V)$?" In this case you start by inverting $S(E,N,V)$ to get a relation for $E(S,N,V)$. Then you can perform the Legendre transform, $$F(T,N,V) = E(S(T,N,V),N,V) - S(T,N,V) \, T \, ,$$ where the function ...

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I am not an expert on this, so I would appreciate if errors are pointed out. To my understanding, the difference between the total energy and the free energy is due to statistical mechanics. Your simulation works on the level of the smallest constituents of the system, and doesn't directly look for the lowest energy configuration, but samples the space of ...

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You may start from the density matrix $\rho$ of the grand canonical ensemble (assuming $k_BT=1$ as the unit of energy), $$\rho=Z^{-1}e^{-H+\mu n},$$ where $Z=\text{Tr}\,e^{-H+\mu n}$ is the partition function. Then $\langle n\rangle$ and $\langle n^2\rangle$ are defined by tracing with the density matrix \langle n\rangle=\text{Tr}\,n\rho\text{, and ...

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