# Tag Info

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A "point" in a macroscopic system is not a geometrical point. It is a volume element that is small on a macroscopic scale and yet has a large number of molecules for entropy and internal energy to be defined. Your temperature probe does not measure its value at a geometrical point but for a small volume of the system in whose contact it is put. The local ...

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The best way to see this is to realize that the zero heat capacity is a quantum effect. Classically, the heat capacity does not go to zero. Quantally what happens is that at low enough temperatures all the particles are in their lowest possible energy states. To get even one particle into a higher energy state requires a small but finite energy ...

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1 - yes the zero temperature limit is not reachable, so you can't measure the heat capacity at zero temperature, what this calculation tell you is that if you measure at smaller and smaller temperatures you will see that C converges towards zero 2- No the reversibility of the path is not important as the entropy is an exact differential

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Length scales are not accounted for properly in your question. When you have a system at local equilibrium where a temperature gradient can be defined then each "point" in this description contains say $10^{10}$ molecules and can be seen as a thermostatistical system at equilibrium. We call that "local" equilibrium because intensive quantities such as ...

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Mad props for a cool question. I'm going to justify essentially the converse of the statement because it doesn't make much sense to talk of the temperature of a system that is in a pure state. Let's assume that we're talking about a quantum system with disrete energy spectrum (with no accumuation points) in thermal equilibrium. Let $\beta = 1/kT$ be the ...

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Since the energy $E$ is a random value, you can define another random variable $c_V=(E-<E>)^2$ such that the heat capacity is the mean value of this quantity: $C_V=<c_V>$ Now we can identificate the mean variance of $c_V$ with the variance of $C_V$, we have: $<\Delta c_V>=\Delta C_V= \sqrt(<E^4>-<E>^4)$ we can compute ...

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Thermal equilibrium between the system $A$ and the reservoir $R$ doesn't mean that the energy of $A$ is fixed. Thermal equilibrium occurs when the ensemble average of the energy of $A$ is fixed. In particular, during thermal equilibrium, the energy of $A$ with generically fluctuate about its ensemble average value. Moreover, knowing the temperature of a ...

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Because of the following. Disorder is usually equated with one's ignorance of the system - the less you know about the outcome of the random variable the more disordered it is. If the system turns out to be in a very unlikely state with low $p(x)$ you will naturally consider yourself to have been more ignorant than when it is in a state you consider very ...

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Okay, my comments are getting too much, so I will answer. If I understand your question correctly it says this: Papers show that the non-planar Ising model (finding its ground state) is NP complete On the other hand, finding the eigenvalues of a matrix is polynomial. So how do these points reconcile? The important point here is in the size of the input. ...

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You have to take into account the differentials. The actual equation is $$f_\text{MB}(\mathbf{v})\,\text{d}v_x\text{d}v_y\text{d}v_z = n\left(\frac{m}{2\pi k_BT}\right)^{3/2}e^{-mv^2/2k_BT}\,\text{d}v_x\text{d}v_y\text{d}v_z.$$ Changing to spherical coordinates, we get $$\text{d}v_x\text{d}v_y\text{d}v_z = ... 0 The distribution function allows to find some kind of probability by summing of f_vdv if it is the distribution over velocities or f_EdE if it concerns energies. One can be transformed to other as:$$f_v(v) dv=f_v(E) v dE,$$since dv=vdE if E is a function of v Taking into account E=\frac{mv^2}{2}, one gets:$$f_v(v)dv=f_v(E) ...

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This kind of exponential decay toward "equilibrium" can be derived when one looks at a Markov process. In this case, if we call $S_t$ the state of the system at time $t$ and $S_{t+1}$ the state at time $t+1$, one has for the evolution: $S_{t+1} = T S_t$ where $T$ is called the transition matrix. This implies that $S_t = T^t S_0$. The idea is then to ...

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This form of $dE/d\tau$ is valid only when the system is not too far from equilibrium and linear response assumption is valid. The fact that $dE/d\tau$ depends on the difference $E - E(0)$ alone is a consequence of assuming a linear response.

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Actually, the Liouville theorem is more general - it is valid even if the distribution function depends on time, and even if the Hamiltonian depends on time. http://en.wikipedia.org/wiki/Liouville%27s_theorem_(Hamiltonian) -> phase space volume preservation but no energy conservation: any Hamiltonian which depends on time, but you already know that. For ...

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Firstly, the logarithm needn't necessarily be to base 2. Changing the base just introduces a (scale) factor, so log10, log2 and ln are all equally useful. Log2 is convenient for people working with binary systems. Let's deconstruct the formula. I will define entropy to be $H = E[-\log(p)]$. You can see that this will reduce to a weighted average which ...

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Liouville's theorem not only depends on the form of Hamilton's equations but also on the fact that $\partial\rho/\partial t = 0$, where $\rho$ is the statistical distribution function of the system. This is strictly true only for closed systems and is approximately true for quasi-closed systems when not observed for too long a time. Energy of a system is ...

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Anyway, consider the following two examples. (1) A random variable always has a single, definite value, i.e., it isn't really random. Then the Shannon entropy is zero. This means that you don't gain any information by being told that my puppy is cute -- puppies are always cute, with probability 1. (2) Suppose a variable is equally likely to take on $n$ ...

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The most physical and understandable definition of Nekrasov's partition function to me uses five-dimensional gauge theories. Namely, any 4d N=2 susy gauge theory has a 5d version with the same matter content, so that compactifying it on a small $S^1$ brings it back to the original 4d theory. Then we put the theory on the so-called Omega background: it is ...

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Symmetric case I'll first explain the definition of the surface tension between two ordered phases of the Potts model, since the symmetry between the phases simplifies things. I'll work in dimension $3$ to be specific, but everything generalizes in a straightforward way to other dimensions. So consider the $q$-states Potts model at inverse temperature ...

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The other answers tackle the statistical/thermodynamic aspect. I will tackle the "falling apple " aspect. Why does the apple fall? From this observation onwards nature was modeled mathematically as interactions between masses, in this case, charges in the electromagnetic case etc. The observations of gravitational interactions led to a mathematical model ...

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Large systems with many degrees of freedom (e.g. a ball consisting of many molecules) tend to settle into low energy states. This is a direct consequence of two fundamental laws, the first and second laws of thermodynamics: energy conservation and entropy increase. A system with many degrees of freedom can be in many different microscopic states (think ...

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I will address such sample systems as a point (or a small metal ball) rolling or bouncing on some hard surface with hills and pits, and an atom which can be either in excited or in basic state. I. If we consider an ideal closed system, then the enegly is conserved. But real systems do not (exactly) behave this way. For a macroscopic mechanic motion we can ...

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The probability of finding a system in a state with energy $E$ is $P(E) = \exp(-\beta E)/Z$, where $\beta = (kT)^{-1}$, $k$ being the Boltzmann constant and $T$ being the absolute temperature. $Z$ in the formula for $P(E)$ is the canonical partition function. For our purpose, we can consider it as a factor introduced to ensure that $0 \le P(E) \le 1$. The ...

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I am not sure if this is what you are up to (it is related to what Xiao-Qi Sun said) to but I'll give it a try too ... At the beginning of Chapter V.2 of his QFT Nutshell, Anthony Zee explains how classical statistical mechanics (characterized by the corresponding partition function involving the Hamilton function) in $d$- dimensional space is related to ...

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Even though the answer you chose is very good I will add my POV Take a box of gas particles. At t=0, the distribution of particles is homogeneous. There is a small probability that at t=1, all particles go to the left side of the box. In this case, entropy is decreasing. Take the statistical mechanics definition of entropy: where kB is the ...

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Microstate:the number of distinct arrangements of partical in cells in phase space , Macrostate:the arrangements of partical in cells in phase space when partical are identical &the number of Microstate in a given Macrostate is called Thermodynamic probability

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Think of entropy as a steady-state quantity related to system dynamics: Wait for a 'long time', smear out the phase space trajectories and measure the resulting volume. This means that even if all gas particles ended up on the left side of the box (unlikely, but not impossible and realized by perfectly valid microstates), entropy only would have decreased ...

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Statistical physics doesn't tell you that entropy will increase all the time. Just that it will increase on average. The maximum-entropy state is the one with the largest number of microstates. This doesn't prevent you from observing an odd state every once in a while -- even one with very low probability -- in fact fluctuations of the state do happen and ...

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Right, there is a small probability that the entropy will decrease. But for the decrease by $-|\Delta S|$, the probability is of the order $\exp(-|\Delta S| / k)$, exponentially small, where $k$ is (in the SI units) the tiny Boltzmann constant. So whenever $|\Delta S|$ is macroscopically large, something like one joule per kelvin, the probability of the ...

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Photons are modeled as bosons with an integer spin, have a symmetry and can occupy the same quantum state. All this means, is that they use the Bose-Einsten distribution instead. Where the Bose Einstein distribution gives the average number of Bosons found in an energy state, $\epsilon$. ...

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I don't want to un-accept @juranga's perfectly good answer, but for future visitors it's worth recording that the macroscopic nature of general relativity is made very clear in this 1995 paper, in which Ted Jacobson derives Einstein's field equations from $dS = \delta Q/T$, together with the Bekenstein bound. (Plus a few other assumptions to do with special ...

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Although summarized as an objection of macroscopic irreversibility when microscopic laws are reversible, Loschmidt's objection originally points that there has to be something breaking the time reversal symmetry in Boltzmann's derivation of the $H$-theorem. I think that Boltzmann's answer was to say that high $H$ states (in absence of external driving) are ...

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Nothing - that is the correct definition. One little caveat is that small systems are usually in contact with a larger system with a temperature that's more easily controlled or measured (the "heat bath"), and $T$ usually stands for the temperature of the heat bath rather than the small system itself. However, this doesn't make a lot of difference in ...

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I think most people would say the paradox is resolved - but, as the answers to this question make clear, they wouldn't necessarily agree about who resolved it or what precisely the resolution is. For my money the paradox was elegantly resolved by Edwin Jaynes in this 1965 paper. In Jaynes' argument, the symmetry is broken by the fact that we, as ...

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Your interpretation is not quite right. One sharp interpretation one can give to this "cutting" of phase space into cubes of size $h^{2N}$ (here $N$ is the dimension of the system's configuration space), is that it allows one to use classical phase space to count the number of energy eigenstates of the corresponding quantum hamiltonian. Instead of trying ...

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Well, you are not specifying the kind of probabilities you are talking about. Therefore, I refer to the tags and assume that we are talking about quantum (not statistical) probabilities. (See the edit at the end) Then, I have to note that your probabilistic definition of independece does not make much sense. Edit Well, according to the comments, I should ...

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How about path integrals? The probability that a system evolves between state $|\phi_1\rangle$ and $|\phi_2\rangle$ is $$\langle \phi_2|\phi_1 \rangle =\int_{\phi_1}^{\phi_2}\mathcal{D}\phi \exp \left(\frac{i}{\hbar}S(\phi)\right)$$ where the measure $\mathcal{D}\phi$ is suitably defined and the action $S(\phi)$ is the integral of the Lagrangian (over ...

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(Upgrading my comment to an answer as per @Nathanial's request but with more detail...and an animation) Mutual induction (e.g. in a transformer) is easier to understand than self-induction (the inductance of a coil). Mutual induction separates the magnetic field source from the motion of the electrons in the pickup coil. A time varying magnetic field ...

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Take a loop of wire (the simplest inductor), carrying a steady current $I_0$ (to start with). Let's say it's a superconductor, so the current is really staying constant. You understand that there is also a steady magnetic field created by this current, $B_0$. So far, there is no force on the electrons--or at any rate, no force that we care about (only a ...

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You do not need neither a coil, nor core. The vacuum is a pretty good conductor of the magnetic field. Cores and coils only complicate the things. Consider a single loop of current. Surprise: It also has a magnetic field! Individual electrons are susceptible to the electric field, that is induced by the changing magnetic field in the usual way: $F = q * ... 2 I assume the system to be$N$particles with potential energy$U(\vec{R})$and kinetic energy$\frac{1}{2}(\dot{\vec{R}},M\dot{\vec{R}})$where$R$is a$3N$-dimensional vector in the configuration space and$M$is the mass matrix. In particular, I assume that the particle numbers are fixed -- there is no interactions that create new particles etc. To avoid ... 7 I prefer to see it in the following way: $$Z_{quantum} \equiv \sum_{m} e^{-\beta E_m}$$ Where$m$is a quantum microstate eigenstate of the Hamiltonian. Now, you can split the sum into two parts; a sum over quantum microstates that yields the same energy eigenvalue$E_n$and a sum over all possible values of$E_n$: ... 3 The only trick here is getting used to how discrete sums are turned into integrals. Suppose you let energy be a function of momentum$p$and position$q$. Then you can rewrite the discrete quantum partition function as$Z_{quantum}=\sum_{p,q}e^{- \beta E(p,q)},$where the sum is over each of the$N$positions and$N$momenta, and the only challenge is how ... 1 Yeah, the example$\sigma=\frac{1}{2} \left | \uparrow \right \rangle \langle \uparrow|+\frac{1}{2} \left | \downarrow \right \rangle \langle \downarrow| $you give describes a completely unpolarized ensemble for single spin-$1/2$system, and the coefficients$1/2$definitely represent the probabilities for the particle be in either up or down spin ... 1 Numerical simulations are a good example to see with your own eyes that statistical mechanics results can be gotten without an infinite number of molecules in general and in particular in the microcanonical ensemble. However, one has to be aware of the finite size effects and see what is the difference with what you would expect from textbooks analytical ... 1 I was wondering about exactly the same question some days ago, reading the seminal paper of Mermin (Rev. Mod. Phys. 51, 591--648 (1979), The topological theory of defects in ordered media), where you find an introductory discussion for the example of spins within the two-dimensional plane. There you find a lot of plots with spins (depicted as arrows in the ... 2 First of all the probability density for a system in a canonical ensemble to be at a given energy (i.e. with respect to a measure$dE$) is not the formula you gave but rather:$\rho(E) = \frac{\Omega(E)e^{-\beta E}}{\int_0^{+\infty}dE\:\Omega(E)e^{-\beta E}} = \frac{e^{S(E)/k_B}e^{-\beta E}}{Q}=\frac{e^{-\beta F(E)}}{Q}$we thus see that weiht associated ... 4 The Maxwell-Boltzmann distributes$N$particles in energy levels$E_i$such that the entropy is maximized for a fixed total energy$E=\sum E_i N_i$. The probability that a particle is in the energy level$E_i$is proportional to the number of particles in the energy level$E_i\$ in this particular arrangement of particles in which entropy is maximized (the ...

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Nice question. Part of this is basic stuff, but the question about why it's at a maximum for low energies is a nice conceptual question that wasn't immediately obvious to me. The Boltzmann factor is the (unnormalized) probability that a specific degree of freedom will be in a specific state. For instance, say we have some helium gas, and we choose one ...

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