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Just because the entropy of the subsystems increases, that doesn't mean that the entropy of the whole system increases. This is possible here because of entanglement: an entangled pure state has zero overall entropy, but the subsystems have non zero entropy. A simple example is the state \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle). ...

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TylerHG: Yes it is easy to calculate the density of states. But what I'm really asking here is "why." Note that a thin circular ring in $\mathbf{k}$-space of thickness $dk$ has area $dA=2\pi k\,dk$ (by elementary geometry). In $E$-space, since $E\propto k^2$, that ring corresponds to a patch of width $dE=2k\,dk$. Thus $$\frac{dA}{dE}=\pi.$$ But ...

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It is easy to show that the total number of electrons in a 3D fermi sphere is : $$N(e)=\frac{V}{3\pi^2}*k_F^3$$ Where $k_F$ is the Fermi wave vector and $V$ is the real space volume of your sphere. Now if you rearrange for $k_F$ in terms of the total number of electrons you'll get a particular equation. It is know that ...

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These sorts of problems are easiest to think about if you build up from simpler problems. Probability that n particles are all of type A: $p_A^{n}$. Probability that, with two particles chosen, the first is of type A, and the second of type B: $p_A p_B$. Probability that, with two particles chosen, one is of type A, and the other of type B (any order): $2 ... 4 I think the following was the argument that the authors used to arrive at their expression: The first particle can choose to be in any one of the g states; the second again. So N particles have a total of g*g*g*... choices - which is the equation given. When they are indistinguishable you divide by N factorial since that is how many times each ... 4 It seems to me that you're looking for the Boltzmann transport equation: $$\frac{\partial f}{\partial t}+\frac{\mathbf p}{m}\cdot\nabla f+\mathbf F\cdot\frac{\partial f}{\partial\mathbf p}=Q+\left(\frac{df}{dt}\right)_{\rm coll}$$ with$f$the distribution in phase-space,$\mathbf p$the particle momentum,$Q$some source term, and the RHS an interaction ... 3 You're running into a tricky property of statistical variables: what is true for an individual particle is not necessarily true when averaged across a distribution. In particular, you can say that the distance one particle travels between collisions, its free path length$\ell$, is equal to that one particle's speed times its free path time$t$: $$\ell = ... 1 A (3d) gas of particles with a gravitational interaction is an example of a system with long range interactions, where the energy is not additive and thus many basic results of classical statistical mechanics are not valid, including the equivalence of the microcanonical, canonical and grand-canonical ensembles. For a general introduction to the subject see ... 0 Why do you say for N particles you have N! distinguishable state? Lets just make sure you got it right. Lets say you have N particles on a lattice having M sites. In the ideal case, no excluded volume therefore each particles can access M sites. The total number of states for distinguishable particles in M^N. Now, if you considere that the particles are ... 0 I'm not sure whether I understood your actual question correctly. But in my opinion chaperones, in the context of statistical mechanics, can act in two ways. 1) They act as a catalyst, i.e. they merely assure that the minimum in free energy is found fast enough. 2) They can keep the polymer in a steady state. The polymer system without the chaperones is ... 0 i will have to disagree with some of the answers posted in this question. First, this involves a matter of interpretation of the quantum formalism (and a prevailing "interpretation", the Copenhagen one) Although this interpretation (which i find unsatisfactory and non-physical) may seem prevailing (and indeed it might be), is not because it offers a better ... 2 I will not directly answer your question, rather I'll try to make plausible the connexion between QFT and statistical physics. To my mind the mathematical details are somehow obscure and confusing, whereas using the theory is worth a deal, and give interesting results, especially in condensed matter and nuclear matter problems. For more details you can have ... 0 I am not sure I understand your question either but will try something else. When you write a probability distribution \mathbb{\pi}: \: \mathbb{R}^3 \rightarrow \mathbb{R}_+ for the cartesian variables x,y,z, you can also look at the induced probability measure for the function r^2 = x^2 + y^2 + z^2. To figure out what is the corresponding probability ... 1 I am not quite sure I understand your question. In your second equation, E is treated as a variable and not as function anymore, so why do you want to find E function? The density of states is basically a function counting the number of state in x\in\mathbb{X} that give the same energy: \Omega(E_0)=|\{x:E(x)=E_0\mbox{ and } x\in\mathbb{X}\}| ... 1 The term you are looking for is premelting or "surface melting." It is an observed phenomenon (which could explain how ice skating works) with some thermodynamic descriptions. Basically what happens is the system is separated into two distinct phases, a solid (ice) and a vapor (air). There is a surface energy associated with this interface. If it happens ... 0 In other words, for, say, an elemental solid, should we expect a portion of its surface to be liquid at any given time, with this portion increasing steadily until the melting point when the whole thing becomes liquid? It is possible for many compounds be part solid and part liquid under the right conditions. As ice melts, you have this condition. ... 0 I found this one: Van der Waals molecules. This answers my question to some extent. 1 An ideal gas in equilibrium cannot be supposed to have reached its equilibrium from a non-equilibrium state by interaction of its particles, because by definition the particles of an ideal gas do not interact: Hard spheres of radius a that collide elastically do interact in the time average - even when they interact "almost never". (Among other things ... 2 Yes, it is possible. The simplest qualitative answer to this is that, at the microscopic level, the electrons in a conductor are dictated by quantum mechanics, which is inherently probabilistic. Velocities and positions are rarely ever totally excluded from a given value; it's just insanely unlikely for a single electron to attain that given value. ... 1 There are multiple ways of looking at the ideal gas model. One is to say we have point particles colliding elastically (as an ideal scenario) and proceed to obtain the exact equation of state, i.e pV=nRT. The other approach, is to state a priori the relevant length scales and time scales at which the system is going to be studied are sufficiently ... 1 Ideal gas does only mean that there are no forces between the particles. They do not have to be point-like. For example 2-atomic gasses could have 3 translatory and 2 rotational degrees of freedom in kinetic gas theory, while still no forces act between the molecules. So ideal but not point-like. For one-atomic gasses the atoms are often taken to be hard ... 0 The notion of a point-like particle needs refinement. An electron is generally thought to be a point particle but still electrons interact through coulomb forces. What matters is that collision cross-sections between the particles constituting an ideal gas vanish. 0 As far as I know, you can still suppose that ideal gas interacts with the wall of the box (particles have to bounce). So, particles with sightly different initial momentum will gain more and more different momentum. This process can redistribute the momentum in agreement with Statistical Mechanics. This is what I remember when I attended my Statistical ... 2 Ideal gasses do not exist. In real gasses, molecules do take up volume. This does affect the pressure-volume curve. Still, an ideal gas model is close enough to reality to be useful. It is more useful if not taken so literally as to include its most unrealistic features. 4 Refs. 1 and 2 define a canonical transformation (CT)$$\tag{1} (q^i,p_i)~\longrightarrow~ (Q^i,P_i)$$[together with choices of Hamiltonian H(q,p,t) and Kamiltonian K(Q,P,t)] as satisfying$$ \tag{2} (p_i\mathrm{d}q^i-H\mathrm{d}t)-(P_i\mathrm{d}Q^i -K\mathrm{d}t) ~=~\mathrm{d}F$$for some generating function F. On the other hand, Wikipedia (March ... 2 The argument goes something like this: Suppose we have a system in contact with a large reservoir at temperature T with which it can exchange energy. The Boltzmann distribution tells us that the probability of finding the system in some small region of phase space \mathrm{d}V is given by$$ \rho(q_i,p_i)\,\mathrm{d}{V}= \frac{1}{Z} ... 1 The short answer is that$\bar E = -\frac{1}{Z}\frac{\partial Z}{\partial \beta}$($\dagger$).$\bar E$is the average energy of a quadratic degree of freedom in this case and$Z$is the partition function which sums over all possible values of$E$associated with your degree of freedom$q$. So it's just how the math works out. There is a short derivation ... 0 The first equation is not correct, the Fermi-level is the chemical potential of the electrons. In semiconductor the carrier density$n$(units of$\text{m}^{-3}$) in the conduction band is the integral of the Fermi-Dirac function over the conduction band density of states$g_c(E)$(units of$\text{m}^{-3}\text{J}^{-1}$), $$n = \int_0^{\infty} g_c(E) f(E, ... 0 The maximality of entropy in mechanical equilibrium requires only the ratio p/T to be equal. So there is a possible equilibrium state where one system has double pressure and double temperature... Where is my mistake? If the systems are allowed to change their volumes, they are allowed to change their internal energies as well. When the wall moves, ... 3 Because if those particles aren't point objects, you must also take into consideration that they take some space in the system and have properties like density and size which have to be taken into consideration when formulating the laws. This makes everything extremely complicated. A huge part of classical mechanics is only true for point objects for the ... 17 If the particles are not point-like, they will take up some volume. As the gas is compressed, the collision frequency will rise more quickly, which will make the pressure-volume curve change. The corrections in the Van der Waals model of a real gas accounts for the volume of the particles. Also if they have internal structure, that structure can have ... 2 This is a classic conundrum and it is called the "problem of the adiabatic piston". You can find it discussed in books on thermodynamics by Landau & Lifhsitz and by Callen. Another very thorough analysis is by Gruber "Thermodynamics of systems with internal adiabatic constraints: time evolution of the adiabatic piston". (You can find Gruber's article ... 0 Firstly, from the middle paragraph it seems you imply that adding, subtracting or changing variables of potentials stands in literal connection to energy transfers and but that's not the case. You can write down all quantities for all situations and in any case, there is only one and the same second law of thermodynamics at work. For any system parameter ... 0 Without magnetism, you have F(T, V, N) and G(T, P, N). So you did a Legendre transform so that your potential depends on P instead of V. It also means that you move from an extensive to an intensive quantity. The free energy F comes via U = TS from the internal energy, which depends purely on extensive quantities. By going from U to F, you ... 1 You are absolutely right that the limit in which this approximation holds is$$\beta(\epsilon - \mu) \gg 1 \,,$$which is not trivially the 'high-temperature limit', and indeed looks rather like the low temperature limit. However, it also looks like the limit of large negative$\mu\$. If we want to know how temperature will affect the exponent, we need to ...

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