# Tag Info

1

You state that: there is literally no way to squeeze more information (entropy) into a given volume than that in a black hole occupying that volume But you must keep in mind that the volume occupied by the radiation+BH system is larger than the volume of the black hole by itself. When the black hole initially forms the horizon has a radius $r$ which ...

1

It's a confusion of terms. The universe is a closed thermodynamic system whether or not it is 'open' or 'closed' in a cosmological sense. In cosmology, open and closed universes refer to the curvature of the universe, whether positive (closed, finite universe), zero (flat, open, infinite universe) or negative (curved, open, infinite universe). In ...

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If the universe is open, there's obviously more universe that you haven't included in your system. The universe, by definition, contains all energy and matter. An open system, by definition, has an outside system to exchange energy and matter with. If that outside system isn't part of the universe, then where is it?

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Most people prefer to do research in the Path Integral Formalism (PIF), instead of the Operator Formalism (OF), because it is "easier." Easier means that whatever property you have in the OF, if you have a PIF for that theory, you can have more properties in the PIF. For example, Noether's theorem is only valid, as far as I know, in a Lagrangian formulation. ...

1

The derivation of ideal gas equation from Hamilton's equations will take the same procedure as what you have seen in Wikipedia. Since you said you haven't understood the way in which the equation is derived I will give you a step by step explanation on it. So we have a system of perfect gas molecules. Of course they are non-interacting. We are going to ...

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What is exactly the canonical ensemble? Thermodynamic ensembles are ensembles in the mathematical sense, so your option no. 2 is the correct one. Consider a system of non-identical particles, this will appear much more clearly. What do "thermal average" and "thermal fluctuation" mean? "Average" is not something per se, one should speak about the thermal ...

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I think, the comment of lucas is the best answer; I'll elaborate a bit If I mix these two containers eventually both the Oxygen and Nitrogen will be at the same temperature. Why is that? By temperature you mean the mean energy per particle. We observe experimentally, that if you mix two heaps of particles with different mean energies, after some time ...

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If you mix two gases, their atoms/molecules exchange with energy, momentum, etc. It is not surprising that a hot gaz heats a cold one. It (heating) can only stop when the mutual energy exchange becomes equal to both gases. So the energy distributions of "subsystems" are equal in the end, but the momentum distributions are still not. P.S. The subsystem ...

1

In a system of many particles, we essentially observe the most probable configuration, and relative fluctuations around it are negligible. Here I will prove that the most probable state of a 2-particle system is this with equal energies. The probability of a state is proportional to the volume of the corresponding part of phase space. If a particle has ...

3

I will be blunt. As fas I know, nobody knows a priori for which systems equilibrium statistical mechanics will work or not. Part of the current effort to determine which systems are fine being described by equilibrium statistical mechanics focuses on various proofs of ergodicity for such systems. For now, they are somewhat limited to either a restrictive ...

0

There are several ways to characterize a system of particles. You can see them as whether they are isolated or in thermal equilibrium or in mechanical or chemical equilibrium. Similarly, you can ask whether a system is made up of identical bosons, identical fermions or distinguishable particle. In each of this cases, different statistics prevails. Simply ...

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The boundary lines are, 1. isolated system 2. equilibrium 3. no composition change such as chemical reaction

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if we assume $E_1$ (zero state) and $E_2$ (state 1), from this question, the system has n units of energy and N particles. The multiplicity is then $$\Omega (N,n) = \frac {(n+N-1)!}{n!(N-1)!}$$ and entropy can be calculated, $$S=k \ln \Omega(N,n)$$

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Pressure is defined as the rate of increase in internal energy to rate of decrease in volume, i.e. $$P=-\frac{\partial U}{\partial V}$$ Assume a particle in a box, for example the classic infinite quantum potential well of width $L$. The quantized energy is $$E_n=\frac{n^2h^2}{8mL^2}$$ In a 3D box this becomes ...

2

The name "Gaussian noise" actually has to do with the higher order correlations in the noise, such as: $$\langle \eta(t) \eta(t+\tau_1) \eta(t+\tau_2) \rangle,$$ $$\langle \eta(t) \eta(t+\tau_1) \eta(t+\tau_2) \eta(t+\tau_3) \rangle,$$ and so on. If the noise is Gaussian then all of these higher order correlations can be rewritten in terms of the two-term ...

3

The Gibbs form $\rho\sim \mathrm{e}^{-\beta H}$ is just a fancy way of writing the standard Boltzmann distribution. A quantum (mixed) state is written in general as $$\rho = \sum_n p_n \lvert \phi_n\rangle \langle \phi_n\rvert, \qquad (1)$$ where $p_n$ is the probability to find the system in the pure state $\lvert \phi_n\rangle$. The thermal equilibrium ...

2

The important point here is that there is no thermodynamic limit for gravitating systems, and thus there is no well-defined temperature. This is, perhaps, not a completely intuitive result, but it comes from work on the stability of matter. This is not as glamorous as it sounds, but revolves around the need to show that the energy of matter is an extensive ...

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A typical velocity dispersion in a globular cluster is 10 km/s. For a typical 1 solar mass subgiant in an old globular, then equating the kinetic energy to $3kT/2$, we get $T = 5\times 10^{60}$ K. Doesn't seem that helpful really... The concept of temperature is only ever applied in a relative sense - i.e. some component is hotter than another. Can't say ...

2

Temperature is not useful concept for describing clusters of stars or other gravitational systems, because such systems are not in the realm described by thermodynamics. There is no way to set up thermodynamic equilibrium - globular clusters partly evaporate and core implodes. Also the velocity distribution can't be Maxwell-Boltzmannian, because very fast ...

3

One benefit of scaling the heat capacity with another extensive variable is that you end up with an intensive property -- heat capacity per # of particles. Similarly specific heat refers to the heat capacity per unit mass so that the value of the intensive property can be compared between samples of the same material but with different sizes or geometries ...

0

It is indeed possible to change between these phases adiabatically. Since, as you noted, the ground state changes between being a superfluid and a Mott insulator, starting in the ground state and making an adiabatic change means that you track that change in state by definition. Note that this diagram is only formally true for the Grand Canonical ensemble, ...

2

While the answer of wbeaty is very interesting in showing points relevant in practice, I think all the answers are still missing an important and simple theoretical point, which you should consider to understand the process. vapour pressure does mean two different things as used above. First, the pressure, the existing water vapour would have (if it were ...

2

When the vapor pressure is equal to the external pressure, there will form a bubble. Not true. Instead, when the vapor pressure is equal to the external pressure, then any existing bubbles will begin growing continuously. And, if no bubbles are already present, then the water will superheat far above the boiling temperature, yet no bubbles will ...

3

You need to be careful about how you go from the full system to the subsystem $A$. You define $\rho^\text{eq}(T) = Z^{-1} \exp(-H/T)$ as the thermal state of the whole system, but then you use $\rho_A^\text{eq}(T)$ without defining how you are reducing the density matrix of the whole system onto just the subsystem. There are two reasonable ways to do so: ...

3

These are conjugate variables.

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Comments to the question (v2): On one hand, the Kuramoto-Sivashinsky (KS) equation is a dissipative differential equation (DE). Each term has an even number of spatial derivatives. It's a non-linear version of the heat equation. Dissipative systems rarely have variational action formulations nor Hamiltonian formulations. On the other hand, in the Korteweg ...

2

Great question. I believe that yes, Liouville's theorem is the key part of the justification for this in classical stat mech. The reason for this is that it leads to a time-invariant equilibrium measure. If you used a volume measure that wasn't time-invariant in this way then it would be very strange, because on the one hand you would say that you had no ...

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This probably more a philosophical than a physical discussion. Let's take a simple everyday example: The air molecules in the room where you are sitting are fairly evenly distributed through the room. Because the molecules are subject to random motion, it's perfectly POSSIBLE to have all molecules bunch up in one half of the room and that there is a perfect ...

0

What justifies identifying this entropy with the one found in thermodynamics? Suppose parameter $X_1$ is contrained to some value $x$. The corresponding number of available microstates is $Ω(X_1=x,X_2,\ldots,X_n)$. This part of phase space is a subspace of the whole one, consisting of all microstates without constraint. So ...

3

We do observe spontaneous symmetry restorations in nature. This is called an emergent symmetry. See e.g. this post. A system posses an emergent symmetry if it appears symmetric at large (coarse-grained) scales although the apparent symmetry is explicitly broken by the microscopic description (typically the Hamiltonian or Lagrangian). I can give two examples ...

2

The key is: Landau theory doesn't assume the order parameter is small. All it assumes is that the free energy is analytic in the order parameter. One then usually expands this free energy up to some order (which is possibly by definition of 'analytic'). It is key to realize that expanding a function in a variable to some order does not mean this variable has ...

1

So why do we say entropy is extensive? It is a convention that is possible and useful for weakly interacting systems. Multiplicity of state of the system made of two systems of the same kind and size and energy $E$ is $$\Omega'(2E) = \Omega(E)\Omega(E)$$ with very good accuracy, the other terms in the sum over $x$ you indicated can be neglected. ...

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The entropy $S$ is extensive as long as you're consistent about what you mean by entropy. In your case you've mixed up two different definitions. One definition of the entropy has the system at fixed energy $E$ -- the other, a fixed temperature $T$. Fixed-E entropy For a system with fixed energy $E$ the entropy is defined to be $$S = \log\Omega(E) ... 1 It may well be that the cross-membrane voltage difference is largely homogeneous, but not because intracellular protons are very mobile. To the contrary, the cytosol mobility (rate of diffusion) of protons, and ions in general, is very different from the high mobility measured in water or dilute solutions of small molecular species. According to this ... 0 The entropy defined by Boltzmann can be applied in systems with fluctuating energy. The only requirement is the sharp definition of \Delta\Gamma, the number of microstate inside the macrostate. When the time is very long, the distribution of probability in phase space tend to be constant at every allowed point. This justifie the microcanonical ensemble. ... 9 I think there is a misunderstanding. You are perfectly right when you write that the total micro canonical entropy of a combined system will be $$S_\textrm{combined}(2E) = k_B\ln \sum_x \Omega(x)\Omega(2E-x)$$ The micro canonical entropy ought to be a function of only the total energy, total amount of matter and total volume of ... 6 Major edit: In @gatsu's answer, it is pointed out that only the amount of energy should matter, which is correct, as there's no such thing as distinguishable microstates with only rearranged energy (think stars-and-bars-type entropy calculations). So, I've edited out that part of the first paragraph and equations (in the first draft, I dropped that part of ... 1 This is because dn_i can be arbitrary. You get an infinite number of equations by choosing different dn_i. For these equations to be statisfied simultaneously, you need the coefficient to be zero. $$\ln n_i + \alpha + \beta \epsilon_i = 0$$ Note because you have included Lagrange multipliers, dn_i can be treated as ... 0 Nifty, The thermal velocity is velocity of thermal motion of particles which make up gas, liquid, and so on. Thermal velocity is a measure of temperature and is also measure of the width of the peak in the Maxwell–Boltzmann particle velocity distribution. 1 First of all, note that one cannot associate a temperature to a single quantum state (cf "vacuum state of the theory is defined as having zero energy and zero temperature"), and having a zero energy vacuum state is just a convention (as it is cut-off dependent, and thus renormalized). Furthermore, the OP is confused. Standard (i.e. zero-temperature) QFTs ... 2 Assuming the functions are well-behaved (continuous and differentiable), you can change the order of differentiation.$$ \left(\frac{\partial T}{\partial V}\right)_S=\frac{\partial}{\partial V}\left(\frac{\partial E}{\partial S}\right) = \frac{\partial}{\partial S}\left(\frac{\partial E}{\partial V} \right) = -\left(\frac{\partial P}{\partial S}\right)_V$$... 1 The total magnetic energy of the system is fixed. This would read (1) E = \sum_{i=1}^N -\vec{\mu}_i \cdot \vec{B} = B\mu_0(-N_+ + N_-)  with the general constraint (2) N_+ + N_- + N_0 = N. Where N_{\pm} is the number of spins with projection m_s = \pm 1, N_0 the number of spins with m_s = 0 and \mu_0 the magnitude of the magnetic dipole. ... 1 A self-contained, careful derivation of (4.10): We consider a thermodynamic system whose state can be characterized by the macroscopic variables (S, V, N), then starting with the fundamental relationship \mathrm dU = T\,\mathrm dS -P\,\mathrm dV + \mu\,\mathrm dN, and noting that \beta = 1/(k T), one can deduce the following useful expression for the ... 0 Let \phi(z)=\sum_{k\geq 1} a_k z^k be absolutely convergent and invertible in a neighborhood of z=0, with a_1\neq 0. Let us denote its (compositional) inverse by \phi^{-1}(u)=\sum_{k\geq 1} b_k u^k. It is not difficult to check that the following relations hold (see below):$$ \sum_{n=1}^m a_n\sum_{\substack{k_1,\dots,k_n\geq 1\\ k_1+\cdots +k_n=m}} ...

0

Ok. After experimenting a bit I think I answered the question, although confirmation would be useful. I will write $N\lambda^3/V = \alpha$ and $1/2 \sqrt{2} = \gamma$ so that we have $$\alpha = z + \gamma z^2$$ or that $$z = \frac{-1 + \sqrt{1+ 4\gamma \alpha}}{2\gamma}$$ Taylor expanding to second order, $$-\frac{1}{2 \gamma} + \frac{1}{2 ... 2 The famous Fermi-Dirac and Bose-Einstein average occupations,$$ \overline{n_i} = \frac{1}{e^{(\epsilon_i-\mu)/kT} \pm 1},$$are only exact in the grand canonical ensemble (GCE) where the total particle number is a flexible (fluctuating) quantity. That flexibility, and the assumption of noninteracting particles, is what allows us to treat each ... 0 In kinetic theory, we can estimate$$ \eta \sim n\bar{p}l_{p} $$and$$ \kappa \sim (c_p/m) \bar{p} l_{q}  where $\bar{p}$ is the mean momentum and $l_p$ and $l_q$ are the mean free paths for momentum and energy transport. The statement that the Prandtl number is of order 1 is then simply the observation that the two mean free paths are comparable. In a ...

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Estimation: I want the two densities of vater and vapour to become approximately equal. the density of water is nearly constant the vapour pressure (you can derive this from the above mentioned Clausius-Clapeiron-equation) is approximately exponential in $1/T$. This means, that if you increase pressure by a factor, the inverse of the evaporation ...

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