# Tag Info

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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 ...

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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 ...

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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 ...

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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 $$S = \log\Omega(E) ... 4 If the universe is open, there's obviously more universe that you haven't included in your system. The universe, by definition, contains all energy and matter. An open system, by definition, has an outside system to exchange energy and matter with. If that outside system isn't part of the universe, then where is it? 3 I will be blunt. As fas I know, nobody knows a priori for which systems equilibrium statistical mechanics will work or not. Part of the current effort to determine which systems are fine being described by equilibrium statistical mechanics focuses on various proofs of ergodicity for such systems. For now, they are somewhat limited to either a restrictive ... 3 The Gibbs form \rho\sim \mathrm{e}^{-\beta H} is just a fancy way of writing the standard Boltzmann distribution. A quantum (mixed) state is written in general as$$ \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 ... 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 ... 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 ... 3 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 ... 3 These are conjugate variables. 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: ... 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 ... 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 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 ... 2 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 The famous Fermi-Dirac and Bose-Einstein average occupations,$$ \overline{n_i} = \frac{1}{e^{(\epsilon_i-\mu)/kT} \pm 1},$$are only exact in the grand canonical ensemble (GCE) where the total particle number is a flexible (fluctuating) quantity. That flexibility, and the assumption of noninteracting particles, is what allows us to treat each ... 2 Assuming the functions are well-behaved (continuous and differentiable), you can change the order of differentiation.$$ \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$$... 2 Temperature is not useful concept for describing clusters of stars or other gravitational systems, because such systems are not in the realm described by thermodynamics. There is no way to set up thermodynamic equilibrium - globular clusters partly evaporate and core implodes. Also the velocity distribution can't be Maxwell-Boltzmannian, because very fast ... 2 The key is: Landau theory doesn't assume the order parameter is small. All it assumes is that the free energy is analytic in the order parameter. One then usually expands this free energy up to some order (which is possibly by definition of 'analytic'). It is key to realize that expanding a function in a variable to some order does not mean this variable has ... 2 The name "Gaussian noise" actually has to do with the higher order correlations in the noise, such as:$$\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, $$and so on. If the noise is Gaussian then all of these higher order correlations can be rewritten in terms of the two-term ... 2 Pressure is defined as the rate of increase in internal energy to rate of decrease in volume, i.e.$$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 ...

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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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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 ...

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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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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. ...

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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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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 ...

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Probably the best place to start classically is with integrable systems. A crude physicist definition is that these are systems that have, in the words of Nandkishore et al, "an infinite set of extensive conserved quantities that are sums of local operators" (1). Roughly speaking, such systems will never approach an equilibrium because none of these ...

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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 ...

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