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13

Your description of critical temperature isn't quite right. If you increase the temperature of a liquid beyond the critical point, the atoms are moving so quickly that persistent structure fails to form and so you have something that behaves a lot like a very dense gas. Similarly, if you increase the pressure of a gas beyond the critical point, it becomes ...


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 \begin{equation} S_\textrm{combined}(2E) = k_B\ln \sum_x \Omega(x)\Omega(2E-x) \end{equation} 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 ...


6

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 ...


5

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) ...


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

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

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

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 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

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

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 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 ...


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

The standard way to put in a temperature is to go to imaginary time (euclidean space) and impose periodic/anti-periodic boundary conditions on bosonic/fermionic fields $$ \phi(x,\tau)=\phi(x,\tau+\beta) , \;\;\;\; \psi(x,\tau)=-\psi(x,\tau+\beta), $$ where $\beta=1/T$. This ensures that the path integral represents the partition function $Z=Tr[\exp(-\beta ...


2

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 ...


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 ...


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 ...


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

(I am not sure this is an answer but it is to long to be a comment) Let us create a simple example of a system of $3$ states, state $1$, state $2$ and state $3$. Let state $1$ and state $2$ both have an energy of $E$ and state $3$ have an energy of $E'\ne E$. Your first summation is summing over individual states. I.e. it is saying 'let us call the energy ...


1

In the second case $$\Omega = \binom{N}{Q}$$ you're picking $Q$ distinguishable balls out of a bag of $N$ and putting each into one of $Q$ cups lined up in a row, but it doesn't matter how they're ordered in the cups. In the first case $$\Omega = \binom{N+Q-1}{Q}$$ you're taking $Q$ indistinguishable balls and putting them into one of $N$ cups, and ...


1

Yes, these are three different values. But why do you call it a problem? This is just so. Take for example three numbers: 1, 1 and 4. The average of the squares (energy) is 6, the square of the average is 4, and the square of the most probable value is 1. There is no problem :)


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 ...


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. \begin{equation} \ln n_i + \alpha + \beta \epsilon_i = 0 \end{equation} Note because you have included Lagrange multipliers, $dn_i$ can be treated as ...



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