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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 ...


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 ...


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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

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

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 ...


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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

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 ...


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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

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 ...


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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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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