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24

The question isn't silly. The speed of each molecule in the liquid is much higher than the speed of either the piston or the water shooting out from the nozzle. At room temperature, for water molecules the average is on the order of 500m/s. And yet, the speed of sound in water is three times higher than that, which implies that pressure can propagate in ...


22

Adjacent molecules in a liquid all repel each other because of the electron clouds that surround the nuclei that they contain. In that sense these molecules never even 'touch' each other (at least not in the intuitive sense of the word). When you apply pressure to the liquid you're squeezing them into a (very slightly) smaller volume, thereby increasing the ...


16

If you consider a system $C$ and two subsystems $A,B$ with associated probability distributions $p_C,p_A,p_B$ and want the entropy to add, you must assume the subsystems are independent in the sense that $p_C(X) = p_A(Y)p_B(Z)$ where $Y$ and $Z$ are a partitioning of the variables $X$ belonging to $C$. Then, the total entropy is \begin{align} S_C & = ...


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 ...


10

$h$ factor The factor of $1/h^{3N}$ is a total hack. The integral over phase space has dimensions, whereas $Z$ only makes sense if it's dimensionless. The $h$ factors are there to make $Z$ dimensionless. Suppose you have a system with only one particle in one dimension. Then the integral in phase space goes over one position variable, $dq$ and one momentum ...


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 ...


7

Both are relevant, and "the misconception that Langevin equation is the universal stochastic differential equation for all kinds of noisy systems is responsible for the difficulties mentioned"* in your post. Take the SDE from Thomas' answer, $$\frac{dy}{dt} = A(y) + C(y)L(t)$$ where $L(t)$ is the noise term. Suppose we can turn the noise off, so we'd only ...


6

In some sense yes. The temperature is defined as an imaginary time in Matsubara Green's functions or some path integrals. Thus, a negative inverse imaginary temperature can be considered as a time. Here is a quotation from Alexander Altland, Ben Simons "Condensed Matter Field Theory": "Thus, real time dynamics and quantum statistical mechanics can be ...


6

It depends on the parameter you consider. The Maxwell-Boltzmann distribution (as the name is applied in the Wikipedia article you link) is the distribution of the particle's speed (absolute value of the velocity) in a gas and this is a chi distribution. You probably were thinking of the distribution in terms of the vectorial velocity and then indeed it ...


6

This phenomenon has been studied by Vella and Mahadevan and written up in the American Journal of Physics (http://scitation.aip.org/content/aapt/journal/ajp/73/9/10.1119/1.1898523). It's called the Cheerios effect. If the cereal pieces clump together away from the edges of the bowl, they gravitate toward a slight concavity in the surface caused by milk ...


6

Let's simplify things down to the barest minimum: one dimension, one particle, and a wall. O | The particle moves to the right, hits the wall, and rebounds, perfectly elastically. If the wall is fixed in place, the particle will leave the collision with exactly the same kinetic energy as it came in with. But what if the wall is moving to ...


6

OP's question (v1) is essentially asking Does the operator identity $$ e^{\frac{it}{\hbar}[\hat{H},~\cdot~]}\hat{A}~ =~ e^{i\hat{H}t/\hbar}\hat{A}e^{-i\hat{H}t/\hbar} \tag{1} $$ have an analog using functions/symbols $H$ and $A$ rather than operators $\hat{H}$ and $\hat{A}$, respectively? The answer is: Yes, in terms of the Groenewold-Moyal star ...


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) ...


5

Both Ito and Stratonovich stochastic PDEs can be used to derive a Fokker-Planck equation. Indeed, for simple one-dimensional processes the Ito process $$ dx = a\, dt + b\, dW(t) $$ is equivalent to the Stratonovich process $$ dx =\left( a\,-\frac{1}{2}b\partial_x b\right) dt + b\, dW(t) $$ The answer is then that both are physically reasonable, for a given ...


5

Yes, neutrinos should obey Fermi-Dirac statistics and yes, the Pauli Exclusion Principle should operate for neutrinos. But let's examine how dense the neutrino population has to be for this to be important. The Fermi momentum is given by $$ p_F = \left( \frac{3}{8\pi}\right) h n_{\nu}^{1/3} $$ where $n_{\nu}$ is the neutrino number density. In order to be ...


5

wsc's answer (i.e., Onsager's computation of the free energy) provides one alternative road to a proof of a phase transition in the Ising model. It implies the existence of a phase transition in dimension 2 (for the nearest-neighbor model). Combined with correlation inequalities, this implies existence of a phase transition in any dimension d≥2, and ...


5

It's true. Special equipment and a long time is required to mix helium and nitrogen. According to one study, a mixture of 2.7% He, 93.3% N at 800 p.s.i.g. required a special cradle to repeatedly upend the cylinder, and 20.5 hours to reach equilibrated gas, which then remained mixed: http://pubs.acs.org/doi/abs/10.1021/je60005a002. The helium repeatedly ...


5

First let me repeat what Yvan Velenik says in the comment: The terminology is somewhat unfortunate, because you don't need that much statistics, rather you'll need some probability theory. To elaborate, quoting Wikipedia, Statistics is the study of the collection, analysis, interpretation, presentation, and organization of data. [...] Statistics deals ...


5

$$\frac{\partial \rho }{\partial t}= -\sum_{i}\left(\frac{\partial \rho}{\partial q_i}\,\dot{q_i}+\frac{\partial\rho}{\partial p_i}\,\dot p_i\right)$$ This means that if we have a function of $t, p, q$ namely $\rho(t,\vec p,\vec q)$ and we have a trajectory that is a curve in $(p,q)$ space, namely $q_i(t), p_i(t), i=1\ldots N,$ then: $$ \frac{\mathrm ...


5

Non-equilibrium systems are most often considered in the approximation where local equilibrium is valid, yielding a hydrodynamic or elasticity description. Local equilibrium means that equilibrium is assumed to hold on a scale large compared to the microscopic scale but small compared with the scale where observations are made. In this case, one considers a ...


4

The von Neumann entropy, written in terms of the quantum mechanical density operator, is a constant of the motion if you keep track of everything (including entanglement with the environment) and don't have any collapse events (which, depending on your favorite interpretation of quantum mechanics, might not exist anyway). The thing is that this fact already ...


4

There are quite a few conceptual confusions in this question. A system is either closed or open. A system is not "equilibrium" or "non-equilibrium". Also, a system is either conservative or dissipative. The ergodic theorem does not apply to open systems, neither to dissipative systems, since they tend to tend to a fixed point or something like that. A ...


4

I think this is just due to the distinction between how scalar field are used in cosmology (where they are assumed to depend only on time) and the more general solutions (superpositions of plane waves, with spatial dependence) that would correspond to a massless Bose gas. The stress-energy tensor for a massless scalar field is $$ T_{ab} = \nabla_a \phi ...


4

To find the global minima of a function in a configuration space using Monte Carlo methods there are two main approaches simulated annealing and parallel tempering. Simulated annealing Simulated annealing is single Markov chain starting at high temperature for global exploration. The system is then evolved via Monte Carlo update whose criteria for ...


4

You have to distinguish between "different states" and "number of states" - or, in the words of @Numrok, between "macrostates" and "microstates". The fundamental theorem refers to "accessible micro states". If I have three white balls and two buckets to put them in, I could put two balls in one and one in the other (that is a macro state); there are in fact ...


4

To determine the upper limit on chemical potential for a gas of $\mathcal N$ bosons, look at the form of the Bose distribution in the grand canonical ensemble with $\langle N \rangle = \mathcal N$. When using the GCE, it's easiest to work at chemical potential $\mu$ and to then choose $\mu(\mathcal N)$ so that $\langle N\rangle(\mu)=\mathcal N$. Each state ...


4

The basic idea is that the term with the highest weight in the exponential series is exactly the desired term. Further, all of the weight is in terms with closeby particle number, and small fluctuations in this number do not matter in many cases. (Note: I will omit the $k=0$ subscript and just write $b^\dagger$ in the following for convenience.) First, we ...


4

In statistical mechanics, at least when you can ignore the spin of the particles you're dealing with, the occupation number of a quantum state (that is, the number of particles in the state, or probability of a particle being found in the state) is proportional to $e^{-E/kT}$, where $E$ is the energy of the state. If you have a harmonic oscillator at ...



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