# Tag Info

## New answers tagged statistical-mechanics

0

In diatomic gases (such as oxygen or nitrogen) where each molecule contains two atoms, energy is stored in the vibration and rotation of these atoms (in between and about each other), but temperature is the average translational kinetic energy of the molecules. This would obviously be the same for molecules with more than two atoms, and as explained above we ...

0

It depends on the setup which constraints are put on the microstate. For example the microcanonical ensemble assumes that the system is entirely isolated from the environmnent so necessarily energy has to be conserved. The microcanonical ensemble doesn't forbid the state where all particles are at the top as long as energy is conserved. So some of the system'...

0

Remember that macro state parameters are derivable from the distribution of micro states. Concretely, macro state parameters like pressure, density etc are averages over micro states of like velocity, number density etc. It is not true that for a gas in a gravitational field, the uniform density micro state does not contribute because it does not reproduce ...

0

Once you define entropy, such as Cort Ammon descibes, add completely symmetric laws of physics (which is the case, as you say), you still need some kind of initial or prior condition to deduce a single emergent arrow of time. The prior condition is the big bang, and it was "large" enough and lower entropy than today. The laws of physics allow for ...

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The question you ask is one of the major open philosophical questions in science today. Why does time appear to move "forward?" Many great minds like Feynman have explored the question (and its dual "what if time doesn't just move forward?") As such, no answer can be completely satisfactory, but some statements can be made and they will ...

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It is actually not, the total entropy can in fact decrease, but it's just highly unlikely. I will make an analogy to the statistical interpretation of entropy as counting the amount of microstates $\Omega$ according to some configuration in phase space. $S = k \ln \Omega$ Consider a uniform gas of indistinguishable particles in a box. Each particle has a ...

-1

If I am correct, one has $$E=\frac{1}{\theta K_2(1/\theta)}\int_1^\infty \gamma^3 \left (\sqrt{1-1/\gamma^2} \right )e^{-\gamma/\theta}d\gamma=\frac{K_1\left(\frac{1}{\theta}\right)}{K_2\left(\frac{1}{\theta}\right)}+3,$$ where $K_n$ is a modified Bessel function of the second kind.

-1

The energy levels used in Boltzmann distribution are normally ranges of energy, when we approximate discrete functions by continuous ones. They are not discrete energy levels in the quantum meaning. For example: each gas molecule has an energy $E_k = \frac{1}{2}mv_k^2 = \frac{p_k^2}{2m}$. The sum for all particles: $\sum_0^N{e^{-\beta E_k}} = \sum_0^N{e^{-\... 1 Yes, Maxwell and Boltzmann produced their theories well before quantum ideas were dreamed of, and they were developed by Gibbs into the form we know today using purely classical ideas. This involves some quite deep assumptions about what macrostates are equally probably in an ensemble. Introducing quantum microstates makes it a lot easier to understand. So ... 1 The standard intuitive path to the classical limit, or, conversely, to quantization, often goes through the phase space formulation of quantum mechanics. The analog of the Liouville probability density—-let's take n=1 for one degree of freedom and easily generalize later—-is the Wigner quasi probability density, normalized $$\int\!\!dqdp~ f(q,p)=1,$$ but ... 0 It is something more fundamental, and the answer is the eigenstate thermalization hypothesis (ETH). Closed quantum systems that fulfill this ansatz will thermalize (so the ETH is a sufficient condition, but it is not proved if it is necessary. So far all systems that thermalize, fulfill the ETH), and this thermalization can be described with the tools of ... 0 According to this text (https://web.math.princeton.edu/~nelson/papers/talk.pdf) gives different correlation values. See on the last pages. Edit: It's about a two HO system. The theory gives different correlation values between the states of the HOs than QM's predictions. 4 A qualitative answer to your question is that a system at higher temperature has higher W, all else being equal. So if you take a certain quantity of heat$Q$out of a system at high temperature$T_h$, without doing anything else, the change in entropy is$-\frac{Q}{T_h} = k \frac{\Delta W_h}{W_h}$(where$\Delta W_h < 0$), and if you put that same heat ... 1 Starting from the entropy expression$dS=\frac{1}{T}dU+\frac{p}{T}dV-\frac{\mu}{T}dN$we can introduce a new variable$L=U=\mu N$, with$dL=dU-Nd\mu -\mu dN$. Insertion gives:$dS=\frac{1}{T}dL+\frac{N}{T}d\mu +\frac{p}{T}dV$This means we can define an ensemble with the variables$L=U-\mu N$,$\mu$and$V$. Most importantly, this is an example of an ... 0 Let's first investigate again the case of the double-well potential$V_{\text{dw}}(x)$, and why a sum over even/odd number of instantons appear. We will normalize the potential such that$V_\text{dw}(\pm a) = 0$. Denote by$s \mapsto x(s;t)$the solution $$m\frac{d^2x(s;t)}{ds^2} = V_{\text{dw}}(x(s;t)) \ ; \\ x(0;t) = -a \ , \ \ x(t;t) = a \ .$$ Note ... -1 Since you are a high school student, I will break it down in simpler terms Heat is a result of combustion, chemical reactions at an atomic or molecular level or radiation like fission/fusion at an atomic and sub-atomic level. At a quantum level particles are not identified with any known elements on periodic table so as to associate temperature. In a known ... 0 In summer, the sun is more directly overhead and more intensely heats the ground surface up. The heated air right next to the surface is more buoyant than the slightly cooler air above it, which is a dynamically unstable condition. The hotter air protrudes up into the cooler air and rises upwards, drawing in more hotter air from around its point of origin ... 2 As it is well known, the equilibrium state of a physical system can be obtained from the extremisation of thermodynamic potential. In the context of phase transitions, the corresponding thermodynamic potential is the (Gibbs) free energy$G(m)$, where I assume for simplicity that the order parameter is constant in space, and is thus just a number$m$. The ... 2 So, critical phenomena can't be addressed via naive mean field. BUT, it is the first simple thing you can do to analyze stuff. Plus, if all you want just for the start is a qualitative description of the phase diagram, MFT is the way to go. Landau's procedure can give a very good qualitative sense of what goes on near criticality. Actually, if you are well ... 1 The statement in question is made in a specific context - that of electrons in condensed matter, which interact via Coulomb interaction, which is quartic in fermion operators. Physically, a quadratic Hamiltonian does not necessarily correspond to non-interacting particles: e.g., bosonization approach reduces Hamiltonians to interactions to quadratic ones. ... 0 The answer is given in the third remark at the end of Section 3 of my paper "Demonstration and resolution of the Gibbs paradox of the first kind" Eur. J. Phys. 35 (2014) 015023 (freely available at arXiv). In short, let's assume you combine two subsystems S1 and S2, each with N indistinguishable particles, by removing a partition between them. As ... 3 The question starts from an incorrect statement. Einstein's theory, but also Debye's theory do not work well at high temperature whatever high the temperature is. The correct statement is that Einstein's law fails to account for the low temperature behavior of specific heat of solids, even qualitatively, but at high temperature it goes to the Dulong and ... 1 From statistical physics, if a system is in contact with a "reservoir" of heat and particles, then the energy of the system$E$and the particle number$N$are allowed to fluctuate. If the set of microstates of the system form a discrete set (which would be the case for a system of particles which can inhabit discrete energy levels), then the ... 2 Q1 As you have shown yourself, this equation does not have a steady-state distribution: if we set$\partial_t P = 0$, i.e., if we assume that the solution is time-independent, we still obtain a solution that depends on time, contradicting our assumption. Q2 and Q3 In some situations one could indeed approximate the solution using form (II). The conditions ... 1 I suppose the answer is no. Consider a transition between the solid-state phase and the gas or liquid phase. Gas and liquid phases are translationally invariant. In a solid crystal, translational invariance is broken due to the presence of lattice. According to the acepted point of view, a continuous transition between phases with different symmetries is ... 0 I will expose a couple of ideas that maybe can help you: -I understand non-equilibrium steady states like those steady states that cannot be predicted by Statistical Mechanics, where your steady state cannot be described by the microcanonical, canonical, etc... ensembles. An example of this is the many-body localization, where local observables of closed ... 0 Consider 𝑂2 or 𝑁2. Then there should be 𝑓=5. And the derivation is by a factor different. Can we still use the law as an approximation? Yes. The so called kinetic temperature of an ideal gas, as normally measured, is based on 3 degrees of freedom and does not (and need not) account for molecular rotation (and vibration). These additional forms of kinetic ... 0 Your comment to @pwf stated that your question actually is My question is, if$T=T_s$, then the expression for a quasi-static process is always the same .......as that for a reversible process, and that is absurd because, every reversible process is quasi-static but not every quasi-static process is reversible. An example of a quasi-static process that is ... 1 As you noted in your question,$\langle \hat{O} \rangle$increases linearly with time... which means that its rate is constant! E.g., if$\langle \hat{O} \rangle$is the electric charge, it gives us a situation with a constant current. I think conceptually the difficulty is that a steady state is more of a theoretical/modeling concept rather than a kind of a ... 2 For me a phase transition is defined (somewhat strictly) as a non-analyticity in the thermodynamic potential. A function$f(x)$is non-analytic at$x=x_0$if it can not be expanded in powers of$x-x_0$. For example$f(x) = \sqrt{x+1}$can be expanded around$x=0$$$f(x) \cong 1 + \frac{x}{2} - \frac{x^2}{8} + \dots$$ but this does not work at$x=-1$because ... 4 A very large class of phase transitions are characterized by the breaking of some symmetry. Usually one finds a quantity called the order parameter, and finds its scaling with respect to an energy scale for the system, like the temperature. Usually for a phase transition one finds that the order parameter is either discontinuous or one of its derivatives is ... 2 Both options ($C_p$or$C_v$) are approximations. In the general situation, a non-rigid solid material will expand non-uniformly and therefore will have a time-dependent stress distribution, which stores internal strain energy. The simplest assumption is that the object is not mechanically constrained and that the non-uniform internal strain energy is small. ... 1 If we have two interacting gases of different temperatures, then it may be possible that a packet of particles(*) which move at high speed... In a "packet of particles" the individual particles will not all move at a "high speed". If the packet of particles is large enough, then the speeds of the individual particles will vary about the ... 1 This Hamiltonian has what is known as a "spin-flip" symmetry. It means that due to the term$\sum S_{z_i}S_{z_j}$, we can simultaneously change the sign of all the$S_{z_i}$operators and we still have the same Hamiltonian (the operator that commutes with the Hamiltonian is$G=\prod_i S_{x_i}$, which produces a global spin-flip over states in the$...

2

What I would like to know is whether ${\rm d}S=\frac{\delta Q_{rev}}{T}$ is just essentially a 'backward engineered formula' In some sense it is. Dividing by temperature is what turns $\delta Q_{rev}$ into the exact differential ${\rm d}S.$ It is what Clausius did (in 1858 I think) when he found that there was such a state quantity, which he called entropy....

1

There is a special relationship between entropy and heat because when heat passes from $A$ to $B$ then entropy comes along for the ride, and this is unavoidable. The entropy of $B$ will go up. The entropy of $A$ may go down or up or stay fixed, but if the process is reversible then it will go down. The only way for $B$ to avoid this entropy increase upon ...

1

At the moment, I don't think there is a special connection since for a Joule expansion there is no heat transfer but there is an entropy increase In the "Joule expansion" the gas cools as it uses its thermal energy to accelerate itself. That is a reversible process. Then the mechanical energy of the gas heats the gas, which is an irreversible ...

1

Caveat: I have not done statistical mechanics. All my knowledge of this subject is based on classical thermodynamics. However, I tried to keep my answer factual by only referencing already well-accepted ideas on the topic while providing references. What I would like to know is whether $dS=\frac{dQ_{rev}}{dT}$ is just essentially a 'backward engineered ...

2

Let the work done on the system be $\delta W$ while its internal energy change be $dU$, assume that the system may also exchange energy with a reservoir that is at temperature $T_r$. Then for an arbitrary process the entropy change $dS$ of the system satisfies $dS \ge \frac{dU-\delta W}{T_r}$. The equality sign holds for a reversible process. When the ...

1

The summation is over all possible mutually exclusive states. The exponential factor is probability of that state.

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So if there are various macrostates that can occur (regardless of how unlikely) what does that mean for our external parameters? How can there be different macrostates with different U, V, N? It means that the quantities which define the macrostate of the system (in this case, $U,V,$ and $N$) are not being held fixed. If your system is in thermal contact ...

2

"But now i read that there is not only a single macrostate of a system, but that there can be various macrostates." This is not in conflict with your previous understanding that a fixed U,V,N defines a macrostate. I think you just misinterpreted the statement. If a particular set of values for U,V,N denotes a particular macrostate, that means that ...

1

First, let's rewrite this in polar coordinates $$\rho = \frac{2}{h^3} \int dp \; 4\pi p^2 e^{-\beta (\frac{p^2}{2m} - \mu)}.$$ Thus we see that the particle densitiy in the intervall $[p, p+dp]$ is $$d\rho = \frac{8\pi}{h^3} p^2 e^{-\beta(p^2/2m - \mu)} dp.$$ Next we calculate $\rho$. This you have ...

0

Multiplicity tells how many microstates have a macrostate. E.g. how many possible multiparticle configurations have a glass of 293K water on surface pressure. Entropy is the logarithm of multiplicity. (multiplied by k)

1

The entropy change of a system is the sum of two parts: Entropy transferred from the surroundings to the system (across the interface with the surroundings) as a result of heat flow, and given by $\int{\frac{dq}{T_B}}$, where dq is the differential heat flow across the boundary interface between the system and surroundings and $T_B$ is the temperature at ...

0

Suppose the entropy associated with the system and the surrounding at the start of thermodynamic process is $S_o$ and the entropy associated with it at the end is$S$. $$∆S=S-S_o$$ The entropy change through any reversible path connecting intial and final state can be given as- $$∆S_{rev}=\int\frac{dq}{T}$$ Here , $T$ is thermodynamic temperature.

19

Negative temperature is mainly to do with (c): a finite number of configurations. It is not a violation of entropy postulates or equilibrium, but I will qualify these statements a little in the following. The heart of this is not to get 'thrown' by the idea of negative temperature. Just follow the ideas and see where they lead. There are two crucial ideas: ...

9

You're pretty much right; in the case of spins, it's the fact that there's an upper bound on the system's energy that causes negative temperature, which is strongly related to the fact that there's a finite number of states. With something like a gas, increasing energy always provides access to an increasingly large set of phase space because the area of a ...

2

Bose-Einstein condensation is not caused by interaction: ideal Bose gas undergo Bose-Einstein condensation. In contrast, superfluidity is due to interaction and one should see its signature in Green functions as a pole associated to phonons.

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Yes, it is a probabilistic statement. But in practical scenarios the number of microstates in the most probable macrostate is so enormously greater than the number in any other macrostate that the system spends almost all of its time in the most probable macrostate, and you would to wait many times longer than the age of the universe before you observed any ...

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