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14

I will try to answer these questions from different views. Macroscopic view The "quantitative" rather than qualitative difference in a liquid-gas phase transition is due to the fact that the molecules arrangement does not change so much (there is no qualitative difference) but the value of the compressibility changes a lot (quantitative difference). This ...


10

You have to be careful to distinguish between microstates and macrostates. Thermodynamic equilibrium is a macrostate which consists of a mixture of all possible microstates of energy $E$ weighted by a Boltzmann weight $e^{- \beta E} / Z$. A state in macroscopic thermal equilibrium can be thought of as "moving through phase space" ergodically (i.e. the ...


9

For a pure substance that can exist in the solid, liquid, and vapor states (i.e., wood is not in this category), let's assume that a closed container is half full of liquid and half full of vapor. As the temperature rises, the liquid expands and the liquid density falls. Also, as the temperature rises, the pressure in the container rises due to the vapor ...


9

Preliminaries: How do we define 'localized?' For a single particle, or for multiple non-entangled particles, it is easy to tell from the expressions for the wavefunctions whether they are localized or delocalized. For example, you might say that if the wavefunction is falling off exponentially or faster for large $x$, that is with a form like $\psi(x)\sim e^...


8

There are two definitions of entropy, which physicists believe to be the same (modulo the dimensional Boltzman scaling constant) and a postulate of their sameness has so far yielded agreement between what is theoretically foretold and what is experimentally observed. There are theoretical grounds, namely most of the subject of statistical mechanics, for our ...


5

$$S=k \ln [\Omega(E)] = k \ln [\omega (E) \delta E] = k \ln [\omega(E)] +k \ln (\delta E)$$ Last term is an arbitrary constant, so that we can set $$S = k \ln[\omega (E)]$$ from which $$\delta S = k \frac{\delta \omega}{\omega}$$ If we can ignore the power contribution and set $\omega (E) \simeq e^{\beta E}$, we get $$\delta S = k \frac{\delta(e^{\beta ...


3

Attempting to answer the "why" question intuitively: In a liquid, the molecules experience significant intermolecular force - so much so, that the average energy of the molecules is insufficient to escape the attractive force of the surrounding materials. The result is that it energetically favorable for them to remain close together, even if that means ...


3

Good question. I don't have my Widom around, but I'll try to answer from memory. I think the consensus is to say a substance is at its gas state if it could be a liquid at the same temperature. This, as opposed to same pressure, same volume, etc. If the temperature is supercritical, there is no transition between liquid and gas, and the generic term "fluid"...


3

It really depends on the boundary conditions. For boundary conditions like a 3D box with reflecting walls, the initial quantum state $\Psi$ will stay a quantum state with the unique wave function depending on variables of each particle: $$\Psi({\bf{r}}_1,...,{\bf{r}}_n, t).$$ If the boundary conditions are such that allow exchange with the environment, then ...


3

To solve your problem exactly, you would have to solve the Schrödinger equation $$i \frac{\partial}{\partial t} \Psi (\vec r_1 \dots \vec r_N,t)= H \ \Psi(\vec r_1 \dots \vec r_N,t)$$ where $\Psi (\vec r_1 \dots \vec r_N,t)$ is the wave function of the $N$ particles and $$H=\sum_i^N \frac{p_i^2}{2 m} + \sum_{i<j}^N u_{ij}+V_{\text{ext}}$$ where $u_{ij}...


2

In the context of solid-state physics, a closely related question has been an area of active research in the past few years. Most interacting systems do indeed thermalize (and thus delocalize) over long time scales. However, certain systems whose disorder is much stronger than their interactions experience "Many-Body Localization," in which the individual ...


2

Quantum effects appear if the concentration of particles satisfies, $$\frac{N}{V} \ge n_q$$ where $N$ is the number of particles, $V$ is the volume, and $n_q$ is the quantum concentration, for which the interparticle distance is equal to the thermal de Broglie wavelength, so that the wavefunctions of the particles are barely overlapping. As the quantum ...


2

The dimension issue is solved easily by defining the probability density function(PDF) as $$P(\{q,p\})=\frac{E_0}{h^{3N}} \ \delta (H(\{q,p\})-E)$$ where $E_0$ is an arbitrary constant which will not affect any thermodynamic quantity or equilibrium property. Actually, this definition is incomplete. We have to take into account the indistinguishability of ...


2

Particle on a rotating ring For further discussion purpose, let's considere the dynamics of a quantum particle on a $r_0$ radius rotating circle at a constant angular velocity $\mathbf{\Omega}=\Omega\,\hat{e}_z$ . In cylindrical coordinates, we fix $z=0$, and we have the azimuthal angle $\theta$, which is the "good" degree of freedom describing the ...


2

I agree that the language is very confusing - I'm a native English speaker, and it also took me a while to understand what they were saying. When they talk about the dimension of "the statistical system itself," they mean the spacetime dimension. So if a system has two spatial dimensions, then it has three dimensions total (including time), and the ...


2

As @valerio92 points out, your mistake is that $S = k \ln (\omega\, \delta E)$, not $\delta S$. To get $\delta S$, you differentiate the right-hand expression to get $\delta S = k \frac{\delta \omega}{\omega}$, and the $\delta E$ drops out and you get an expression with the right dimensions. The notation is a bit misleading, because the $\delta$ in the $\...


2

Why does a critical point even exist? I think this question is equal to this one: "Why the width of the two phase region is bigger at lower temperatures and pressures?" Specific volume of liquids mostly depends on the temperature of them in comparing with their pressure. This means, for a well-defined increment of the pressure, we can neglect its effect ...


1

Corresponding to a macroscopic equilibrium state of a system, there exists a set of microscopic states, such that at any time the system can be in any of those microscopic states with equal probability. In other words, the system passes through every micro-state in that set, without preference for one micro-state rather than the other. This $\textit{set}$ is ...


1

The molecule has only 3*N (number of atoms) positional degrees of freedom, since every atom can move in 3 dimensions. Three of of the 3N are translations of the whole molecule, 3 are rotational and 3N-6 are vibrational (symmetry will even break these into invariant subspaces) Thus the partition function includes all relevant microstates in this definition.


1

I read in the documentation that came with this simulator that when you change the number of particles "N", the total energy of the system "E" or the number of dimensions, the field will turn yellow (which indicates that the simulator has not incorporated the changes) until you hit enter. I made changes to these values, hit enter with the cursor still in ...


1

I'll try to explain why there could be a critical line and not just a critical point, and hopefully that will answer your question. If you think about the Ising model, we have the standard Hamiltonian: \begin{equation} -\beta H = J_1\sum_{<i,j>}s_i s_j + h\sum_{i}s_i \end{equation} where $\sum_{<i,j>}$ is a sum over nearest neighbors. This model ...



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