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0

As you correctly point out, the solution follows by integrating the differential equation. The general idea is to expand the left hand side of your differential equation by realizing that $$\text{d}\ln n^c(T) = \frac{1}{n^c(T)} \text{d}n^c(T) = \frac{1}{n^c(T)} \frac{\partial n^c(T)}{\partial T}\text{d}T = \frac{T}{n^c(T)} \frac{\partial n^c(T)}{\partial ...


4

I can see different subtleties in Landau's argument. First of all, it isn't entirely clear what is meant by "there are only seven additive constants of motion". To give an example, consider a single particle hamiltonian: $$H=\frac{\mathbf p^2}{2m}+m\omega ^2\frac{\mathbf q ^2}{2}.$$ For this hamiltonian there are several conserved quantities: $$e(\mathbf ...


0

If entropy can be explained probabilistically, does that mean we can describe a system's entropy as a random variable? No, it does not. Thermodynamic entropy can be given additional explanation in probabilistic terms applied to mechanics, but it is still thermodynamic entropy - a function of macroscopic variables like internal energy, volume, number of ...


1

Here's a way to get a decent distance estimate for solids, liquids or gases.: for solids or liquids, you can get the number density, $n$, of atoms or molecules (as needed), from the density, Avogadro's number ($6.02E23$) and the molecular weight ($\rho,N_a,W$: $$n = \frac{\rho N_a}{W}$$ for a gas with pressure, temperature and the boltzmann constant ...


0

I took a quantum statistical mechanics course as an undergraduate using Kittel's book "Thermal Physics"s this made sense to me because he used simple models. In graduate school we used Landau and Liftshitz book which was confusing. My understanding of the ideas was that Bolzman made an identification of classical thermodynamic quantities Temperature, ...


0

Brionius has the right answer, but there is more to be said. Water at room temperature in air will slowly evaporate. Water at room temperature in a vacuum will boil, as is shown here. So these mini torpedos can prevent damage to chemical bonds. Water molecules are polar. The O's are a little negatively charged. The H's are a little positive. The H's and ...


0

I think the $\Omega_N$ in your question is the number of states inside an energy sphere $E$; Now we need to consider micro-canonical ensemble, which means we need to get the number of states in an energy shell between $E$ and $E+\Delta$: $\Omega'=\frac{1}{N!h^{3N}} V^N \frac{\pi^{3N/2}}{(3N/2)!}\{[2m(E+\Delta)]^{3N/2}-(2mE)^{3N/2}\}$ Which is equal to: ...


-1

In fact, they do!! Watch what happens to an ice cube that is left in the air... trillions of particles of its exterior are torn out of their stable arrangement, and soon they cascade down the sides—a microscopic waterfall! So in this case you are right, but it is just the very exterior surface of an object that is exposed to the air and thus affected ...


3

The thermal energy $k_{B} T$ is really referring to the probability of finding a system in a state of energy $E$, given that it is in a surrounding enviroment at temperature $T$. This probability is proportional to $e^{-E/(k_{B} T)}$. Using this you can derive a great many things, including the Boltzmann/Fermi distributions. The proportionality constant is ...


5

Giving the value simply of $k_B T$ is generally more useful, because I can plug that into anything. Sure, I might need to know the ideal gas energy, and multiply by $3/2$. But maybe I need to put it into a partition function, and I just need $k_B T$. Maybe I'm worried about a harmonic oscillator and I just have the two degrees of freedom. The 3/2 is ...


-1

Another way of looking at this is that things that would be destroyed by the environment (be it heat, light, etc) have already been destroyed (like ice on a hot summer day). The things that you see around you are the ones where the bond energy was high enough that they survived.


1

You are considering a system of fermions (SC hamiltonian) and a system of bosons (weakly interacting BEC). In order for the density to be finite, fermions must have a positive chemical potential $\mu>0$. On the other hand, the chemical potential of a system of bosons is less or equal to the energy of the lowest-energy state. In a non-interacting system ...


1

I found it: $$\bar{n}_{BE}=-\frac{1}{\mathcal{Z}}\frac{\partial \mathcal{Z}}{\partial x}$$ where $x =(\epsilon -\mu)/kT$ $$\mathcal{Z} = \sum\limits_{n}^\infty e^{-n(\epsilon -\mu)/kT}$$ which converges if $\epsilon -\mu<0$


2

Things actually do get destroyed by what those air molecules pick up and throw around. Take look at this example [image from here: http://en.wikipedia.org/wiki/File:Arbol_de_Piedra.jpg ] Just like their bigger sized brothers, it's the load of those mini-torpedos that brings the destruction.


60

When you say "why aren't things being destroyed", you presumably mean "why aren't the chemical bonds that hold objects together being broken". Now, we can determine the energy it takes to break a bond - that's called the "bond energy". Let's take, for example, a carbon-carbon bond, since it's a common one in our bodies. The bond energy of a carbon-carbon ...


1

You state the second law as : The entropy of the universe always increases. In my college textbook it is stated as : Processes in which the entropy of an isolated system would decrease do not occur, or, in every process taking place in an isolated system, the entropy of the system either increases or remains constant.( F.W.Sears an introduction ...


3

There are of course many books out there, quantum statistics is a really well-established field, so regardless of suggestions here you should really look further on your own as well and find one that suits you best. But here are a few that I've used in the past that you may find useful: Statistical mechanics: A survival guide by Mike Glazer and Justin ...


1

Define $H' \equiv H/kT$ and $F' \equiv F / kT$. Then you get $$Z = \sum e^{-H / kT} = \sum e^{-H'}$$ and $$F' = F / kT = -\ln Z \, .$$ The author is probably dropping the $'$ symbol to keep the notation compact. This is annoying but common.


1

I will only focus on the "measure" aspect of the problem you seem having trouble with. When it comes to the $N!$, let's say the argument given by By Symmetry is essentially correct. Now, one way to look at it is to start directly from the actual quantum partition function of a system of $N$ distinguishable particles: \begin{equation} Q_q(\beta, N, V) ...


1

I think you're almost there. The only major point you are missing is that your $N$ particles are identical. This means if you exchange the position and momentum of two particles you get back the same state and your final equation is, therefore, double counting certain states. Now since we are dealing with classical particles we can assume that there are ...


0

You have to express $dk$ in terms of $\omega$ and $d\omega$ as well. This is just a matter of using the method of u-substitution: Just reexpress the integral in terms of $\omega$.


0

There is a simple way of looking at this. Would a container of gas have a change in temperature if the container was given a different velocity? For your second question, the vibrating membrane acts like a spring pendulum which transfers energy into the surroundings. The membrane does not have a change in temperature until it absorbs the energy back from ...


2

This may be the argument: you have $N$ particles, and for each one you can put it on the left side or on the right side. Each of these choices, for each particle, leads to a different microstate. There are $2^N$ possible choices you can make for how to distribute the $N$ particles between the left and right halves of the box (assuming the particles are ...


7

The ratio between avaible microstates is given by the following ratio: $$ \frac{n_{B}}{n_{A}}=\frac{\int_{V/2}d^{3N}q d^{3N}p}{\int_{V}d^{3N}q d^{3N}p}=\frac{\int_{V/2}d^{3N}q}{\int_{V}d^{3N}q}=\frac{\int_{V/2}d^{3}q_1 \int_{V/2}d^3 q_{2}\dots }{\int_{V}d^{3}q_1 \int_{V}d^3 q_{2}\dots}=\left( V/2 \right)^N/V^N=\frac{1}{2^N}$$ where $N$ is the number of ...


0

In first place, the temperature is a quantity that measure thermal equilibrium by the zeroth law of thermodynamics. We have the contact with this quantity by with a thermal equilibrium can do. For example, the Celsius units is constructed by define $0°C$ as the volume of mercury in contact with freezing water and $100 °C$ as the volume of mercury in contact ...


0

Can anyone help with this derivation? There are $N!$ ways to arrange $N$ different objects, but you don't have $N$ different objects, you have $n_0$ indistinguishable objects in the ground state, $n_1$ indistinguishable objects in the first state, and so on. So that means, from your $N!$ different permutations (assuming all the objects are ...


2

The chemical potential is sort of the potential energy needed to add another particle from the surrounding reservoir to the system. Thus to add another particle to a particular single particle level requires energy if the chemical potential is larger than the energy of single-particle level. If the chemical potential was smaller than the energy level, then ...


1

So the first law for an system where we don't have mass flows in or out ( a closed system ), is $\Delta Q + \Delta W = \Delta U$ Where $Q$ is your net heat added, and $W$ is your net work added, and $U$ is your net internal energy change: internal energy being like the sum of all the different kinetic energies of the molecules. This is why when we add ...


0

The equipartition theorem says that if the system is in contact with thermal reservoir of temperature $T$ and has Hamiltonian description with Hamiltonian being a sum of terms, one of which is quadratic function of some canonical variable $q_c$ ($aq_c^2$, where $a$ is constant), then the expected average value of $aq_c^2$ is $k_B T/2$. This theorem follows ...


3

TL;DR: The Gross-Pitaevskii equation is only applicable for very weakly-interacting bosons. At $a=\infty$ the gas displays universal physics. Strictly speaking, the Gross-Pitaevskii equation (GPE) is only valid for $$na^3 \ll 1,$$ where $n$ is the density of particles and $a$ is the $s$-wave scattering length. As it is a mean-field theory, one has to look ...


1

The partition function $\mathcal Z$ is the sum over all those these exponential indexed by the state index. In German it's called "$\mathcal Z$ustandssumme", literally "state sum". You could device a scheme of indexing the possible overall energies so that you could write it as $\sum_n$, i.e. finding an injection ${\mathbb N}\to{\mathbb N}^2$, but if the ...


2

You should start by the definition of temperature $$\frac{1}{T} = \frac{\partial s}{\partial E }= k \frac{\partial \ln \Omega}{\partial E} $$ Also you have the partition function $$Z_1=\sum \exp(-bE_r) \text{ where } b=1/kT $$ Let's now assume a single magnetic dipole in a magnetic field inside a heat tank( we can assume the heat tank as the rest of the ...


0

Short answers: no, they are not the same; they are somewhat related. A more detailed discussion follows. Indeed, in most QFT books zero temperature is usually assumed. However, if one is interested in energy scales that are way beyond the temperature of the system, the zero-temperature approximation is a valid one. For example, the thermal energy at room ...


1

Quantum mechanically the general expression you want for the partition function is $$ Z = \mathrm{Tr} \left( \mathrm{e}^{-\beta H} \right),$$ where $\mathrm{Tr}$ means the trace (i.e. sum over micro-states). Now you can use the fact that the modes are independent, so that quantum Boltzmann operator $\mathrm{e}^{-\beta H}$ factorises into a product. This ...


1

My original answer was incorrect and Mark gave a good answer above. Here is a slightly different approach by considering the total energy of the system. Say we have $\alpha$ total harmonic oscillators. We can write the total energy of the whole system as $$ E = \sum_\alpha \frac{1}{2}\hbar\omega_\alpha + \sum_\alpha n_\alpha \hbar \omega_\alpha .$$ For ...


1

Consider the general diffusion equation of the form, $$ \frac{\partial \psi}{\partial t}=D\frac{\partial\Phi}{\partial x}=D\frac{\partial^2\psi}{\partial x^2}\tag{1} $$ where $\Phi=\partial_x\psi$ is the probability current (flux). In order to replicate a reflecting boundary, we establish an infinite potential at the boundaries, ...


3

You can think of the whole thing as a "fluid of systems" and each one of them can be in any of the states $i$ available. $\pi_{ij}$ tells you what is the speed at which a system in state $i$ will go to state $j$ $p_i$ tells you how likely it is for a system (in this fluid or ensemble of systems) to be in state $i$ and is proportional to the number of ...


0

Kinetics are focused on the rate and mechanism of chemical processes, so you are definitely right to say that you gain a lot of insight about mechanism from kinetics. Many kinetic theory make extensive use of statistical thermodynamics methods, and that's why you perceive a resemblance. However, keep in mind that in kinetics the system is not in ...


2

response function = susceptibility = (pure or mixed) second derivative of a (Helmholtz, Gibbs, etc.) free energy. Magnetization is not a response function as the free energy is not observable, so one cannot observe the response to a change of some variable.


2

The answer lies in probability theory. Roughly, the probability of an event or macro state $A$ to happen is the number of instances $\Omega(A)$ in which it is fulfilled divided by the total number of possible instances or micro states $\Omega$ i.e. $p(A) = \frac{\Omega(A)}{\Omega}$. So the reason why you want to maximize $\Omega(A)$ is because you seek ...


0

Widespread beliefs are not laws. Look to the genesis of thermodynamics: The short range interaction field, the electromagnetic one, that allows the atomic structure and the transfer of energy, heat, in the collisions of atoms and molecules was, for sure, present in the analysis of the thermodynamics since its inception. In this original context the ...


3

For gravitational systems one has to be careful making statements about entropy and the second law of thermodynamics. Your example is similar to the gravitational collapse of a gas cloud if you think carefully about it. In that case and in yours, the shrinking of the gas will raise it's heat. Now even though the increase of entropy due to the increased ...


0

I might be able to answer your question in the context of linear response theory: Response function: the power series expansion of the applied field generated by a weak external perturbation. Mathematically speaking, we can relate the average value of an observable $X$_i to the response function $\chi$ via \begin{align} \langle X_i(t)\rangle=\int_0^t dt'' ...


1

While lemon's answer is of course correct, it is not the only way to calculate the temperature from a molecular dynamics simulation: it can also be obtained from the configurations, that is, the particle coordinates, of the system. This is called "configurational temperature" (see, e.g., this article; pay-walled). The key identity is (for a canonical ...


3

The instantaneous temperature of a system of $N$ particles of masses $m_i$ and velocities $v_i$ is $$ T(t) = \sum_{i=1}^N \frac{m_iv_i^2(t)}{k_BN_f} $$ where $N_f$ equals the number of degrees of freedom, typically $N_f=3N-3$ for fixed total momentum. Note that the instantaneous temperature will fluctuate 5-10% about the true (thermodynamic) temperature ...


2

"The double summation (first over the numbers $n_\epsilon$ constrained by a fixed value of the total number N, and then over all possible values of N) is equivalent to a summation over all possible values of the numbers $n_\epsilon$ independent of one another" Why is this true? I have been stuck on this argument for hours... The ...


1

A short answer to questions 2 and 3: In Mermin-Wagner's paper the short-range condition is stated as $\sum_{\bf R} {\bf R}^2 |J_{\bf R}|<+\infty$. For interactions with (or more precisely majorized by a) power law decay $|J_{\bf R}| \sim R^{-\alpha}$, this requires $\alpha > D+2$, where $D$ is the space dimensionality (i.e., $\alpha >4$ for $D=2$ ...


2

I think your concern is why to use the fundamental "symmetrization postulate" and not only the hamiltonian symmetry. The thing is that doesn't matter, suppose i'm trying to describe two particles (fermions for example) with Hilbert space $\mathfrak{H}_1$ and $\mathfrak{H}_2$, the Hilbert space of the two particles is $$ \mathfrak{H} = ...


1

Basically, you need a higher dimensional version of the Leibniz rule to differentiate under the integral sign. Suppose you have an integral of this kind $$ \int_{\Omega(s)} f(s,\vec x) \, d\vec x $$ That's it, your region of integration and your function depends of a parameter $s$. (in your case $s = x_i$). Then if you want to calculate $$ ...


3

For the Laplacian $$ \Delta ~:=~ -\frac{d^2}{dx^2} ~\geq~ 0, $$ the corresponding HS transformation reads $$\exp\left(-\frac{a}{2}\Delta\right) f(x)~=~\int_{\mathbb{R}} \!\frac{dy}{\sqrt{2\pi a}}\exp\left(-\frac{y^2}{2a}\right) f(x+y), \qquad a~>~0.$$ Proof: Use Fourier transformation.



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