# Tag Info

61

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 ...

17

$\hbar$ does not need to appear in classical statistical mechanics. You are free to replace it with any quantity with units of angular momentum, say $\hbar_{\mathrm{C}}$. As long as this is choosen smaller than the size you can experimentally probe (i.e., as long as you don't ask questions of the theory that contain structure on this length scale or below) ...

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 ...

6

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: $$e(\mathbf ... 5 Giving the value simply of k_B T is generally more useful, because I can plug that into anything. Sure, I might need to know the ideal gas energy, and multiply by 3/2. But maybe I need to put it into a partition function, and I just need k_B T. Maybe I'm worried about a harmonic oscillator and I just have the two degrees of freedom. The 3/2 is ... 4 There are of course many books out there, quantum statistics is a really well-established field, so regardless of suggestions here you should really look further on your own as well and find one that suits you best. But here are a few that I've used in the past that you may find useful: Statistical mechanics: A survival guide by Mike Glazer and Justin ... 4 Characteristically, a critical point occurs somewhere anytime you have a continuous phase transistion. That is, if you have two phases of a substance that themselves share their intrinsic symmetries. The classic example is the critical point associated with the liquid gas transition, as you note. Liquids are isotropic and homogenous, gases are isotropic ... 3 The thermal energy k_{B} T is really referring to the probability of finding a system in a state of energy E, given that it is in a surrounding enviroment at temperature T. This probability is proportional to e^{-E/(k_{B} T)}. Using this you can derive a great many things, including the Boltzmann/Fermi distributions. The proportionality constant is ... 3 For the Laplacian$$ \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.$$Proof: Use Fourier transformation. 3 The instantaneous temperature of a system of N particles of masses m_i and velocities v_i is$$ T(t) = \sum_{i=1}^N \frac{m_iv_i^2(t)}{k_BN_f} $$where N_f equals the number of degrees of freedom, typically N_f=3N-3 for fixed total momentum. Note that the instantaneous temperature will fluctuate 5-10% about the true (thermodynamic) temperature ... 3 For gravitational systems one has to be careful making statements about entropy and the second law of thermodynamics. Your example is similar to the gravitational collapse of a gas cloud if you think carefully about it. In that case and in yours, the shrinking of the gas will raise it's heat. Now even though the increase of entropy due to the increased ... 3 You can think of the whole thing as a "fluid of systems" and each one of them can be in any of the states i available. \pi_{ij} tells you what is the speed at which a system in state i will go to state j p_i tells you how likely it is for a system (in this fluid or ensemble of systems) to be in state i and is proportional to the number of ... 3 TL;DR: The Gross-Pitaevskii equation is only applicable for very weakly-interacting bosons. At a=\infty the gas displays universal physics. Strictly speaking, the Gross-Pitaevskii equation (GPE) is only valid for$$na^3 \ll 1,$$where n is the density of particles and a is the s-wave scattering length. As it is a mean-field theory, one has to look ... 2 You should start by the definition of temperature$$\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 $$Let's now assume a single magnetic dipole in a magnetic field inside a heat tank( we can assume the heat tank as the rest of the ... 2 The answer lies in probability theory. Roughly, the probability of an event or macro state A to happen is the number of instances \Omega(A) in which it is fulfilled divided by the total number of possible instances or micro states \Omega i.e. p(A) = \frac{\Omega(A)}{\Omega}. So the reason why you want to maximize \Omega(A) is because you seek ... 2 response function = susceptibility = (pure or mixed) second derivative of a (Helmholtz, Gibbs, etc.) free energy. Magnetization is not a response function as the free energy is not observable, so one cannot observe the response to a change of some variable. 2 I think your concern is why to use the fundamental "symmetrization postulate" and not only the hamiltonian symmetry. The thing is that doesn't matter, suppose i'm trying to describe two particles (fermions for example) with Hilbert space \mathfrak{H}_1 and \mathfrak{H}_2, the Hilbert space of the two particles is$$ \mathfrak{H} = ...

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.

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 ...

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 ...

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 ...

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 ...

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: Q_q(\beta, N, V) ...

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

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 ...

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 $\epsilon -\mu<0$

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

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 ...

1

Without analyzing your (or Landau's) argument in depth, I'd like to point you to other alternatives to Laplacian Insufficient Reason. The idea of getting rid of Insufficient Reason is not new and was one of the drivers that motivated the great probability theorist and physicist Edwin T Jaynes, who was unsatisfied with Insufficient Reason unless clear ...

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 $D=2$ ...

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