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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 ...


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 ...


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

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 ...


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.


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

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 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: \begin{equation} 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

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 ...



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