Hot answers tagged statistical-mechanics
8
I'll state one version of the theorem, valid for classical systems. I'll not give the most general framework, as things become messy, but this should still give you an idea of how general the result is.
We need the following ingredients:
Spins: to each vertex of the lattice $\mathbb{Z}^2$, we attach a spin $\phi_x$ taking values in some compact ...
5
Good question. I think that you're incorrect to assume the chemical potential is independent of the zero point.
I think it's easiest to see that the chemical potential must change by using $\mu = (\partial U/\partial N)_{S,V}$. Suppose I first define the zero point as 0 so that the energy is $U(N)$. Then I redefine the zero-point energy (for a ...
4
Length scales are not accounted for properly in your question. When you have a system at local equilibrium where a temperature gradient can be defined then each "point" in this description contains say $10^{10}$ molecules and can be seen as a thermostatistical system at equilibrium. We call that "local" equilibrium because intensive quantities such as ...
4
Mad props for a cool question. I'm going to justify essentially the converse of the statement because it doesn't make much sense to talk of the temperature of a system that is in a pure state.
Let's assume that we're talking about a quantum system with disrete energy spectrum (with no accumuation points) in thermal equilibrium. Let $\beta = 1/kT$ be the ...
3
The maximum entropy probability distribution for a system of fixed expected total energy is the Boltzmann distribution. This means that Boltzmann distribution is appropriate for systems where we know the total energy but little else. The distribution is given by $p(E) = Z^{-1} e^{-E/kT}$, where $Z$ is the normalization constant (making sure probabilities add ...
3
A "point" in a macroscopic system is not a geometrical point. It is a volume element that is small on a macroscopic scale and yet has a large number of molecules for entropy and internal energy to be defined. Your temperature probe does not measure its value at a geometrical point but for a small volume of the system in whose contact it is put.
The local ...
2
The role of the atomic structure of the black body has its importance, as it will determine how the light is absorbed. And a black body is defined as matter that absorbs everything. However, matter will always absorb a little bit, and reflect or disperse the rest. Having a little hole allows for light to bounce many times after entering, every time absorbing ...
2
Let us compare thermal equilibrium radiation you have described and the black body radiation.
Take an object B at thermal equilibrium (not necessarily a black body), and denote with $\varepsilon(\lambda)$ the fraction of radiation of wavelength $\lambda$ it absorbs.
Thus if $a(\lambda)$ of radiation of the wavelength $\lambda$ falls on B, it absorbs
...
2
This site might help in clearing up the concepts, which are not simple. In the link, click "heat and thermodynamics" and then " radiation" and then "black body" on the right column to get the options shown in the image. Unfortunately the site does not link back to individual pages. :(.
Clicking on the main line,
Blackbody radiation" or "cavity ...
2
The integral carries units of $[momentum]^{3N}[space]^{3N}$ which is exactley the same as $1/h^{3N}$, so this is the factor you need for making the phase space volume dimensionless. I don't understand why you say that it has to be $[momentum]^{3N-1}$, just look directly at the integral.
Furthermore, and maybe this is the main problem, the phase space VOLUME ...
2
I think it is a mistake, as often happens in popularizations of science.
A water or any molecule may lose kinetic energy and acquire potential energy, but it is the kinetic energy distribution that gives the temperature of an ensemble of molecules. The shape of the distribution shows that there will always be individual molecules at very high energy , in ...
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Let $|\mathbf{n}\rangle = |n_0, n_1, n_2, \dots\rangle$ denote a state with $n_i$ particles in the $i^\mathrm{th}$ energy state of the single-particle hamiltonian. Let $\epsilon_i$ denote the energy of the state with label $i$. The $n_i$ are "occupation numbers" in the standard terminology. Note that my notation is such that $i=0,1,2,\dots$ labels ...
1
As the other answers have said, temperature is a collective property and can only be defined when you have an assemblage of particles. However by definition in a molecule you have an assemblage of atoms, and they have relative motions described by the vibrational excitations of the molecule.
So if you have a large enough molecule you can look at the ...
1
Thermodynamics makes sense when you have large numbers of particles. For example, the second law of thermodynamics has an extremely low probability of being violated when you have Avogadro's number's worth of particles. However, if you have a very small number of particles, the second law will frequently be violated.
This comes up in nuclear physics, where ...
1
It makes sense if all you know about the molecule is its expected energy. Then you can show that it's energy distribution is the Boltzmann distribution $p(E) = e^{-E/kT}$ for some constant $T$, which is related to the expected energy.
So the question reduces to a philosophical view of probabilities. Does it make sense to assign probabilities to a ...
1
Integrability of the inexact differential $\delta Q$ is a law of nature. Although in general Pfaffian differential forms like $\delta Q$ are not integrable, second law of thermodynamics guarantees that an integrating factor always exists and it is $1/T$ in all cases, $T$ being the absolute temperature.
1
The best way to see this is to realize that the zero heat capacity is a quantum effect. Classically, the heat capacity does not go to zero.
Quantally what happens is that at low enough temperatures all the particles are in their lowest possible energy states. To get even one particle into a higher energy state requires a small but finite energy ...
1
1 - yes the zero temperature limit is not reachable, so you can't measure the heat capacity at zero temperature, what this calculation tell you is that if you measure at smaller and smaller temperatures you will see that C converges towards zero
2- No the reversibility of the path is not important as the entropy is an exact differential
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