Hot answers tagged statistical-mechanics
13
The realization of non-Abelian statistics in condensed matter systems was first proposed in the following two papers.
G. Moore and N. Read, Nucl. Phys. B 360, 362 (1991)
X.-G. Wen, Phys. Rev. Lett. 66, 802 (1991)
Zhenghan Wang and I wrote a review article to explain FQH state (include non-Abelian FQH state)
to mathematicians, which include the explanations ...
13
If you were to surround the atmosphere by an adiabatic envelope and allow it to come to equilibrium, it probably would settle into such a state. However, the atmosphere is not a static place. It is actively mixed due to heating of the ground by the sun, and by cooling of the upper atmosphere by radiation into space. This makes the surface air less dense than ...
9
Here is a video of the film's science advisor explaining what the equation is and how he came up with it: http://www.youtube.com/watch?v=WjfT6MqTCqQ
It is based on the Gompertz equation, which is a model of mortality rates, with some added "mathematical glitter."
9
Long wires are real macroscopic bodies, kilometers of superconducting wires are used at the LHC of CERN and the currents can be described by quantum mechanical equations.
Crystals also can be described by quantum mechanical equations, and can be quite large, maybe not as large as a table. Superfluids too are in the realm of macroscopic quantum mechanics.
...
9
Hannesh, you are correct that the second law of thermodynamics only describes what is most likely to happen in macroscopic systems, rather than what has to happen. It is true that a system may spontaneously decrease its entropy over some time period, with a small but non-zero probability. However, the probability of this happening over and over again tends ...
8
Yes. In general relativity and cosmology, the collection of galaxies is often even treated as an ideal fluid, in the thermodynamic sense, with temperature and pressure.
The standard textbook on general relativity, the book Gravitation by Misner, Thorne, and Wheeler, discusses this model in Section 27.2, though only on the thermodynamic level.
For the ...
8
My understanding is that this question is being asked in the context of the kinetic theory of classical gases. In that context, here is the argument:
If the system is rotationally invariant, then we should have $\langle v_x^2 \rangle = \langle v_y^2 \rangle = \langle v_z^2 \rangle$. Thus $\langle v^2 \rangle = \langle (v_x^2 + v_y^2 + v_z^2 )\rangle $ ...
8
From a fluid dynamics standpoint, as a body moves through a fluid, a small region of fluid is dragged along with it. This is what forms the boundary layer. In the near-body region, odor will be dragged along with the body.
Likewise, behind a moving person is a turbulent wake and a low pressure region. The low pressure reason will "suck" the odor along with ...
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 ...
7
I think you are both wrong.
"The lowest energy state still has non-zero energy" does not mean that the temperature cannot be zero. If the system is in the ground state with 100% probability, then the temperature is zero. It doesn't matter what the ground state energy is.
It's true that all molecules in the substance would stand perfectly still at absolute ...
7
This answer is somewhat hand-wavy, but I do believe it should help to grasp the concepts on an intuitive level.
First of all, entropy is not a measure of randomness. For an isolated system in equilibrium under the fundamental assumption of statistical mechanics, the entropy is just
$$
S=k\ln\Omega
$$
where $\Omega$ is the number of microstates - microscopic ...
7
Negative temparature makes sense in statistical mechanics only if the associated Hamiltonian is bounded from above; otherwise the trace in the definition of the partition function does not exist.
In particular, a gas cannot have negative temperature since its Hamiltonian contains a kinetic energy term, which is not bounded from above. (Edit: Since the ...
7
I prefer to see it in the following way:
\begin{equation}
Z_{quantum} \equiv \sum_{m} e^{-\beta E_m}
\end{equation}
Where $m$ is a quantum microstate eigenstate of the Hamiltonian. Now, you can split the sum into two parts; a sum over quantum microstates that yields the same energy eigenvalue $E_n$ and a sum over all possible values of $E_n$:
...
7
Right, there is a small probability that the entropy will decrease. But for the decrease by $-|\Delta S|$, the probability is of the order $\exp(-|\Delta S| / k)$, exponentially small, where $k$ is (in the SI units) the tiny Boltzmann constant. So whenever $|\Delta S|$ is macroscopically large, something like one joule per kelvin, the probability of the ...
6
The basic idea is that statistical properties of complex physical systems
fall into a small number of universal classes. A very known example of
this phenomenon is the universal law implied by the central limit theorem
where the sum of a large number of random variables belonging to a large
class of distrubutions converges to the normal distribution. Please ...
6
David Bar Moshe's answer is fine, but I wanted to go into more detail. The main reason that random matrices show up in dynamical systems is because they describe the level statistics of classically chaotic motions. In classically integrable systems, there is a semiclassical formula for the level-spacing, determined by the Bohr-Sommerfeld rule. If you know ...
6
By the third law of thermodynamics, a quantum system has temperature absolute zero if and only if its entropy is zero, i.e., if it is in a pure state.
Because of the unavoidable interaction with the environment this is impossible to achieve.
But it has nothing to do with all molecules standing still, which is impossible for a quantum system as the mean ...
6
In my opinion, it isn't strictly correct to say that entropy is "randomness" or "disorder". The entropy is defined in statistical mechanics as $-k_B \sum_i p_i \log p_i$, where $k_B$ is Boltzmann's constant (which is only there to put it into physically convenient units) and $p_i$ is the probability that the system is in state $i$. These probabilities do not ...
6
If I create an electron on earth and someone else creates an electron on Andromeda, they're identical particles. They have the same quantum numbers, they're both excitations of the electron field. However they're distinguishable by means of their spatial separation. Their wavefunctions don't overlap.
Edit: perhaps I should add that not everyone uses the ...
6
Assuming that the $N$ argon atoms are being treated as a classical system, then there are 3 degrees of freedom per atom. That assumes that we are neglecting electronic degrees of freedom, which is OK since one needs a fairly high temperature to thermally excite the electrons in argon.
Now if there are $N$ atoms, then there are $3N$ degrees of freedom. No ...
6
I think that the most prominent example of "prediction before observation" in statistical physics is the Bose-Einstein condensate.
It was predicted in ~1925 by, well, Bose and Einstein, obviously. Then after more than ten years it was proposed as an explanation for superfluidity and superconductivity. And the actual BEC of atoms (as a new state of matter) ...
6
Your interpretation is not quite right. One sharp interpretation one can give to this "cutting" of phase space into cubes of size $h^{2N}$ (here $N$ is the dimension of the system's configuration space), is that it allows one to use classical phase space to count the number of energy eigenstates of the corresponding quantum hamiltonian. Instead of trying ...
5
Yes mean-field theory is wrong for the one-dimensional case (and wrong for the two and three dimensional cases as well, where the transition exists but the mean-field approximation gets the wrong critical temperature and exponents). In fact it's a typical first year exercise to solve the 1D Ising model exactly using transfer matrices, and I suggest you look ...
5
There are several different notions of microstates or distinguishability that might be relevant to your question.
Coarse-graining of phase space into Planck cells.
Consider two classical variables $x$ and $p$ with $x \sim x+x_0$ and $p \sim p+p_0$. You can think of this system as describing a particle that lives on a circle of radius $x_0$ and where ...
5
The answer to your question is yes, your intuition is 100% correct. It all boils down to the topology of the configuration space $\mathcal C$, mainly the first homotopy group $\pi_1(\mathcal C)$ (which is non-zero in your example). See problem set 1, problem 3 from this course at Oxford. This exercise is precisely about loops in 3+1D! One has to argue that ...
5
Because by convention, we want to write $\beta$ as the coefficient in front of energy $E$ in the exponent $\exp(-\beta E)$. Exponents have to be dimensionless so $\beta$ has to have units of inverse energy. That's why it has to be objects such as $\beta=1/kT$ because $kT$ has units of energy. The latter statement holds because the energy per degree of ...
5
The von Neumann entropy is the analogue of the Boltzmann entropy in quantum mechanics. It really is exactly the same thing. Any density matrix $\rho$ can be written as $\rho = \sum_i p_i |i\rangle\langle i|$, where $p_i = \mbox{probability}(\mbox{state}_i)$ is a probability distribution on state vectors. The von Neumann entropy is the Boltzmann entropy of ...
5
For the partition sum, you have so sum $e^{-E}$ ($T=1$) over all possible eigenstates of the system where $E$ is the energy of the corresponding state.
Two bosons can be in the states 10
$|kl\rangle$, with $1\leq k \leq l \leq 4$ where we accounted for the degeneracy by introducing an additional state with $E_4 =2E$. The corresponding partition sum reads ...
5
Qualitatively, the Morse potential has two competing effects. The first is at small separations, where the potential becomes (infinitely) large; this effect is roughly due to the electrostatic repulsion between the two atoms, and it increases as the atoms get closer together. On the other hand, two atoms may covalently bond, and generally speaking, the ...
5
Off the top of my head, the example I can think of is the whole work that Boltzmann did. He based his entire theory of statistical mechanics on the concept of indivisible particles (i.e. that all matter is made up out of atoms). Doing this, his theory (using theoretical mathematical methods as you said) was able to predict how the atoms determine the visible ...
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