26
Yes, one can observe a decrease in entropy of an isolated system.
The statistics of these observation are quantified by the fluctuation theorem. The logic of it is based on what you (the OP) suggest: since statistical mechanics is about statistical laws, one would expect there to be fluctuations.
The first paragraph of wikipedia states this well, so I'll ...
6
My question, then, is: have we ever observed a natural example of a
"spontaneous" reduction in entropy?
Yes, of course, but as @Mark_Bell pointed out, it depends on if you are talking about the entropy of the system only (or the surroundings only) or the total entropy change (system + surroundings).
If heat spontaneously transfers from the system ...
4
The partition function $Z$ is defined as
$$Z(T,\mathbf{x})=\sum_{\{\mu\}}e^{-\beta \mathcal{H}(\mu)}$$
where sum is over microstate ($\mu$).
For an ideal gas,
$$Z(T,V,N)=\int \frac{1}{N!} \prod_{i=1}^N\frac{d^3q_id^3p_i}{h^3}\exp\left[-\beta \sum_{i=1}^N\frac{p_i^2}{2m}\right]$$
$$Z(T,V,N)=\frac{1}{N!}\left(\frac{V}{\lambda(T)^3}\right)^N$$
$$\Rightarrow \...
3
If you want to write $\exp\{-\beta \hat H\}$ in closed form in the position representation, you can use Mehler's formula:
$$
\sum_{n=0}^\infty s^n \varphi_n(x)\varphi_n(y) =\\ \frac
1{\sqrt{\pi (1-s^2)}} \exp\left\{\frac{4xys
-(x^2+y^2)(1+s^2)}{2(1-s^2)}\right\}, \quad 0\le |s|<1.
$$
with
$$
s= e^{-\beta(n+1/2)}.
$$
Here
$$
\varphi_n(x)\equiv \frac{1}{\...
3
OP seems to basically have this but is missing a complex conjugation somewhere. We want $(C^\dagger)_{ij}=C^*_{ji}$ that is:
$$(C^{\dagger})_{ij} = \langle c_j^\dagger c_i \rangle^*$$
the inner product simply returns a number and so complex conjugating is equivalent to taking the adjoint of the operator in the middle:
$$(C^\dagger)_{ij} = \langle (c_j^\...
3
Assuming a cubic sample of side $L$ and periodic boundary conditions.
The wavevector $\mathrm{k}$ can assume any value on the grid
\begin{equation}
\mathrm{k} = (k_x, k_y, k_z) = \left(\frac{2\pi}{L}n_x, \frac{2\pi}{L}n_y, \frac{2\pi}{L}n_z\right)
\end{equation}
where $n_x, n_y, n_z$ are integers. The Fermi momentum is the maximum momentum allowed for ...
3
The 3-dimensional Maxwell velocity distribution function in the rest frame of the container is
$$f(v_x,v_y,v_z) = \left(\frac{m}{2 \pi kT}\right)^{3/2}\, exp(-\frac{m(v_x^2+v_y^2+v_z^2)}{2kT})$$
If the container is moving with velocity $V$ say in the x-direction, this becomes
$$f(v_x,v_y,v_z) = \left(\frac{m}{2 \pi kT}\right)^{3/2}\, exp(-\frac{m((v_x-V)^2+...
3
From the Hyperphysics website:
The Maxwell-Boltzmann distribution is the classical distribution function for distribution of an amount of energy between identical but distinguishable particles
$$f(E)=\frac{1}{Ae^{E/kT}}$$
Where $f(E)$ is the probability of that a particle will have energy E, A is a constant, $k$ is Boltzmann’s constant, and $T$ is the ...
2
So, let's start with the fundamental vagueness of the question. The reason behind this vagueness is the lack of clear unpacking of what is meant by "computation" or "prediction" or "modelling".
Let's suppose for a second that you actually do have an infinitely powerful computer that is able to simulate the dynamics of any number ...
2
Let's first be clear about some basics then we go on to questions:
The periodic boundary condition requires that any wave in the sample $e^{ikr}$ have the same value for a position $r$ as it has for $r+L$ (after one round around the circle). This imposes quantization of $k$
$$k=\frac{2\pi n}{L} \ \ \ n\ \text{is integer}.$$
In three dimensions,
$$\mathbf{k}=\...
2
Does Bose-Einstein condensation depend on boundary conditions?
No.
This can be shown rigorously in 'On the Bose-Einstein condensation of an ideal gas' by L. J. Landau and I. F. Wilde (1979).
The proof lies in computing the fugacity $z = \mathrm{e}^{\beta \mu}$ (called activity in the paper) and showing that it exhibits a non-analytic behaviour at some $T = ...
2
This gets exactly into what is called "coarse graining." The idea goes as follows. Say you have some computer simulation where you know the position and momentum of every particle in a large box. There is indeed no way to assign an entropy to this microstate. If you know the microstate, then $\Omega = 1$, and $S = k \ln(\Omega) = k \ln(1) = 0$.
...
1
Using Transition state theory approximation for the rate of water molecule to cross over the bonding barrier, $E_b = 8KT \approx 21 KJ/mole$:
$$ \tag{1}
\frac{1}{\tau} = \omega \exp\{-\frac{E_b}{KT}\}.
$$
Now, we have to estimate the frequency prefactor $\omega$ in Eq.(1). Since there is no data available for the vibration between water-water molecule ...
1
The following may not necessarily be true
$$
\langle c^{\dagger}_ic_j\rangle=\langle c^{\dagger}_jc_i\rangle
$$
Actually, there is a whole branch of condensed matter models that depend on
$$
\langle c^{\dagger}_ic_j\rangle \neq \langle c^{\dagger}_jc_i\rangle
$$
meaning
$$
\langle c^{\dagger}_ic_j\rangle \sim e^{\theta_{ij}i}
$$
and
$$
\langle c^{\dagger}...
1
Classical Boltzmann distribution gives the probability density of finding a particle in a particular point of the phase space. E.g., If we have Hamiltonian
$$
H= \frac{p^2}{2m} + V(x)
$$
the distribution is given by
$$
W(p,x) = Ce^{-\frac{H(p,x)}{k_BT}}, C=\int dp dx e^{-H(p,x)}
$$
The momentum can be integrated out, so the distribution that you need is
$$
w(...
1
Quoting from this directly related post,
According to the fluctuation theorem the second law of thermodynamics is a statistical law. Violations at the micro scale, therefore, certainly have a non-zero probability.
We can calculate the entropy using the Boltzmann formula $S=k_B\log N$, where $N$ is the number of states. From the initial state $i$ to the the ...
1
For completeness sake and later reference i'll add the actual result here. The derivation requires besides the evaluation of the sum also a fair amount of hyperbolic trig function manipulation. Here it is,
$$\begin{aligned}
\rho(x,x') &\equiv \langle x | \hat \rho |x'\rangle \\[1.0em]
&= \frac{\langle x|\exp(-\beta \hat H)| x'\rangle }{Z}\\[1.0em]
&...
1
I think you are mixing up spontaneous process vs unspontaneous process with isolated system vs non-isolated system.
If you state the 2nd law of thermodynamics for isolated systems then you are right in stating that the entropy will not decrease for any processes, spontaneous or not.
If you state the law for non-isolated systems then you can only state that ...
1
What you are describing seems to me like a system of non-intarcting two-level particles. The temperature for such system is derived here.
As you can see, the temperature for such systems can be pretty wild. It can be even negative!
1
I think we should speak here not of semiclassical model, but of semiclassical approximation, which preserves some of the quantum features in favor making things more tractable.
Indeed, if we were to demand that the position and momentum can be measured simulatenously, we could not have quantum effects. However, if we demand that they are measured ...
1
First of all, it's important to remember that we are dealing with models here, which are mathematical expressions that are derived from certain assumptions. Generally speaking, in materials systems we need to work with very large numbers of particles that can interact in non-trivial ways, so there is a tendency to simplify things to make the mathematics ...
1
Equal probability is a base concept for the microcanonical ensemble: all microstates have eqaul accessing probability.
The microcanonical ensemble is based on this concept, but also the Boltzmann hypothesis of entropy formula:
$$
S(E, V, N) = K_B \ln \Gamma(E, V, N).
$$
Where $\Gamma$ is the total number of microscopic state with the given condition of $E$,...
1
Let me point to one example of $\Bbb Z_2$ symmetry breaking. It is contained in my paper
S. Shlosman: Phase transitions for two-dimensional models with isotropic short-range interactions and continuous symmetry, Comm. Math. Phys. 71(1980), 207-212.
There I consider a 2D system with $O(2)$ symmetry group (which is disconnected). In agreement with the Mermin-...
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