Mixture Dynamics And Shells — Examples

For each integer $n\ge 2$, the set $E_n=[1\mathrm{/}n,1)$ is Borel, and ${\bigcup }_{n\ge 2}{E}_n=(0,1)$. Since $P((0,1))=1$, at least one $E_n$ has positive probability: otherwise the disjoint sets $E_2,E_3\setminus E_2,E_4\setminus E_3,\dots$ all have probability zero, and countable additivity would give $P((0,1))=0$. On $E_n$, $x\ge 1\mathrm{/}n$, while elsewhere $x\ge 0$, so ${\int }_{(0,1)}xdP\ge \frac{1}{n}P(E_n)>0.$ Therefore every admissible probability has strictly positive objective.

The point mass ${\delta }_{1\mathrm{/}n}$ at $1\mathrm{/}n$ is a Borel probability on $(0,1)$ and has objective $1\mathrm{/}n$. These values tend to zero, while 1.1 shows all objectives are positive. Hence the infimum is zero and is not attained. The generator set is noncompact, so this conclusion does not contradict the compact result [F2]; it shows why an attainment theorem must retain its compactness/coercivity hypotheses.

Since $e_{N,k}\ge 1$, $e^{N{e}_{N,k}^2}\ge e^N$. For $x\ge e^N$, $x-1\le \lfloor x\rfloor \le x$, so $e_{N,k}^2+\frac{1}{N}\log (1-e^{-N})\le \frac{1}{N}\log g_{N,k}\le e_{N,k}^2\mathrm{.}$ The error tends to zero uniformly in $k$. The grid mesh is $1\mathrm{/}N$; for a continuous function on $[1,2]$, uniform continuity makes its grid maxima converge to its interval maximum.

Let $Z_N(b)={\sum }_{k=0}^N{g}_{N,k}{e}^{-Nb{e}_{N,k}}$ and $m_N={\max }_k[N^{-1}\log g_{N,k}-b{e}_{N,k}]$. The largest summand is $e^{N{m}_N}$ and there are $N+1$ summands, so $m_N\le N^{-1}\log Z_N(b)\le m_N+N^{-1}\log (N+1)\mathrm{.}$ By 1.1, $m_N\to {\max }_{e\in [1,2]}(e^2-be)$. If $e=\theta \cdot 1+(1-\theta )\cdot 2$ with $0<\theta <1$, direct expansion gives $e^2-be=\theta (1-b)+(1-\theta )(4-2b)-\theta (1-\theta ),$ strictly below the weighted endpoint values. Thus the maximum is at an endpoint and equals $\max (1-b,4-2b)$.

If $K=\mathrm{\varnothing }$, the probability is zero. Otherwise, for compact $K\subset (1,2)$ strict convexity gives $\eta :={\min }_{e\in K}[\max (1-b,4-2b)-(e^2-be)]>0$ by [F3]. Uniform approximation in 1.1 makes every grid exponent in $K$ at most the maximum minus $\eta \mathrm{/}2$ for all sufficiently large $N$. On the other hand, an endpoint group has exponent tending to the maximum. By [F2], $\log (N+1)\mathrm{/}N\le 2\sqrt{N+1}\mathrm{/}N\to 0$, so the polynomial factor is dominated by the exponential gap and the canonical probability of $K$, at most $(N+1)e^{-N\eta \mathrm{/}2+o(N)}$, tends to zero.

Choose integers $k_N$ with $e_{N,k_N}\to e\in (1,2)$. The microcanonical exact-group law has energy density exactly $e_{N,k_N}$ at every state. By 1.1 its entropy density $N^{-1}\log g_{N,k_N}$ tends to $e^2$. Thus, for example at the crossing $b=3$, canonical mass escapes every compact neighborhood of any interior target while the microcanonical group remains there. Finally $s(3\mathrm{/}2)=9\mathrm{/}4<5\mathrm{/}2=(s(1)+s(2))\mathrm{/}2$, so $s(e)=e^2$ is not concave. The example is a constructed model, not an observation about a material.

Write $X_H=a{\mathrm{\partial }}_q+b{\mathrm{\partial }}_p$. Then ${\iota }_{X_H}(dq\wedge dp)=adp-bdq$, while $dH=-\sin qdq$. Equating coefficients gives $a=0$ and $b=\sin q$. The explicit solution is ${\varphi }_t(q,p)=(q,p+t\sin q(\mathrm{m}\mathrm{o}\mathrm{d}2\pi )),$ which is defined for every real $t$ and is a diffeomorphism with inverse ${\varphi }_{-t}$.

Since $H\circ {\varphi }_t=H$, the set $A$ is invariant under every ${\varphi }_t$. The torus has total area $(2\pi {)}^2=4{\pi }^2$. The allowed $q$ values in $A$ have total length $\pi$, while $p$ ranges over length $2\pi$, so $m(A)=2{\pi }^2$, strictly between zero and total area. The map preserves area by [F2], and the invariant set violates the criterion [F3]; hence it is not ergodic for any nonzero sampling time.

The unitary evolution sends $\psi$ to $(e^{-i{E}_{1}t\mathrm{/}\mathrm{\hslash }}{u}_1+e^{-i{E}_{2}t\mathrm{/}\mathrm{\hslash }}{u}_2)\mathrm{/}\sqrt{2}$, which remains a unit vector by [F2]. Its rank-one density has eigenvalues $1,0$, so its entropy is $-k_B(1\log 1+0\log 0)=0$ for every $t$.

The two energy eigenspaces are one-dimensional. The matrix entry between them is an exponential with nonzero frequency; by [F1] its average has magnitude at most $\mathrm{\hslash }\mathrm{/}(T\mathrm{\mid }{E}_1-E_2\mathrm{\mid })$ and tends to zero. The diagonal entries remain $1\mathrm{/}2$, so the direct finite-dimensional average yields ${\rho }_{\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}}=\frac{1}{2}{P}_1+\frac{1}{2}{P}_2$. Its eigenvalues are $(1\mathrm{/}2,1\mathrm{/}2)$, so $S({\rho }_{\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}})=-k_B(\frac{1}{2}\log \frac{1}{2}+\frac{1}{2}\log \frac{1}{2})=k_B\log 2.$ Since every $S({\rho }_t)=0$, the time average of instantaneous entropy is exactly zero.

In the basis $(u_1,u_2)$, ${\bar{\rho }}_T$ has diagonal entries $1\mathrm{/}2$ and off-diagonal entry $c_T=\frac{1}{2T}{\int }_0^T{e}^{-i(E_1-E_2)t\mathrm{/}\mathrm{\hslash }}dt,$ whose magnitude is at most $\mathrm{\hslash }\mathrm{/}(T\mathrm{\mid }{E}_1-E_2\mathrm{\mid })$ by direct integration. Its eigenvalues are $1\mathrm{/}2\pm \mathrm{\mid }{c}_T\mathrm{\mid }$, so both tend to $1\mathrm{/}2$. Continuity of $x\mapsto -x\log x$ at $1\mathrm{/}2$ gives $S({\bar{\rho }}_T)\to k_B\log 2$. Thus entropy of the averaged state has a nonzero limit even though the averaged instantaneous entropy remains zero.