The Vanishing of Excess Heat for Nonequilibrium Processes Reaching Zero Ambient Temperature

Abstract

We present the mathematical ingredients for an extension of the Third Law of Thermodynamics (Nernst heat postulate) to nonequilibrium processes. The central quantity is the excess heat which measures the quasistatic addition to the steady dissipative power when a parameter in the dynamics is changed slowly. We prove for a class of driven Markov jump processes that it vanishes at zero environment temperature. Furthermore, the nonequilibrium heat capacity goes to zero with temperature as well. Main ingredients in the proof are the matrix-forest theorem for the relaxation behavior of the heat flux, and the matrix-tree theorem giving the low-temperature asymptotics of the stationary probability. The main new condition for the extended Third Law requires the absence of major (low-temperature induced) delays in the relaxation to the steady dissipative structure.

Appendices

Appendix A: Matrix-Forest Theorem

As it may seem less clear how (4.4) and (4.5) arise, we give here their origin from the matrix-forest theorem.

Consider a continuous-time irreducible Markov process on a finite state space $\mathcal {V}$ characterized by transition rates $k(x,y)\ge 0$ for $x \,, y \in \mathcal {V}$. Let L be the backward generator,

$$\begin{aligned} L_{xy}&=k(x,y), \quad x\ne y \,\, \text {and } \, x,y\in \mathcal {V}\nonumber \\ L_{xx}&=-\sum _y k(x,y). \end{aligned}$$

(A.1)

Consider $e^{tL}g (x) = \langle g(X_t)\,|\,X_0=x\rangle$.

$$\begin{aligned} V_g(x) = \int _0^\infty [\langle g(X_t)\,|\,X_0=x\rangle - \langle g\rangle ^s ] \text{ d }t \end{aligned}$$

(A.2)

where we subtract the asymptotic stationary value and

$$\begin{aligned} \langle g(x_t)\,|\,x_0=x\rangle \xrightarrow []{t\uparrow \infty }\langle g\rangle ^s . \end{aligned}$$

Putting $f(x)=g(x)-\left\langle g\right\rangle$,

$$\begin{aligned} V_g(x) = V_f(x) = \int _0^\infty e^{Lt}f(x) \,\text {d}t \end{aligned}$$

(A.3)

so then

$$\begin{aligned} V_f(x) = \lim _{b \rightarrow 0}\dfrac{-1}{(L - b I)}f(x) \end{aligned}$$

(A.4)

where $\dfrac{1}{L-b I}$ is the resolvent-inverse of the backward generator L (see [22]).

For our purposes, the set $\mathcal {V}$ is the vertex set of a connected graph G, and the transitions happen over its edges. We use [33,34,35] to obtain a graphical representation of $V_f$; put $V:= V_f$. A spanning forest is a collection of trees that forms a spanning subgraph. Define the set $\mathcal {F}^{x \rightarrow y}_m$ to be the set of all spanning forests in G with m edges having the properties: every tree in the forest is a rooted tree, y is the root of one of the trees and x and y are in the same tree (so there is a path from x to y). $F_m^{xy}$ denotes an element from the set $\mathcal {F}^{x \rightarrow y}_m$. Define $\mathcal {F}_m$ as the union of sets $\mathcal {F}^{x \rightarrow x}_m$ in graph G.

Proposition A.1

$$\begin{aligned} V(x) = \sum _{y} \dfrac{ w\left( \mathcal {F}_{n-2}^{x\rightarrow y}\right) }{w\left( \mathcal {F}_{n-1}\right) }f(y).\end{aligned}$$

(A.5)

Proof

From [34], for all $c>0$,

$$\begin{aligned} \bigg ( \dfrac{1}{I - c L}\bigg )_{xy} = \frac{\sum _{m=0}^{n-1} c ^m w(\mathcal {F}^{x \rightarrow y}_m)}{\sum _{m=0}^{n-1}c ^m w(\mathcal {F}_m)} \end{aligned}$$

(A.6)

where $w(\mathcal {F}^{x \rightarrow y}_k)$ is the weight of the set $\mathcal {F}^{x \rightarrow y}_k$

$$\begin{aligned} w(\mathcal {F}^{x \rightarrow y}_m)= \sum _{F_m^{xy}\in \mathcal {F}^{x \rightarrow y}_m }\prod _{(z,z')\in F_k^{xy}}k(z,z'). \end{aligned}$$

(A.7)

Thus,

$$\begin{aligned} V(x)&=\sum _{y} \left( \lim _{c \rightarrow \infty } \dfrac{c}{ I-c L}\right) _{xy} f(y) \nonumber \\&=\sum _{y} \lim _{c \rightarrow \infty } \frac{\sum _{m=0}^{n-1} c^{m+1}w(\mathcal {F}^{x \rightarrow y}_m)f(y)}{\sum _{m=0}^{n-1}c^m w(\mathcal {F}_m)}\nonumber \\&=\sum _{y} \lim _{c \rightarrow \infty } \frac{\sum _{m=0}^{n-2} c^{m+1}w(\mathcal {F}^{x \rightarrow y}_m)f(y)}{\sum _{m=0}^{n-1}c^m w(\mathcal {F}_m)} + \sum _{y} \lim _{c \rightarrow \infty } \frac{ c^{n}w(\mathcal {F}^{x \rightarrow y}_{n-1})f(y)}{\sum _{m=0}^{n-1}c^mw(\mathcal {F}_m)}. \end{aligned}$$

(A.8)

Next, use that $\left<f\right>^s = 0$ and hence, $\sum _{y} \rho ^s (y)f(y) = 0$. For the stationary distribution $\rho ^s$ we use the Kirchhoff formula (2.16). Therefore, $\sum _y w(\mathcal {F}^{x \rightarrow y}_{n-1})f(y) = 0$, and the second term in the last line above is equal to zero. To continue the calculation,

$$\begin{aligned} V(x)&= \sum _{y} \lim _{c \rightarrow \infty } \frac{\sum _{m=0}^{n-2} c^{m+1}w(\mathcal {F}^{x \rightarrow y}_m)f(y)}{\sum _{m=0}^{n-1}c^m w(\mathcal {F}_m)} \nonumber \\&= \sum _{y} \dfrac{ w\left( \mathcal {F}_{n-2}^{x\rightarrow y}\right) }{ w\left( \mathcal {F}_{n-1}\right) }f(y). \end{aligned}$$

(A.9)

$\square$

Define $\mathcal {F}^{x\rightarrow y}:= \mathcal {F}_{n-2}^{x\rightarrow y}$ as the set of all spanning forests consisting of two trees. Remark that $\mathcal {F}_{n-1}$ is the set of all rooted spanning trees and W is the sum over the weights of all rooted spanning trees, so then, $w(\mathcal {F}_{n-1})=W$.

Corollary A.2

The solution V with $\langle V\rangle ^s =0$ of $LV = -f$ with $\langle f \rangle ^s=0$ is given by

$$\begin{aligned} V(x) = \frac{1}{W} \sum _{y} w\left( \mathcal {F}^{x\rightarrow y}\right) f(y). \end{aligned}$$

(A.10)

1.1 1. Quasipotential for a Specific Source

Here we consider the quasipotential of (A.10) for a specific source. We focus on the case $f(y)=\mathcal {P}(y)-\left\langle \mathcal {P}\right\rangle ^s$ in (A.10), where $\mathcal {P}(y)=\sum _x k(y,x)q(y,x)$ and $q(x,y)=-q(y,x)$. We define

$$\begin{aligned} V^\mathcal {P}(x):=\dfrac{\sum _y w(\mathcal {F}^{x\rightarrow y})\mathcal {P}(y)}{W}, \quad V^{\left\langle \mathcal {P} \right\rangle }(x):= \left\langle \mathcal {P} \right\rangle ^s \dfrac{ \sum _y w(\mathcal {F}^{x\rightarrow y})}{ W}. \end{aligned}$$

(A.11)

So then, the quasipotential can be written as

$$\begin{aligned} V(x)=V^\mathcal {P}(x)-V^{\left\langle \mathcal {P}\right\rangle }(x) \end{aligned}$$

(A.12)

Proposition A.6 shows that the quasipotential in (A.12) can be decomposed into two terms, one related to spanning trees only and the other containing loops.

$F^{xy}$ denotes a forest in the set of $\mathcal {F}^{x \rightarrow y}$. Write $k_{yx}:= k(y,x)$ and $q(y,x):=q_{yx}$.

Lemma A.3

For each x on the graph G,

$$\begin{aligned} \sum _y w(\mathcal {F}^{x\rightarrow y} )\mathcal {P}(y) = \sum _{z,y} \sum _{\begin{array}{c} F^{xy} \in \mathcal {F}^{x\rightarrow y} \\ (z,y)\notin F^{xy} \end{array}}\,w(F^{xy})\, k_{yz}\,q_{yz}. \end{aligned}$$

(A.13)

where $\sum _{\begin{array}{c} F^{xy} \in \mathcal {F}^{x\rightarrow y}(z,y)\notin F^{xy} \end{array}}$ is sum over all forests in set $\mathcal {F}^{x\rightarrow y}$ such that the edge (z, y) is not in the forest.

Proof

The product $w(F^{xy})\, k_{yz}$ is the weight when adding edge (y, z) to the forest $F^{xy}$. Let us consider forests $F^{xy}$ and $F^{xz}$ which have different directions for the edge $\{z,y\}$: (z, y) is in the forest $F^{xy}\in \mathcal {F}^{x\rightarrow y}$ and the edge (y, z) is in the forest $F^{xz}\in \mathcal {F}^{x\rightarrow z}$. Adding the edge (y, z) to the forest $F^{xy}$ is the same as adding the edge (z, y) to the forest $F^{xz}$; see Fig. 9.

Fig. 9

Adding the edge (y, z) to $F^{xy}$ and adding the edge (z, y) to the forest $F^{xz}$

and

$$\begin{aligned} w(F^{xy})\,k(y,z)q(y,z)\, +\, w(F^{xz})\,k(z,y)q(z,y)=0. \end{aligned}$$

(A.14)

So then

$$\begin{aligned} \sum _y w(\mathcal {F}^{x\rightarrow y} )\mathcal {P}(y)&=\sum _y w(\mathcal {F}^{x\rightarrow y} )\sum _zk_{yz}q_{yz}\nonumber \\&=\sum _{z,y} \sum _{\begin{array}{c} F^{xy} \in \mathcal {F}^{x\rightarrow y} \\ (z,y)\notin F^{xy} \end{array}}\,w(F^{xy})\, k_{yz}\,q_{yz}. \end{aligned}$$

(A.15)

$\square$

Lemma A.4

By adding the edge $(y,z) \in G$ to the forest $F^{xy} \in \mathcal {F}^{x\rightarrow y}$ where $(z,y)\notin F^{xy}$, the new graph is either a rooted spanning tree or an oriented tree-loop-tree.

Proof

Consider the forest $F^{xy}$. It has two trees: one is rooted in y and the other tree is rooted in some vertex r. Vertices x and y are located in a same tree and the edge (z, y) is not in the forest. If the vertex z is on the same tree with y, adding the edge (y, z) creates an oriented tree-loop-tree, Fig. 10a, and if z is in another tree then a rooted spanning tree is created, Fig. 10b. $\square$

Fig. 10

The edge (y, z) is added to the forest $F^{xy}$, either an oriented tree-loop-tree is created (a) or a rooted spanning tree (b)

We need extra notations for the next Lemma. Put $\mathcal {K}_x$ for the set of all tree-loop-trees such that x is located on the tree-loop part. If $\kappa \in \mathcal {K} _x$, then $O(\kappa )$ is the set of oriented tree-loop-trees for all possible directions in $\kappa$ and $\overline{\kappa }$ denotes an element from the set $O(\kappa )$. $\mathcal {L}$ is the set of all loops in the graph G and $\mathcal {K}_{\ell ,x}$ denotes the set of all tree-loop-trees including the loop $\ell$ and x in the tree-loop part.

Lemma A.5

$V^\mathcal {P}(x)$ splits into two parts; one where we only sum over spanning trees and one where loops are present:

$$\begin{aligned} V^\mathcal {P}(x)= V^\mathcal {P}_{\text {tree}}(x) + V^\mathcal {P}_{\text {loop}}(x). \end{aligned}$$

(A.16)

Here, the first part is equal to

$$\begin{aligned} V^\mathcal {P}_{\text {tree}}(x) = \dfrac{1}{W}\; \sum _{y\in \mathcal {V}}\sum _{T\in \mathcal {T} } w(T_y)\,q_{T }(x\rightarrow y ) \end{aligned}$$

(A.17)

and

$$\begin{aligned} q_{T }(x\rightarrow y ):= \sum _{(u,u')\in (x\xrightarrow {T }y)}q(u,u') \end{aligned}$$

(A.18)

where $(x\xrightarrow {T }y)$ is the path from x to y in the spanning tree T. The second part is equal to

$$\begin{aligned} V^\mathcal {P}_{\text {loop}}(x) =\dfrac{1}{W}\,\sum _{\ell \in \mathcal {L}}\sum _{\kappa \in \mathcal {K}_{\ell ,x}}\,\sum _{\overline{\kappa } \in O(\kappa ) }w(\overline{\kappa })q(\overline{\ell } ) \end{aligned}$$

(A.19)

where $\overline{\ell }$ is an oriented loop the same as the loop in $\overline{\kappa }$ and $q(\overline{\ell })\!=\!\sum _{(u,u')\in \overline{\ell }}q(u,u')$.

Proof

Consider the first case in Lemma A.4; there are different possible tree-loop-trees for every loop in the graph G and it follows there are different possible oriented tree-loop-tree for every oriented loop. Take an edge $(u,u')$ located on the oriented loop of the oriented tree-loop-tree, removing the edge $(u,u')$ from the oriented tree-loop-tree gives a forest where u and $u'$ are in the same connected component. There is a forest $F^{xu}$ such that $u'\in F^{xu}$ and $(u,u') \notin F^{xu}$, and

$$\begin{aligned} \sum _{(u,u')\in \ell }w(F^{xu})k_{uu'}q_{uu'}=w(\overline{\kappa })q(\overline{\ell }). \end{aligned}$$

(A.20)

Let us go back to Lemma A.4. For a rooted spanning tree $T_y$ in graph G, there is a unique path $(x\xrightarrow {T }y)$ from x to y on this tree. Then,

$$\begin{aligned} w(T_y)q_{T}(x\rightarrow y)=\sum _{(u,u')\,\in (x\xrightarrow {T }y)}w(F^{xu})k_{uu'}q_{uu'} \end{aligned}$$

(A.21)

where $F^{xu}\in \mathcal {F}^{x\rightarrow u}$. We have used the fact that by removing an edge $(u,u')$ located on the path $(x\xrightarrow {T }y)$ in spanning tree $T_y$, a forest including two trees, $\tau _u$ (toward u) and $\tau _y$, is created. Obviously, the edge $(u,u') \notin F^{xu}$ and there is no other path between u and $u'$ in the new forest. So then, the sum over all possible oriented spanning trees in the left-hand side of relation (A.21) will give the weight of all possible forests such that

$$\begin{aligned} \sum _{y\in \mathcal {V}}\sum _{T \in \mathcal {T}}w(T_y)q_{T}(x\rightarrow y)&=\sum _{y\in \mathcal {V}}\sum _{z\in A_{F^{xy}}}w(\mathcal {F}^{x\rightarrow y})k_{yz}q_{yz} \end{aligned}$$

(A.22)

where

$$\begin{aligned} A_{F^{xy}} = \{z \in \mathcal {V} ; \quad \not \exists \, (z\xrightarrow {F^{x y}}y) \in F^{x y}\} \end{aligned}$$

(A.23)

is the set of all vertices from which there is no path to y in the forest $F^{xy}$. Hence, according to the second case in Lemma A.4, the tree-term of $V^\mathcal {P}$ is

$$\begin{aligned} V^\mathcal {P}_\text {tree}(x) = \dfrac{1}{W}\; \sum _{y\in \mathcal {V}}\sum _{T\in \mathcal {T}} w(T_y)\,q_{T}(x\rightarrow y ). \end{aligned}$$

(A.24)

$\square$

We are ready for the main result of this Appendix.

Proposition A.6

The graphical representation of the quasipotential in (A.10) for

$$\begin{aligned} f(y)=\mathcal {P}(y)-\left\langle \mathcal {P} \right\rangle ^s , \quad \mathcal {P}(y)=\sum _x k(y,x)q(y,x), \quad q(x,y)=-q(y,x) \end{aligned}$$

(A.25)

consists of two distinct classes of terms, trees and loops:

$$\begin{aligned} V(x)&= V_{\text {tree}}(x)+V_{\text {loop}}(x) \end{aligned}$$

where

$$\begin{aligned} V_\text {tree}(x)&= \dfrac{1}{W}\; \sum _{y\in \mathcal {V}}\sum _{T\in \mathcal {T}} w(T_y)\,q_{T}(x\rightarrow y ) \end{aligned}$$

(A.26)

$$\begin{aligned} V_{\text{ loop }}(x)&=\dfrac{1}{W}\sum _{\ell \in \mathcal {L}}\,\sum _{\kappa \in \mathcal {K}_{\ell ,x}}\,\sum _{\overline{\kappa } \in O(\kappa ) }w(\overline{\kappa })q(\overline{\ell } ) -\frac{1}{W}\sum _y w(\mathcal {F}^{x\rightarrow y})\left\langle \mathcal {P} \right\rangle ^s . \end{aligned}$$

(A.27)

Proof

From Eq. (A.12) we get that

$$\begin{aligned} V(x) = V^\mathcal {P}(x) - V^{\left<\mathcal {P}\right>}(x). \end{aligned}$$

(A.28)

Now use the representation of $V^\mathcal {P}(x)$ from Lemma A.5. The graphical representation of $\left\langle \mathcal {P} \right\rangle$ is given in Lemma 5.2. $\square$

Lemma A.7

Consider a connected graph G and the edge $(x,x')\in \mathcal {E}(G)$ then

$$\begin{aligned} \sum _z w(\mathcal {F}^{x\rightarrow z})-\sum _y w(\mathcal {F}^{x'\rightarrow y})=0. \end{aligned}$$

(A.29)

Proof

Take arbitrary z and $F \in \mathcal {F}^{x \rightarrow z}$. Then F consists of two disconnected trees $\tau _1$ and $\tau _2$, where $\tau _1$ is a tree rooted in z, $x \in \tau _1$ and $\tau _2$ is a tree rooted in some vertex r. There are two possibilities, $x'\in \tau _1$ or $x' \in \tau _2$. If $x' \in \tau _1$, then also $F \in \mathcal {F}^{x' \rightarrow z}$. If $x' \in \tau _2$, then $F \in \mathcal {F}^{x'\rightarrow r}$. In the same way, if we fix y, every forest $F \in \mathcal {F}^{x' \rightarrow y}$ corresponds to a forest $F \in \mathcal {F}^{x \rightarrow z}$ for some z. This is a one-to-one correspondence. $\square$

1.2 2. Difference-Quasipotential on an Edge

Consider an edge $e:=\{x,y\}$. We write $\overline{e}:=(x,y)$ when it has a direction. Put $V(\overline{e}):=V(x)-V(y)$. From Proposition A.6,

$$\begin{aligned} V(\overline{e})= V_{\text{ tree }}(\overline{e})+ V_{\text {loop}}(\overline{e}). \end{aligned}$$

(A.30)

Recall (A.26).

$$\begin{aligned} V_\text {tree}(\overline{e})&= \dfrac{1}{W}\; \sum _{u\in \mathcal {V}}\sum _{T\in \mathcal {T}} w(T_u)\,q_{T}(x\rightarrow u )-\dfrac{1}{W}\; \sum _{u\in \mathcal {V}}\sum _{T\in \mathcal {T}} w(T_u)\,q_{T}(y\rightarrow u )\nonumber \\&=\dfrac{1}{W}\; \sum _{u\in \mathcal {V}}\sum _{T\in \mathcal {T}} w(T_u)\big (q_{T}(x\rightarrow u )+q_{T}(u\rightarrow y )\big )\nonumber \\&=\dfrac{1}{W}\; \sum _{u\in \mathcal {V}}\sum _{T\in \mathcal {T}} w(T_u)\,q_{T}(x\rightarrow y ). \end{aligned}$$

(A.31)

According to Proposition A.6 and Lemma A.7 the difference of loop terms is given as

$$\begin{aligned} V_{\text {loop}}(\overline{e})=\dfrac{1}{W}\sum _{\ell \in \mathcal {L}}\sum _{\kappa \in \mathcal {K}_{\ell ,x}}\,\sum _{\overline{\kappa } \in O(\kappa ) } w(\overline{\kappa } )q(\overline{\ell } )-\dfrac{1}{W}\sum _{\ell \in \mathcal {L}}\sum _{\kappa \in \mathcal {K}_{\ell ,y}}\,\sum _{\overline{\kappa }\in O(\kappa ) }w(\overline{\kappa } )q(\overline{\ell } ) \end{aligned}$$

(A.32)

where $\mathcal {K}_{\ell ,x}$ denotes the set of all tree-loop-trees such that x is located in the tree-loop part. Take one tree-loop-tree from graph G. If x and y both are in the tree-loop part, then the tree-loop appears in both terms of the right-hand side of (A.32). We continue with the case that in tree-loop-trees, x and y are located in different parts. One of them is located in the tree-loop part and the other is located in the tree part:

$$\begin{aligned} V_{\text {loop}}(\overline{e})=\dfrac{1}{W}\sum _{\ell \in \mathcal {L}}&\bigg (\sum _{\kappa \in \mathcal {K}_{\ell ,x|y}}\,\sum _{\overline{\kappa } \in O(\kappa ) }w(\overline{\kappa } )q(\overline{\ell } )-\sum _{\kappa \in \mathcal {K}_{\ell ,y|x}}\,\sum _{\overline{\kappa } \in O(\kappa ) }w(\overline{\kappa } )q(\overline{\ell } )\bigg ). \end{aligned}$$

(A.33)

Here, $\mathcal {K}_{\ell ,x|y}$ denotes a tree-loop-tree with loop $\ell$ such that x is located in the tree-loop part and y is located in the tree part. Put $\mathcal {H}_{\ell ,e}$ as the set of all spanning tree-loops (including the loop $\ell$) such that the edge e is located on a tree and $\mathcal {H}_{\ell }^{(e)}$ is the set of all tree-loop-trees which are created by removing the edge e from a spanning tree-loop of $\mathcal {H}_{\ell ,e}$. Corresponding to what state of the edge e is closer to the loop, the set of $\mathcal {H}_{\ell ,e}$ splits into two groups. The set of spanning tree-loops where x is closer to the loop which is denoted by $\mathcal {H}_{\ell _x, e }$ and the set of all spanning tree-loops where y is closer to the loop which is denoted by $\mathcal {H}_{\ell _y, e }$. $\mathcal {H}^{(e)}_{\ell _x }$ is the set of all tree-loop-trees such that x is located on the tree-loop part. Rewrite (A.33) as

$$\begin{aligned} V_{\text{ loop }}(\overline{e})=\dfrac{1}{W}\bigg (\sum _{\ell \in \mathcal {L}}\sum _{H \in \mathcal {H}^{(e)}_{\ell _x }}\sum _{\overline{ H}\in O(H) }w(\overline{ H})q(\overline{\ell } )-\sum _{\ell \in \mathcal {L}}\sum _{H \in \mathcal {H}^{(e)}_{\ell _y }}\sum _{\overline{ H}\in O(H) }w(\overline{ H})q(\overline{\ell } )\bigg ). \end{aligned}$$

(A.34)

Notice that for every spanning tree-loop including the edge e on a tree either $\mathcal {H}^{(e)}_{\ell _x }$ or $\mathcal {H}^{(e)}_{\ell _y }$ happens.

As an example, consider the graph in Fig. 5 which has three loops. The possible tree-loops are shown in Fig. 6.

To find the differences of quasipotentials over the edge $\overline{e}=(x,w)$ the tree-loops including the edge $e:\{x,w\}$ in a tree are engaging, see $H_2$ and $H_3$ in Fig. 6.

We first look at a (non-oriented) tree-loop $H_{\ell ,e}\in \mathcal {H}_{\ell ,e}$ and we remove the edge e. In that way a tree-loop-tree is created. Secondly, we consider different possible orientations for the created tree-loop-trees. We rewrite (A.33),

$$\begin{aligned} V_{\text {loop}}(x,y)=\frac{1}{W}\sum _{\ell \in \mathcal {L}}\,\sum _{H\in \mathcal {H}_{\ell ,e}}\,\sum _{\overline{H}\in O(H)}\sigma _H(\overline{e})\, q(\overline{\ell })\,w(\overline{H}) \end{aligned}$$

(A.35)

where $\sigma _H(\overline{e})=\pm 1$. If the edge $\overline{e}$ is oriented toward the loop of H, then $\sigma _H(\overline{e})=-1$; otherwise it is positive. So then,

$$\begin{aligned} V_{\text {loop}}(\overline{e})=\frac{1}{W}\,\sum _{H\in \mathcal {H}_{e}}\, \sum _{\overline{H}\in O(H^{(e)})}\sigma _H(\overline{e})\, q(\overline{\ell })\,w(\overline{H}),\quad e\in \text {trees} \end{aligned}$$

(A.36)

where $\mathcal {H}_e$ is the set of all spanning tree-loops including the edge e in a tree. If $H\in \mathcal {H}_e$, then $H^{(e)}$ denotes a tree-loop-tree which is created by removing the edge e from the tree-loop H. $O(H^{(e)})$ is the set of all oriented tree-loop-trees made by giving all possible orientations to $H^{(e)}$ (remember Definition 3.2). Finally, the difference of the quasipotential over a directed edge $\overline{e}$ is

$$\begin{aligned} V(\overline{e})&= V_{\text{ tree }}(\overline{e})+ V_{\text {loop}}(\overline{e})\nonumber \\&=\dfrac{1}{W}\; \sum _{u\in \mathcal {V}}\sum _{T\in \mathcal {T}} w(T_u)\,q_{T}(x\rightarrow y )\nonumber \\&\quad +\frac{1}{W}\,\sum _{H\in \mathcal {H}_{e}}\, \sum _{\overline{H}\in O(H^{(e)})}\sigma _H(\overline{e})\, q(\overline{\ell })\,w(\overline{H}),\quad e\in \text {trees}. \end{aligned}$$

(A.37)

Appendix B: Proof of Lemma 5.2

Lemma B.1

The average of $\mathcal {P}(x)=\sum _{y} k(x,y)q(x,y)$ is

$$\begin{aligned} \left\langle \mathcal {P} \right\rangle ^s =\sum _{x,y\in \mathcal {V}}k(x,y)q(x,y)\rho ^s (x)=\frac{1}{W}\sum _{\overline{H} \in O(\mathcal {H})}w(\overline{H})q(\overline{\ell }) \end{aligned}$$

(B.1)

where $w(\overline{H})=\prod _{(z,z')\in \overline{H}}k(z,z')$.

Proof

Take a rooted spanning tree $T_{x}$ and the edge $\overline{e}=(x,y)$. We consider two cases:

First case, if the edge $\overline{e}'=(y,x)\in T_{x}$, then

$$\begin{aligned} k(x,y)w(T_{x})q(x,y)=k(y,x)w(T_{y})q(x,y)=-k(y,x)w(T_{y})q(y,x). \end{aligned}$$

(B.2)

From here we can conclude that

$$\begin{aligned} \sum _{x,y\in \mathcal {V}}\sum _{T_x\in \mathcal {T}_x}k(x,y)w(T_{x})q(x,y)=0. \end{aligned}$$

(B.3)

Second case, if the edge $\overline{e}'=(y,x)\notin T_{x}$, then $k(x,y)w(T_{x})$ is equal to the weight of a graphical object made by adding the edge (x, y) to the rooted tree $T_{x}$. That graphical object is a spanning oriented tree-loop. As a consequence,

$$\begin{aligned} k(x,y)w(T_{x})q(x,y)=w(\overline{H})q(\overline{e}) \end{aligned}$$

(B.4)

where $\overline{H}$ is made by tree $T_x$ and the edge (x, y) and its loop is in the same direction as (x, y). Summing over the edges in the oriented loop $\overline{\ell }$ (in clockwise or counter clockwise direction) in the underlying graph G gives

$$\begin{aligned} \sum _{ (x,y)\in \overline{\ell }} k(x,y)w(T_{x})q(x,y)=w(\overline{H}_{\ell })q(\overline{\ell }). \end{aligned}$$

(B.5)

Finally, we obtain

$$\begin{aligned} \sum _{x,y\in \mathcal {V}}k(x,y)q(x,y)\rho (x)&=\frac{1}{W} \sum _{x,y\in \mathcal {V}}\sum _{T_x\in \mathcal {T}} k(x,y)q(x,y) w(T_x)\nonumber \\&=\frac{1}{W} \sum _{x,y\in \mathcal {V}}\sum _{T_x\in \mathcal {T}_x} k(x,y)q(x,y) w(T_{x})\nonumber \\&= \frac{1}{W} \sum _{x,y\in \mathcal {V}}\sum _{T_x \ni (y,x)} k(x,y)q(x,y) w(T_{x})\nonumber \\&\quad + \sum _{x,y\in \mathcal {V}}\sum _{T_x\not \ni (y,x) } k(x,y)q(x,y) w(T_{x})\nonumber \\&=\frac{1}{W} \sum _\ell w(\overline{H}_{\ell })q(\overline{\ell })\nonumber \\&=\frac{1}{W}\sum _{\overline{H} \in O(\mathcal {H})}w(\overline{H})q(\overline{\ell }). \end{aligned}$$

(B.6)

$\square$

Appendix C: Examples and Illustrations

This section is meant to clarify the graphical conditions of the main Theorems 4.1–4.2. It is not essential for the proofs and it can be skipped at first reading. We first illustrate the notations with some simple examples.

Example C.1

Consider the graph G in Fig. 11.

Fig. 11

Graph G made by three loops

The transition rates are

$$\begin{aligned} k(y,x)=k(y,z)=k(x,u)=k(u,z)=e^{-\beta } \end{aligned}$$

(C.1)

and the rates over all other edges are equal to one. Definition 2.19 gives

$$\begin{aligned} \phi (y,x)&=\phi (y,z)=\phi (x,u)=\phi (u,z)=-1, \nonumber \\ \phi (x,y)&=\phi (z,y)=\phi (u,x)=\phi (z,u)=\phi (z,x)=\phi (x,z)=0. \end{aligned}$$

(C.2)

To find $\phi ^*$ we need to look at all the rooted spanning trees in the graph. The spanning trees are in Fig. 12.

Fig. 12

Spanning trees of the graph in Fig. 11. See [31] for more details

We find

$$\begin{aligned} \phi (x)=\phi (z)=\phi (u)=-1 , \quad \phi (y)=0 \end{aligned}$$

(C.3)

and thus $\phi ^*=0$. Denote

$$\begin{aligned} e_1:=\{x,y\}, \quad e_2:=\{y,z\}, \quad e_3:=\{z,u\}, \quad e_4:=\{x,u\}, \quad e_5:=\{x,z\}. \end{aligned}$$

(C.4)

To check Condition 1a we also need to find all spanning tree-loops in the graph which are shown in Fig. 13.

Fig. 13

All tree-loops in the graph of Fig. 11

For every edge e, we need to look at set of all the spanning tree-loops that include edge e on a tree,

$$\begin{aligned} \mathcal {H}_{e_1}=\{H_1\}, \quad \mathcal {H}_{e_2}=\{H_2\}, \quad \mathcal {H}_{e_3}=\{H_3\}, \quad \mathcal {H}_{e_4}=\{H_4\}, \quad \mathcal {H}_{e_5}=\emptyset . \end{aligned}$$

(C.5)

The next step is to construct for every possible edge the set of all oriented tree-loop-trees $O(\mathcal {H}^{(e)})$ (but that is not possible for the edge $e_5$). For example, if we look at the edge $e_1$, we need the set $O(H^{(e_1)}_1)$, which consists of two elements depending on the two possible orientations in the loop of $H_1$ see Fig. 14.

Fig. 14

The possible oriented tree-loop-trees created by removing the edge $e_1$ from $H_{1}$

Therefore, $O(H_1^{(e_1)})=\{\overline{H}_{1,1},\overline{H}_{1,2}\}$, where $\phi (\overline{H}_{1,1})=0$ and $\phi (\overline{H}_{1,2})=-2$. We now do the same for the other edges and get,

$$\begin{aligned} O(H_2^{(e_2)})&=\{\overline{H}_{2,1},\overline{H}_{2,2}\}, \quad \phi (\overline{H}_{2,1})=0,\, \phi (\overline{H}_{2,2})=-2\nonumber \\ O(H_3^{(e_3)})&=\{\overline{H}_{3,1},\overline{H}_{3,2}\}, \quad \phi (\overline{H}_{3,1})=-1,\, \phi (\overline{H}_{3.2})=-1\nonumber \\ O(H_4^{(e_4)})&=\{\overline{H}_{4,1},\overline{H}_{4,2}\}, \quad \phi (\overline{H}_{4,1})=-1,\, \phi (\overline{H}_{4,2})=-1. \end{aligned}$$

(C.6)

We see that Condition 1a is satisfied.

To check Condition 1b, we can use the Kirchhoff formula which expresses the stationary distribution in terms of weights of rooted spanning trees; see, e.g., [31]. Here, the state y is the unique dominant state: $x^*=y$, unique maximizer of $\phi (z)$ in (2.19).

Next we give a counterexample to Condition 2, which can be interpreted as a “no-delay condition”; for details of such a physical interpretation, see the earlier discussion in [16] and the recent [23].

Example C.2

Consider the graph in Fig. 15 made by a centered triangle such that each state is symmetrically connected to the state in the center.

Fig. 15

The graph has four loops

A random walker is moving on the graph with transition rates

$$\begin{aligned} k(x,y)&=k(y,z)=k(z,x)= \nu , \quad k(y,x)=k(x,z)=k(z,y)= \nu e^{-\beta \varepsilon }\nonumber \\ k(x,u)&=k(y,u)=k(z,u)=\mu \,e^{-\beta a}, \quad k(u,x)=k(u,y)=k(u,z)=\mu \,e^{-(a+\Delta )\beta } \end{aligned}$$

(C.7)

where $\varepsilon , \, a, \, \Delta \ge 0$. The spanning trees with the largest weight are rooted in u; they start from a state on the triangle and visit the next state on the outer triangle in clockwise direction before going to the center. Therefore, $\phi ^*=-a$. There are 12 spanning tree-loops, shown in Fig. 16.

Fig. 16

All spanning tree-loops in the graph of Fig. 15

Putting

$$\begin{aligned}&e_1:=\{x,y\}, \quad e_2:=\{y,z\}, \quad e_3:=\{z,x\}, \quad e_4:=\{x,u\},\nonumber \\ {}&\quad e_5:=\{y,u\},\quad e_6:=\{z,u\} \end{aligned}$$

(C.8)

we have

$$\begin{aligned} \mathcal {H}_{e_1}&=\{H_3,H_6\}, \quad \mathcal {H}_{e_2}=\{H_4,H_7\}, \quad \mathcal {H}_{e_3}=\{H_1,H_9\}, \nonumber \\ \mathcal {H}_{e_4}&=\{H_2,H_{12}\}, \quad \mathcal {H}_{e_5}=\{H_5,H_{10}\}, \quad \mathcal {H}_{e_6}=\{H_8,H_{11}\}. \end{aligned}$$

(C.9)

The $\phi$’s for all small loops in the clockwise direction are equal to $- (2\,a+\Delta )$ and in counter-clockwise direction equal to $-(2\,a+\Delta +\varepsilon )$. The set $\mathcal {H}^{(e_1)}$ contains two tree-loop-trees made by removing the edge $\{x,y\}$ from $H_3$ and $H_6$. If we give orientations to both, we get four oriented tree-loop-trees for which the $\phi$ value is either equal to $-(2\,a+\Delta )$ or to $-(2\,a+\Delta +\varepsilon )$. That scenario repeats itself for $\mathcal {H}^{(e_2)}$ and $\mathcal {H}^{(e_3)}$. Next, we take an edge connecting the outer triangle to the center and look at the sets $\mathcal {H}^{(e_4)}=\mathcal {H}^{(e_5)}=\mathcal {H}^{(e_6)}$ all containing two elements. Giving orientations to these two tree-loop-trees we get four oriented tree-loop-trees. The $\phi$ of the large loop in clockwise direction is zero and in counter-clockwise direction $-3 \varepsilon$. We see that Condition 1a is satisfied for the edges $e_1, e_2, e_3$ but not for the edges $e_4, e_5, e_6$. In fact, it is easy to convince oneself that V(x), V(y) and V(z) are diverging for $a>0$ while V(u) is uniformly bounded only when $\Delta \, >\, a$. That is interesting because it shows that the conditions of our Theorems are not necessary: for $a<\Delta$ the heat capacity and the excess heat do go to zero at absolute zero. The reason is that the divergence of the quasipotential $V(x)= V(y) = V(z)$ in $\beta \uparrow \infty$ is slower than how their stationary probabilities go to zero. For $a> \Delta$ also V(u) is diverging and the heat capacity as well. That again shows the physical content of the condition 1a: if $a > \Delta$, there is a high barrier between the states x, y, z on the one hand and z on the other hand. Starting say from state x shows much delay in reaching the dominant state u.