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Orient each edge both ways. A unit flow $j$ from $x$ to $y$ is antisymmetric, $j(u,v)=-j(v,u)$, with divergence of a flow equal to $+1$ at $x$, $-1$ at $y$ and zero elsewhere. Its energy iswhere each unoriented edge is counted once. Put $c_e=1/r_e$. The electrical unit current is $i(u,v)=c_{uv}(V(u)-V(v))$, where the potential $V$ solves the Kirchhoff node law with those source and sink divergences. Because $G$ is a connected graph, this potential is unique up to an additive constant. The effective resistance is the electric potential difference needed for unit current,Discrete summation by parts gives ${\mathcal{E}}_r(i)={\sum }_{u}V(u)\mathrm{div}i(u)=R(x,y)$.

The Thomson principle says that this current minimizes energy among all unit flows. Indeed write any other unit flow as $j=i+k$, where $\mathrm{div}k=0$. ThenConsequentlyThis also proves uniqueness of the minimizing current.

If $r_e^{\prime }\ge r_e$ for every edge, then ${\mathcal{E}}_{r^{\prime }}(j)\ge {\mathcal{E}}_r(j)$ for every admissible flow. The admissible set is unchanged, so taking its minimum proves the Rayleigh monotonicity principle:Deletion is the limiting operation $r_e\to \infty$: finite-energy flows must then carry zero current on that edge. If the terminals become disconnected, the effective resistance is infinite.

Let $e=\{a,b\}$ be an edge of $G$. Apply the preceding result with source $a$ and sink $b$. A tree contains $e$ exactly when its unique path from $a$ to $b$ is the single edge $e$, in the positive direction. Consequently the unit electrical current on $e$ is $\mathbb{P}(e\in T)$. By Ohm's law and the definition of effective resistance,since this edge has unit resistance. This is the unit-conductance case of the edge-inclusion formula for a uniform spanning tree.

The general principle needed is Rayleigh monotonicity principle: adding edges, or decreasing their resistances, cannot increase the effective resistance between fixed terminals. One precise variational formulation is the Thomson principle:with one orientation chosen for each unoriented edge. To see this, write any unit flow as $i+k$, where $i$ is the electrical unit current and $k$ has zero divergence. Since $r_e{i}_e$ is a potential difference, summation by parts gives ${\sum }_e{r}_e{i}_e{k}_e=0$. Thus its energy is ${\sum }_e{r}_e{i}_e^2+{\sum }_e{r}_e{k}_e^2$, minimized by $i$. The energy of $i$ equals its terminal voltage drop, hence the effective resistance.

Extend any unit flow on $G$ to $G^{\prime }$ by zero on all new edges. It is still a unit flow, including at new graph vertices, and its energy is unchanged. The variational minimum on $G^{\prime }$ is therefore no larger. Using the edge-inclusion formula in both graphs givesThe graphs are understood to be finite and connected, as required for their uniform spanning-tree laws. The comparison allows added graph vertices as well as added edges.

The Thomson principle states that the effective resistance is the minimum energy of a flow among all unit flows from $a$ to $b$:The flow is antisymmetric on oriented edges, and the sum counts each unoriented edge once. The Rayleigh monotonicity principle states that increasing edge resistances, including deleting edges by setting their resistance to infinity, cannot decrease the effective resistance. Decreasing resistances or identifying vertices cannot increase it.

Take a shortest path in a graph from $a$ to $b$, of length $d=d_G(a,b)$, and send one unit of flow along it, with zero flow on all other edges. Its energy of a flow is $d$ because each edge resistance is one. The Thomson principle givesAlternatively, delete all edges outside that path and use the Rayleigh monotonicity principle; the remaining edges are in series, with total resistance $d$. When $a=b$, both quantities are zero.

Let $d=d_G(a,b)>0$, and for $i=1,\dots ,d$ putEach edge cutset ${\Pi }_i$ separates $a$ from $b$. Adjacent vertices have graph distances from $a$ differing by at most one. Hence an edge can cross at most one of these level boundaries, so the edge cutsets are pairwise edge-disjoint. Write $k_i=|{\Pi }_i|\ge 1$; then ${\sum }_i{k}_i\le |E|$.

The Nash-Williams inequality for unit conductances giveswhere the middle inequality is the Cauchy-Schwarz inequality. One can also derive the first inequality directly: every unit flow has signed net flux one through each ${\Pi }_i$, so its energy of a flow on that cut is at least $1/k_i$; sum over the disjoint cuts and use the Thomson principle.

Combining with the commute time identity yieldsThe case $a=b$ is trivial. A path graph, with $a,b$ its endpoints, attains equality.

Thomson principle states that the effective resistance is the minimum energy of a unit flow from $a$ to $z$:An edge cutset separating $a$ and $z$ is a set of edges whose removal disconnects them. Every unit flow has net flux one across each such cutset ${\Pi }_k$. By Cauchy-Schwarz inequality,For disjoint cutsets, summing these energy lower bounds and applying Thomson's principle gives the Nash-Williams inequalityThe commute time identity is

Increasing edge resistances, including deleting edges, cannot decrease effective resistance. Decreasing resistances or identifying vertices cannot increase effective resistance. These statements also follow from the Thomson principle.

Thomson principle says that effective resistance is the minimum energy among unit flows from $a$ to $z$: