Resistor model
$
\def \J {\Delta}
$
Let us devise for example a tetrahedron, built up from electrical resistors.
Six (6) of these should be associated with the six sides of the tetrahedron.
Let the vertices of the tetrahedron be numbered $0,1,2,3$ and the accompanying
admittances be named accordingly. Assemble the accompanying element-matrices:
$$
\left[ \begin{array}{cccc}
+ A_{01} & - A_{01} & 0 & 0 \\
- A_{01} & + A_{01} & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \end{array} \right] +
\left[ \begin{array}{cccc}
+ A_{02} & 0 & - A_{02} & 0 \\
0 & 0 & 0 & 0 \\
- A_{02} & 0 & + A_{02} & 0 \\
0 & 0 & 0 & 0 \end{array} \right] +
\left[ \begin{array}{cccc}
+ A_{03} & 0 & 0 & - A_{03} \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
- A_{03} & 0 & 0 & - A_{03} \end{array} \right] +
$$ $$
\left[ \begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & + A_{12} & - A_{12} & 0 \\
0 & - A_{12} & + A_{12} & 0 \\
0 & 0 & 0 & 0 \end{array} \right] +
\left[ \begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & + A_{13} & 0 & - A_{13} \\
0 & 0 & 0 & 0 \\
0 & - A_{13} & 0 & + A_{13} \end{array} \right] +
\left[ \begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & + A_{23} & - A_{23} \\
0 & 0 & - A_{23} & + A_{23} \end{array} \right]
$$
Giving:
$$
\left[ \begin{array}{cccc}
A_{01} + A_{02} + A_{03} & - A_{01} & - A_{02} & - A_{03} \\
- A_{01} & A_{01} + A_{12} + A_{13} & - A_{12} & - A_{13} \\
- A_{02} & - A_{12} & A_{02} + A_{12} + A_{23} & - A_{23} \\
- A_{03} & - A_{13} & - A_{23} & A_{03} + A_{13} + A_{23}
\end{array} \right]
$$
It is trivially seen that each row of the matrix sums up to zero.
Let's compare this with the finite element matrix for 3-D diffusion, as has
been found in the preceding paragraph:
$$
\left[ \begin{array}{cccc}
\vec{N}_{123} \cdot \vec{N}_{123} &
\vec{N}_{123} \cdot \vec{N}_{230} &
\vec{N}_{123} \cdot \vec{N}_{301} &
\vec{N}_{123} \cdot \vec{N}_{012} \\
\vec{N}_{230} \cdot \vec{N}_{123} &
\vec{N}_{230} \cdot \vec{N}_{230} &
\vec{N}_{230} \cdot \vec{N}_{301} &
\vec{N}_{230} \cdot \vec{N}_{012} \\
\vec{N}_{301} \cdot \vec{N}_{123} &
\vec{N}_{301} \cdot \vec{N}_{230} &
\vec{N}_{301} \cdot \vec{N}_{301} &
\vec{N}_{301} \cdot \vec{N}_{012} \\
\vec{N}_{012} \cdot \vec{N}_{123} &
\vec{N}_{012} \cdot \vec{N}_{230} &
\vec{N}_{012} \cdot \vec{N}_{301} &
\vec{N}_{012} \cdot \vec{N}_{012}
\end{array}\right] / \J
$$
Now it is possible to identify terms in the diffusion matrix and the resistor
matrix respectively:
$$
\begin{array}{c}
A_{01} = - \vec{N}_{123} \cdot \vec{N}_{230} / \J \\
A_{02} = - \vec{N}_{123} \cdot \vec{N}_{301} / \J \\
A_{03} = - \vec{N}_{123} \cdot \vec{N}_{012} / \J \\
A_{12} = - \vec{N}_{230} \cdot \vec{N}_{301} / \J \\
A_{13} = - \vec{N}_{230} \cdot \vec{N}_{012} / \J \\
A_{23} = - \vec{N}_{301} \cdot \vec{N}_{012} / \J
\end{array}
$$
Thus we can consider the $4 \times 4$ matrix for diffusion at a tetrahedron as
a superposition of one-dimensional resistor-like elements, where any resistor
corresponds with the inner product of the normals of the two triangles which
are not adjacent to the side corresponding with the resistor (divided by
the volume of the tetrahedron).
Resistors are positive if (and only if) the corresponding inner products of the
normals are negative. This means that the accompanying boundary triangle planes
have to make sharp angles which each other.
It is seen that for a (rectangular) Finite Difference grid the normals on the
triangles $(012)$,$(230)$ and $(301)$ will be perpendicular to each other and
therefore the accompanying inner products will be zero. Hence in this special
case: $ A_{12} = A_{13} = A_{23} = 0 $.
When assembling these element matrices into the global system, most resistors
have to be replaced by parallel resistors, one resistor for each side of a
tetrahedron, according to the law of superposition. Exceptions are at the
boundary.