Left lateral, right lateral, normal, reverse

Assume we are given a reference point r and a normal n pointing from the “+”-side to the “-“-side of a fault segment. How does one describe the motion of the fault?

Left-lateral and right-lateral

“In a left-lateral (right-lateral) fault, an observer on one of the walls will see the other wall moving to the left (right).” [J. Pujol, Elastic Wave Propagation and Generation in Seismology]

Assume we stand on the “-“-side and look towards the “+”-side, then if the strike vector s points to left, we have a left-lateral motion (for a positive slip-rate). We formalize “points to the left” with:


where u is the unit vector which points up (e.g. (0,0,1) for enu or (0,0,-1) for ned).

In SeisSol, the strike vector is (not normalized)

\(s:=(-e_3)\times n\)

So, e.g., for enu we always have a left-lateral motion, as s and l are parallel, and for ned we always have a right-lateral motion as s and l are anti-parallel.

Normal and reverse

“The foot wall (hanging wall) is defined as the block below (above) the fault plane. (…) the hanging wall moves up with respect to the foot wall and the fault is known as reverse. (…) the opposite happens and the fault is said to be normal.” [J. Pujol, Elastic Wave Propagation and Generation in Seismology]

In SeisSol, the dip vector is (not normalized)

\(d:=n\times s=n\times(-e_3\times n)=-e_3+n_zn\)

We used Grassmann’s Identity for the last step. In particular, we observe that the dip vector d is independent of the reference point, as we obtain the same vector for -n and, as n is normalized and


the dip vector always points in -z direction. That is, the “+”-side moves down for enu and the “+”-side moves up for ned (assuming positive slip rate).

Normal or reverse depends also on the reference point. If the reference point is inside the hanging wall (foot wall) the “+”-side corresponds to the hanging wall (foot wall).

In summary, we obtain the following table

foot wall= + hanging wall = +
z = up reverse normal
z = down normal reverse

or logically

\(\text{isNormal}:=(+=\text{hanging wall})\leftrightarrow(z=\text{up})\)


We have a 60° dipping normal fault with 90° strike and zero rake with enu convention. (That is, the strike is east.) The normal of the fault plane, which points from the foot wall to the hanging wall, is given by

\(N:=\frac{1}{2}\begin{pmatrix}1 & 0 & \sqrt{3}\end{pmatrix}\)

Hence, we set the reference point to x + a N, where a > 0 and x is an arbitrary point on the fault. In this case, the reference point is inside the hanging wall and we obtain a normal fault.