Auxiliary Plane Calculation

1. The Nodal Plane Ambiguity

Every earthquake focal mechanism has two nodal planes that produce the exact same beachball pattern. One is the actual fault plane where slip occurred; the other is called the auxiliary plane. Seismological data alone (P-wave first motions, waveform modeling) cannot distinguish which is which — additional geological evidence (aftershock distribution, surface rupture, mapped faults) is needed.

Given Nodal Plane 1 defined by strike ($\varphi_1$), dip ($\delta_1$), and rake ($\lambda_1$), we can compute Nodal Plane 2 using a fundamental geometric relationship:

Key relationship: The slip vector of Plane 1 is the normal vector of Plane 2, and the normal vector of Plane 1 is the slip vector of Plane 2. The two nodal planes are mutually perpendicular, and both produce an identical seismic moment tensor.

2. Coordinate Convention

We use the Aki & Richards (2002) convention with axes oriented as (North, East, Down):

$x_1$ = North
$x_2$ = East
$x_3$ = Down (into the Earth)

Fault plane parameters:

Strike ($\varphi$): Azimuth (0–360°) of the fault trace, measured clockwise from North, with the fault dipping to the right when looking along the strike direction.
Dip ($\delta$): Angle (0–90°) of the fault plane from horizontal, measured in the down-dip direction.
Rake ($\lambda$): Angle (-180–180°) of the slip vector within the fault plane, measured from the strike direction. Positive = thrust component, negative = normal component.

3. Step-by-Step Calculation

Adjust the sliders to see all intermediate values update in real time:

Strike ($\varphi_1$)

30°

Dip ($\delta_1$)

45°

Rake ($\lambda_1$)

90°

Normal Vector of Plane 1

The outward normal to the fault plane, pointing into the upper hemisphere (by convention $n_3 \leq 0$ in NED, meaning the normal points upward):

$\hat{n} = \begin{pmatrix} -\sin\delta\;\sin\varphi \\ \sin\delta\;\cos\varphi \\ -\cos\delta \end{pmatrix}_{\text{NED}}$

$n_N$ (North)

$n_E$ (East)

$n_D$ (Down)

Slip Vector of Plane 1

The direction of hanging wall motion relative to the footwall, lying within the fault plane:

$\hat{d} = \begin{pmatrix} \cos\lambda\;\cos\varphi + \cos\delta\;\sin\lambda\;\sin\varphi \\ \cos\lambda\;\sin\varphi - \cos\delta\;\sin\lambda\;\cos\varphi \\ -\sin\delta\;\sin\lambda \end{pmatrix}_{\text{NED}}$

$d_N$ (North)

$d_E$ (East)

$d_D$ (Down)

✓ Perpendicularity: $\hat{n} \cdot \hat{d}$ = 0.000 (must be 0)

Normal of Plane 2 = Slip Vector of Plane 1

The fundamental relationship: the slip vector $\hat{d}$ of Plane 1 is the normal vector of Plane 2. We enforce the convention that the normal points upward ($n_3^{(2)} \leq 0$ in NED):

$\hat{n}^{(2)} = \begin{cases} +\hat{d} & \text{if } d_D \leq 0 \\ -\hat{d} & \text{if } d_D > 0 \end{cases}$

$n_N^{(2)}$

$n_E^{(2)}$

$n_D^{(2)}$

Extract Strike and Dip of Plane 2

From the normal vector components, using the inverse of the formulas in Step 1:

$\delta_2 = \arccos(-n_D^{(2)})$

$\varphi_2 = \text{atan2}\!\left(\dfrac{-n_N^{(2)}}{\sin\delta_2},\;\dfrac{n_E^{(2)}}{\sin\delta_2}\right)$

$\delta_2$ (Dip)

0°

$\varphi_2$ (Strike)

0°

Slip Vector of Plane 2 = Normal of Plane 1

Symmetrically, the normal $\hat{n}$ of Plane 1 becomes the slip vector of Plane 2. We orient it consistently with the flipping done in Step 3:

$\hat{d}^{(2)} = \begin{cases} +\hat{n} & \text{if } d_D \leq 0 \\ -\hat{n} & \text{if } d_D > 0 \end{cases}$

$d_N^{(2)}$

$d_E^{(2)}$

$d_D^{(2)}$

Extract Rake of Plane 2

Project the slip vector onto the strike direction and use the vertical component:

Along-strike component: $A = d_N^{(2)}\cos\varphi_2 + d_E^{(2)}\sin\varphi_2$

$\lambda_2 = \text{atan2}\!\left(\dfrac{-d_D^{(2)}}{\sin\delta_2},\;A\right)$

Along-strike $A$

$\lambda_2$ (Rake)

0°

Final Result

Nodal Plane 1 (Input)

Strike: 30° | Dip: 45° | Rake: 90°

Nodal Plane 2 (Calculated)

Strike: — | Dip: — | Rake: —

4. Verification

Both nodal planes must produce the identical seismic moment tensor. This is the definitive test that the auxiliary plane was computed correctly.

Moment Tensor Comparison

The moment tensor $M_{ij}$ in spherical coordinates $(r, \theta, \phi)$ from Aki & Richards:

From Plane 1

From Plane 2

✓ Moment tensors match: max |difference| = 0.000

Perpendicularity Check

The two nodal planes must be perpendicular, meaning their normal vectors are orthogonal:

✓ $\hat{n}^{(1)} \cdot \hat{n}^{(2)}$ = 0.000 (must be 0)

5. Interactive 3D Visualization

Both nodal planes on the focal sphere, with their normal and slip vectors. Drag to rotate.

Legend

Plane 1 great circle

Plane 2 great circle

Normal vectors ($\hat{n}$)

Slip vectors ($\hat{d}$)

Notice how the normal of one plane is parallel to the slip vector of the other — this is the fundamental relationship.

The sphere shows the lower hemisphere (standard focal mechanism projection).

References

Aki, K. & Richards, P.G. (2002). Quantitative Seismology, 2nd edition. University Science Books.
Cronin, V. (2010). A Primer on Focal Mechanism Solutions for Geologists. Baylor University.
Shearer, P.M. (2019). Introduction to Seismology, 3rd edition. Cambridge University Press.