To calculate the strain (denoted as $e$) and rotation (denoted as $\omega$) tensors for a point on the fault, we need to understand the relationship between displacement and velocity changes along the $x$- and $y$-axes.

Given:

• Change in displacement along the x-axis, $\Delta x = 70 \, \text{km} = 70,000 \, \text{m}$
• Change in velocity along the y-axis, $\Delta v_y = 75 \, \text{mm/year} = 0.075 \, \text{m/year}$

The strain tensor $e$ and rotation tensor $\omega$ are defined in terms of the displacement gradients (strain) and velocity gradients (rotation).

1. Strain Tensor $e$:

Strain is related to the change in displacement per unit length. The strain tensor in 2-D is given by:

\epsilon_{xx} & \epsilon_{xy} \\ \epsilon_{yx} & \epsilon_{yy} \end{pmatrix}$$ Where: - $$\epsilon_{xx}$$ is the normal strain in the x-direction. - $$\epsilon_{yy}$$ is the normal strain in the y-direction. - $$\epsilon_{xy} = \epsilon_{yx}$$ is the shear strain. Now, assuming these displacements are constant, the strain components can be calculated as: - $$\epsilon_{xx} = \frac{\Delta x}{L_x}$$, where $$L_x$$ is the length over which the displacement occurs. - $$\epsilon_{yy} = \frac{\Delta v_y}{L_y}$$, where $$L_y$$ is the length over which the velocity change occurs. - The shear strain $$\epsilon_{xy}$$ will generally depend on the rate of change of displacement across both axes. Since we don’t have specific lengths over which the changes happen, I will assume $$L_x = 1 \, \text{m}$$ and $$L_y = 1 \, \text{m}$$ for simplicity, yielding: $$\epsilon_{xx} = \frac{70,000}{1} = 70,000 \, \text{m/m}$$ $$\epsilon_{yy} = \frac{0.075}{1} = 0.075 \, \text{m/m}$$ $$\epsilon_{xy} = \epsilon_{yx} = 0 \, \text{m/m} \quad (\text{since there is no specified shear displacement})$$ Thus, the strain tensor $$e$$ is: $$e = \begin{pmatrix} 70,000 & 0 \\ 0 & 0.075 \end{pmatrix}$$ ### 2. **Rotation Tensor $$\omega$$:** The rotation tensor is related to the change in the direction of the displacement and is expressed in terms of the velocity gradient. The rotation tensor $$\omega$$ is given by: $$\omega = \frac{1}{2} \begin{pmatrix} \omega_{xx} & \omega_{xy} \\ \omega_{yx} & \omega_{yy} \end{pmatrix}$$ Where: - $$\omega_{xy} = -\omega_{yx}$$ describes the shear rotation between the axes. Since there is no specified angular velocity or rotational component, we assume $$\omega_{xx} = \omega_{yy} = 0$$, meaning no pure rotation is considered along the axes. For simplicity, the rotation tensor $$\omega$$ will have only off-diagonal components: $$\omega = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$ ### Final Answer: - **Strain Tensor $$e$$:** $$e = \begin{pmatrix} 70,000 & 0 \\ 0 & 0.075 \end{pmatrix}$$ - **Rotation Tensor $$\omega$$:** $$\omega = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$ --- Would you like further clarification on any of the steps or concepts? Here are some related questions you might find helpful: 1. What is the physical interpretation of the strain tensor components? 2. How can we calculate shear strain if there was displacement in both directions? 3. What are the units of strain and rotation in this context? 4. How does the rotation tensor relate to angular velocity in geophysical problems? 5. What assumptions did I make in this calculation, and how would changing them affect the results? **Tip:** When calculating strain or rotation in geophysics or material science, it’s important to ensure the consistency of units across all quantities involved.