Question 5.1.2: An isotropic elastic material occupies the half-space z >......

In a general case, both longitudinal (wave 2) and transverse (wave 3) reflected waves will be present. Let us denote their displacement amplitudes as \vec{u}_{2} and \vec{u}_{3}, when \vec{u}_{1} is the amplitude of the incident wave (wave 1) (the directions of these vectors is shown in Figure 5.2). For a given \vec{u}_{1}, the vectors \vec{u}_{2} and \vec{u}_{3} have to be found from the boundary condition at the plane z = 0. In the present case of an absolutely rigid attachment it means zero total displacement there. Thus, one comes to the following relations:

u_{z} = u_{1} \cos θ_{l} − u_{2} \cos θ_{l} + u_{3} \sin θ_{t} = 0,

u_{x} = u_{1} \sin θ_{l} + u_{2} \sin θ_{l} + u_{3} \cos θ_{t} = 0,

which yield u_{2} = u_{1} \cos(θ_{l} + θ_{t})/  \cos(θ_{l} − θ_{t}) and, hence, the reflection coefficient is equal to

R_{l} = (u_{2}/u_{1})^{2} = \cos^{2}(θ_{l} + θ_{t})/  \cos^{2}(θ_{l} − θ_{t})        (5.12)

The angle, θ_{t}, of the reflected transverse wave can be expressed in terms of the angle θ_{l}, the longidutinal sound speed, c_{l}, and the transverse sound speed, c_{t}, in the following way. Since all three waves have the same frequency and the same k_{x} (x-component of their wave vectors \vec{k}_{l,t}), one gets that c_{l}k_{l} = c_{t}k_{t} and k_{l} \sin θ_{l} = k_{t} \sin θ_{t}. These yield \sin θ_{t} = (c_{t}/c_{l}) \sin θ_{l}, which together with expression (5.12) completes the answer. Finally, one may notice from (5.12) that the longitudinal reflection coefficient, R_{l}, becomes equal to zero when θ_{l} + θ_{t} = π/2, which has a simple geometrical explanation. Indeed, in this case the vectors \vec{u}_{1} and \vec{u}_{3} are collinear, and, thus, the required boundary condition \vec{u}_{z=0} = 0 is satisfied with \vec{u}_{3} = −\vec{u}_{1} and \vec{u}_{2} = 0.