# Longitudinal wave propagation in 3D

When a longitudinal wave is sent through a body, there is a strain in the emitted direction (x). What about the three-dimensional body with the strains in the direction perpendicular to the emitting direction (y, z). This strains must inevitably occur due to the "pressure wave". how can these be calculated?

We can write the displacement vector as

$$\mathbf{u} = A \cos(kx)\cos(\omega t)(1, 0, 0)\, .$$

Thus, the strain tensor that is defined as

$$\epsilon = \frac{1}{2}[\nabla \mathbf{u} + (\nabla\mathbf{u})^T]\, ,$$

is

$$\epsilon = -Ak\sin(kx)\cos(\omega t) \begin{bmatrix} 1 &0&0\\ 0 &0 &0\\ 0 &0 &0\end{bmatrix}\, .$$

Furthermore, the stress tensor, that for a linear elastic material isotropic is defined as

$$\sigma = \lambda \mathrm{tr}(\epsilon) + 2\mu\epsilon\, ,$$

looks like

$$\sigma = -Ak\sin(kx)\cos(\omega t)\left[\begin{matrix}\lambda + 2 \mu & 0 & 0\\0 & \lambda & 0\\0 & 0 & \lambda\end{matrix}\right]\, .$$

• That was very helpful! What would the strains eyy and ezz be (or also the strain tensor)? Jul 25 at 19:05
• @Frank, I have updated the answer. Jul 25 at 20:51
• Thank you for the renewed answer, @nicoguaro!!! I don't quite understand the strain tensor yet. The pressure wave of the longitudinal wave would inevitably also have to generate -small- strains in eyy and ezz, since the definition of Hook's law means that the compressive stresses in 3D also influence all vectors, right? Jul 26 at 6:41
• One more question about the stress tensor. Shouldn't it be like that: <br/> \sigma = -Ak\sin(kx)\cos(\omega t)\left[\begin{matrix}\lambda + 2 \mu & 0 & 0\\0 & \mu & 0\\0 & 0 & \mu\end{matrix}\right]\, . Jul 26 at 7:10
• @Frank, I don't think that your intuition is correct. I have added some details and hyperlinks for reference. Jul 26 at 14:31