7.2. Helmholtz decomposition and sum of waves equation for \(p\) and \(s\) waves
Displacement field can be written using Helmholtz decomposition as the sum of a potential \(\mathbf{s}_p = \nabla \phi\) (s.t. \(\nabla \times \nabla \phi = \mathbf{0}\)) and a divergence-free \(\mathbf{s}_s = \nabla \times \mathbf{a}\) (s.t. \(\nabla \cdot \nabla \times \mathbf{a} = \mathbf{0}\)) part,
\[\mathbf{s} = \mathbf{s}_p + \mathbf{s}_s = \nabla \phi + \nabla \times \mathbf{a} \ .\]
Introducing the last expression in the momentum equation, using vector identity
\[\nabla^2 \mathbf{v} = \nabla \nabla \cdot \mathbf{v} - \nabla \times \nabla \times \mathbf{v} \ , \]
the equation can be written as
\[\begin{split}\begin{aligned}
\mathbf{0}
& = \rho_0 \partial_{tt} \nabla \phi - (2 \mu + \lambda) \nabla^2 \nabla \phi + \rho_0 \partial_{tt} \nabla \times \mathbf{a} + \mu \nabla^2 \nabla \times \mathbf{a} = \\
& = \rho_0 \partial_{tt} \mathbf{s}_p - (2 \mu + \lambda) \nabla^2 \mathbf{s}_p + \rho_0 \partial_{tt} \mathbf{s}_s - \mu \nabla^2 \mathbf{s}_s \ ,
\end{aligned}\end{split}\]
i.e. as the “sum of two wave equations” for the potential part \(\mathbf{s}_p\) and the divergence-free part \(\mathbf{s}_s\) of the displacement. Speed of propagation of \(p\)- and \(s\)-displacement read
\[c_p = \sqrt{\dfrac{2 \mu + \lambda}{\rho_0}} \qquad , \qquad c_s = \sqrt{\dfrac{\mu}{\rho_0}} \ .\]
7.3. Fourier decomposition: \(p\) is longitudinal, \(s\) is transverse
Using Fourier decomposition of fields as sum of harmonic plane waves,
\[\mathbf{s}(\mathbf{r},t) = \sum_{\mathbf{k},\omega} \mathbf{s}_{\mathbf{k},\omega} e^{i\left( \mathbf{k} \cdot \mathbf{r}- \omega t \right)} \ ,\]
it’s immediate to prove that the potential part can be associated to a longitudinal perturbation (i.e. with displacement in the same direction of the wave vector \(\mathbf{k}\), representing the direction of propagation of the perturbation, while the divergence-free part can be associated to a transverse perturnation (i.e. with displacement orthogonal to the wave vector \(\mathbf{k}\)). Helmholtz’s decomposition of the field in Fourier domain reads
\[\begin{split}\begin{aligned}
\mathbf{s}(\mathbf{r},t)
& = \sum_{\mathbf{k},\omega} \mathbf{s}_{\mathbf{k},\omega} e^{i\left( \mathbf{k} \cdot \mathbf{r}- \omega t \right)} = \\
& = \sum_{\mathbf{k},\omega} \left( \mathbf{s}^p_{\mathbf{k},\omega} + \mathbf{s}^s_{\mathbf{k}, \omega} \right) e^{i\left( \mathbf{k} \cdot \mathbf{r}- \omega t \right)} = \\
& = \sum_{\mathbf{k},\omega} \left( i \, \mathbf{k} \phi_{\mathbf{k},\omega} + i \, \mathbf{k} \times \mathbf{a}_{\mathbf{k}, \omega} \right) e^{i\left( \mathbf{k} \cdot \mathbf{r}- \omega t \right)} = \ .
\end{aligned}\end{split}\]
For each individual harmonic contribution, the potential part is thus proportional, i.e. aligned, to wave vector \(\mathbf{k}\),
\[\mathbf{s}^p_{\mathbf{k},\omega} = i \,\mathbf{k} \phi_{\mathbf{k},\omega} \ ,\]
while the divergence-free is orthogonal, and thus transverse, w.r.t. the direction of wave propagation,
\[\mathbf{k} \cdot \mathbf{s}^s_{\mathbf{k},\omega} = i \, \mathbf{k} \times \left( \mathbf{k} \times \mathbf{a}_{\mathbf{k}, \omega} \right) = \mathbf{0} \ .\]
