P wave

A P-wave is longitudinal, carries nonzero dilatation, and has speed $c_P=\sqrt{(\lambda +2\mu )/\rho }$ in an isotropic solid.

Let $t=T(\mathbf{x})$ be a smooth wavefront and suppose the first jump in time derivatives of displacement field is $[{\partial }_t^k\mathbf{u}]=a^{[k]}\mathbf{n}$, with $k\ge 2$. Substitution of $\mathbf{u}=a^{[k]}\mathbf{n}(t-T{)}_{+}^k/k!+\cdots$ into the elastic equation first gives the eikonal equation $|\nabla T|=1/\alpha$ and longitudinal polarization $\mathbf{n}=\alpha \nabla T$. Projecting the next-order compatibility condition onto $\mathbf{n}$ gives $2\rho \alpha \mathbf{n}\cdot \nabla a^{[k]}+a^{[k]}\nabla \cdot (\rho \alpha \mathbf{n})=0$, hence the displayed conservation law. Coefficients and wavefront must be sufficiently differentiable for this classical transport equation; its local form fails at a caustic.

The spherical source potential $f(t-r/\alpha )/r$ gives leading longitudinal displacement field $-f^{\prime }(t-r/\alpha )/(\alpha r)$. Apply the local solid-fluid P-wave transmission coefficient to its potential at a smooth interface. In hyperbolic interface collimation of P-waves, transmitted rays are parallel, so ray-tube conservation for a P-wave jump makes the leading coefficient constant along each vertical ray. This yields the displayed jump for the first nonzero derivative discontinuity of the waveform. Smooth transverse derivatives multiply $f$ rather than $f^{\prime }$ and are one order more regular. The construction presupposes the indicated one-sided derivative jump exists; a source smooth to every order may have no jump.

A tube of neighboring elastic rays follows the wavefront normal. If $J$ is its infinitesimal transverse area, its expansion is $d\log J/ds=\nabla \cdot \mathbf{n}$ along arclength. The P-wavefront discontinuity transport law therefore gives the displayed conserved flux. Geometrical spreading, mass density and wave speed determine the leading jump wave amplitude together; the rule is valid before the ray tube reaches a caustic.

For a stable isotropic solid in $y<0$, use $u_x={\varphi }_x+{\psi }_y$, $u_y={\varphi }_y-{\psi }_x$. An incident P wave with frequency $\omega$, tangential wavenumber $\kappa$ and upward normal wavenumber $a$ reflects into a P potential $A{e}^{i(\kappa x-ay-\omega t)}$ and an SV potential $B{e}^{i(\kappa x-by-\omega t)}$, where $b^2={\omega }^2/c_s^2-{\kappa }^2$. Vanishing of both displacements at the rigid plane gives $\kappa (1+A)-bB=0$, $a(1-A)-\kappa B=0$. Hence $A=(ab-{\kappa }^2)/(ab+{\kappa }^2)$ and $B=2a\kappa /(ab+{\kappa }^2)$. Because $c_p>c_s$ and ${\kappa }^2\le {\omega }^2/c_p^2$, $b^2>0$ and the converted S wave propagates rather than being evanescent.

In a longitudinal plane wave, displacement is parallel to the wavevector and the curl vanishes.