Past exam of the physics of the Earth as a Planet course of the University of Cambridge / 2016 / Paper 1 / 3 / a

For uniform density $\rho >0$ and stiffness, linear momentum balance without body force is $\rho {\ddot{u}}_i={\partial }_j{\sigma }_{ij}$. The minor elastic symmetries make the two symmetric-strain terms equal, so ${\partial }_j{\sigma }_{ij}=c_{ijkl}{u}_{k,lj}$. Write $s=ct-{\hat{p}}_j{x}_j$ and $u_i=a_{i}f(s)$. Then ${\ddot{u}}_i=c^2{a}_i{f}^{\prime \prime }$ and $u_{k,lj}=a_k{\hat{p}}_l{\hat{p}}_j{f}^{\prime \prime }$. Thus a nontrivial plane-wave profile satisfiesThe density-normalized elastic acoustic tensor is consequentlyThe alternative dummy-index form $B_{ik}={\rho }^{-1}{c}_{ijkl}{\hat{p}}_j{\hat{p}}_l$ is identical. In ordinary conservative elasticity, major symmetry $c_{ikjl}=c_{jlik}$ gives $B_{ji}=B_{ij}$ after interchanging the dummy indices. This symmetry comes from a strain-energy density; it is an assumption of the ordinary elastic tensor, not a consequence of homogeneity alone.

For any such eigenpair the profile $f$ is arbitrary, since both sides carry the same $f^{\prime \prime }$. This nondispersive result uses linear, local, time-independent elasticity without attenuation or spatial dispersion. A profile with $f^{\prime \prime }=0$ is a trivial uniform or affine displacement field and cannot be used by itself to infer the wave-speed eigenvalue condition.