 Solution

The equilibrium quantities $\rho ,p,\mathbf{B},\Phi$ are real, and the gravitational Hessian ${\Phi }_{,ij}={\partial }_i{\partial }_j\Phi$ is symmetric. The crucial tensor property isThis follows directly from the displayed expression for $V$: the products ${\delta }_{ij}{\delta }_{kl}$ and ${\delta }_{il}{\delta }_{jk}$ are unchanged by interchanging the pairs $(i,j)$ and $(k,l)$; $B_j{B}_l{\delta }_{ik}$ is also unchanged; and the two terms involving $B_i{B}_j{\delta }_{kl}$ and $B_k{B}_l{\delta }_{ij}$ exchange with each other. All coefficients are real.

Multiply $F\mathbf{\xi }$ by $\rho {\mathbf{\eta }}^{\ast }$ and integrate by parts over all space. For admissible displacement fields, the boundary contribution at infinity vanishes: the fluid pressure and density are confined to the body, and the magnetic stress decays in the surrounding conducting medium. The conducting exterior is included in the integration. For a piecewise description, the usual displacement and traction matching conditions cancel the two opposite interface contributions; alternatively, take a smooth transition and then its limit. ThusThe first term is a Hermitian pairing because ${\Phi }_{,ij}={\Phi }_{,ji}$, and the second is Hermitian because $V_{ijkl}=V_{klij}$. Taking the complex conjugate of the same expression with $\mathbf{\eta }$ and $\mathbf{\xi }$ interchanged and relabelling $(i,j)\leftrightarrow (k,l)$ reproduces it. HenceThis establishes the required self-adjoint operator identity in the mass-weighted inner product, on the physical displacement domain with the stated decay and interface conditions. The negligible-density exterior can be treated as a small-positive-density conducting limit when defining this weighted pairing; its magnetic contribution is retained even as its inertia becomes negligible.