 Solution

Under a Cartesian frame change $B_i^{\prime }=R_{ik}{B}_k$, where $R$ is an orthogonal matrix. Orthogonality gives $B_k^{\prime }{B}_k^{\prime }=B_k{B}_k$ and $R_{ik}{R}_{j\ell }{\delta }_{k\ell }={\delta }_{ij}$. Thereforewhich is the tensor component transformation law. Also $T_{ij}=T_{ji}$ directly, proving that it is a symmetric Cartesian second-rank tensor for every choice of the two scalars.

For $\mathbf{B}\ne 0$, the direction of $\mathbf{B}$ is an eigenvector with eigenvalue $(\alpha +\beta )|\mathbf{B}{|}^2$. Every vector perpendicular to $\mathbf{B}$ is an eigenvector with eigenvalue $\beta |\mathbf{B}{|}^2$, a two-dimensional eigenspace. If $\alpha =0$, these eigenvalues coincide and every nonzero vector is an eigenvector. If $\mathbf{B}=0$, the tensor is zero and again every nonzero vector is an eigenvector, now with eigenvalue zero.

For the position-dependent field with zero divergence, the product rule givesContraction of the two Levi-Civita symbols in the curl and cross product givesConsequentlyThese constants give the magnetic part of the Maxwell stress tensor, with its physical prefactor suppressed. They are forced if the identity is to hold for every divergence-free field: $\mathbf{B}=(y,0,0)$ forces $2\beta =-1$; at the origin $\mathbf{B}=(y,1,0)$ then forces $\alpha =1$.

Apply the divergence theorem separately to each component:Since the whole vector field vanishes on $S$, all components of $T$ vanish there, and hence the vector volume integral is zero. The argument assumes a differentiable field and a boundary for which the divergence theorem applies.