Volume integrals of vector quantities

A tensor divergence analogy of the above trick can easily be found. Using Einstein summation convention, one can write tensors on the LHS whose divergence gives vectors on the RHS,

$$\partial_i \left( \begin{array}{c} \delta_{ij} a^* b \\ r_j ... ...\partial_m b ) \partial_m \partial_j a^* \end{array} \right) .$$ (H.15)

Notice that no dependence on $d$ has entered. The relation can be written

$$\partial_i t_{ij}^{(\alpha)} \; = \; {\mathcal{T}}_{\alpha \beta} \, w_j^{(\beta)}.$$ (H.16)

Remember that Roman indices $i,j,m = 1\cdots d$ are spatial, whereas Greek indices $\alpha,\beta = 1\cdots6$ label coefficients. Note that the vector of vectors ${\mathbf w}^{(\beta)}$ does contain some different functions than ${\mathbf v}_\beta$ from Section H.2.1. The matrix determinant is det ${\mathcal{T}} = \varepsilon ^2$, the same as for $\mathcal{M}$. The tensor Gauss' theorem is

$$\int_{\mathcal{D}} \!\! d{\mathbf r} \, \partial_i A_{ij} \; = \; \oint_\Gamma \!\! d{\mathbf s} \, n_i A_{ij} .$$ (H.17)

Applying this to the domain integral of (H.16) gives

$$\int_{\mathcal{D}} \!\! d{\mathbf r} \, w_j^{(\alpha)} \; =... ...ta} \oint_\Gamma \!\! d{\mathbf s} \,n_i \, t_{ij}^{(\beta)} ,$$ (H.18)

which are the desired closed-form expressions for volume integrals of (the $j$ component of) the vector functions ${\mathbf w}^{(\alpha)}$. The symbolic inverse is found to be,

$${\mathcal{T}}^{-1} \; \; = \; \; \frac{1}{\varepsilon } \lef... ...k_a^2 k_b^2 &0&0& -k_a^2 & -k_b^2 &k_a^2 \end{array} \right) .$$ (H.19)

Again, singular solutions can still be found when $k_a = k_b$. The full consequences of these types of matrix relations have not been explored.