= Symmetric gradient and curl decomposition
{title2=$\partial_iF_j=\frac12(\partial_iF_j+\partial_jF_i)+\frac12\epsilon_{ijk}(\nabla\times F)_k$}
For a differentiable three-dimensional <vector field> $F$, split its gradient <tensor> into symmetric and antisymmetric parts. The symmetric part is $S_{ij}=(\partial_iF_j+\partial_jF_i)/2$. Contracting the antisymmetric part with the <Levi-Civita symbol> gives the <axial vector> $u_k=\epsilon_{kij}\partial_iF_j/2=(\nabla\times F)_k/2$, proving the decomposition. If $\nabla\cdot F=0$, the divergence of the symmetric part is $\partial_jS_{ij}=\Delta F_i/2$. This lets a vector <Laplacian> be converted into a boundary integral of a symmetric <tensor> by the <divergence theorem>.
Back to article page
