= Euclidean scalar-gradient stress identity
{c}
{title2=$\partial_k\Sigma_{kj}=(\Delta\psi)\partial_j\psi$}
For a sufficiently smooth real <scalar field> on <Euclidean space>, set $\Sigma_{kj}=\partial_k\psi\partial_j\psi-\delta_{kj}|\nabla\psi|^2/2$. The <product rule> and equality of mixed <partial derivatives> give
$$
\partial_k\Sigma_{kj}=(\Delta\psi)\partial_j\psi+\partial_k\psi\partial_k\partial_j\psi-\partial_k\psi\partial_j\partial_k\psi=(\Delta\psi)\partial_j\psi.
$$
The <divergence theorem>, applied twice, consequently gives the useful pure-boundary identity
$$
\int_V\mathbf r\,\nabla\cdot[(\Delta\psi)\nabla\psi]\,dV=\int_{\partial V}\left[\mathbf r(\Delta\psi)\partial_n\psi-(\partial_n\psi)\nabla\psi+\tfrac12|\nabla\psi|^2\hat{\mathbf n}\right]dS.
$$
Here $\partial_n\psi=\hat{\mathbf n}\cdot\nabla\psi$ uses the outward <unit normal>. This is the Euclidean zero-potential counterpart of the <scalar stress-energy divergence identity>.
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