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My question is basically the same as this, but the answer in that page was not clear to me.

Let me restate the question here: let $\Omega\subset\mathbb{R}^3$ be a domain with boundary $\Gamma$, and let $\mathbf{u}$ be a vector field defined on all $\Omega$.

We define the surface stress as

$$ \mathbf{s}_t = \nabla \mathbf{u}\cdot\mathbf{n} - (\mathbf{n}\cdot\nabla \mathbf{u}\cdot\mathbf{n})\mathbf{n} $$

Question: can I compute $\mathbf{s}_t$ using only $\nabla_\mathbf{v}\mathbf{u}$, for one or more suitable choices of $\mathbf{v}$? Here, $\nabla_\mathbf{v} \mathbf{u}$ denotes the covariant derivative of $\mathbf{u}$ along the vector $\mathbf{v}$.

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bartgol
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1 Answers1

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If you mean the covariant derivative of $\mathbb R^3$, then yes, but since the covariant derivative coincides with the Euclidean derivative it's pretty much just rewriting the formula you already have as $$ {\bf s}_t = {\bf \nabla_n u - n\cdot(\nabla_n u)\;n}.$$

If you mean the covariant derivative of $\Gamma$, the answer is no: ${\bf s}_t$ depends explicitly upon the normal derivative while the covariant derivative of $\Gamma$ only measures changes in tangential directions.

If you define the "$\Gamma$-tangential derivative" of $u$ by $\nabla^\Gamma_v u = v\cdot\nabla u - ({\bf n} \cdot v\cdot \nabla u){\bf n}$ then it's true that ${\bf s}_t = \nabla_{\bf n} \bf u$ and it's also true that $\nabla^\Gamma$ agrees with the covariant derivative where the latter is defined. The issue is that $\bf n$ is not tangent to $\Gamma$, so it's not correct to call $\nabla^\Gamma_\bf nu$ a covariant derivative.

Anthony Carapetis
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