Micromechanics of composites/Proof 6

Curl of the gradient of a vector - 1

Let $𝐯$ be a vector field. Show that

$\mathrm{\nabla }\times (\mathrm{\nabla }𝐯)=0.$

Proof:

For a second order tensor field $𝑺$, we can define the curl as

$(\mathrm{\nabla }\times 𝑺)\cdot 𝐚=\mathrm{\nabla }\times ({𝑺}^T\cdot 𝐚)$

where $𝐚$ is an arbitrary constant vector. Substituting $\mathrm{\nabla }𝐯$ into the definition, we have

$[\mathrm{\nabla }\times (\mathrm{\nabla }𝐯)]\cdot 𝐚=\mathrm{\nabla }\times (\mathrm{\nabla }{𝐯}^T\cdot 𝐚).$

Since $𝐚$ is constant, we may write

$\mathrm{\nabla }{𝐯}^T\cdot 𝐚=\mathrm{\nabla }(𝐯\cdot 𝐚)=\mathrm{\nabla }\phi$

where $\phi =𝐯\cdot 𝐚$ is a scalar. Hence,

$[\mathrm{\nabla }\times (\mathrm{\nabla }𝐯)]\cdot 𝐚=\mathrm{\nabla }\times (\mathrm{\nabla }\mathit{\phi }\mathit{)}.$

Since the curl of the gradient of a scalar field is zero (recall potential theory), we have

$\mathrm{\nabla }\times (\mathrm{\nabla }\mathit{\phi }\mathit{)}=\mathrm{𝟎}.$

Hence,

$[\mathrm{\nabla }\times (\mathrm{\nabla }𝐯)]\cdot 𝐚=\mathrm{𝟎}\mathrm{\forall }𝐚.$

The arbitrary nature of $𝐚$ gives us

$\mathrm{\nabla }\times (\mathrm{\nabla }𝐯)=\mathrm{𝟎}◻$

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