Let
be a vector field. Show that
![{\displaystyle {\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {v} )=0~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4fdb30972301aef3923213cdbcd3a8fb75927f61)
Proof:
For a second order tensor field
, we can define the curl as
![{\displaystyle ({\boldsymbol {\nabla }}\times {\boldsymbol {S}})\cdot \mathbf {a} ={\boldsymbol {\nabla }}\times ({\boldsymbol {S}}^{T}\cdot \mathbf {a} )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e97aae219d52220e7f83d084249851b1bc628b7e)
where
is an arbitrary constant vector. Substituting
into
the definition, we have
![{\displaystyle [{\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {v} )]\cdot \mathbf {a} ={\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {v} ^{T}\cdot \mathbf {a} )~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dff3f2f359c9d16745995076e648bba3ceaa69d1)
Since
is constant, we may write
![{\displaystyle {\boldsymbol {\nabla }}\mathbf {v} ^{T}\cdot \mathbf {a} ={\boldsymbol {\nabla }}(\mathbf {v} \cdot \mathbf {a} )={\boldsymbol {\nabla }}\varphi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ea17a0084f6c0c88792023af7890e0e76a1630c)
where
is a scalar. Hence,
![{\displaystyle [{\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {v} )]\cdot \mathbf {a} ={\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla \varphi )}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dccad809072128b9fd5823a3460cee25c2afd17d)
Since the curl of the gradient of a scalar field is zero (recall potential
theory), we have
![{\displaystyle {\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla \varphi )}}=\mathbf {0} ~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8763a6a73ff312a37e834c56a217b0859103a498)
Hence,
![{\displaystyle [{\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {v} )]\cdot \mathbf {a} =\mathbf {0} \qquad \qquad \forall ~~\mathbf {a} ~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/07493460ded81a758844b493c8aa8fd23299a467)
The arbitrary nature of
gives us
![{\displaystyle {{\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {v} )=\mathbf {0} \qquad \square }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5b46b36e71762f574b9eae72739104e483bd93a)
Let
be a vector field. Show that
![{\displaystyle {\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {v} ^{T})={\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\times \mathbf {v} )~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5faf33a1df9219de9a0c2cbdf0b394c9d66182ab)
Proof:
The curl of a second order tensor field
is defined as
![{\displaystyle ({\boldsymbol {\nabla }}\times {\boldsymbol {S}})\cdot \mathbf {a} ={\boldsymbol {\nabla }}\times ({\boldsymbol {S}}^{T}\cdot \mathbf {a} )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e97aae219d52220e7f83d084249851b1bc628b7e)
where
is an arbitrary constant vector.
If we write the right hand side in index notation with respect to a
Cartesian basis, we have
![{\displaystyle [{\boldsymbol {S}}^{T}\cdot \mathbf {a} ]_{k}=[\mathbf {b} ]_{k}=b_{k}=S_{pk}~a_{p}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d691445dbdf4b3f66eacd88098e412c9c5f5e524)
and
![{\displaystyle [{\boldsymbol {\nabla }}\times \mathbf {b} ]_{i}=e_{ijk}{\frac {\partial b_{k}}{\partial x_{j}}}=e_{ijk}{\frac {\partial (S_{pk}~a_{p})}{\partial x_{j}}}=e_{ijk}{\frac {\partial S_{pk}}{\partial x_{j}}}~a_{p}=[({\boldsymbol {\nabla }}\times {\boldsymbol {S}})]_{ip}~a_{p}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a8eba99b2e076a199cc0380b3d2e70d76491477)
In the above a quantity
represents the
-th component of a
vector, and the quantity
represents the
-th components of
a second-order tensor.
Therefore, in index notation, the curl of a second-order tensor
can
be expressed as
![{\displaystyle [{\boldsymbol {\nabla }}\times {\boldsymbol {S}}]_{ip}=e_{ijk}{\frac {\partial S_{pk}}{\partial x_{j}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b786a08dc372c47adb4994a682efb79d89c8802e)
Using the above definition, we get
![{\displaystyle [{\boldsymbol {\nabla }}\times {\boldsymbol {S}}^{T}]_{ip}=e_{ijk}{\frac {\partial S_{kp}}{\partial x_{j}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0127ac19bf1d019dc7b1187d5d4ed8e206357ea0)
If
, we have
![{\displaystyle [{\boldsymbol {\nabla }}\times {\boldsymbol {\nabla }}\mathbf {v} ^{T}]_{ip}=e_{ijk}{\frac {\partial }{\partial x_{j}}}\left({\frac {\partial v_{k}}{\partial x_{p}}}\right)={\frac {\partial }{\partial x_{p}}}\left(e_{ijk}{\frac {\partial v_{k}}{\partial x_{j}}}\right)={\frac {\partial }{\partial x_{p}}}\left([{\boldsymbol {\nabla }}\times \mathbf {v} ]_{i}\right)=[{\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\times \mathbf {v} )]_{ip}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd50999dad5938bddeeaca49629213bee3905dd3)
Therefore,
![{\displaystyle {{\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {v} ^{T})={\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\times \mathbf {v} )\qquad \square }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/75edb0a278eb529c844dcb82723b60f6f1730d49)