Proof:
The infinitesimal strain tensor is given by
ε
=
1
2
(
∇
u
+
∇
u
T
)
.
{\displaystyle {\boldsymbol {\varepsilon }}={\frac {1}{2}}({\boldsymbol {\nabla }}\mathbf {u} +{\boldsymbol {\nabla }}\mathbf {u} ^{T})~.}
Therefore,
∇
×
ε
=
1
2
[
∇
×
(
∇
u
)
+
∇
×
(
∇
u
T
)
]
.
{\displaystyle {\boldsymbol {\nabla }}\times {\boldsymbol {\varepsilon }}={\frac {1}{2}}[{\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {u} )+{\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {u} ^{T})]~.}
Recall that
∇
×
(
∇
u
)
=
0
and
∇
×
(
∇
u
T
)
=
∇
(
∇
×
u
)
.
{\displaystyle {\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {u} )=\mathbf {0} \qquad {\text{and}}\qquad {\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {u} ^{T})={\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\times \mathbf {u} )~.}
Hence,
∇
×
ε
=
1
2
[
∇
(
∇
×
u
)
]
.
{\displaystyle {\boldsymbol {\nabla }}\times {\boldsymbol {\varepsilon }}={\frac {1}{2}}[{\boldsymbol {\nabla }}({\boldsymbol {\nabla }}\times \mathbf {u} )]~.}
Also recall that
θ
=
1
2
∇
×
u
.
{\displaystyle {\boldsymbol {\theta }}={\frac {1}{2}}{\boldsymbol {\nabla }}\times \mathbf {u} ~.}
Therefore,
∇
×
ε
=
∇
θ
◻
{\displaystyle {{\boldsymbol {\nabla }}\times {\boldsymbol {\varepsilon }}={\boldsymbol {\nabla }}{\boldsymbol {\theta }}\qquad \square }}