Proof:
The axial vector
w
{\displaystyle \mathbf {w} }
W
{\displaystyle {\boldsymbol {W}}}
W
⋅
a
=
w
×
a
{\displaystyle {\boldsymbol {W}}\cdot \mathbf {a} =\mathbf {w} \times \mathbf {a} }
for all vectors
a
{\displaystyle \mathbf {a} }
W
i
p
a
p
=
e
i
j
k
w
j
a
k
{\displaystyle W_{ip}~a_{p}=e_{ijk}~w_{j}~a_{k}}
Since
e
i
j
k
=
−
e
i
k
j
{\displaystyle e_{ijk}=-e_{ikj}}
W
i
p
a
p
=
−
e
i
k
j
w
j
a
k
≡
−
e
i
p
q
w
q
a
p
{\displaystyle W_{ip}~a_{p}=-e_{ikj}~w_{j}~a_{k}\equiv -e_{ipq}~w_{q}~a_{p}}
or,
W
i
p
=
−
e
i
p
q
w
q
.
{\displaystyle W_{ip}=-e_{ipq}~w_{q}~.}
Therefore, the relation between the components of
ω
{\displaystyle {\boldsymbol {\omega }}}
θ
{\displaystyle {\boldsymbol {\theta }}}
ω
i
j
=
−
e
i
j
k
θ
k
.
{\displaystyle \omega _{ij}=-e_{ijk}~\theta _{k}~.}
Multiplying both sides by
e
p
i
j
{\displaystyle e_{pij}}
e
p
i
j
ω
i
j
=
−
e
p
i
j
e
i
j
k
θ
k
=
−
e
p
i
j
e
k
i
j
θ
k
.
{\displaystyle e_{pij}~\omega _{ij}=-e_{pij}~e_{ijk}~\theta _{k}=-e_{pij}~e_{kij}~\theta _{k}~.}
Recall the identity
e
i
j
k
e
p
q
k
=
δ
i
p
δ
j
q
−
δ
i
q
δ
j
p
.
{\displaystyle e_{ijk}~e_{pqk}=\delta _{ip}~\delta _{jq}-\delta _{iq}~\delta _{jp}~.}
Therefore,
e
i
j
k
e
p
j
k
=
δ
i
p
δ
j
j
−
δ
i
j
δ
j
p
=
3
δ
i
p
−
δ
i
p
=
2
δ
i
p
{\displaystyle e_{ijk}~e_{pjk}=\delta _{ip}~\delta _{jj}-\delta _{ij}~\delta _{jp}=3\delta _{ip}-\delta _{ip}=2\delta _{ip}}
Using the above identity, we get
e
p
i
j
ω
i
j
=
−
2
δ
p
k
θ
k
=
−
2
θ
p
.
{\displaystyle e_{pij}~\omega _{ij}=-2\delta _{pk}~\theta _{k}=-2\theta _{p}~.}
Rearranging,
θ
p
=
−
1
2
e
p
i
j
ω
i
j
{\displaystyle \theta _{p}=-{\frac {1}{2}}~e_{pij}~\omega _{ij}}
Now, the components of the tensor
ω
{\displaystyle {\boldsymbol {\omega }}}
ω
i
j
=
1
2
(
∂
u
i
∂
x
j
−
∂
u
j
∂
x
i
)
{\displaystyle \omega _{ij}={\frac {1}{2}}\left({\frac {\partial u_{i}}{\partial x_{j}}}-{\frac {\partial u_{j}}{\partial x_{i}}}\right)}
Therefore, we may write
θ
p
=
−
1
4
e
p
i
j
(
∂
u
i
∂
x
j
−
∂
u
j
∂
x
i
)
{\displaystyle \theta _{p}=-{\cfrac {1}{4}}~e_{pij}\left({\frac {\partial u_{i}}{\partial x_{j}}}-{\frac {\partial u_{j}}{\partial x_{i}}}\right)}
Since the curl of a vector
v
{\displaystyle \mathbf {v} }
∇
×
v
=
e
i
j
k
∂
u
k
∂
x
j
e
i
{\displaystyle {\boldsymbol {\nabla }}\times \mathbf {v} =e_{ijk}~{\frac {\partial u_{k}}{\partial x_{j}}}~\mathbf {e} _{i}}
we have
e
p
i
j
∂
u
j
∂
x
i
=
[
∇
×
u
]
p
and
e
p
i
j
∂
u
i
∂
x
j
=
−
e
p
j
i
∂
u
i
∂
x
j
=
−
[
∇
×
u
]
p
{\displaystyle e_{pij}~{\frac {\partial u_{j}}{\partial x_{i}}}~=[{\boldsymbol {\nabla }}\times \mathbf {u} ]_{p}\qquad {\text{and}}\qquad e_{pij}~{\frac {\partial u_{i}}{\partial x_{j}}}~=-e_{pji}{\frac {\partial u_{i}}{\partial x_{j}}}=-[{\boldsymbol {\nabla }}\times \mathbf {u} ]_{p}}
where
[
]
p
{\displaystyle [~]_{p}}
p
{\displaystyle p}
Hence,
θ
p
=
−
1
4
(
−
[
∇
×
u
]
p
−
[
∇
×
u
]
p
)
=
1
2
[
∇
×
u
]
p
.
{\displaystyle \theta _{p}=-{\cfrac {1}{4}}~\left(-[{\boldsymbol {\nabla }}\times \mathbf {u} ]_{p}-[{\boldsymbol {\nabla }}\times \mathbf {u} ]_{p}\right)={\frac {1}{2}}~[{\boldsymbol {\nabla }}\times \mathbf {u} ]_{p}~.}
Therefore,
θ
=
1
2
∇
×
u
◻
{\displaystyle {{\boldsymbol {\theta }}={\frac {1}{2}}{\boldsymbol {\nabla }}\times \mathbf {u} \qquad \square }}
![{\displaystyle e_{pij}~{\frac {\partial u_{j}}{\partial x_{i}}}~=[{\boldsymbol {\nabla }}\times \mathbf {u} ]_{p}\qquad {\text{and}}\qquad e_{pij}~{\frac {\partial u_{i}}{\partial x_{j}}}~=-e_{pji}{\frac {\partial u_{i}}{\partial x_{j}}}=-[{\boldsymbol {\nabla }}\times \mathbf {u} ]_{p}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca92c913ee52df02591563417503d3b009fb510e)
![{\displaystyle [~]_{p}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3a3d86bc6670572df5ac68710c4cde9b6e547bc)
![{\displaystyle \theta _{p}=-{\cfrac {1}{4}}~\left(-[{\boldsymbol {\nabla }}\times \mathbf {u} ]_{p}-[{\boldsymbol {\nabla }}\times \mathbf {u} ]_{p}\right)={\frac {1}{2}}~[{\boldsymbol {\nabla }}\times \mathbf {u} ]_{p}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e5a5a6865dc9ebb0f7cb3a48e287ce569e35dad3)
