Relation between axial vector and displacement
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Proof:
The axial vector
of a skew-symmetric tensor
satisfies the
condition
![{\displaystyle {\boldsymbol {W}}\cdot \mathbf {a} =\mathbf {w} \times \mathbf {a} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/86b1e01bc2c3b5f8b5e06b5adea91a06c96e4711)
for all vectors
. In index notation (with respect to a Cartesian
basis), we have
![{\displaystyle W_{ip}~a_{p}=e_{ijk}~w_{j}~a_{k}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c24958923cbfdf86fc078df80475232efbfefd6a)
Since
, we can write
![{\displaystyle W_{ip}~a_{p}=-e_{ikj}~w_{j}~a_{k}\equiv -e_{ipq}~w_{q}~a_{p}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/845869c94221c36898407627e5eff31c6d4149c1)
or,
![{\displaystyle W_{ip}=-e_{ipq}~w_{q}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aad7cc5a59d74642b047a46e6d3ed871a6e037f0)
Therefore, the relation between the components of
and
is
![{\displaystyle \omega _{ij}=-e_{ijk}~\theta _{k}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae024e606c9a5149802439f5ce94766ef7e426be)
Multiplying both sides by
, we get
![{\displaystyle e_{pij}~\omega _{ij}=-e_{pij}~e_{ijk}~\theta _{k}=-e_{pij}~e_{kij}~\theta _{k}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8cbc184d860219e10307986658036d270d536cfe)
Recall the identity
![{\displaystyle e_{ijk}~e_{pqk}=\delta _{ip}~\delta _{jq}-\delta _{iq}~\delta _{jp}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0100c701dd4e104bd5b31b0f55c1f16a38a8bbd7)
Therefore,
![{\displaystyle e_{ijk}~e_{pjk}=\delta _{ip}~\delta _{jj}-\delta _{ij}~\delta _{jp}=3\delta _{ip}-\delta _{ip}=2\delta _{ip}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a2df176e5dda640e3fdea77a608ef2855fd4636)
Using the above identity, we get
![{\displaystyle e_{pij}~\omega _{ij}=-2\delta _{pk}~\theta _{k}=-2\theta _{p}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e45a5c92d3e661031e24ba54c252d38f1474f7c4)
Rearranging,
![{\displaystyle \theta _{p}=-{\frac {1}{2}}~e_{pij}~\omega _{ij}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2fe45dbe9f8f376fc5e0220691dccfa95005ea89)
Now, the components of the tensor
with respect to a Cartesian
basis are given by
![{\displaystyle \omega _{ij}={\frac {1}{2}}\left({\frac {\partial u_{i}}{\partial x_{j}}}-{\frac {\partial u_{j}}{\partial x_{i}}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/72747ba25ed4a7cc2fc839591ef277e7917b1839)
Therefore, we may write
![{\displaystyle \theta _{p}=-{\cfrac {1}{4}}~e_{pij}\left({\frac {\partial u_{i}}{\partial x_{j}}}-{\frac {\partial u_{j}}{\partial x_{i}}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/477139e39504d2fc4f336077e3573fde70e9139d)
Since the curl of a vector
can be written in index notation as
![{\displaystyle {\boldsymbol {\nabla }}\times \mathbf {v} =e_{ijk}~{\frac {\partial u_{k}}{\partial x_{j}}}~\mathbf {e} _{i}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b27430d0ef3b706e56f40618f0e8bcad5c470d9)
we have
![{\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)
where
indicates the
-th component of the vector inside the
square brackets.
Hence,
![{\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)
Therefore,
![{\displaystyle {{\boldsymbol {\theta }}={\frac {1}{2}}{\boldsymbol {\nabla }}\times \mathbf {u} \qquad \square }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/abf7bb7c9d6f9f94fd9a501760f46ea25b5d38e1)