I have learnt that in order to prove $A$ is a second order Cartesian tensor its components have to obey the following transformation rule: $$ \bar{A}_{pq} = l_{ip}l_{jq}A_{ij} $$ where $l_{mn}$ are the components of the orthogonal transformation matrix and $\bar{A}_{pq}$ the transformed components of the tensor. Summation is implied on the $i,j$ indices. This is quite simple to show for a case such as $A=ab$ where $a,b$ are vectors.
Is there a way to prove that $A_{ij}=e_{ijk}a_k$ is a cartesian tensor using the above idea? $a_k$ is any vector and a summation is implied on $k$. According to my textbook this is a definitely a cartesian tensor but I can't prove it no matter what i tried
Best Answer
By definition, a second order Cartesian tensor $\boldsymbol{T}$ is an arbitrary linear combination of the second order basis tensors $\boldsymbol{e}_i\otimes\boldsymbol{e}_j\,.$ Each such basis tensor is the Kronecker product of the two basis vectors that is: the $3\times 3$ matrix that has a single 1 at row $i$ and column $j$ and zeros otherwise.
To make a long story short: a 2nd order Cartesian tensor is just a matrix.
The second order tensor $\epsilon_{ijk}\,a_k$ can be written as \begin{align} \left(\begin{matrix}0&a_3&-a_2\\-a_3&0&a_1\\a_2&-a_1&0\end{matrix}\right)\,. \end{align} In tensor notation this is \begin{align} a_3(\boldsymbol{e}_1\otimes\boldsymbol{e}_2)-a_3(\boldsymbol{e}_2\otimes\boldsymbol{e}_1)+ a_2(\boldsymbol{e}_3\otimes\boldsymbol{e}_1)-a_2(\boldsymbol{e}_1\otimes\boldsymbol{e}_3)+ a_1(\boldsymbol{e}_2\otimes\boldsymbol{e}_3)-a_1(\boldsymbol{e}_3\otimes\boldsymbol{e}_2)\,. \end{align} A new basis is related to the old basis by $$ \overline{\boldsymbol{e}}_i=l_{ik}\,\boldsymbol{e}_k\,,\quad\boldsymbol{e}_k=l_{ki}\,\overline{\boldsymbol{e}}_i\,. $$ The transformation rule $l_{ip}\,l_{jq}\,\epsilon_{ijk}\,a_k$ therefore follows automatically.