User:Maschen/Cartesian tensor

(This article attempts to be the transition from basic vector algebra to tensor algebra.)

In geometry and linear algebra, a Cartesian tensor is a tensor in Euclidean space represented by the standard basis. In 3d, Cartesian coordinates are employed (hence the name), and higher dimensional generalizations are easily possible. Dyadic tensors were historically the first approach to formulating second order tensors, Cartesian tensors are more general.

First order Cartesian tensors in 3d

Basis vectors in 3d

In 3d the standard basis is (e_x, e_y, e_z) = (e₁, e₂, e₃). Each basis vector points along the x, y, and z axes, and the vectors are all unit vectors (or "normalized"), so the basis is orthonormal.

Dot product and Kronecker delta

For the dot product

${\begin{array}{cccc}\mathbf {e} _{x}\cdot \mathbf {e} _{y}&=\mathbf {e} _{y}\cdot \mathbf {e} _{z}&=\mathbf {e} _{z}\cdot \mathbf {e} _{x}\\\mathbf {e} _{y}\cdot \mathbf {e} _{x}&=\mathbf {e} _{z}\cdot \mathbf {e} _{y}&=\mathbf {e} _{x}\cdot \mathbf {e} _{z}&=0\end{array}}$

while

$\mathbf {e} _{x}\cdot \mathbf {e} _{x}=\mathbf {e} _{y}\cdot \mathbf {e} _{y}=\mathbf {e} _{z}\cdot \mathbf {e} _{z}=1$

which can be summarized by

$\mathbf {e} _{i}\cdot \mathbf {e} _{j}=\left\langle {\begin{array}{cc}1&i=j\\0&i\neq j\end{array}}\right.$

where i and j are placeholders for x, y, z. This is so important and useful it's given a name: the Kronecker delta.

Cross and product and the Levi-Civita symbol

For the cross product it's (almost) the other way round:

${\begin{array}{lllc}\mathbf {e} _{x}\times \mathbf {e} _{y}&=\mathbf {e} _{y}\times \mathbf {e} _{z}&=\mathbf {e} _{z}\times \mathbf {e} _{x}\\-\mathbf {e} _{y}\times \mathbf {e} _{x}&=-\mathbf {e} _{z}\times \mathbf {e} _{y}&=-\mathbf {e} _{x}\times \mathbf {e} _{z}&=1\end{array}}$

while

$\mathbf {e} _{x}\times \mathbf {e} _{x}=\mathbf {e} _{y}\times \mathbf {e} _{y}=\mathbf {e} _{z}\times \mathbf {e} _{z}=0$

which can be summarized by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \mathbf{e}_{i}\times\mathbf{e}_{j}=\left\langle \begin{array}{cc} +1 & \text{cyclic permutations of }i,j\in\left\{ x,y,z\right\} :i\neq j\\ -1 & \text{anticyclic permutations of }i,j\in\left\{ x,y,z\right\} :i\neq j\\ 0 & i=j \end{array}\right. }

where again i and j are placeholders for x, y, z. The cyclic permutations of $i,j\in \left\{x,y,z\right\}:i\neq j$ are $\left(x,y\right),\left(y,z\right),\left(z,x\right)$, and the anticyclic permutations of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle i,j\in\left\{ x,y,z\right\} :i\neq j} are $\left(y,x\right),\left(z,y\right),\left(x,z\right)$.

Similarly $\mathbf {e} _{k}\cdot \mathbf {e} _{i}\times \mathbf {e} _{j}=\left\langle {\begin{array}{cc}+1&{\text{cyclic permutations of }}i,j,k\in \left\{x,y,z\right\}:i\neq j\\-1&{\text{anticyclic permutations of }}i,j,k\in \left\{x,y,z\right\}:i\neq j\\0&i=j{\text{ or }}j=k{\text{ or }}k=i\end{array}}\right.$

Again this is so important and useful it's given a name: the Levi-Civita symbol.

• The cyclic permutations of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle i,j,k\in\left\{ x,y,z\right\} :i\neq j} are $\left(x,y,z\right),\left(y,z,x\right),\left(z,x,y\right)$ and

• the anticyclic permutations of $i,j,k\in \left\{x,y,z\right\}:i\neq j$ are $\left(y,x,z\right),\left(z,y,x\right),\left(x,z,y\right)$.

Transformations of Cartesian vectors (any number of dimensions)

A scalar is a quantity which does not change from one coordinate system to another, it's invariant under change of coordinates.

The standard example the rotation of a vector. The position vector x should take the same form in any coordinate system. In one coordinate system it has components x_i and basis e_i, so

$\mathbf {x} =x_{i}\mathbf {e} _{i}$

In another coordinate system it has components x_i and basis e_i, so

$\mathbf {x} ={\bar {x}}_{i}{\bar {\mathbf {e} }}_{i}$

The dependence of coordinates is

${\bar {x}}{}_{i}={\bar {x}}{}_{i}\left(x_{1},x_{2},\cdots \right)$

and dependence of basis is

${\bar {\mathbf {e} }}{}_{j}=\left(\mathbf {e} _{1},\mathbf {e} _{2}\cdots \right)$

for each i, j.

The components can be linearly transformed by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \bar{x}{}_{j}=L_{ij}x_{i}}

where L_ij represents the transformation matrix, and basis by

${\bar {\mathbf {e} }}{}_{j}=\left(L_{ik}\right)^{-1}\mathbf {e} _{k}$

where (L_ij)⁻¹ denotes the inverse matrix of L_ij, so that

${\bar {x}}_{j}{\bar {\mathbf {e} }}_{j}=L_{ij}x_{i}\left(L_{ik}\right)^{-1}\mathbf {e} _{k}=L_{ij}\left(L_{ik}\right)^{-1}x_{i}\mathbf {e} _{k}=\delta _{ik}x_{i}\mathbf {e} _{k}=x_{i}\mathbf {e} _{i}$

Exactly the same is neccersarily true for any vector to be invariant under changes of coordinate systems; a = a_ie_i. If a does not transform according to this rule, it's not a vector.

The linear transformation can be interpreted as a matrix; L_ij are the entries if the matrix (row number is i and column number is j).

If L is an orthogonal transformation (orthogonal matrix) then there are considerable simplifications, the matrix transpose is the inverse (by definition):

$\mathbf {L} ^{\mathrm {T} }=\mathbf {L} ^{-1}\Rightarrow \left(L_{ij}\right)^{-1}=\left(L_{ij}\right)^{\mathrm {T} }=L_{ji}$

and moreover, det(L) = ±1, (+1) for rotations and (−1) for reflections. To illustrate all possible symmetries in the indices:

${\begin{array}{c}\left(L_{jk}\right)^{-1}L_{ki}=\left(L_{jk}\right)^{\mathrm {T} }L_{ki}=L_{kj}L_{ki}=\delta _{ji}\\L_{ik}\left(L_{kj}\right)^{-1}=L_{ik}\left(L_{kj}\right)^{\mathrm {T} }=L_{ik}L_{jk}=\delta _{ij}\\\left(L_{jk}\right)^{-1}L_{ik}=\left(L_{jk}\right)^{\mathrm {T} }L_{ik}=L_{kj}L_{ik}=\delta _{ji}\\L_{ik}\left(L_{jk}\right)^{-1}=L_{ik}\left(L_{jk}\right)^{\mathrm {T} }=L_{ik}L_{kj}=\delta _{ij}\end{array}}$

The inverse transformation of coordinates is simply

$\left(L_{ki}\right)^{-1}{\bar {x}}_{i}=\left(L_{ki}\right)^{-1}L_{ij}x_{j}=\delta _{kj}x_{j}=x_{k}$

and the inverse transformation of basis is:

$L_{ji}{\bar {\mathbf {e} }}{}_{i}=L_{ji}\left(L_{ik}\right)^{-1}\mathbf {e} _{k}=\delta _{jk}\mathbf {e} _{k}=\mathbf {e} _{j}$

The components of L are partial derivatives of the coordinates.

Differentiating x_i with respect to x_k

${\frac {\partial {\bar {x}}_{i}}{\partial x_{k}}}={\dfrac {\partial }{\partial x_{k}}}\left(L_{ij}x_{j}\right)=L_{ij}{\dfrac {\partial x_{j}}{\partial x_{k}}}=L_{ij}\delta _{jk}=L_{ik}$

so

$L_{ij}={\dfrac {\partial {\bar {x}}_{i}}{\partial x_{j}}}$

is an element of the Jacobian matrix;

$J_{ij}={\dfrac {\partial \left({\bar {x}}_{1},{\bar {x}}_{2},{\bar {x}}_{3}\right)}{\partial \left(x_{1},x_{2},x_{3}\right)}}={\dfrac {\partial {\bar {x}}_{i}}{\partial x_{j}}}$

The first index i for the top of the derivative, second j for the bottom). Many sources state transformations in terms of contractions with this partial derivative.

Conversely differentiating x_j with respect to x_i:

${\frac {\partial x_{j}}{\partial {\bar {x}}_{k}}}={\dfrac {\partial }{\partial {\bar {x}}_{k}}}\left(\left(L_{ij}\right)^{-1}{\bar {x}}_{i}\right)=\left(L_{ij}\right)^{-1}{\dfrac {\partial {\bar {x}}_{i}}{\partial {\bar {x}}_{k}}}=\left(L_{ij}\right)^{-1}\delta _{ik}=\left(L_{kj}\right)^{-1}$

so

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \left(L_{ij}\right)^{-1}=\dfrac{\partial x_{j}}{\partial\bar{x}_{i}}}

is an element of the inverse Jacobian matrix

$\left(J^{-1}\right)_{ij}={\dfrac {\partial \left(x_{1},x_{2},x_{3}\right)}{\partial \left({\bar {x}}_{1},{\bar {x}}_{2},{\bar {x}}_{3}\right)}}={\dfrac {\partial x_{i}}{\partial {\bar {x}}_{j}}}$

Since L is orthogonal:

$L_{ij}=\left(L_{ij}\right)^{-1}={\dfrac {\partial x_{j}}{\partial {\bar {x}}_{i}}}$

Note the first index j for the bottom of the derivative, second i for top).

Contracting them gives the Kronecker delta:

${\dfrac {\partial {\bar {x}}_{i}}{\partial {\bar {x}}_{k}}}={\dfrac {\partial {\bar {x}}_{i}}{\partial x_{j}}}{\dfrac {\partial x_{j}}{\partial {\bar {x}}_{k}}}=L_{ij}\left(L_{kj}\right)^{-1}=L_{ij}L_{jk}=\delta _{ik}$

also

${\dfrac {\partial x_{i}}{\partial x_{k}}}={\dfrac {\partial x_{i}}{\partial {\bar {x}}_{j}}}{\dfrac {\partial {\bar {x}}_{j}}{\partial x_{k}}}=\left(L_{ji}\right)^{-1}L_{jk}=L_{ij}L_{jk}=\delta _{ik}$

which parallels the matrix multiplication of the Jacobian and its inverse:

$\mathbf {J} \mathbf {J} ^{-1}=\mathbf {J} ^{-1}\mathbf {J} =\mathbf {I}$

As a special case;

$L_{ij}\left(L_{ij}\right)^{-1}=L_{ij}L_{ji}={\dfrac {\partial {\bar {x}}_{i}}{\partial x_{j}}}{\dfrac {\partial x_{j}}{\partial {\bar {x}}_{i}}}=1$

As with all linear transformations, L depends on the basis chosen. Since ${\bar {\mathbf {x} }}={\bar {x}}_{j}{\bar {\mathbf {e} }}_{j},\mathbf {x} =x_{j}\mathbf {e} _{j}$, for two orthonormal bases

${\bar {\mathbf {e} }}_{i}\cdot {\bar {\mathbf {e} }}_{j}=\mathbf {e} _{i}\cdot \mathbf {e} _{j}=\delta _{ij},\left|\mathbf {e} _{i}\right|=\left|{\bar {\mathbf {e} }}_{i}\right|=1$,

• projecting x to the x axes${\bar {x}}_{i}={\bar {\mathbf {e} }}_{i}\cdot \mathbf {x} ={\bar {\mathbf {e} }}_{i}\cdot \mathbf {e} _{j}x_{j}=L_{ij}x_{j}$,

• projecting x to the x axes: $x_{i}=\mathbf {e} _{i}\cdot \mathbf {x} =\mathbf {e} _{i}\cdot {\bar {\mathbf {e} }}_{j}{\bar {x}}_{j}=\left(L_{ij}\right)^{-1}{\bar {x}}_{j}=L_{ji}{\bar {x}}_{j}$.

Hence the components reduce to direction cosines between the x_i and x_j axes:

$L_{ij}={\bar {\mathbf {e} }}_{i}\cdot \mathbf {e} _{j}=\cos \theta _{ij}$

$\left(L_{ij}\right)^{-1}=\mathbf {e} _{i}\cdot {\bar {\mathbf {e} }}_{j}=\cos \theta _{ji}$

where θ_ij, θ_ji are the angles between the x_i and x_j axes.

NB: $\mathbf {e} _{i}\cdot {\bar {\mathbf {e} }}_{j}\neq \delta _{ij}$ unless the basis vectors are identical: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \mathbf{e}_{i}=\bar{\mathbf{e}}_{i}}

Accumulating all results obtained, from equation (??):

The transformation components are

${\begin{array}{c}L_{ij}={\dfrac {\partial {\bar {x}}_{i}}{\partial x_{j}}}={\bar {\mathbf {e} }}_{i}\cdot \mathbf {e} _{j}=\cos \theta _{ij}\\\upharpoonleft \downharpoonright \\L_{ji}={\dfrac {\partial x_{i}}{\partial {\bar {x}}_{j}}}=\mathbf {e} _{i}\cdot {\bar {\mathbf {e} }}_{j}=\cos \theta _{ji}\end{array}}$

in matrix form

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \mathbf{L}=\begin{pmatrix}\dfrac{\partial\bar{x}_{1}}{\partial x_{1}} & \dfrac{\partial\bar{x}_{1}}{\partial x_{2}} & \dfrac{\partial\bar{x}_{1}}{\partial x_{3}}\\ \dfrac{\partial\bar{x}_{2}}{\partial x_{1}} & \dfrac{\partial\bar{x}_{2}}{\partial x_{2}} & \dfrac{\partial\bar{x}_{2}}{\partial x_{3}}\\ \dfrac{\partial\bar{x}_{3}}{\partial x_{1}} & \dfrac{\partial\bar{x}_{3}}{\partial x_{2}} & \dfrac{\partial\bar{x}_{3}}{\partial x_{3}} \end{pmatrix}=\begin{pmatrix}\bar{\mathbf{e}}_{1}\cdot\mathbf{e}_{1} & \bar{\mathbf{e}}_{1}\cdot\mathbf{e}_{2} & \bar{\mathbf{e}}_{1}\cdot\mathbf{e}_{3}\\ \bar{\mathbf{e}}_{2}\cdot\mathbf{e}_{1} & \bar{\mathbf{e}}_{2}\cdot\mathbf{e}_{2} & \bar{\mathbf{e}}_{2}\cdot\mathbf{e}_{3}\\ \bar{\mathbf{e}}_{3}\cdot\mathbf{e}_{1} & \bar{\mathbf{e}}_{3}\cdot\mathbf{e}_{2} & \bar{\mathbf{e}}_{3}\cdot\mathbf{e}_{3} \end{pmatrix}=\begin{pmatrix}\cos\theta_{11} & \cos\theta_{12} & \cos\theta_{13}\\ \cos\theta_{21} & \cos\theta_{22} & \cos\theta_{23}\\ \cos\theta_{31} & \cos\theta_{32} & \cos\theta_{33} \end{pmatrix},\quad\mathbf{L}^{-1}=\mathbf{L}^{\mathrm{T}} }

The transformation of components can be fully written:

${\bar {\mathbf {x} }}=\mathbf {L} \mathbf {x} ,\quad {\begin{pmatrix}{\bar {x}}_{1}\\{\bar {x}}_{2}\\{\bar {x}}_{3}\end{pmatrix}}={\begin{pmatrix}{\dfrac {\partial {\bar {x}}_{1}}{\partial x_{1}}}&{\dfrac {\partial {\bar {x}}_{1}}{\partial x_{2}}}&{\dfrac {\partial {\bar {x}}_{1}}{\partial x_{3}}}\\{\dfrac {\partial {\bar {x}}_{2}}{\partial x_{1}}}&{\dfrac {\partial {\bar {x}}_{2}}{\partial x_{2}}}&{\dfrac {\partial {\bar {x}}_{2}}{\partial x_{3}}}\\{\dfrac {\partial {\bar {x}}_{3}}{\partial x_{1}}}&{\dfrac {\partial {\bar {x}}_{3}}{\partial x_{2}}}&{\dfrac {\partial {\bar {x}}_{3}}{\partial x_{3}}}\end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}={\begin{pmatrix}{\bar {\mathbf {e} }}_{1}\cdot \mathbf {e} _{1}&{\bar {\mathbf {e} }}_{1}\cdot \mathbf {e} _{2}&{\bar {\mathbf {e} }}_{1}\cdot \mathbf {e} _{3}\\{\bar {\mathbf {e} }}_{2}\cdot \mathbf {e} _{1}&{\bar {\mathbf {e} }}_{2}\cdot \mathbf {e} _{2}&{\bar {\mathbf {e} }}_{2}\cdot \mathbf {e} _{3}\\{\bar {\mathbf {e} }}_{3}\cdot \mathbf {e} _{1}&{\bar {\mathbf {e} }}_{3}\cdot \mathbf {e} _{2}&{\bar {\mathbf {e} }}_{3}\cdot \mathbf {e} _{3}\end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}={\begin{pmatrix}\cos \theta _{11}&\cos \theta _{12}&\cos \theta _{13}\\\cos \theta _{21}&\cos \theta _{22}&\cos \theta _{23}\\\cos \theta _{31}&\cos \theta _{32}&\cos \theta _{33}\end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}$

similarly for

$\mathbf {x} =\mathbf {L} ^{-1}{\bar {\mathbf {x} }}=\mathbf {L} ^{\mathrm {T} }{\bar {\mathbf {x} }}$

The geometric interpretation is the x_i components equal to the sum of projecting the x_j components onto the x_j axes.

Transformations of Cartesian tensors (any number of dimensions)

Tensors are defined as quantities which transform in a certain way under linear transformations of coordinates.

Let $\mathbf {a} =a_{i}\mathbf {e} _{i},\mathbf {b} =b_{i}\mathbf {e} _{i}$ be two vectors, so that they transform according to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \bar{a}_{i}=L_{ij}a_{j},\bar{b}_{i}=L_{ij}b_{j}} .

Taking the tensor product gives:

$\mathbf {a} \otimes \mathbf {b} =a_{i}\mathbf {e} _{i}\otimes b_{j}\mathbf {e} _{j}=a_{i}b_{j}\mathbf {e} _{i}\otimes \mathbf {e} _{j}$

then applying the transformation to the components

${\bar {a}}_{p}{\bar {b}}_{q}=L_{pi}a_{i}L_{qj}b_{j}=L_{pi}L_{qj}a_{i}b_{j}$

and to the bases

${\bar {\mathbf {e} }}_{p}\otimes {\bar {\mathbf {e} }}_{q}=\left(L_{pi}\right)^{-1}\mathbf {e} _{i}\otimes \left(L_{qj}\right)^{-1}{\bar {\mathbf {e} }}_{j}=\left(L_{pi}\right)^{-1}\left(L_{qj}\right)^{-1}\mathbf {e} _{i}\otimes {\bar {\mathbf {e} }}_{j}=L_{ip}L_{jq}\mathbf {e} _{i}\otimes {\bar {\mathbf {e} }}_{j}$

gives the transformation law of an order-2 tensor. The tensor $\mathbf {a} \otimes \mathbf {b}$ is invariant under this tranformation:

${\begin{array}{cl}{\bar {a}}_{p}{\bar {b}}_{q}{\bar {\mathbf {e} }}_{p}\otimes {\bar {\mathbf {e} }}_{q}&=L_{pk}L_{q\ell }a_{k}b_{\ell }\left(L_{pi}\right)^{-1}\left(L_{qj}\right)^{-1}\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\&=a_{k}b_{\ell }L_{pk}\left(L_{pi}\right)^{-1}L_{q\ell }\left(L_{qj}\right)^{-1}\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\&=a_{k}b_{\ell }\delta _{ki}\delta _{\ell j}\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\&=a_{i}b_{j}\mathbf {e} _{i}\otimes \mathbf {e} _{j}\end{array}}$

More generally, for any order-2 tensor $\mathbf {R} =R_{ij}\mathbf {e} _{i}\otimes \mathbf {e} _{j}$, the components transform according to;

${\bar {R}}_{pq}=L_{pi}L_{qj}R_{ij}$,

and the basis transforms by:

${\bar {\mathbf {e} }}_{p}\otimes {\bar {\mathbf {e} }}_{q}=\left(L_{pi}\right)^{-1}\mathbf {e} _{i}\otimes \left(L_{qj}\right)^{-1}\mathbf {e} _{j}$

If R does not transform according to this rule - whatever quantity R may be, it's not an order 2 tensor.

Now suppose we have an additional vector Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \mathbf{c}=c_{i}\mathbf{e}_{i}} which transforms according to ${\bar {c}}_{i}=L_{ij}c_{j}$.

Taking the tensor product with the other two vectors a and b above gives:

$\mathbf {a} \otimes \mathbf {b} \otimes \mathbf {c} =a_{i}\mathbf {e} _{i}\otimes b_{j}\mathbf {e} _{j}\otimes c_{k}\mathbf {e} _{k}=a_{i}b_{j}c_{k}\mathbf {e} _{i}\otimes \mathbf {e} _{j}\otimes \mathbf {e} _{k}$

then applying the transformation to the components gives the transformation law of an order-3 tensor:

${\bar {a}}_{p}{\bar {b}}_{q}{\bar {c}}_{r}=L_{ip}a_{i}L_{jq}b_{j}L_{kr}b_{k}=L_{ip}L_{jq}L_{kr}a_{i}b_{j}c_{k}$

and to the bases

${\begin{array}{cl}{\bar {\mathbf {e} }}_{p}\otimes {\bar {\mathbf {e} }}_{q}\otimes {\bar {\mathbf {e} }}_{r}&=\left(L_{pi}\right)^{-1}\mathbf {e} _{i}\otimes \left(L_{qj}\right)^{-1}{\bar {\mathbf {e} }}_{j}\otimes \left(L_{rk}\right)^{-1}{\bar {\mathbf {e} }}_{r}\\&=\left(L_{pi}\right)^{-1}\left(L_{qj}\right)^{-1}\left(L_{rk}\right)^{-1}\mathbf {e} _{i}\otimes {\bar {\mathbf {e} }}_{j}\otimes {\bar {\mathbf {e} }}_{k}\\&=L_{ip}L_{jq}L_{kr}\mathbf {e} _{i}\otimes {\bar {\mathbf {e} }}_{j}\otimes {\bar {\mathbf {e} }}_{k}\end{array}}$

gives the transformation law of an order-3 tensor. The tensor a⊗b⊗c is invariant under this transformation:

${\begin{array}{cl}{\bar {a}}_{p}{\bar {b}}_{q}{\bar {c}}_{r}{\bar {\mathbf {e} }}_{p}\otimes {\bar {\mathbf {e} }}_{q}\otimes {\bar {\mathbf {e} }}_{r}&=L_{p\ell }L_{qm}L_{rn}a_{\ell }b_{m}c_{n}\left(L_{pi}\right)^{-1}\left(L_{qj}\right)^{-1}\left(L_{rk}\right)^{-1}\mathbf {e} _{i}\otimes \mathbf {e} _{j}\otimes \mathbf {e} _{k}\\&=a_{\ell }b_{m}c_{n}L_{p\ell }\left(L_{pi}\right)^{-1}L_{qm}\left(L_{qj}\right)^{-1}L_{rn}\left(L_{rk}\right)^{-1}\mathbf {e} _{i}\otimes \mathbf {e} _{j}\otimes \mathbf {e} _{k}\\&=a_{\ell }b_{m}c_{n}\delta _{\ell i}\delta _{mj}\delta _{nk}\mathbf {e} _{i}\otimes \mathbf {e} _{j}\otimes \mathbf {e} _{k}\\&=a_{i}b_{j}c_{k}\mathbf {e} _{i}\otimes \mathbf {e} _{j}\otimes \mathbf {e} _{k}\end{array}}$

For any order-3 tensor $\mathbf {S} =S_{ijk}\mathbf {e} _{i}\otimes \mathbf {e} _{j}\otimes \mathbf {e} _{k}$, the components transform according to;${\bar {S}}_{pqr}=L_{ip}L_{jq}L_{kr}S_{ijk}$ and the basis transforms by:

${\bar {\mathbf {e} }}_{p}\otimes {\bar {\mathbf {e} }}_{q}=\left(L_{pi}\right)^{-1}\mathbf {e} _{i}\otimes \left(L_{qj}\right)^{-1}\mathbf {e} _{j}\otimes \left(L_{rk}\right)^{-1}\mathbf {e} _{k}$.

If S does not transform according to this rule - whatever quantity S may be, it's not an order-3 tensor.

More generally, for any order p tensor

$\mathbf {T} =T_{i_{1}i_{2}\cdots -i_{p}}\mathbf {e} _{i_{1}}\otimes \mathbf {e} _{i_{2}}\otimes \cdots \mathbf {e} _{i_{p}}$

the components transform according to;

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \bar{T}_{j_{1}j_{2}\cdots-j_{p}}=L_{i_{1}j_{1}}L_{i_{2}j_{2}}\cdots L_{i_{p}j_{p}}T_{i_{1}i_{2}\cdots i_{p}}}

and the basis transforms by:

${\bar {\mathbf {e} }}_{j_{1}}\otimes {\bar {\mathbf {e} }}_{j_{2}}\cdots \otimes {\bar {\mathbf {e} }}_{i_{p}}=\left(L_{i_{1}j_{1}}\right)^{-1}\mathbf {e} _{i_{1}}\otimes \left(L_{i_{2}j_{2}}\right)^{-1}\mathbf {e} _{i_{2}}\cdots \otimes \left(L_{i_{p}j_{p}}\right)^{-1}\mathbf {e} _{i_{p}}$.

If T does not transform according to this rule - whatever quantity T may be it's not an order-3 tensor.

Second order Cartesian tensors in 3d

Exterior product of Cartesian basis in 3d

For the exterior product we have:

$\mathbf {e} _{x}\wedge \mathbf {e} _{y}=-\mathbf {e} _{y}\wedge \mathbf {e} _{x}$

$\mathbf {e} _{y}\wedge \mathbf {e} _{z}=-\mathbf {e} _{z}\wedge \mathbf {e} _{y}$

$\mathbf {e} _{z}\wedge \mathbf {e} _{x}=-\mathbf {e} _{x}\wedge \mathbf {e} _{z}$

and

$\mathbf {e} _{x}\wedge \mathbf {e} _{x}=\mathbf {e} _{y}\wedge \mathbf {e} _{y}=\mathbf {e} _{z}\wedge \mathbf {e} _{z}=0$

Tensor product of Cartesian basis in 3d

For tensors of order order 1 a Cartesian vector a can be written algebraically as a linear combination of the basis vectors or equivalently as a column vector of the coordinates with respect to the basis:

$\mathbf {a} =a_{x}\mathbf {e} _{x}+a_{y}\mathbf {e} _{y}+a_{z}\mathbf {e} _{z}={\begin{pmatrix}a_{x}&a_{y}&a_{z}\end{pmatrix}}$

A dyadic tensor T is an order 2 tensor formed by the tensor product (designated the symbol ⊗) of two Cartesian vectors a and b, again it can be written as a linear combination of the tensor basis e_x ⊗ e_y, e_x ⊗ e_z ... e_y ⊗ e_z, e_z ⊗ e_z, or more systematically as a matrix:

$\mathbf {T} =\mathbf {a} \otimes \mathbf {b} =\left(a_{x}\mathbf {e} _{x}+a_{y}\mathbf {e} _{y}+a_{z}\mathbf {e} _{z}\right)\otimes \left(b_{x}\mathbf {e} _{x}+b_{y}\mathbf {e} _{y}+b_{z}\mathbf {e} _{z}\right)={\begin{pmatrix}a_{x}b_{x}&a_{x}b_{y}&a_{x}b_{z}\\a_{y}b_{x}&a_{y}b_{y}&a_{y}b_{z}\\a_{z}b_{x}&a_{z}b_{y}&a_{z}b_{z}\\\end{pmatrix}}$

See matrix multiplication for the notational correspondence between matrices and the dot and tensor products.

Cartesian dyadic tensors of second order are reducible, which means they can be re-expressed in terms of the two vectors as follows:

$\mathbf {T} =\mathbf {T} ^{(1)}+\mathbf {T} ^{(2)}+\mathbf {T} ^{(3)}$

$T_{ij}^{(1)}={\frac {a_{k}b_{k}}{3}}\delta _{ij}$

$T_{ij}^{(2)}={\frac {1}{2}}[a_{i}b_{j}-a_{j}b_{i}]=a_{[i}b_{j]}$

$T_{ij}^{(3)}={\frac {1}{2}}(a_{i}b_{j}+a_{j}b_{i})-{\frac {a_{k}b_{k}}{3}}\delta _{ij}=a_{(i}b_{j)}-T_{ij}^{(1)}$

where δ_ij is the Kronecker delta, the components of the identity matrix. These three terms are irreducible representations, which means they cannot be decomposed further and still be tensors satisfying the defining transformation laws under which they must be invariant. Each of the irreducible representations transform like angular momentum according to the number of independent components.

References

• Pei Chi Chou (1992). Elasticity: Tensor, Dyadic, and Engineering Approaches. Courier Dover Publications. ISBN 048-666-958-0.

• T.W. Körner (2012). Vectors, Pure and Applied: A General Introduction to Linear Algebra. Cambridge University Press. p. 216. ISBN 11070-3356-X.

• R. Torretti (1996). Relativity and Geometry. Courier Dover Publications. p. 103. ISBN 0-4866-90466.

• J.J.L. Synge, A. Schild (1978). Tensor Calculus. Courier Dover Publications. p. 128. ISBN 0-4861-4139-X.

• C.A. Balafoutis, R.V. Patel (1991). Dynamic Analysis of Robot Manipulators: A Cartesian Tensor Approach. The Kluwer International Series in Engineering and Computer Science: Robotics: vision, manipulation and sensors. Vol. 131. Springer. ISBN 0792-391-454.

• S. G. Tzafestas (1992). Robotic systems: advanced techniques and applications. Springer. ISBN 0-792-317-491.

• T. Dass, S.K. Sharma (1998). Mathematical Methods In Classical And Quantum Physics. Universities Press. p. 144. ISBN 817-371-0899.

External links

• R. E. Hunt (2002) Cartesian Tensors