thumb|Two different 3d [[Orthonormal basis|orthonormal bases: each basis consists of unit vectors that are mutually perpendicular.]]

In geometry and linear algebra, a Cartesian tensor uses an orthonormal basis to represent a tensor in a Euclidean space in the form of components. Converting a tensor's components from one such basis to another is done through an orthogonal transformation.

The most familiar coordinate systems are the two-dimensional and three-dimensional Cartesian coordinate systems. Cartesian tensors may be used with any Euclidean space, or more technically, any finite-dimensional vector space over the field of real numbers that has an inner product.

Use of Cartesian tensors occurs in physics and engineering, such as with the Cauchy stress tensor and the moment of inertia tensor in rigid body dynamics. Sometimes general curvilinear coordinates are convenient, as in high-deformation continuum mechanics, or even necessary, as in general relativity. While orthonormal bases may be found for some such coordinate systems (e.g. tangent to spherical coordinates), Cartesian tensors may provide considerable simplification for applications in which rotations of rectilinear coordinate axes suffice. The transformation is a passive transformation, since the coordinates are changed and not the physical system.

Vectors in three dimensions

In 3D Euclidean space, <math>\R^3</math>, the standard basis is , , . 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.

Throughout, when referring to Cartesian coordinates in three dimensions, a right-handed system is assumed and this is much more common than a left-handed system in practice, see orientation (vector space) for details.

For Cartesian tensors of order 1, a Cartesian vector can be written algebraically as a linear combination of the basis vectors , , :

<math display="block">\mathbf{a} = a_\text{x}\mathbf{e}_\text{x} + a_\text{y}\mathbf{e}_\text{y} + a_\text{z}\mathbf{e}_\text{z} </math>

where the coordinates of the vector with respect to the Cartesian basis are denoted , , . It is common and helpful to display the basis vectors as column vectors

<math display="block">

\mathbf{e}_\text{x} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \,,\quad

\mathbf{e}_\text{y} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \,,\quad

\mathbf{e}_\text{z} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}

</math>

when we have a coordinate vector in a column vector representation:

<math display="block">\mathbf{a} = \begin{pmatrix} a_\text{x} \\ a_\text{y} \\ a_\text{z} \end{pmatrix}</math>

A row vector representation is also legitimate, although in the context of general curvilinear coordinate systems the row and column vector representations are used separately for specific reasons – see Einstein notation and covariance and contravariance of vectors for why.

The term "component" of a vector is ambiguous: it could refer to:

  • a specific coordinate of the vector such as (a scalar), and similarly for and , or
  • the coordinate scalar-multiplying the corresponding basis vector, in which case the "-component" of is (a vector), and similarly for and .

A more general notation is tensor index notation, which has the flexibility of numerical values rather than fixed coordinate labels. <!---PLEASE DON'T DELETE THIS ANCHOR, IT'S USED LATER--->The Cartesian labels are replaced by tensor indices in the basis vectors , , and coordinates , , . In general, the notation , , refers to any basis, and , , refers to the corresponding coordinate system; although here they are restricted to the Cartesian system. Then:

<math display="block">\mathbf{a} = a_1\mathbf{e}_1 + a_2\mathbf{e}_2 + a_3\mathbf{e}_3 = \sum_{i=1}^3 a_i\mathbf{e}_i</math>

It is standard to use the Einstein notation—the summation sign for summation over an index that is present exactly twice within a term may be suppressed for notational conciseness:

<math display="block">\mathbf{a} = \sum_{i=1}^3 a_i\mathbf{e}_i \equiv a_i\mathbf{e}_i </math>

An advantage of the index notation over coordinate-specific notations is the independence of the dimension of the underlying vector space, i.e. the same expression on the right hand side takes the same form in higher dimensions (see below). Previously, the Cartesian labels x, y, z were just labels and not indices. (It is informal to say "i = x, y, z").

Second-order tensors in three dimensions

A dyadic tensor T is an order-2 tensor formed by the tensor product of two Cartesian vectors and , written . Analogous to vectors, it can be written as a linear combination of the tensor basis , , ..., (the right-hand side of each identity is only an abbreviation, nothing more):

<math display="block">\begin{align}

\mathbf{T} =\quad

&\left(a_\text{x}\mathbf{e}_\text{x} + a_\text{y}\mathbf{e}_\text{y} + a_\text{z}\mathbf{e}_\text{z}\right)\otimes\left(b_\text{x}\mathbf{e}_\text{x} + b_\text{y}\mathbf{e}_\text{y} + b_\text{z}\mathbf{e}_\text{z}\right) \\[5pt]

{}=\quad &a_\text{x} b_\text{x} \mathbf{e}_\text{x} \otimes \mathbf{e}_\text{x} + a_\text{x} b_\text{y}\mathbf{e}_\text{x} \otimes \mathbf{e}_\text{y} + a_\text{x} b_\text{z}\mathbf{e}_\text{x} \otimes \mathbf{e}_\text{z} \\[4pt]

{}+{} &a_\text{y} b_\text{x}\mathbf{e}_\text{y} \otimes \mathbf{e}_\text{x} + a_\text{y} b_\text{y}\mathbf{e}_\text{y} \otimes \mathbf{e}_\text{y} + a_\text{y} b_\text{z}\mathbf{e}_\text{y} \otimes \mathbf{e}_\text{z} \\[4pt]

{}+{} &a_\text{z} b_\text{x} \mathbf{e}_\text{z} \otimes \mathbf{e}_\text{x} + a_\text{z} b_\text{y}\mathbf{e}_\text{z} \otimes \mathbf{e}_\text{y} + a_\text{z} b_\text{z}\mathbf{e}_\text{z} \otimes \mathbf{e}_\text{z}

\end{align}</math>

Representing each basis tensor as a matrix:

<math display="block">\begin{align}

\mathbf{e}_\text{x} \otimes \mathbf{e}_\text{x} &\equiv \mathbf{e}_\text{xx} = \begin{pmatrix}

1 & 0 & 0\\

0 & 0 & 0\\

0 & 0 & 0

\end{pmatrix}\,,&

\mathbf{e}_\text{x} \otimes \mathbf{e}_\text{y} &\equiv \mathbf{e}_\text{xy} = \begin{pmatrix}

0 & 1 & 0\\

0 & 0 & 0\\

0 & 0 & 0

\end{pmatrix}\,,&

\mathbf{e}_\text{z} \otimes \mathbf{e}_\text{z} &\equiv \mathbf{e}_\text{zz} = \begin{pmatrix}

0 & 0 & 0\\

0 & 0 & 0\\

0 & 0 & 1

\end{pmatrix}

\end{align}</math>

then can be represented more systematically as a matrix:

<math display="block">\mathbf{T} = \begin{pmatrix}

a_\text{x} b_\text{x} & a_\text{x} b_\text{y} & a_\text{x} b_\text{z} \\

a_\text{y} b_\text{x} & a_\text{y} b_\text{y} & a_\text{y} b_\text{z} \\

a_\text{z} b_\text{x} & a_\text{z} b_\text{y} & a_\text{z} b_\text{z}

\end{pmatrix}</math>

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

More generally, whether or not is a tensor product of two vectors, it is always a linear combination of the basis tensors with coordinates , , ..., :

<math display="block">\begin{align}

\mathbf{T} =\quad

&T_\text{xx}\mathbf{e}_\text{xx} + T_\text{xy}\mathbf{e}_\text{xy} + T_\text{xz}\mathbf{e}_\text{xz} \\[4pt]

{}+{} &T_\text{yx}\mathbf{e}_\text{yx} + T_\text{yy}\mathbf{e}_\text{yy} + T_\text{yz}\mathbf{e}_\text{yz} \\[4pt]

{}+{} &T_\text{zx}\mathbf{e}_\text{zx} + T_\text{zy}\mathbf{e}_\text{zy} + T_\text{zz}\mathbf{e}_\text{zz}

\end{align}</math>

while in terms of tensor indices:

<math display="block">\mathbf{T} = T_{ij} \mathbf{e}_{ij} \equiv \sum_{ij} T_{ij} \mathbf{e}_i \otimes \mathbf{e}_j \,,</math>

and in matrix form:

<math display="block">\mathbf{T} = \begin{pmatrix}

T_\text{xx} & T_\text{xy} & T_\text{xz} \\

T_\text{yx} & T_\text{yy} & T_\text{yz} \\

T_\text{zx} & T_\text{zy} & T_\text{zz}

\end{pmatrix}</math>

Second-order tensors occur naturally in physics and engineering when physical quantities have directional dependence in the system, often in a "stimulus-response" way. This can be mathematically seen through one aspect of tensors – they are multilinear functions. A second-order tensor T which takes in a vector u of some magnitude and direction will return a vector v; of a different magnitude and in a different direction to u, in general. The notation used for functions in mathematical analysis leads us to write , while the same idea can be expressed in matrix and index notations (including the summation convention), respectively:

<math display="block">\begin{align}

\begin{pmatrix}

v_\text{x} \\

v_\text{y} \\

v_\text{z}

\end{pmatrix} &= \begin{pmatrix}

T_\text{xx} & T_\text{xy} & T_\text{xz} \\

T_\text{yx} & T_\text{yy} & T_\text{yz} \\

T_\text{zx} & T_\text{zy} & T_\text{zz}

\end{pmatrix}\begin{pmatrix}

u_\text{x} \\

u_\text{y} \\

u_\text{z}

\end{pmatrix}\,, &

v_i &= T_{ij}u_j

\end{align}</math>

By "linear", if for two scalars and and vectors and , then in function and index notations:

<math display="block">\begin{align}

\mathbf{v} &=&& \mathbf{T}(\rho\mathbf{r} + \sigma\mathbf{s}) &=&& \rho\mathbf{T}(\mathbf{r}) + \sigma\mathbf{T}(\mathbf{s}) \\[1ex]

v_i &=&& T_{ij}(\rho r_j + \sigma s_j) &=&& \rho T_{ij} r_j + \sigma T_{ij} s_j

\end{align}</math>

and similarly for the matrix notation. The function, matrix, and index notations all mean the same thing. The matrix forms provide a clear display of the components, while the index form allows easier tensor-algebraic manipulation of the formulae in a compact manner. Both provide the physical interpretation of directions; vectors have one direction, while second-order tensors connect two directions together. One can associate a tensor index or coordinate label with a basis vector direction.

The use of second-order tensors are the minimum to describe changes in magnitudes and directions of vectors, as the dot product of two vectors is always a scalar, while the cross product of two vectors is always a pseudovector perpendicular to the plane defined by the vectors, so these products of vectors alone cannot obtain a new vector of any magnitude in any direction. (See also below for more on the dot and cross products). The tensor product of two vectors is a second-order tensor, although this has no obvious directional interpretation by itself.

The previous idea can be continued: if takes in two vectors and , it will return a scalar . In function notation we write , while in matrix and index notations (including the summation convention) respectively:

<math display="block">r = \begin{pmatrix}

p_\text{x} &

p_\text{y} &

p_\text{z}

\end{pmatrix}\begin{pmatrix}

T_\text{xx} & T_\text{xy} & T_\text{xz} \\

T_\text{yx} & T_\text{yy} & T_\text{yz} \\

T_\text{zx} & T_\text{zy} & T_\text{zz}

\end{pmatrix}\begin{pmatrix}

q_\text{x} \\

q_\text{y} \\

q_\text{z}

\end{pmatrix} = p_i T_{ij} q_j

</math>

The tensor T is linear in both input vectors. When vectors and tensors are written without reference to components, and indices are not used, sometimes a dot ⋅ is placed where summations over indices (known as tensor contractions) are taken. For the above cases:

<math display="block">\mathbf{c} \cdot \mathbf{a} \times \mathbf{b} = \begin{vmatrix} c_\text{x} & a_\text{x} & b_\text{x} \\ c_\text{y} & a_\text{y} & b_\text{y} \\ c_\text{z} & a_\text{z} & b_\text{z} \end{vmatrix} </math>

This in turn can be used to rewrite the cross product of two vectors as follows:

<math display="block">\begin{align}

(\mathbf{a} \times \mathbf{b})_i = {\mathbf{e}_i \cdot \mathbf{a} \times \mathbf{b

&= \varepsilon_{\ell jk} {(\mathbf{e}_i)}_\ell a_j b_k = \varepsilon_{\ell jk} \delta_{i \ell} a_j b_k = \varepsilon_{ijk} a_j b_k \\

\Rightarrow\quad {\mathbf{a} \times \mathbf{b = (\mathbf{a} \times \mathbf{b})_i \mathbf{e}_i

&= \varepsilon_{ijk} a_j b_k \mathbf{e}_i

\end{align}</math>

Contrary to its appearance, the Levi-Civita symbol is not a tensor, but a pseudotensor, the components transform according to:

<math display="block">\bar{\varepsilon}_{pqr} = \det(\boldsymbol{\mathsf{L) \varepsilon_{ijk} \mathsf{L}_{ip}\mathsf{L}_{jq}\mathsf{L}_{kr} \,.</math>

Therefore, the transformation of the cross product of and is:

<!---PLEASE LEAVE THE SPACES \;\; FOR CLARITY, AND LEAVE IN ALL THE STEPS - CLARITY IS NOT "TEXTBOOKY" - THANKS--->

<math display="block">\begin{align}

&\left(\bar{\mathbf{a \times \bar{\mathbf{b\right)_i \\[1ex]

{}={} &\bar{\varepsilon}_{ijk} \bar{a}_j \bar{b}_k \\[1ex]

{}={} &\det(\boldsymbol{\mathsf{L) \;\; \varepsilon_{pqr} \;\; \mathsf{L}_{pi}\mathsf{L}_{qj} \mathsf{L}_{rk} \;\; a_m \mathsf{L}_{mj} \;\; b_n \mathsf{L}_{nk} \\[1ex]

{}={} &\det(\boldsymbol{\mathsf{L) \;\; \varepsilon_{pqr} \;\; \mathsf{L}_{pi} \;\; \mathsf{L}_{qj} \left(\boldsymbol{\mathsf{L^{-1}\right)_{jm} \;\; \mathsf{L}_{rk} \left(\boldsymbol{\mathsf{L^{-1}\right)_{kn} \;\; a_m \;\; b_n \\[1ex]

{}={} &\det(\boldsymbol{\mathsf{L) \;\; \varepsilon_{pqr} \;\; \mathsf{L}_{pi} \;\; \delta_{qm} \;\; \delta_{rn} \;\; a_m \;\; b_n \\[1ex]

{}={} &\det(\boldsymbol{\mathsf{L) \;\; \mathsf{L}_{pi} \;\; \varepsilon_{pqr} a_q b_r \\[1ex]

{}={} &\det(\boldsymbol{\mathsf{L) \;\; (\mathbf{a}\times\mathbf{b})_p \mathsf{L}_{pi}

\end{align}</math>

and so transforms as a pseudovector, because of the determinant factor.

The tensor index notation applies to any object which has entities that form multidimensional arrays – not everything with indices is a tensor by default. Instead, tensors are defined by how their coordinates and basis elements change under a transformation from one coordinate system to another.

Note the cross product of two vectors is a pseudovector, while the cross product of a pseudovector with a vector is another vector.

Applications of the tensor and pseudotensor

Other identities can be formed from the tensor and pseudotensor, a notable and very useful identity is one that converts two Levi-Civita symbols adjacently contracted over two indices into an antisymmetrized combination of Kronecker deltas:

<math display="block">\varepsilon_{ijk}\varepsilon_{pqk} = \delta_{ip}\delta_{jq} - \delta_{iq}\delta_{jp} </math>

The index forms of the dot and cross products, together with this identity, greatly facilitate the manipulation and derivation of other identities in vector calculus and algebra, which in turn are used extensively in physics and engineering. For instance, it is clear the dot and cross products are distributive over vector addition:

<math display="block">\begin{align}

\mathbf{a}\cdot(\mathbf{b} + \mathbf{c}) &= a_i ( b_i + c_i ) = a_i b_i + a_i c_i = \mathbf{a}\cdot\mathbf{b} + \mathbf{a}\cdot\mathbf{c} \\[1ex]

\mathbf{a}\times(\mathbf{b} + \mathbf{c}) &= \mathbf{e}_i\varepsilon_{ijk} a_j ( b_k + c_k ) = \mathbf{e}_i \varepsilon_{ijk} a_j b_k + \mathbf{e}_i \varepsilon_{ijk} a_j c_k = \mathbf{a}\times\mathbf{b} + \mathbf{a}\times\mathbf{c}

\end{align}</math>

without resort to any geometric constructions – the derivation in each case is a quick line of algebra. Although the procedure is less obvious, the vector triple product can also be derived. Rewriting in index notation:

<math display="block"> \left[ \mathbf{a}\times(\mathbf{b}\times\mathbf{c})\right]_i = \varepsilon_{ijk} a_j ( \varepsilon_{k \ell m} b_\ell c_m ) = (\varepsilon_{ijk} \varepsilon_{k \ell m} ) a_j b_\ell c_m </math>

and because cyclic permutations of indices in the symbol does not change its value, cyclically permuting indices in to obtain allows us to use the above - identity to convert the symbols into tensors:

<math display="block">\begin{align}

\left[ \mathbf{a}\times(\mathbf{b}\times\mathbf{c})\right]_i

{}={} &\left(\delta_{i\ell} \delta_{jm} - \delta_{im} \delta_{j\ell}\right) a_j b_\ell c_m \\

{}={} &\delta_{i\ell} \delta_{jm} a_j b_\ell c_m - \delta_{im} \delta_{j\ell} a_j b_\ell c_m \\

{}={} &a_j b_i c_j - a_j b_j c_i \\

{}={} &\left[(\mathbf{a}\cdot\mathbf{c})\mathbf{b} - (\mathbf{a}\cdot\mathbf{b})\mathbf{c}\right]_i

\end{align}</math>

thusly:

<math display="block">\mathbf{a}\times(\mathbf{b}\times\mathbf{c}) = (\mathbf{a}\cdot\mathbf{c})\mathbf{b} - (\mathbf{a}\cdot\mathbf{b})\mathbf{c}</math>

Note this is antisymmetric in and , as expected from the left hand side. Similarly, via index notation or even just cyclically relabelling , , and in the previous result and taking the negative:

<math display="block">(\mathbf{a}\times \mathbf{b})\times\mathbf{c} = (\mathbf{c}\cdot\mathbf{a})\mathbf{b} - (\mathbf{c}\cdot\mathbf{b})\mathbf{a} </math>

and the difference in results show that the cross product is not associative. More complex identities, like quadruple products;

<math display="block">(\mathbf{a}\times \mathbf{b})\cdot(\mathbf{c}\times\mathbf{d}),\quad (\mathbf{a}\times \mathbf{b})\times(\mathbf{c}\times\mathbf{d}),\quad \ldots</math>

and so on, can be derived in a similar manner.

Transformations of Cartesian tensors (any number of dimensions)

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

Second order

Let and be two vectors, so that they transform according to , .

Taking the tensor product gives:

<math display="block">\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</math>

then applying the transformation to the components

<math display="block">\bar{a}_p\bar{b}_q = a_i \mathsf{L}_i{}_p b_j \mathsf{L}_j{}_q = \mathsf{L}_i{}_p\mathsf{L}_j{}_q a_i b_j </math>

and to the bases

<math display="block">\bar{\mathbf{e_p\otimes\bar{\mathbf{e_q = \left(\boldsymbol{\mathsf{L^{-1}\right)_{pi}\mathbf{e}_i\otimes\left(\boldsymbol{\mathsf{L^{-1}\right)_{qj}\mathbf{e}_j = \left(\boldsymbol{\mathsf{L^{-1}\right)_{pi}\left(\boldsymbol{\mathsf{L^{-1}\right)_{qj}\mathbf{e}_i\otimes\mathbf{e}_j = \mathsf{L}^{-1}_{ip} \mathsf{L}^{-1}_{jq} \mathbf{e}_i\otimes\mathbf{e}_j</math>

gives the transformation law of an order-2 tensor. The tensor is invariant under this transformation:

<math display="block">\begin{align}

\bar{a}_p\bar{b}_q\bar{\mathbf{e_p\otimes\bar{\mathbf{e_q

{}={} &\mathsf{L}_{kp} \mathsf{L}_{\ell q} a_k b_{\ell} \, \left(\boldsymbol{\mathsf{L^{-1}\right)_{pi} \left(\boldsymbol{\mathsf{L^{-1}\right)_{qj} \mathbf{e}_i\otimes\mathbf{e}_j \\[1ex]

{}={} &\mathsf{L}_{kp} \left(\boldsymbol{\mathsf{L^{-1}\right)_{pi} \mathsf{L}_{\ell q} \left(\boldsymbol{\mathsf{L^{-1}\right)_{q j} \, a_k b_{\ell} \mathbf{e}_i\otimes\mathbf{e}_j \\[1ex]

{}={} &\delta_k{}_i \delta_{\ell j} \, a_k b_{\ell} \mathbf{e}_i\otimes\mathbf{e}_j \\[1ex]

{}={} &a_ib_j\mathbf{e}_i\otimes\mathbf{e}_j

\end{align}</math>

More generally, for any order-2 tensor

<math display="block">\mathbf{R} = R_{ij}\mathbf{e}_i\otimes\mathbf{e}_j\,,</math>

the components transform according to;

<math display="block">\bar{R}_{pq}=\mathsf{L}_i{}_p\mathsf{L}_j{}_q R_{ij},</math>

and the basis transforms by:

<math display="block">\bar{\mathbf{e_p\otimes\bar{\mathbf{e_q = \left(\boldsymbol{\mathsf{L^{-1}\right)_{ip}\mathbf{e}_i\otimes \left(\boldsymbol{\mathsf{L^{-1}\right)_{jq}\mathbf{e}_j</math>

If does not transform according to this rule – whatever quantity may be – it is not an order-2 tensor.

Any order

More generally, for any order tensor

<math display="block">\mathbf{T} = T_{j_1 j_2 \cdots j_p} \mathbf{e}_{j_1}\otimes\mathbf{e}_{j_2}\otimes\cdots\mathbf{e}_{j_p}</math>

the components transform according to;

<math display="block">\bar{T}_{j_1j_2\cdots j_p} = \mathsf{L}_{i_1 j_1} \mathsf{L}_{i_2 j_2}\cdots \mathsf{L}_{i_p j_p} T_{i_1 i_2\cdots i_p}</math>

and the basis transforms by:

<math display="block">\bar{\mathbf{e_{j_1}\otimes\bar{\mathbf{e_{j_2}\cdots\otimes\bar{\mathbf{e_{j_p}=\left(\boldsymbol{\mathsf{L^{-1}\right)_{j_1 i_1}\mathbf{e}_{i_1}\otimes\left(\boldsymbol{\mathsf{L^{-1}\right)_{j_2 i_2}\mathbf{e}_{i_2}\cdots\otimes\left(\boldsymbol{\mathsf{L^{-1}\right)_{j_p i_p}\mathbf{e}_{i_p}</math>

For a pseudotensor of order , the components transform according to;

<math display="block">\bar{S}_{j_1j_2\cdots j_p} = \det(\boldsymbol{\mathsf{L) \mathsf{L}_{i_1 j_1} \mathsf{L}_{i_2 j_2}\cdots \mathsf{L}_{i_p j_p} S_{i_1 i_2\cdots i_p}\,.</math>

Pseudovectors as antisymmetric second order tensors

The antisymmetric nature of the cross product can be recast into a tensorial form as follows.