Cartesian tensor

In 3D Euclidean space, ${\displaystyle \mathbb {R} ^{3}}$, the standard basis is e_x, e_y, e_z. 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 a can be written algebraically as a linear combination of the basis vectors e_x, e_y, e_z:

$${\displaystyle \mathbf {a} =a_{\text{x}}\mathbf {e} _{\text{x}}+a_{\text{y}}\mathbf {e} _{\text{y}}+a_{\text{z}}\mathbf {e} _{\text{z}}}$$

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

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

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

$${\displaystyle \mathbf {a} ={\begin{pmatrix}a_{\text{x}}\\a_{\text{y}}\\a_{\text{z}}\end{pmatrix}}}$$

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_z (a scalar), and similarly for x and y, or
• the coordinate scalar-multiplying the corresponding basis vector, in which case the "y-component" of a is a_ye_y (a vector), and similarly for x and z.

A more general notation is tensor index notation, which has the flexibility of numerical values rather than fixed coordinate labels. The Cartesian labels are replaced by tensor indices in the basis vectors e_x ↦ e₁, e_y ↦ e₂, e_z ↦ e₃ and coordinates a_x ↦ a₁, a_y ↦ a₂, a_z ↦ a₃. In general, the notation e₁, e₂, e₃ refers to any basis, and a₁, a₂, a₃ refers to the corresponding coordinate system; although here they are restricted to the Cartesian system. Then:

$${\displaystyle \mathbf {a} =a_{1}\mathbf {e} _{1}+a_{2}\mathbf {e} _{2}+a_{3}\mathbf {e} _{3}=\sum _{i=1}^{3}a_{i}\mathbf {e} _{i}}$$

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:

$${\displaystyle \mathbf {a} =\sum _{i=1}^{3}a_{i}\mathbf {e} _{i}\equiv a_{i}\mathbf {e} _{i}}$$

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").

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

$${\displaystyle {\begin{aligned}\mathbf {T} =\quad &\left(a_{\text{x}}\mathbf {e} _{\text{x}}+a_{\text{y}}\mathbf {e} _{\text{y}}+a_{\text{z}}\mathbf {e} _{\text{z}}\right)\otimes \left(b_{\text{x}}\mathbf {e} _{\text{x}}+b_{\text{y}}\mathbf {e} _{\text{y}}+b_{\text{z}}\mathbf {e} _{\text{z}}\right)\\[5pt]{}=\quad &a_{\text{x}}b_{\text{x}}\mathbf {e} _{\text{x}}\otimes \mathbf {e} _{\text{x}}+a_{\text{x}}b_{\text{y}}\mathbf {e} _{\text{x}}\otimes \mathbf {e} _{\text{y}}+a_{\text{x}}b_{\text{z}}\mathbf {e} _{\text{x}}\otimes \mathbf {e} _{\text{z}}\\[4pt]{}+{}&a_{\text{y}}b_{\text{x}}\mathbf {e} _{\text{y}}\otimes \mathbf {e} _{\text{x}}+a_{\text{y}}b_{\text{y}}\mathbf {e} _{\text{y}}\otimes \mathbf {e} _{\text{y}}+a_{\text{y}}b_{\text{z}}\mathbf {e} _{\text{y}}\otimes \mathbf {e} _{\text{z}}\\[4pt]{}+{}&a_{\text{z}}b_{\text{x}}\mathbf {e} _{\text{z}}\otimes \mathbf {e} _{\text{x}}+a_{\text{z}}b_{\text{y}}\mathbf {e} _{\text{z}}\otimes \mathbf {e} _{\text{y}}+a_{\text{z}}b_{\text{z}}\mathbf {e} _{\text{z}}\otimes \mathbf {e} _{\text{z}}\end{aligned}}}$$

Representing each basis tensor as a matrix:

$${\displaystyle {\begin{aligned}\mathbf {e} _{\text{x}}\otimes \mathbf {e} _{\text{x}}&\equiv \mathbf {e} _{\text{xx}}={\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}}\,,&\mathbf {e} _{\text{x}}\otimes \mathbf {e} _{\text{y}}&\equiv \mathbf {e} _{\text{xy}}={\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}}\,,&\mathbf {e} _{\text{z}}\otimes \mathbf {e} _{\text{z}}&\equiv \mathbf {e} _{\text{zz}}={\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}}\end{aligned}}}$$

then T can be represented more systematically as a matrix:

$${\displaystyle \mathbf {T} ={\begin{pmatrix}a_{\text{x}}b_{\text{x}}&a_{\text{x}}b_{\text{y}}&a_{\text{x}}b_{\text{z}}\\a_{\text{y}}b_{\text{x}}&a_{\text{y}}b_{\text{y}}&a_{\text{y}}b_{\text{z}}\\a_{\text{z}}b_{\text{x}}&a_{\text{z}}b_{\text{y}}&a_{\text{z}}b_{\text{z}}\end{pmatrix}}}$$

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

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

$${\displaystyle {\begin{aligned}\mathbf {T} =\quad &T_{\text{xx}}\mathbf {e} _{\text{xx}}+T_{\text{xy}}\mathbf {e} _{\text{xy}}+T_{\text{xz}}\mathbf {e} _{\text{xz}}\\[4pt]{}+{}&T_{\text{yx}}\mathbf {e} _{\text{yx}}+T_{\text{yy}}\mathbf {e} _{\text{yy}}+T_{\text{yz}}\mathbf {e} _{\text{yz}}\\[4pt]{}+{}&T_{\text{zx}}\mathbf {e} _{\text{zx}}+T_{\text{zy}}\mathbf {e} _{\text{zy}}+T_{\text{zz}}\mathbf {e} _{\text{zz}}\end{aligned}}}$$

while in terms of tensor indices:

$${\displaystyle \mathbf {T} =T_{ij}\mathbf {e} _{ij}\equiv \sum _{ij}T_{ij}\mathbf {e} _{i}\otimes \mathbf {e} _{j}\,,}$$

and in matrix form:

$${\displaystyle \mathbf {T} ={\begin{pmatrix}T_{\text{xx}}&T_{\text{xy}}&T_{\text{xz}}\\T_{\text{yx}}&T_{\text{yy}}&T_{\text{yz}}\\T_{\text{zx}}&T_{\text{zy}}&T_{\text{zz}}\end{pmatrix}}}$$

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 v − T(u),[1] while the same idea can be expressed in matrix and index notations[2] (including the summation convention), respectively:

$${\displaystyle {\begin{aligned}{\begin{pmatrix}v_{\text{x}}\\v_{\text{y}}\\v_{\text{z}}\end{pmatrix}}&={\begin{pmatrix}T_{\text{xx}}&T_{\text{xy}}&T_{\text{xz}}\\T_{\text{yx}}&T_{\text{yy}}&T_{\text{yz}}\\T_{\text{zx}}&T_{\text{zy}}&T_{\text{zz}}\end{pmatrix}}{\begin{pmatrix}u_{\text{x}}\\u_{\text{y}}\\u_{\text{z}}\end{pmatrix}}\,,&v_{i}&=T_{ij}u_{j}\end{aligned}}}$$

By "linear", if u = ρr + σs for two scalars ρ and σ and vectors r and s, then in function and index notations:

$${\displaystyle {\begin{aligned}\mathbf {v} &=&&\mathbf {T} (\rho \mathbf {r} +\sigma \mathbf {s} )&=&&\rho \mathbf {T} (\mathbf {r} )+\sigma \mathbf {T} (\mathbf {s} )\\[1ex]v_{i}&=&&T_{ij}(\rho r_{j}+\sigma s_{j})&=&&\rho T_{ij}r_{j}+\sigma T_{ij}s_{j}\end{aligned}}}$$

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 T takes in two vectors p and q, it will return a scalar r. In function notation we write r = T(p, q), while in matrix and index notations (including the summation convention) respectively:

$${\displaystyle r={\begin{pmatrix}p_{\text{x}}&p_{\text{y}}&p_{\text{z}}\end{pmatrix}}{\begin{pmatrix}T_{\text{xx}}&T_{\text{xy}}&T_{\text{xz}}\\T_{\text{yx}}&T_{\text{yy}}&T_{\text{yz}}\\T_{\text{zx}}&T_{\text{zy}}&T_{\text{zz}}\end{pmatrix}}{\begin{pmatrix}q_{\text{x}}\\q_{\text{y}}\\q_{\text{z}}\end{pmatrix}}=p_{i}T_{ij}q_{j}}$$

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:[1][2]

$${\displaystyle {\begin{aligned}\mathbf {v} &=\mathbf {T} \cdot \mathbf {u} \\r&=\mathbf {p} \cdot \mathbf {T} \cdot \mathbf {q} \end{aligned}}}$$

motivated by the dot product notation:

$${\displaystyle \mathbf {a} \cdot \mathbf {b} \equiv a_{i}b_{i}}$$

More generally, a tensor of order m which takes in n vectors (where n is between 0 and m inclusive) will return a tensor of order m − n, see Tensor § As multilinear maps for further generalizations and details. The concepts above also apply to pseudovectors in the same way as for vectors. The vectors and tensors themselves can vary within throughout space, in which case we have vector fields and tensor fields, and can also depend on time.

Following are some examples:

An applied or given...	...to a material or object of...	...results in...	...in the material or object, given by:
unit vector n	Cauchy stress tensor σ	a traction force t	${\displaystyle \mathbf {t} ={\boldsymbol {\sigma }}\cdot \mathbf {n} }$
angular velocity ω	moment of inertia I	an angular momentum J	${\displaystyle \mathbf {J} =\mathbf {I} \cdot {\boldsymbol {\omega }}}$
		a rotational kinetic energy T	${\displaystyle T={\tfrac {1}{2}}{\boldsymbol {\omega }}\cdot \mathbf {I} \cdot {\boldsymbol {\omega }}}$
electric field E	electrical conductivity σ	a current density flow J	${\displaystyle \mathbf {J} ={\boldsymbol {\sigma }}\cdot \mathbf {E} }$
	polarizability α (related to the permittivity ε and electric susceptibility χE)	an induced polarization field P	${\displaystyle \mathbf {P} ={\boldsymbol {\alpha }}\cdot \mathbf {E} }$
magnetic H field	magnetic permeability μ	a magnetic B field	${\displaystyle \mathbf {B} ={\boldsymbol {\mu }}\cdot \mathbf {H} }$

For the electrical conduction example, the index and matrix notations would be:

$${\displaystyle {\begin{aligned}J_{i}&=\sigma _{ij}E_{j}\equiv \sum _{j}\sigma _{ij}E_{j}\\{\begin{pmatrix}J_{\text{x}}\\J_{\text{y}}\\J_{\text{z}}\end{pmatrix}}&={\begin{pmatrix}\sigma _{\text{xx}}&\sigma _{\text{xy}}&\sigma _{\text{xz}}\\\sigma _{\text{yx}}&\sigma _{\text{yy}}&\sigma _{\text{yz}}\\\sigma _{\text{zx}}&\sigma _{\text{zy}}&\sigma _{\text{zz}}\end{pmatrix}}{\begin{pmatrix}E_{\text{x}}\\E_{\text{y}}\\E_{\text{z}}\end{pmatrix}}\end{aligned}}}$$

while for the rotational kinetic energy T:

$${\displaystyle {\begin{aligned}T&={\frac {1}{2}}\omega _{i}I_{ij}\omega _{j}\equiv {\frac {1}{2}}\sum _{ij}\omega _{i}I_{ij}\omega _{j}\,,\\&={\frac {1}{2}}{\begin{pmatrix}\omega _{\text{x}}&\omega _{\text{y}}&\omega _{\text{z}}\end{pmatrix}}{\begin{pmatrix}I_{\text{xx}}&I_{\text{xy}}&I_{\text{xz}}\\I_{\text{yx}}&I_{\text{yy}}&I_{\text{yz}}\\I_{\text{zx}}&I_{\text{zy}}&I_{\text{zz}}\end{pmatrix}}{\begin{pmatrix}\omega _{\text{x}}\\\omega _{\text{y}}\\\omega _{\text{z}}\end{pmatrix}}\,.\end{aligned}}}$$

See also constitutive equation for more specialized examples.

In n-dimensional Euclidean space over the real numbers, ${\displaystyle \mathbb {R} ^{n}}$, the standard basis is denoted e₁, e₂, e₃, ... e_n. Each basis vector e_i points along the positive x_i axis, with the basis being orthonormal. Component j of e_i is given by the Kronecker delta:

$${\displaystyle (\mathbf {e} _{i})_{j}=\delta _{ij}}$$

A vector in ${\displaystyle \mathbb {R} ^{n}}$ takes the form:

$${\displaystyle \mathbf {a} =a_{i}\mathbf {e} _{i}\equiv \sum _{i}a_{i}\mathbf {e} _{i}\,.}$$

Similarly for the order-2 tensor above, for each vector a and b in ${\displaystyle \mathbb {R} ^{n}}$:

$${\displaystyle \mathbf {T} =a_{i}b_{j}\mathbf {e} _{ij}\equiv \sum _{ij}a_{i}b_{j}\mathbf {e} _{i}\otimes \mathbf {e} _{j}\,,}$$

or more generally:

$${\displaystyle \mathbf {T} =T_{ij}\mathbf {e} _{ij}\equiv \sum _{ij}T_{ij}\mathbf {e} _{i}\otimes \mathbf {e} _{j}\,.}$$

The position vector x in ${\displaystyle \mathbb {R} ^{n}}$ is a simple and common example of a vector, and can be represented in any coordinate system. Consider the case of rectangular coordinate systems with orthonormal bases only. It is possible to have a coordinate system with rectangular geometry if the basis vectors are all mutually perpendicular and not normalized, in which case the basis is orthogonal but not orthonormal. However, orthonormal bases are easier to manipulate and are often used in practice. The following results are true for orthonormal bases, not orthogonal ones.

In one rectangular coordinate system, x as a contravector has coordinates xⁱ and basis vectors e_i, while as a covector it has coordinates x_i and basis covectors eⁱ, and we have:

$${\displaystyle {\begin{aligned}\mathbf {x} &=x^{i}\mathbf {e} _{i}\,,&\mathbf {x} &=x_{i}\mathbf {e} ^{i}\end{aligned}}}$$

In another rectangular coordinate system, x as a contravector has coordinates xⁱ and basis e_i, while as a covector it has coordinates x_i and basis eⁱ, and we have:

$${\displaystyle {\begin{aligned}\mathbf {x} &={\bar {x}}^{i}{\bar {\mathbf {e} }}_{i}\,,&\mathbf {x} &={\bar {x}}_{i}{\bar {\mathbf {e} }}^{i}\end{aligned}}}$$

Each new coordinate is a function of all the old ones, and vice versa for the inverse function:

$${\displaystyle {\begin{aligned}{\bar {x}}{}^{i}={\bar {x}}{}^{i}\left(x^{1},x^{2},\ldots \right)\quad &\rightleftharpoons \quad x{}^{i}=x{}^{i}\left({\bar {x}}^{1},{\bar {x}}^{2},\ldots \right)\\{\bar {x}}{}_{i}={\bar {x}}{}_{i}\left(x_{1},x_{2},\ldots \right)\quad &\rightleftharpoons \quad x{}_{i}=x{}_{i}\left({\bar {x}}_{1},{\bar {x}}_{2},\ldots \right)\end{aligned}}}$$

and similarly each new basis vector is a function of all the old ones, and vice versa for the inverse function:

$${\displaystyle {\begin{aligned}{\bar {\mathbf {e} }}{}_{j}={\bar {\mathbf {e} }}{}_{j}\left(\mathbf {e} _{1},\mathbf {e} _{2},\ldots \right)\quad &\rightleftharpoons \quad \mathbf {e} {}_{j}=\mathbf {e} {}_{j}\left({\bar {\mathbf {e} }}_{1},{\bar {\mathbf {e} }}_{2},\ldots \right)\\{\bar {\mathbf {e} }}{}^{j}={\bar {\mathbf {e} }}{}^{j}\left(\mathbf {e} ^{1},\mathbf {e} ^{2},\ldots \right)\quad &\rightleftharpoons \quad \mathbf {e} {}^{j}=\mathbf {e} {}^{j}\left({\bar {\mathbf {e} }}^{1},{\bar {\mathbf {e} }}^{2},\ldots \right)\end{aligned}}}$$

for all i, j.

A vector is invariant under any change of basis, so if coordinates transform according to a transformation matrix L, the bases transform according to the matrix inverse L⁻¹, and conversely if the coordinates transform according to inverse L⁻¹, the bases transform according to the matrix L. The difference between each of these transformations is shown conventionally through the indices as superscripts for contravariance and subscripts for covariance, and the coordinates and bases are linearly transformed according to the following rules:

Vector elements	Contravariant transformation law	Covariant transformation law
Coordinates	${\displaystyle {\bar {x}}^{j}=x^{i}({\boldsymbol {\mathsf {L}}})_{i}{}^{j}=x^{i}{\mathsf {L}}_{i}{}^{j}}$	${\displaystyle {\bar {x}}_{j}=x_{k}\left({\boldsymbol {\mathsf {L}}}^{-1}\right)_{j}{}^{k}}$
Basis	${\displaystyle {\bar {\mathbf {e} }}_{j}=\left({\boldsymbol {\mathsf {L}}}^{-1}\right)_{j}{}^{k}\mathbf {e} _{k}}$	${\displaystyle {\bar {\mathbf {e} }}^{j}=({\boldsymbol {\mathsf {L}}})_{i}{}^{j}\mathbf {e} ^{i}={\mathsf {L}}_{i}{}^{j}\mathbf {e} ^{i}}$
Any vector	${\displaystyle {\bar {x}}^{j}{\bar {\mathbf {e} }}_{j}=x^{i}{\mathsf {L}}_{i}{}^{j}\left({\boldsymbol {\mathsf {L}}}^{-1}\right)_{j}{}^{k}\mathbf {e} _{k}=x^{i}\delta _{i}{}^{k}\mathbf {e} _{k}=x^{i}\mathbf {e} _{i}}$	${\displaystyle {\bar {x}}_{j}{\bar {\mathbf {e} }}^{j}=x_{i}\left({\boldsymbol {\mathsf {L}}}^{-1}\right)_{j}{}^{i}{\mathsf {L}}_{k}{}^{j}\mathbf {e} ^{k}=x_{i}\delta ^{i}{}_{k}\mathbf {e} ^{k}=x_{i}\mathbf {e} ^{i}}$

where L_i^j represents the entries of the transformation matrix (row number is i and column number is j) and (L⁻¹)_i^k denotes the entries of the inverse matrix of the matrix L_i^k.

If L is an orthogonal transformation (orthogonal matrix), the objects transforming by it are defined as Cartesian tensors. This geometrically has the interpretation that a rectangular coordinate system is mapped to another rectangular coordinate system, in which the norm of the vector x is preserved (and distances are preserved).

The determinant of L is det(L) = ±1, which corresponds to two types of orthogonal transformation: (+1) for rotations and (−1) for improper rotations (including reflections).

There are considerable algebraic simplifications, the matrix transpose is the inverse from the definition of an orthogonal transformation:

$${\displaystyle {\boldsymbol {\mathsf {L}}}^{\textsf {T}}={\boldsymbol {\mathsf {L}}}^{-1}\Rightarrow \left({\boldsymbol {\mathsf {L}}}^{-1}\right)_{i}{}^{j}=\left({\boldsymbol {\mathsf {L}}}^{\textsf {T}}\right)_{i}{}^{j}=({\boldsymbol {\mathsf {L}}})^{j}{}_{i}={\mathsf {L}}^{j}{}_{i}}$$

From the previous table, orthogonal transformations of covectors and contravectors are identical. There is no need to differ between raising and lowering indices, and in this context and applications to physics and engineering the indices are usually all subscripted to remove confusion for exponents. All indices will be lowered in the remainder of this article. One can determine the actual raised and lowered indices by considering which quantities are covectors or contravectors, and the relevant transformation rules.

Exactly the same transformation rules apply to any vector a, not only the position vector. If its components a_i do not transform according to the rules, a is not a vector.

Despite the similarity between the expressions above, for the change of coordinates such as x^j = L_i^jxⁱ, and the action of a tensor on a vector like b_i = T_ij a_j, L is not a tensor, but T is. In the change of coordinates, L is a matrix, used to relate two rectangular coordinate systems with orthonormal bases together. For the tensor relating a vector to a vector, the vectors and tensors throughout the equation all belong to the same coordinate system and basis.

The entries of L are partial derivatives of the new or old coordinates with respect to the old or new coordinates, respectively.

Differentiating x_i with respect to x_k:

$${\displaystyle {\frac {\partial {\bar {x}}_{i}}{\partial x_{k}}}={\frac {\partial }{\partial x_{k}}}(x_{j}{\mathsf {L}}_{ji})={\mathsf {L}}_{ji}{\frac {\partial x_{j}}{\partial x_{k}}}=\delta _{kj}{\mathsf {L}}_{ji}={\mathsf {L}}_{ki}}$$

so

$${\displaystyle {{\mathsf {L}}_{i}}^{j}\equiv {\mathsf {L}}_{ij}={\frac {\partial {\bar {x}}_{j}}{\partial x_{i}}}}$$

is an element of the Jacobian matrix. There is a (partially mnemonical) correspondence between index positions attached to L and in the partial derivative: i at the top and j at the bottom, in each case, although for Cartesian tensors the indices can be lowered.

Conversely, differentiating x_j with respect to x_i:

$${\displaystyle {\frac {\partial x_{j}}{\partial {\bar {x}}_{k}}}={\frac {\partial }{\partial {\bar {x}}_{k}}}\left({\bar {x}}_{i}\left({\boldsymbol {\mathsf {L}}}^{-1}\right)_{ij}\right)={\frac {\partial {\bar {x}}_{i}}{\partial {\bar {x}}_{k}}}\left({\boldsymbol {\mathsf {L}}}^{-1}\right)_{ij}=\delta _{ki}\left({\boldsymbol {\mathsf {L}}}^{-1}\right)_{ij}=\left({\boldsymbol {\mathsf {L}}}^{-1}\right)_{kj}}$$

so

$${\displaystyle \left({\boldsymbol {\mathsf {L}}}^{-1}\right)_{i}{}^{j}\equiv \left({\boldsymbol {\mathsf {L}}}^{-1}\right)_{ij}={\frac {\partial x_{j}}{\partial {\bar {x}}_{i}}}}$$

is an element of the inverse Jacobian matrix, with a similar index correspondence.

Many sources state transformations in terms of the partial derivatives:

and the explicit matrix equations in 3d are:

$${\displaystyle {\begin{aligned}{\bar {\mathbf {x} }}&={\boldsymbol {\mathsf {L}}}\mathbf {x} \\{\begin{pmatrix}{\bar {x}}_{1}\\{\bar {x}}_{2}\\{\bar {x}}_{3}\end{pmatrix}}&={\begin{pmatrix}{\frac {\partial {\bar {x}}_{1}}{\partial x_{1}}}&{\frac {\partial {\bar {x}}_{1}}{\partial x_{2}}}&{\frac {\partial {\bar {x}}_{1}}{\partial x_{3}}}\\{\frac {\partial {\bar {x}}_{2}}{\partial x_{1}}}&{\frac {\partial {\bar {x}}_{2}}{\partial x_{2}}}&{\frac {\partial {\bar {x}}_{2}}{\partial x_{3}}}\\{\frac {\partial {\bar {x}}_{3}}{\partial x_{1}}}&{\frac {\partial {\bar {x}}_{3}}{\partial x_{2}}}&{\frac {\partial {\bar {x}}_{3}}{\partial x_{3}}}\end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}\end{aligned}}}$$

similarly for

$${\displaystyle \mathbf {x} ={\boldsymbol {\mathsf {L}}}^{-1}{\bar {\mathbf {x} }}={\boldsymbol {\mathsf {L}}}^{\textsf {T}}{\bar {\mathbf {x} }}}$$

Top: Angles from the x_i axes to the x_i axes. Bottom: Vice versa.

As with all linear transformations, L depends on the basis chosen. For two orthonormal bases

$${\displaystyle {\begin{aligned}{\bar {\mathbf {e} }}_{i}\cdot {\bar {\mathbf {e} }}_{j}&=\mathbf {e} _{i}\cdot \mathbf {e} _{j}=\delta _{ij}\,,&\left|\mathbf {e} _{i}\right|&=\left|{\bar {\mathbf {e} }}_{i}\right|=1\,,\end{aligned}}}$$

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

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

$${\displaystyle {\begin{aligned}{\mathsf {L}}_{ij}&={\bar {\mathbf {e} }}_{i}\cdot \mathbf {e} _{j}=\cos \theta _{ij}\\\left({\boldsymbol {\mathsf {L}}}^{-1}\right)_{ij}&=\mathbf {e} _{i}\cdot {\bar {\mathbf {e} }}_{j}=\cos \theta _{ji}\end{aligned}}}$$

where θ_ij and θ_ji are the angles between the x_i and x_j axes. In general, θ_ij is not equal to θ_ji, because for example θ₁₂ and θ₂₁ are two different angles.

The transformation of coordinates can be written:

and the explicit matrix equations in 3d are:

$${\displaystyle {\begin{aligned}{\bar {\mathbf {x} }}&={\boldsymbol {\mathsf {L}}}\mathbf {x} \\{\begin{pmatrix}{\bar {x}}_{1}\\{\bar {x}}_{2}\\{\bar {x}}_{3}\end{pmatrix}}&={\begin{pmatrix}{\bar {\mathbf {e} }}_{1}\cdot \mathbf {e} _{1}&{\bar {\mathbf {e} }}_{1}\cdot \mathbf {e} _{2}&{\bar {\mathbf {e} }}_{1}\cdot \mathbf {e} _{3}\\{\bar {\mathbf {e} }}_{2}\cdot \mathbf {e} _{1}&{\bar {\mathbf {e} }}_{2}\cdot \mathbf {e} _{2}&{\bar {\mathbf {e} }}_{2}\cdot \mathbf {e} _{3}\\{\bar {\mathbf {e} }}_{3}\cdot \mathbf {e} _{1}&{\bar {\mathbf {e} }}_{3}\cdot \mathbf {e} _{2}&{\bar {\mathbf {e} }}_{3}\cdot \mathbf {e} _{3}\end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}={\begin{pmatrix}\cos \theta _{11}&\cos \theta _{12}&\cos \theta _{13}\\\cos \theta _{21}&\cos \theta _{22}&\cos \theta _{23}\\\cos \theta _{31}&\cos \theta _{32}&\cos \theta _{33}\end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}\end{aligned}}}$$

similarly for

$${\displaystyle \mathbf {x} ={\boldsymbol {\mathsf {L}}}^{-1}{\bar {\mathbf {x} }}={\boldsymbol {\mathsf {L}}}^{\textsf {T}}{\bar {\mathbf {x} }}}$$

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

The numbers e_i⋅e_j arranged into a matrix would form a symmetric matrix (a matrix equal to its own transpose) due to the symmetry in the dot products, in fact it is the metric tensor g. By contrast e_i⋅e_j or e_i⋅e_j do not form symmetric matrices in general, as displayed above. Therefore, while the L matrices are still orthogonal, they are not symmetric.

Apart from a rotation about any one axis, in which the x_i and x_i for some i coincide, the angles are not the same as Euler angles, and so the L matrices are not the same as the rotation matrices.

The dot product ⋅ of each possible pairing of the basis vectors follows from the basis being orthonormal. For perpendicular pairs we have

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

while for parallel pairs we have

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

Replacing Cartesian labels by index notation as shown above, these results can be summarized by

$${\displaystyle \mathbf {e} _{i}\cdot \mathbf {e} _{j}=\delta _{ij}}$$

where δ_ij are the components of the Kronecker delta. The Cartesian basis can be used to represent δ in this way.

In addition, each metric tensor component g_ij with respect to any basis is the dot product of a pairing of basis vectors:

$${\displaystyle g_{ij}=\mathbf {e} _{i}\cdot \mathbf {e} _{j}.}$$

For the Cartesian basis the components arranged into a matrix are:

$${\displaystyle \mathbf {g} ={\begin{pmatrix}g_{\text{xx}}&g_{\text{xy}}&g_{\text{xz}}\\g_{\text{yx}}&g_{\text{yy}}&g_{\text{yz}}\\g_{\text{zx}}&g_{\text{zy}}&g_{\text{zz}}\\\end{pmatrix}}={\begin{pmatrix}\mathbf {e} _{\text{x}}\cdot \mathbf {e} _{\text{x}}&\mathbf {e} _{\text{x}}\cdot \mathbf {e} _{\text{y}}&\mathbf {e} _{\text{x}}\cdot \mathbf {e} _{\text{z}}\\\mathbf {e} _{\text{y}}\cdot \mathbf {e} _{\text{x}}&\mathbf {e} _{\text{y}}\cdot \mathbf {e} _{\text{y}}&\mathbf {e} _{\text{y}}\cdot \mathbf {e} _{\text{z}}\\\mathbf {e} _{\text{z}}\cdot \mathbf {e} _{\text{x}}&\mathbf {e} _{\text{z}}\cdot \mathbf {e} _{\text{y}}&\mathbf {e} _{\text{z}}\cdot \mathbf {e} _{\text{z}}\\\end{pmatrix}}={\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\\\end{pmatrix}}}$$

so are the simplest possible for the metric tensor, namely the δ:

$${\displaystyle g_{ij}=\delta _{ij}}$$

This is not true for general bases: orthogonal coordinates have diagonal metrics containing various scale factors (i.e. not necessarily 1), while general curvilinear coordinates could also have nonzero entries for off-diagonal components.

The dot product of two vectors a and b transforms according to

$${\displaystyle \mathbf {a} \cdot \mathbf {b} ={\bar {a}}_{j}{\bar {b}}_{j}=a_{i}{\mathsf {L}}_{ij}b_{k}\left({\boldsymbol {\mathsf {L}}}^{-1}\right)_{jk}=a_{i}\delta _{i}{}_{k}b_{k}=a_{i}b_{i}}$$

which is intuitive, since the dot product of two vectors is a single scalar independent of any coordinates. This also applies more generally to any coordinate systems, not just rectangular ones; the dot product in one coordinate system is the same in any other.

Cyclic permutations of index values and positively oriented cubic volume.

Anticyclic permutations of index values and negatively oriented cubic volume.

Non-zero values of the Levi-Civita symbol ε_ijk as the volume e_i ⋅ e_j × e_k of a cube spanned by the 3d orthonormal basis.

For the cross product (×) of two vectors, the results are (almost) the other way round. Again, assuming a right-handed 3d Cartesian coordinate system, cyclic permutations in perpendicular directions yield the next vector in the cyclic collection of vectors:

$${\displaystyle {\begin{aligned}\mathbf {e} _{\text{x}}\times \mathbf {e} _{\text{y}}&=\mathbf {e} _{\text{z}}&\mathbf {e} _{\text{y}}\times \mathbf {e} _{\text{z}}&=\mathbf {e} _{\text{x}}&\mathbf {e} _{\text{z}}\times \mathbf {e} _{\text{x}}&=\mathbf {e} _{\text{y}}\\[1ex]\mathbf {e} _{\text{y}}\times \mathbf {e} _{\text{x}}&=-\mathbf {e} _{\text{z}}&\mathbf {e} _{\text{z}}\times \mathbf {e} _{\text{y}}&=-\mathbf {e} _{\text{x}}&\mathbf {e} _{\text{x}}\times \mathbf {e} _{\text{z}}&=-\mathbf {e} _{\text{y}}\end{aligned}}}$$

while parallel vectors clearly vanish:

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

and replacing Cartesian labels by index notation as above, these can be summarized by:

$${\displaystyle \mathbf {e} _{i}\times \mathbf {e} _{j}={\begin{cases}+\mathbf {e} _{k}&{\text{cyclic permutations: }}(i,j,k)=(1,2,3),(2,3,1),(3,1,2)\\[2pt]-\mathbf {e} _{k}&{\text{anticyclic permutations: }}(i,j,k)=(2,1,3),(3,2,1),(1,3,2)\\[2pt]{\boldsymbol {0}}&i=j\end{cases}}}$$

where i, j, k are indices which take values 1, 2, 3. It follows that:

$${\displaystyle {\mathbf {e} _{k}\cdot \mathbf {e} _{i}\times \mathbf {e} _{j}}={\begin{cases}+1&{\text{cyclic permutations: }}(i,j,k)=(1,2,3),(2,3,1),(3,1,2)\\[2pt]-1&{\text{anticyclic permutations: }}(i,j,k)=(2,1,3),(3,2,1),(1,3,2)\\[2pt]0&i=j{\text{ or }}j=k{\text{ or }}k=i\end{cases}}}$$

These permutation relations and their corresponding values are important, and there is an object coinciding with this property: the Levi-Civita symbol, denoted by ε. The Levi-Civita symbol entries can be represented by the Cartesian basis:

$${\displaystyle \varepsilon _{ijk}=\mathbf {e} _{i}\cdot \mathbf {e} _{j}\times \mathbf {e} _{k}}$$

which geometrically corresponds to the volume of a cube spanned by the orthonormal basis vectors, with sign indicating orientation (and not a "positive or negative volume"). Here, the orientation is fixed by ε₁₂₃ = +1, for a right-handed system. A left-handed system would fix ε₁₂₃ = −1 or equivalently ε₃₂₁ = +1.

The scalar triple product can now be written:

$${\displaystyle \mathbf {c} \cdot \mathbf {a} \times \mathbf {b} =c_{i}\mathbf {e} _{i}\cdot a_{j}\mathbf {e} _{j}\times b_{k}\mathbf {e} _{k}=\varepsilon _{ijk}c_{i}a_{j}b_{k}}$$

with the geometric interpretation of volume (of the parallelepiped spanned by a, b, c) and algebraically is a determinant:[3]^: 23

$${\displaystyle \mathbf {c} \cdot \mathbf {a} \times \mathbf {b} ={\begin{vmatrix}c_{\text{x}}&a_{\text{x}}&b_{\text{x}}\\c_{\text{y}}&a_{\text{y}}&b_{\text{y}}\\c_{\text{z}}&a_{\text{z}}&b_{\text{z}}\end{vmatrix}}}$$

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

$${\displaystyle {\begin{aligned}(\mathbf {a} \times \mathbf {b} )_{i}={\mathbf {e} _{i}\cdot \mathbf {a} \times \mathbf {b} }&=\varepsilon _{\ell jk}{(\mathbf {e} _{i})}_{\ell }a_{j}b_{k}=\varepsilon _{\ell jk}\delta _{i\ell }a_{j}b_{k}=\varepsilon _{ijk}a_{j}b_{k}\\\Rightarrow \quad {\mathbf {a} \times \mathbf {b} }=(\mathbf {a} \times \mathbf {b} )_{i}\mathbf {e} _{i}&=\varepsilon _{ijk}a_{j}b_{k}\mathbf {e} _{i}\end{aligned}}}$$

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

$${\displaystyle {\bar {\varepsilon }}_{pqr}=\det({\boldsymbol {\mathsf {L}}})\varepsilon _{ijk}{\mathsf {L}}_{ip}{\mathsf {L}}_{jq}{\mathsf {L}}_{kr}\,.}$$

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

$${\displaystyle {\begin{aligned}&\left({\bar {\mathbf {a} }}\times {\bar {\mathbf {b} }}\right)_{i}\\[1ex]{}={}&{\bar {\varepsilon }}_{ijk}{\bar {a}}_{j}{\bar {b}}_{k}\\[1ex]{}={}&\det({\boldsymbol {\mathsf {L}}})\;\;\varepsilon _{pqr}\;\;{\mathsf {L}}_{pi}{\mathsf {L}}_{qj}{\mathsf {L}}_{rk}\;\;a_{m}{\mathsf {L}}_{mj}\;\;b_{n}{\mathsf {L}}_{nk}\\[1ex]{}={}&\det({\boldsymbol {\mathsf {L}}})\;\;\varepsilon _{pqr}\;\;{\mathsf {L}}_{pi}\;\;{\mathsf {L}}_{qj}\left({\boldsymbol {\mathsf {L}}}^{-1}\right)_{jm}\;\;{\mathsf {L}}_{rk}\left({\boldsymbol {\mathsf {L}}}^{-1}\right)_{kn}\;\;a_{m}\;\;b_{n}\\[1ex]{}={}&\det({\boldsymbol {\mathsf {L}}})\;\;\varepsilon _{pqr}\;\;{\mathsf {L}}_{pi}\;\;\delta _{qm}\;\;\delta _{rn}\;\;a_{m}\;\;b_{n}\\[1ex]{}={}&\det({\boldsymbol {\mathsf {L}}})\;\;{\mathsf {L}}_{pi}\;\;\varepsilon _{pqr}a_{q}b_{r}\\[1ex]{}={}&\det({\boldsymbol {\mathsf {L}}})\;\;(\mathbf {a} \times \mathbf {b} )_{p}{\mathsf {L}}_{pi}\end{aligned}}}$$

and so a × b 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.

Other identities can be formed from the δ tensor and ε 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:

$${\displaystyle \varepsilon _{ijk}\varepsilon _{pqk}=\delta _{ip}\delta _{jq}-\delta _{iq}\delta _{jp}}$$

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:

$${\displaystyle {\begin{aligned}\mathbf {a} \cdot (\mathbf {b} +\mathbf {c} )&=a_{i}(b_{i}+c_{i})=a_{i}b_{i}+a_{i}c_{i}=\mathbf {a} \cdot \mathbf {b} +\mathbf {a} \cdot \mathbf {c} \\[1ex]\mathbf {a} \times (\mathbf {b} +\mathbf {c} )&=\mathbf {e} _{i}\varepsilon _{ijk}a_{j}(b_{k}+c_{k})=\mathbf {e} _{i}\varepsilon _{ijk}a_{j}b_{k}+\mathbf {e} _{i}\varepsilon _{ijk}a_{j}c_{k}=\mathbf {a} \times \mathbf {b} +\mathbf {a} \times \mathbf {c} \end{aligned}}}$$

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:

$${\displaystyle \left[\mathbf {a} \times (\mathbf {b} \times \mathbf {c} )\right]_{i}=\varepsilon _{ijk}a_{j}(\varepsilon _{k\ell m}b_{\ell }c_{m})=(\varepsilon _{ijk}\varepsilon _{k\ell m})a_{j}b_{\ell }c_{m}}$$

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

$${\displaystyle {\begin{aligned}\left[\mathbf {a} \times (\mathbf {b} \times \mathbf {c} )\right]_{i}{}={}&\left(\delta _{i\ell }\delta _{jm}-\delta _{im}\delta _{j\ell }\right)a_{j}b_{\ell }c_{m}\\{}={}&\delta _{i\ell }\delta _{jm}a_{j}b_{\ell }c_{m}-\delta _{im}\delta _{j\ell }a_{j}b_{\ell }c_{m}\\{}={}&a_{j}b_{i}c_{j}-a_{j}b_{j}c_{i}\\{}={}&\left[(\mathbf {a} \cdot \mathbf {c} )\mathbf {b} -(\mathbf {a} \cdot \mathbf {b} )\mathbf {c} \right]_{i}\end{aligned}}}$$

thusly:

$${\displaystyle \mathbf {a} \times (\mathbf {b} \times \mathbf {c} )=(\mathbf {a} \cdot \mathbf {c} )\mathbf {b} -(\mathbf {a} \cdot \mathbf {b} )\mathbf {c} }$$

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

$${\displaystyle (\mathbf {a} \times \mathbf {b} )\times \mathbf {c} =(\mathbf {c} \cdot \mathbf {a} )\mathbf {b} -(\mathbf {c} \cdot \mathbf {b} )\mathbf {a} }$$

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

$${\displaystyle (\mathbf {a} \times \mathbf {b} )\cdot (\mathbf {c} \times \mathbf {d} ),\quad (\mathbf {a} \times \mathbf {b} )\times (\mathbf {c} \times \mathbf {d} ),\quad \ldots }$$

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

Let a = a_ie_i and b = b_ie_i be two vectors, so that they transform according to a_j = a_iL_ij, b_j = b_iL_ij.

Taking the tensor product gives:

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

then applying the transformation to the components

$${\displaystyle {\bar {a}}_{p}{\bar {b}}_{q}=a_{i}{\mathsf {L}}_{i}{}_{p}b_{j}{\mathsf {L}}_{j}{}_{q}={\mathsf {L}}_{i}{}_{p}{\mathsf {L}}_{j}{}_{q}a_{i}b_{j}}$$

and to the bases