Dyadics

A dyad is a tensor of order two and rank one, and is the dyadic product of two vectors (complex vectors in general), whereas a dyadic is a general tensor of order two (which may be full rank or not).

There are several equivalent terms and notations for this product:

• the dyadic product of two vectors ${\displaystyle \mathbf {a} }$ and ${\displaystyle \mathbf {b} }$ is denoted by ${\displaystyle \mathbf {a} \mathbf {b} }$ (juxtaposed; no symbols, multiplication signs, crosses, dots, etc.)
• the outer product of two column vectors ${\displaystyle \mathbf {a} }$ and ${\displaystyle \mathbf {b} }$ is denoted and defined as ${\displaystyle \mathbf {a} \otimes \mathbf {b} }$ or ${\displaystyle \mathbf {a} \mathbf {b} ^{\mathsf {T}}}$, where ${\displaystyle {\mathsf {T}}}$ means transpose,
• the tensor product of two vectors ${\displaystyle \mathbf {a} }$ and ${\displaystyle \mathbf {b} }$ is denoted ${\displaystyle \mathbf {a} \otimes \mathbf {b} }$,

In the dyadic context they all have the same definition and meaning, and are used synonymously, although the tensor product is an instance of the more general and abstract use of the term.

Dirac's bra–ket notation makes the use of dyads and dyadics intuitively clear, see Cahill (2013).

To illustrate the equivalent usage, consider three-dimensional Euclidean space, letting:

${\displaystyle {\begin{aligned}\mathbf {a} &=a_{1}\mathbf {i} +a_{2}\mathbf {j} +a_{3}\mathbf {k} \\\mathbf {b} &=b_{1}\mathbf {i} +b_{2}\mathbf {j} +b_{3}\mathbf {k} \end{aligned}}}$

be two vectors where i, j, k (also denoted e₁, e₂, e₃) are the standard basis vectors in this vector space (see also Cartesian coordinates). Then the dyadic product of a and b can be represented as a sum:

${\displaystyle {\begin{aligned}\mathbf {ab} =\qquad &a_{1}b_{1}\mathbf {ii} +a_{1}b_{2}\mathbf {ij} +a_{1}b_{3}\mathbf {ik} \\{}+{}&a_{2}b_{1}\mathbf {ji} +a_{2}b_{2}\mathbf {jj} +a_{2}b_{3}\mathbf {jk} \\{}+{}&a_{3}b_{1}\mathbf {ki} +a_{3}b_{2}\mathbf {kj} +a_{3}b_{3}\mathbf {kk} \end{aligned}}}$

or by extension from row and column vectors, a 3×3 matrix (also the result of the outer product or tensor product of a and b):

${\displaystyle \mathbf {ab} \equiv \mathbf {a} \otimes \mathbf {b} \equiv \mathbf {ab} ^{\mathsf {T}}={\begin{pmatrix}a_{1}\\a_{2}\\a_{3}\end{pmatrix}}{\begin{pmatrix}b_{1}&b_{2}&b_{3}\end{pmatrix}}={\begin{pmatrix}a_{1}b_{1}&a_{1}b_{2}&a_{1}b_{3}\\a_{2}b_{1}&a_{2}b_{2}&a_{2}b_{3}\\a_{3}b_{1}&a_{3}b_{2}&a_{3}b_{3}\end{pmatrix}}.}$

A dyad is a component of the dyadic (a monomial of the sum or equivalently an entry of the matrix) — the dyadic product of a pair of basis vectors scalar multiplied by a number.

Just as the standard basis (and unit) vectors i, j, k, have the representations:

${\displaystyle {\begin{aligned}\mathbf {i} &={\begin{pmatrix}1\\0\\0\end{pmatrix}},&\mathbf {j} &={\begin{pmatrix}0\\1\\0\end{pmatrix}},&\mathbf {k} &={\begin{pmatrix}0\\0\\1\end{pmatrix}}\end{aligned}}}$

(which can be transposed), the standard basis (and unit) dyads have the representation:

${\displaystyle {\begin{aligned}\mathbf {ii} &={\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}},&\mathbf {ij} &={\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}},&\mathbf {ik} &={\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}}\\\mathbf {ji} &={\begin{pmatrix}0&0&0\\1&0&0\\0&0&0\end{pmatrix}},&\mathbf {jj} &={\begin{pmatrix}0&0&0\\0&1&0\\0&0&0\end{pmatrix}},&\mathbf {jk} &={\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix}}\\\mathbf {ki} &={\begin{pmatrix}0&0&0\\0&0&0\\1&0&0\end{pmatrix}},&\mathbf {kj} &={\begin{pmatrix}0&0&0\\0&0&0\\0&1&0\end{pmatrix}},&\mathbf {kk} &={\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}}\end{aligned}}}$

For a simple numerical example in the standard basis:

${\displaystyle {\begin{aligned}\mathbf {A} &=2\mathbf {ij} +{\frac {\sqrt {3}}{2}}\mathbf {ji} -8\pi \mathbf {jk} +{\frac {2{\sqrt {2}}}{3}}\mathbf {kk} \\[2pt]&=2{\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}}+{\frac {\sqrt {3}}{2}}{\begin{pmatrix}0&0&0\\1&0&0\\0&0&0\end{pmatrix}}-8\pi {\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix}}+{\frac {2{\sqrt {2}}}{3}}{\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}}\\[2pt]&={\begin{pmatrix}0&2&0\\{\frac {\sqrt {3}}{2}}&0&-8\pi \\0&0&{\frac {2{\sqrt {2}}}{3}}\end{pmatrix}}\end{aligned}}}$

If the Euclidean space is N-dimensional, and

${\displaystyle {\begin{aligned}\mathbf {a} &=\sum _{i=1}^{N}a_{i}\mathbf {e} _{i}=a_{1}\mathbf {e} _{1}+a_{2}\mathbf {e} _{2}+{\ldots }+a_{N}\mathbf {e} _{N}\\\mathbf {b} &=\sum _{j=1}^{N}b_{j}\mathbf {e} _{j}=b_{1}\mathbf {e} _{1}+b_{2}\mathbf {e} _{2}+\ldots +b_{N}\mathbf {e} _{N}\end{aligned}}}$

where e_i and e_j are the standard basis vectors in N-dimensions (the index i on e_i selects a specific vector, not a component of the vector as in a_i), then in algebraic form their dyadic product is:

${\displaystyle \mathbf {ab} =\sum _{j=1}^{N}\sum _{i=1}^{N}a_{i}b_{j}\mathbf {e} _{i}\mathbf {e} _{j}.}$

This is known as the nonion form of the dyadic. Their outer/tensor product in matrix form is:

${\displaystyle \mathbf {ab} =\mathbf {ab} ^{\mathsf {T}}={\begin{pmatrix}a_{1}\\a_{2}\\\vdots \\a_{N}\end{pmatrix}}{\begin{pmatrix}b_{1}&b_{2}&\cdots &b_{N}\end{pmatrix}}={\begin{pmatrix}a_{1}b_{1}&a_{1}b_{2}&\cdots &a_{1}b_{N}\\a_{2}b_{1}&a_{2}b_{2}&\cdots &a_{2}b_{N}\\\vdots &\vdots &\ddots &\vdots \\a_{N}b_{1}&a_{N}b_{2}&\cdots &a_{N}b_{N}\end{pmatrix}}.}$

A dyadic polynomial A, otherwise known as a dyadic, is formed from multiple vectors a_i and b_j:

${\displaystyle \mathbf {A} =\sum _{i}\mathbf {a} _{i}\mathbf {b} _{i}=\mathbf {a} _{1}\mathbf {b} _{1}+\mathbf {a} _{2}\mathbf {b} _{2}+\mathbf {a} _{3}\mathbf {b} _{3}+\ldots }$

A dyadic which cannot be reduced to a sum of less than N dyads is said to be complete. In this case, the forming vectors are non-coplanar, see Chen (1983).

The following table classifies dyadics:

	Determinant	Adjugate	Matrix and its rank
Zero	= 0	= 0	= 0; rank 0: all zeroes
Linear	= 0	= 0	≠ 0; rank 1: at least one non-zero element and all 2 × 2 subdeterminants zero (single dyadic)
Planar	= 0	≠ 0 (single dyadic)	≠ 0; rank 2: at least one non-zero 2 × 2 subdeterminant
Complete	≠ 0	≠ 0	≠ 0; rank 3: non-zero determinant

The following identities are a direct consequence of the definition of the tensor product:[1]

There are four operations defined on a vector and dyadic, constructed from the products defined on vectors.

	Left	Right
Dot product	${\displaystyle \mathbf {c} \cdot \left(\mathbf {a} \mathbf {b} \right)=\left(\mathbf {c} \cdot \mathbf {a} \right)\mathbf {b} }$	${\displaystyle \left(\mathbf {a} \mathbf {b} \right)\cdot \mathbf {c} =\mathbf {a} \left(\mathbf {b} \cdot \mathbf {c} \right)}$
Cross product	${\displaystyle \mathbf {c} \times \left(\mathbf {ab} \right)=\left(\mathbf {c} \times \mathbf {a} \right)\mathbf {b} }$	${\displaystyle \left(\mathbf {ab} \right)\times \mathbf {c} =\mathbf {a} \left(\mathbf {b} \times \mathbf {c} \right)}$

There are five operations for a dyadic to another dyadic. Let a, b, c, d be real vectors. Then:

		Dot	Cross
Dot	Dot product ${\displaystyle {\begin{aligned}\left(\mathbf {a} \mathbf {b} \right)\cdot \left(\mathbf {c} \mathbf {d} \right)&=\mathbf {a} \left(\mathbf {b} \cdot \mathbf {c} \right)\mathbf {d} \\&=\left(\mathbf {b} \cdot \mathbf {c} \right)\mathbf {a} \mathbf {d} \end{aligned}}}$	Double-dot product ${\displaystyle {\begin{aligned}\left(\mathbf {ab} \right){}_{\,\centerdot }^{\,\centerdot }\left(\mathbf {cd} \right)&=\mathbf {c} \cdot \left(\mathbf {ab} \right)\cdot \mathbf {d} \\&=\left(\mathbf {a} \cdot \mathbf {c} \right)\left(\mathbf {b} \cdot \mathbf {d} \right)\end{aligned}}}$ and ${\displaystyle \mathbf {ab} {\underline {{}_{\,\centerdot }^{\,\centerdot }}}\mathbf {cd} =\left(\mathbf {a} \cdot \mathbf {d} \right)\left(\mathbf {b} \cdot \mathbf {c} \right)}$	Dot–cross product ${\displaystyle \left(\mathbf {ab} \right){}_{\,\centerdot }^{\times }\left(\mathbf {c} \mathbf {d} \right)=\left(\mathbf {a} \cdot \mathbf {c} \right)\left(\mathbf {b} \times \mathbf {d} \right)}$
Cross		Cross–dot product ${\displaystyle \left(\mathbf {ab} \right){}_{\times }^{\,\centerdot }\left(\mathbf {cd} \right)=\left(\mathbf {a} \times \mathbf {c} \right)\left(\mathbf {b} \cdot \mathbf {d} \right)}$	Double cross product ${\displaystyle \left(\mathbf {ab} \right){}_{\times }^{\times }\left(\mathbf {cd} \right)=\left(\mathbf {a} \times \mathbf {c} \right)\left(\mathbf {b} \times \mathbf {d} \right)}$

Letting

${\displaystyle \mathbf {A} =\sum _{i}\mathbf {a} _{i}\mathbf {b} _{i},\quad \mathbf {B} =\sum _{j}\mathbf {c} _{j}\mathbf {d} _{j}}$

be two general dyadics, we have:

		Dot	Cross
Dot	Dot product ${\displaystyle \mathbf {A} \cdot \mathbf {B} =\sum _{i,j}\left(\mathbf {b} _{i}\cdot \mathbf {c} _{j}\right)\mathbf {a} _{i}\mathbf {d} _{j}}$	Double dot product ${\displaystyle {\begin{aligned}\mathbf {A} {}_{\,\centerdot }^{\,\centerdot }\mathbf {B} &=\sum _{i,j}\left(\mathbf {a} _{i}\cdot \mathbf {c} _{j}\right)\left(\mathbf {b} _{i}\cdot \mathbf {d} _{j}\right)\end{aligned}}}$ and ${\displaystyle {\begin{aligned}\mathbf {A} {\underline {{}_{\,\centerdot }^{\,\centerdot }}}\mathbf {B} &=\sum _{i,j}\left(\mathbf {a} _{i}\cdot \mathbf {d} _{j}\right)\left(\mathbf {b} _{i}\cdot \mathbf {c} _{j}\right)\end{aligned}}}$	Dot–cross product ${\displaystyle \mathbf {A} {}_{\,\centerdot }^{\times }\mathbf {B} =\sum _{i,j}\left(\mathbf {a} _{i}\cdot \mathbf {c} _{j}\right)\left(\mathbf {b} _{i}\times \mathbf {d} _{j}\right)}$
Cross		Cross–dot product ${\displaystyle \mathbf {A} {}_{\times }^{\,\centerdot }\mathbf {B} =\sum _{i,j}\left(\mathbf {a} _{i}\times \mathbf {c} _{j}\right)\left(\mathbf {b} _{i}\cdot \mathbf {d} _{j}\right)}$	Double cross product ${\displaystyle \mathbf {A} {}_{\times }^{\times }\mathbf {B} =\sum _{i,j}\left(\mathbf {a} _{i}\times \mathbf {c} _{j}\right)\left(\mathbf {b} _{i}\times \mathbf {d} _{j}\right)}$

The first definition of the double-dot product is the Frobenius inner product,

${\displaystyle {\begin{aligned}\operatorname {tr} \left(\mathbf {A} \mathbf {B} ^{\mathsf {T}}\right)&=\sum _{i,j}\operatorname {tr} \left(\mathbf {a} _{i}\mathbf {b} _{i}^{\mathsf {T}}\mathbf {d} _{j}\mathbf {c} _{j}^{\mathsf {T}}\right)\\&=\sum _{i,j}\operatorname {tr} \left(\mathbf {c} _{j}^{\mathsf {T}}\mathbf {a} _{i}\mathbf {b} _{i}^{\mathsf {T}}\mathbf {d} _{j}\right)\\&=\sum _{i,j}(\mathbf {a} _{i}\cdot \mathbf {c} _{j})(\mathbf {b} _{i}\cdot \mathbf {d} _{j})\\&=\mathbf {A} {}_{\centerdot }^{\centerdot }\mathbf {B} \end{aligned}}}$

Furthermore, since,

${\displaystyle {\begin{aligned}\mathbf {A} ^{\mathsf {T}}&=\sum _{i,j}\left(\mathbf {a} _{i}\mathbf {b} _{j}^{\mathsf {T}}\right)^{\mathsf {T}}\\&=\sum _{i,j}\mathbf {b} _{i}\mathbf {a} _{j}^{\mathsf {T}}\end{aligned}}}$

we get that,

${\displaystyle \mathbf {A} {}_{\centerdot }^{\centerdot }\mathbf {B} =\mathbf {A} {\underline {{}_{\centerdot }^{\centerdot }}}\mathbf {B} ^{\mathsf {T}}}$

so the second possible definition of the double-dot product is just the first with an additional transposition on the second dyadic. For these reasons, the first definition of the double-dot product is preferred, though some authors still use the second.

We can see that, for any dyad formed from two vectors a and b, its double cross product is zero.

${\displaystyle \left(\mathbf {ab} \right){}_{\times }^{\times }\left(\mathbf {ab} \right)=\left(\mathbf {a} \times \mathbf {a} \right)\left(\mathbf {b} \times \mathbf {b} \right)=0}$

However, by definition, a dyadic double-cross product on itself will generally be non-zero. For example, a dyadic A composed of six different vectors

${\displaystyle \mathbf {A} =\sum _{i=1}^{3}\mathbf {a} _{i}\mathbf {b} _{i}}$

has a non-zero self-double-cross product of

${\displaystyle \mathbf {A} {}_{\times }^{\times }\mathbf {A} =2\left[\left(\mathbf {a} _{1}\times \mathbf {a} _{2}\right)\left(\mathbf {b} _{1}\times \mathbf {b} _{2}\right)+\left(\mathbf {a} _{2}\times \mathbf {a} _{3}\right)\left(\mathbf {b} _{2}\times \mathbf {b} _{3}\right)+\left(\mathbf {a} _{3}\times \mathbf {a} _{1}\right)\left(\mathbf {b} _{3}\times \mathbf {b} _{1}\right)\right]}$

The spur or expansion factor arises from the formal expansion of the dyadic in a coordinate basis by replacing each dyadic product by a dot product of vectors:

${\displaystyle {\begin{aligned}|\mathbf {A} |=\qquad &A_{11}\mathbf {i} \cdot \mathbf {i} +A_{12}\mathbf {i} \cdot \mathbf {j} +A_{13}\mathbf {i} \cdot \mathbf {k} \\{}+{}&A_{21}\mathbf {j} \cdot \mathbf {i} +A_{22}\mathbf {j} \cdot \mathbf {j} +A_{23}\mathbf {j} \cdot \mathbf {k} \\{}+{}&A_{31}\mathbf {k} \cdot \mathbf {i} +A_{32}\mathbf {k} \cdot \mathbf {j} +A_{33}\mathbf {k} \cdot \mathbf {k} \\[6pt]=\qquad &A_{11}+A_{22}+A_{33}\end{aligned}}}$

in index notation this is the contraction of indices on the dyadic:

${\displaystyle |\mathbf {A} |=\sum _{i}A_{i}{}^{i}}$

In three dimensions only, the rotation factor arises by replacing every dyadic product by a cross product

${\displaystyle {\begin{aligned}\langle \mathbf {A} \rangle =\qquad &A_{11}\mathbf {i} \times \mathbf {i} +A_{12}\mathbf {i} \times \mathbf {j} +A_{13}\mathbf {i} \times \mathbf {k} \\{}+{}&A_{21}\mathbf {j} \times \mathbf {i} +A_{22}\mathbf {j} \times \mathbf {j} +A_{23}\mathbf {j} \times \mathbf {k} \\{}+{}&A_{31}\mathbf {k} \times \mathbf {i} +A_{32}\mathbf {k} \times \mathbf {j} +A_{33}\mathbf {k} \times \mathbf {k} \\[6pt]=\qquad &A_{12}\mathbf {k} -A_{13}\mathbf {j} -A_{21}\mathbf {k} \\{}+{}&A_{23}\mathbf {i} +A_{31}\mathbf {j} -A_{32}\mathbf {i} \\[6pt]=\qquad &\left(A_{23}-A_{32}\right)\mathbf {i} +\left(A_{31}-A_{13}\right)\mathbf {j} +\left(A_{12}-A_{21}\right)\mathbf {k} \\\end{aligned}}}$

In index notation this is the contraction of A with the Levi-Civita tensor

${\displaystyle \langle \mathbf {A} \rangle =\sum _{jk}{\epsilon _{i}}^{jk}A_{jk}.}$

${\displaystyle {\begin{aligned}\left(\mathbf {a} \times \mathbf {I} \right)\cdot \left(\mathbf {b} \times \mathbf {I} \right)&=\mathbf {ba} -\left(\mathbf {a} \cdot \mathbf {b} \right)\mathbf {I} \\\mathbf {I} {}_{\times }^{\,\centerdot }\left(\mathbf {ab} \right)&=\mathbf {b} \times \mathbf {a} \\\mathbf {I} {}_{\times }^{\times }\mathbf {A} &=(\mathbf {A} {}_{\,\centerdot }^{\,\centerdot }\mathbf {I} )\mathbf {I} -\mathbf {A} ^{\mathsf {T}}\\\mathbf {I} {}_{\,\centerdot }^{\,\centerdot }\left(\mathbf {ab} \right)&=\left(\mathbf {I} \cdot \mathbf {a} \right)\cdot \mathbf {b} =\mathbf {a} \cdot \mathbf {b} =\mathrm {tr} \left(\mathbf {ab} \right)\end{aligned}}}$

where "tr" denotes the trace.

A nonzero vector a can always be split into two perpendicular components, one parallel (‖) to the direction of a unit vector n, and one perpendicular (⊥) to it;

${\displaystyle \mathbf {a} =\mathbf {a} _{\parallel }+\mathbf {a} _{\perp }}$

The parallel component is found by vector projection, which is equivalent to the dot product of a with the dyadic nn,

${\displaystyle \mathbf {a} _{\parallel }=\mathbf {n} (\mathbf {n} \cdot \mathbf {a} )=(\mathbf {nn} )\cdot \mathbf {a} }$

and the perpendicular component is found from vector rejection, which is equivalent to the dot product of a with the dyadic I − nn,

${\displaystyle \mathbf {a} _{\perp }=\mathbf {a} -\mathbf {n} (\mathbf {n} \cdot \mathbf {a} )=(\mathbf {I} -\mathbf {nn} )\cdot \mathbf {a} }$

The dyadic

${\displaystyle \mathbf {J} =\mathbf {ji} -\mathbf {ij} ={\begin{pmatrix}0&-1\\1&0\end{pmatrix}}}$

is a 90° anticlockwise rotation operator in 2d. It can be left-dotted with a vector r = xi + yj to produce the vector,

${\displaystyle (\mathbf {ji} -\mathbf {ij} )\cdot (x\mathbf {i} +y\mathbf {j} )=x\mathbf {ji} \cdot \mathbf {i} -x\mathbf {ij} \cdot \mathbf {i} +y\mathbf {ji} \cdot \mathbf {j} -y\mathbf {ij} \cdot \mathbf {j} =-y\mathbf {i} +x\mathbf {j} ,}$

in summary

${\displaystyle \mathbf {J} \cdot \mathbf {r} =\mathbf {r} _{\mathrm {rot} }}$

or in matrix notation

${\displaystyle {\begin{pmatrix}0&-1\\1&0\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}-y\\x\end{pmatrix}}.}$

For any angle θ, the 2d rotation dyadic for a rotation anti-clockwise in the plane is

${\displaystyle \mathbf {R} =\mathbf {I} \cos \theta +\mathbf {J} \sin \theta =(\mathbf {ii} +\mathbf {jj} )\cos \theta +(\mathbf {ji} -\mathbf {ij} )\sin \theta ={\begin{pmatrix}\cos \theta &-\sin \theta \\\sin \theta &\;\cos \theta \end{pmatrix}}}$

where I and J are as above, and the rotation of any 2d vector a = a_xi + a_yj is

${\displaystyle \mathbf {a} _{\mathrm {rot} }=\mathbf {R} \cdot \mathbf {a} }$

A general 3d rotation of a vector a, about an axis in the direction of a unit vector ω and anticlockwise through angle θ, can be performed using Rodrigues' rotation formula in the dyadic form

${\displaystyle \mathbf {a} _{\mathrm {rot} }=\mathbf {R} \cdot \mathbf {a} \,,}$

where the rotation dyadic is

${\displaystyle \mathbf {R} =\mathbf {I} \cos \theta +{\boldsymbol {\Omega }}\sin \theta +{\boldsymbol {\omega \omega }}(1-\cos \theta )\,,}$

and the Cartesian entries of ω also form those of the dyadic

${\displaystyle {\boldsymbol {\Omega }}=\omega _{x}(\mathbf {kj} -\mathbf {jk} )+\omega _{y}(\mathbf {ik} -\mathbf {ki} )+\omega _{z}(\mathbf {ji} -\mathbf {ij} )\,,}$

The effect of Ω on a is the cross product

${\displaystyle {\boldsymbol {\Omega }}\cdot \mathbf {a} ={\boldsymbol {\omega }}\times \mathbf {a} }$

which is the dyadic form the cross product matrix with a column vector.

In special relativity, the Lorentz boost with speed v in the direction of a unit vector n can be expressed as

${\displaystyle t'=\gamma \left(t-{\frac {v\mathbf {n} \cdot \mathbf {r} }{c^{2}}}\right)}$

${\displaystyle \mathbf {r} '=[\mathbf {I} +(\gamma -1)\mathbf {nn} ]\cdot \mathbf {r} -\gamma v\mathbf {n} t}$

where

${\displaystyle \gamma ={\frac {1}{\sqrt {1-{\dfrac {v^{2}}{c^{2}}}}}}}$

is the Lorentz factor.