每日彙整: 2018-04-07

樹莓派、樹莓派之學習、樹莓派之教育

STEM 隨筆︰古典力學︰運動學【二．六．三】

即使為著閱讀科技文獻之目的，也得了解一些張量代數！

Dyadic algebra

Product of dyadic and vector

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)

Product of dyadic and dyadic

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

    Dot Cross
Dot Dot product\displaystyle \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} Double dot product\displaystyle \mathbf {ab} \colon \mathbf {cd} =\left(\mathbf {a} \cdot \mathbf {d} \right)\left(\mathbf {b} \cdot \mathbf {c} \right)

or

\displaystyle \left(\mathbf {ab} \right):\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)

 Dot–cross product\displaystyle \left(\mathbf {ab} \right)\!\!\!{\begin{array}{c}_{\cdot }\\^{\times }\end{array}}\!\!\!\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)\!\!\!{\begin{array}{c}_{\times }\\^{\cdot }\end{array}}\!\!\!\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)\!\!\!{\begin{array}{c}_{\times }\\^{\times }\end{array}}\!\!\!\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 _{j}\sum _{i}\left(\mathbf {b} _{i}\cdot \mathbf {c} _{j}\right)\mathbf {a} _{i}\mathbf {d} _{j}

 

 Double dot product\displaystyle \mathbf {A} \colon \mathbf {B} =\sum _{j}\sum _{i}\left(\mathbf {a} _{i}\cdot \mathbf {d} _{j}\right)\left(\mathbf {b} _{i}\cdot \mathbf {c} _{j}\right)

or

\displaystyle \mathbf {A} \colon \mathbf {B} =\sum _{j}\sum _{i}\left(\mathbf {a} _{i}\cdot \mathbf {c} _{j}\right)\left(\mathbf {b} _{i}\cdot \mathbf {d} _{j}\right)

 Dot–cross product\displaystyle \mathbf {A} \!\!\!{\begin{array}{c}_{\cdot }\\^{\times }\end{array}}\!\!\!\mathbf {B} =\sum _{j}\sum _{i}\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} \!\!\!{\begin{array}{c}_{\times }\\^{\cdot }\end{array}}\!\!\!\mathbf {B} =\sum _{j}\sum _{i}\left(\mathbf {a} _{i}\times \mathbf {c} _{j}\right)\left(\mathbf {b} _{i}\cdot \mathbf {d} _{j}\right) Double cross product\displaystyle \mathbf {A} \!\!\!{\begin{array}{c}_{\times }\\^{\times }\end{array}}\!\!\!\mathbf {B} =\sum _{i,j}\left(\mathbf {a} _{i}\times \mathbf {c} _{j}\right)\left(\mathbf {b} _{i}\times \mathbf {d} _{j}\right)

Double-dot product

There are two ways to define the double dot product, one must be careful when deciding which convention to use. As there are no analogous matrix operations for the remaining dyadic products, no ambiguities in their definitions appear.

The double-dot product is commutative due to commutativity of the normal dot-product:

\displaystyle \mathbf {A} \colon \!\mathbf {B} =\mathbf {B} \colon \!\mathbf {A}

There is a special double dot product with a transpose
\displaystyle \mathbf {A} \colon \!\mathbf {B} ^{\mathrm {T} }=\mathbf {A} ^{\mathrm {T} }\colon \!\mathbf {B}
Another identity is:
\displaystyle \mathbf {A} \colon \mathbf {B} =\left(\mathbf {A} \cdot \mathbf {B} ^{\mathrm {T} }\right)\colon \mathbf {I} =\left(\mathbf {B} \cdot \mathbf {A} ^{\mathrm {T} }\right)\colon \mathbf {I}

Double-cross product

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)\!\!\!{\begin{array}{c}_{\times }\\^{\times }\end{array}}\!\!\!\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} \!\!\!{\begin{array}{c}_{\times }\\^{\times }\end{array}}\!\!\!\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]

Tensor contraction

Main article: Tensor contraction

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{array}{llll}|\mathbf {A} |&=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} \\\\&=A_{11}+A_{22}+A_{33}\\\end{array}}

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{array}{llll}\langle \mathbf {A} \rangle &=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} \\\\&=A_{12}\mathbf {k} -A_{31}\mathbf {j} -A_{21}\mathbf {k} \\&+A_{23}\mathbf {i} +A_{31}\mathbf {j} -A_{32}\mathbf {i} \\\\&=(A_{23}-A_{32})\mathbf {i} +(A_{31}-A_{13})\mathbf {j} +(A_{12}-A_{21})\mathbf {k} \\\end{array}}
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}.

 

但是又如何能靠短短維基百科幾行摘要文本而得到呢？

如果輔以功能強大之 SymPy 電腦代數系統，借著練習範例，可否自學乎？？

Rotation dyadic

Further information: Axis–angle representation

2d rotations

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 = axi + ayj is
\displaystyle \mathbf {a} _{\mathrm {rot} }=\mathbf {R} \cdot \mathbf {a}

※ 自習

 

一位追求自問？

3d rotations

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 +\sin \theta {\boldsymbol {\Omega }}+(1-\cos \theta ){\boldsymbol {\omega \omega }}\,,
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.

自答者！自得心證，自能知之耶！！