| Version 6 |
Version 5 |
| A third kind of ``products'' between two Euclidean vectors $\vec{a}$ and $\vec{b}$, besides the scalar product $\vec{a}\!\cdot\!\vec{b}$ and the vector product $\vec{a}\!\times\!\vec{b}$, is the\, {\em dyad product}\, $\vec{a}\,\vec{b}$,\, which is usually denoted without any multiplication symbol.\, The dyad products and the finite sums |
A third kind of ``products'' between two Euclidean vectors $\vec{a}$ and $\vec{b}$, besides the scalar product $\vec{a}\!\cdot\!\vec{b}$ and the vector product $\vec{a}\!\times\!\vec{b}$, is the\, {\em dyad product}\, $\vec{a}\,\vec{b}$,\, which is usually denoted without any multiplication symbol.\, The dyad products and the finite sums |
| \begin{align} |
$$\Phi := \sum_\mu \vec{a}_\mu \vec{b}_\mu$$ |
| \Phi := \sum_\mu \vec{a}_\mu \vec{b}_\mu |
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| \end{align} |
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| of them are called {\em dyads}. |
of them are called {\em dyads}. |
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| A dyad is not a vector, but an operator.\, It \PMlinkescapetext{functions} on any vector $\vec{v}$ producing from it new vectors or new dyads according to the definitions |
A dyad is neither a scalar nor a vector, but an operator.\, It \PMlinkescapetext{functions} on any vector $\vec{v}$ producing from it new vectors or new dyads according to the definitions |
| \begin{align} |
$$\Phi*\vec{v} := \sum_\mu \vec{a}_\mu(\vec{b}_\mu*\vec{v}), \quad |
| \Phi*\vec{v} := \sum_\mu \vec{a}_\mu(\vec{b}_\mu*\vec{v}), \quad |
\vec{v}*\Phi := \sum_\mu (\vec{v}*\vec{a}_\mu)\vec{b}_\mu.$$ |
| \vec{v}*\Phi := \sum_\mu (\vec{v}*\vec{a}_\mu)\vec{b}_\mu. |
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| \end{align} |
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| Here the asterisks \PMlinkescapetext{mean either dots (producing two vectors) or crosses (producing two dyads).\, One can also allow the asterisks to mean} empty, in which case the vector $\vec{v}$ must be replaced by a scalar $v$; the products $\Phi v$ and $v\Phi$ are dyads. |
Here the asterisks \PMlinkescapetext{mean either dots (producing two vectors) or crosses (producing two dyads).\, One can also allow the asterisks to mean} empty, in which case the vector $\vec{v}$ must be replaced by a scalar $v$; the products $\Phi v$ and $v\Phi$ are dyads. |
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| The dyad product obeys the distributive laws |
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| $$\vec{a}(\vec{b}+\vec{c}) = \vec{a}\vec{b}+\vec{a}\vec{c}, \quad |
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| (\vec{b}+\vec{c})\vec{a} = \vec{b}\vec{a}+\vec{c}\vec{a},$$ |
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| which can be verified by multiplying an arbitrary vector $\vec{v}$ and both sides and then comparing the results.\, Likewise, the scalar factor transfer rule is valid.\, It follows that if\, |
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| $\vec{a} = a_1\vec{e_1}+a_2\vec{e_2}+a_3\vec{e_3}$\, and\, |
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| $\vec{b} = b_1\vec{e_1}+b_2\vec{e_2}+b_3\vec{e_3}$\, in the orthonormal basis\, $\{\vec{e_1},\,\vec{e_2},\,\vec{e_3}\}$ (for the brevity, we confine us to vectors of $\mathbb{R}^3$), their dyad product is the sum |
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| $$a_1b_1\vec{e_1}\vec{e_1}+a_1b_2\vec{e_1}\vec{e_2}+a_1b_3\vec{e_1}\vec{e_3}\\ |
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| +a_2b_1\vec{e_2}\vec{e_1}+a_2b_2\vec{e_2}\vec{e_2}+a_2b_3\vec{e_2}\vec{e_3}\\ |
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| +a_3b_1\vec{e_3}\vec{e_1}+a_3b_2\vec{e_3}\vec{e_2}+a_3b_3\vec{e_3}\vec{e_3},$$ |
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| which shows that the dyad product has been formed similarly as the matrix product of the vectors\, $(a_1,\,a_2,\,a_3)^T$\, and $(b_1,\,b_2,\,b_3)$. |
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| The {\em unit dyad}\, |
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| $I := \vec{e_1}\vec{e_1}+\vec{e_2}\vec{e_2}+\vec{e_3}\vec{e_3}$\, |
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| \PMlinkescapetext{satisfies} |
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| $$I\cdot\vec{v} = \vec{v}\cdot I = \vec{v}$$ |
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| and |
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| $$\Phi\cdot I = I\cdot\Phi = \Phi$$ |
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| for all vectors $\vec{v}$ and dyads $\Phi$. |
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| \begin{thebibliography}{9} |
\begin{thebibliography}{9} |
| \bibitem{VV}{\sc K. V\"ais\"al\"a:} {\em Vektorianalyysi}. \,Werner S\"oderstr\"om Osakeyhti\"o, Helsinki (1961). |
\bibitem{VV}{\sc K. V\"ais\"al\"a:} {\em Vektorianalyysi}. \,Werner S\"oderstr\"om Osakeyhti\"o, Helsinki (1961). |
| \end{thebibliography} |
\end{thebibliography} |
