In the paper http://www.goshen.edu/physix/mathphys/gco/TensorGuideAJP.pdf at page 499, the authors defined the dot product of two vectors in a weird way: In the 3D Euclidian space let $\textbf{A}$ and $\textbf{B}$ two vectors and let ${\{\hat{x_a}:a=1,2,3\}}$ the orthonormal basis vectors. $\textbf{A}$ and $\textbf{B}$ are defined as: $$ \textbf{A}=\sum_{a=1}^3A_a\hat{x_a} \;\;\;\;\;\;\;\;\;\; \textbf{B}=\sum_{b=1}^3B_b\hat{x_b} $$

They defined the dot product of $\textbf{A}$ and $\textbf{B}$ as: $$\textbf{A} \cdot \textbf{B} = \sum_{a=1}^3\sum_{b=1}^3(A_a\hat{x_a})\cdot (B_b\hat{x_b})=\sum_{a=1}^3\sum_{b=1}^3(A_aB_b)(\hat{x_a}\cdot \hat{x_b})=\sum_{a=1}^3\sum_{b=1}^3(A_aB_b)\delta_{ab}=\sum_{a=1}^3A_aB_a$$ Where $\delta_{ab}$ is the Kronecker delta.

As far as I know the dot product is only defined as the multiplication of the components belonging to the same axes and then summing them. But in this definition dot product is shown as the regular, element wise multiplication of $(A_1\hat{x_1}+A_2\hat{x_2}+A_3\hat{x_3})$ and $(B_1\hat{x_1}+B_2\hat{x_2}+B_3\hat{x_3})$. So, what is the actual definition of dot product here? Which one is correct?

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    $\begingroup$ The final part of the equality looks precisely like the definition you want? $\endgroup$ – James Jun 24 '14 at 12:33
  • $\begingroup$ Well, but in the beginning they used a completely different form for it? It is written as the dot product is the element wise product of two vectors, not as the product of same axes. $\endgroup$ – Ufuk Can Bicici Jun 24 '14 at 12:38
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    $\begingroup$ Well, yes you have all these extra terms but they all disappear if you have $\hat{x}_1 = (1,0,0),\hat{x}_2 = (0,1,0),\hat{x}_3 = (0,0,1)$. I imagine the definition on the left is more general than the classic one. If you pick the standard basis then it turns out to be the same, but you could, say, take a different basis (not necessarily orthonormal) and the definition on the left is still ok. Or perhaps this makes sense in settings in which the dot product does not. $\endgroup$ – James Jun 24 '14 at 12:48
  • $\begingroup$ It is the 'usual' dot product in the case of the standard basis. Otherwise it is calculated in the same way (as in the case of the 'usual' dot product) by summing the products of corresponding vector components. $\endgroup$ – AnyAD Jun 24 '14 at 12:48
  • $\begingroup$ @Any Does this mean that the actual "usual" dot product is in the form of $\textbf{A}.\textbf{B} = \sum_{a=1}^3\sum_{b=1}^3(A_a\hat{x_a})(B_b\hat{x_b})$? If the basis is not orthonormal this becomes clearly different than the $\sum_{a=1}^3A_aB_b$ formula? $\endgroup$ – Ufuk Can Bicici Jun 24 '14 at 13:38

All of the paper is using the usual dot product.

It starts with the two vectors being expressed as a linear combination of some basis vectors, and applies the calculation rules of the dot product. $X=aU+bV$ and $Y=u'U'+b'V'$ can be multiplied together: $$X\cdot Y=(uU+vV)\cdot(u'U'+v'V')= uu'U\cdot U'+uv'U\cdot V'+vu'V\cdot U'+vv'U'\cdot V',$$ by distributivity of the dot product over vector addition, and associativity of the scalar product.

When the basis vectors are biorthogonal (hence the Kronecker delta), this simplifies to $uu'+vv'$.

In particular, when $U=U'=1_x$ and $V=V'=1_y$, you find the familiar $A\cdot B=a_xb_x+a_yb_y.$


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