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Say we have a basis $\mathbf{S}$ for $\Bbb{R}^n$ $$\mathbf{S}={\{\vec{e_1}; \vec{e_2};..;\vec{e_n}\}}$$

where $i \in[1;n];\ i \land n\in\Bbb{N^+}$. A vector $\vec{x}$ is thus written as

$$\vec{x}=\sum_{i=1}^na_i\vec{e_i} =a_i\vec{e_i},\ \ \ a_i\in\Bbb{R}$$

using Einstein's summation convention. Denote the components of the $i^\text{th}$ basis vector as

$$\begin{bmatrix}e_{1i} \\ e_{2i} \\ . \\ .\\e_{ni}\\\end{bmatrix}_{[\mathbf{S}]}.$$ Therefore, the modulus $|\vec{x}|$ takes the form

$$|\vec{x}|=\sqrt{\sum_{j=1}^n\left(\sum_{i=1}^na_ie_{ji}\right)^2}.$$

Question: How can one represent this double sum in Einstein's shorthand?

I am only able to drop one of the captital sigmas, i.e.,

$$|\vec{x}|=\sqrt{\sum_{j=1}^n\left(a_ie_{ji}\right)^2}.$$

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  • $\begingroup$ When you write the components of the $i^th$ basis vector, is that the components of the basis with respect to itself? $\endgroup$ – snulty Sep 20 '16 at 9:33
  • $\begingroup$ @snulty: the components of the $i^\text{th}$ basis vector are indeed meant with respect to itself. But other answers are welcome, too. The more general the answer, the better. $\endgroup$ – Linear Christmas Sep 20 '16 at 9:38
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Given a vector $v^i$ the square root of $v_iv^i=v^iv^jg_{ij}$ is the modulus, where $g_{ij}$ is the metric. In the Euclidean case the metric is just $\delta_{ij}$.

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