Draft:Linear algebra/Rotation of axes

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Notice: Tensors/Transformation rule under a change of basis gets about 200 pageviews per month. This draft might need to be moved, or at least linked.

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Linear algebra/Rotation of axes, w:Rotation matrix, w:Rotation of axes

Alternative notation: $(x,y)=(x_1,x_2),$ $(x,y)=(x_1^{\prime },x_2^{\prime }),$ and $(\hat{x},\hat{y})=({\hat{e}}_1,{\hat{e}}_2).$

These days most introductory physics textbooks display the position (displacement) vector as:

$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$

This use of seven letters is peculiar in a field where the combined Latin, Greek, and Cyrillic alphabets sometimes seems insufficient. Here, we shall replace the unit vectors

$\{\hat{i},\hat{j},...\}$

by

$\{\hat{x},\hat{y}\dots \}$

,

$\{{\hat{e}}_1,{\hat{e}}_2,\dots \}$

, or even simply

${\hat{e}}_j$

, where

$j=\{1,2,3\}$

. Even though matter exists in three spatial dimensions, one dimension can often be "ignored". The two-dimensional figure shown can also model rotations rotations of a cube or long wire of uniform cross sectino. Two different coordinate systems can be used to represent the point

$P$

:

$\underset{\_}{x}=\sum _j{x}_j{\hat{e}}_j=\sum _j{x}_j^{\prime }{\hat{e}}_j^{\prime }$

We shall see that underline convention for vectors is also useful for tensors. The unit vectors reminds readers that the coordinate system has been rotated.

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Quizbank ideas: Does the value of

$\underset{\_}{x}$

change if the index Template:Math is changed? Can the figure be applied to a three dimensional object? Why is

${\underset{\_}{x}}^{\prime }$

absent from this discussion? Is

$\hat{x}\cdot {\hat{y}}^{\prime }$

positive, negative, or zero? Though it might seem trivial, the following question serves a purpose: It is the author's opinion that three dimensional objects should generally be drawn in perspective by displaying the x,y, and z axes (True or False)?

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Deriving the rotation tensor

The dot product of any vector with a unit vector yields the vector's component in that direction. Pick any value for $j$ and consider the dot product with and vector consider for example the dot product of $\underset{\_}{x}$ with the unit vector ${\hat{e}}_k$:

${\hat{e}}_k\cdot \underset{\_}{x}={\hat{e}}_k\cdot \sum _j{x}_j{\hat{e}}_j=\sum _j\underset{{\delta }_{kj}=\text{0 or 1}}{\underbrace{({\hat{e}}_k\cdot {\hat{e}}_j)}}{x}_j=\sum _j{x}_j{\delta }_{jk}=x_k,$

where ${\delta }_{jk}$ is the Kronecker delta function. Here we took the dot product with the unprimed unit vector ${\hat{e}}_k$. Something entirely different happens if we instead multiply by the primed unit vector ${\hat{e}}_k^{\prime }$:

${\hat{e}}_k^{\prime }\cdot \underset{\_}{x}={\hat{e}}_k^{\prime }\cdot \sum _j{x}_j{\hat{e}}_j=\sum _j({\hat{e}}_k^{\prime }\cdot {\hat{e}}_j)x_j\equiv \underset{\_}{\underset{\_}{R}}\cdot \underset{\_}{x},$

where,

$R_{jk}={\hat{e}}_k^{\prime }\cdot {\hat{e}}_j$

,

$(\text{in dyad notation, }\underset{\_}{\underset{\_}{R}}={\hat{e}}_k^{\prime }{\hat{e}}_j)$

is the rotation tensor. To understand why, note that since ${\hat{e}}_k^{\prime }\cdot \underset{\_}{x}=x_k^{\prime },$ we have an expression for the k-th component of $\underset{\_}{x}$ in the primed $(x_1^{\prime },x_2^{\prime }\dots )$ coordinate system.

${\hat{e}}_k^{\prime }\cdot \underset{\_}{x}=x_k^{\prime }\Rightarrow \sum _j{\hat{e}}_j{x}_j^{\prime }=\sum _{j,k}{\hat{e}}_j{R}_{jk}{x}_k\Rightarrow x_j^{\prime }=\sum _k{R}_{jk}{x}_k,$

The latter form can be written as $x_j^{\prime }=R_{jk}{x}_k$ if it is understood that repeated and adjacent subscripts are always summed.

Scratchwork

Our first tensor will of course be the rotation tensor:

$x_j^{\prime }=\sum _k{R}_{jk}{x}_k.$

To appreciate the value of unit vectors, we calculate the rotation tensor for the two dimensional rotation shown above. Let $\underset{\_}{x}$ be the displacement vector from the origin to point $P.$

$\underset{\_}{x}=\sum _j{x}_j{\hat{e}}_j=\sum _j{x}_j^{\prime }{\hat{e}}_j^{\prime },$

where, for example, $x_2^{\prime }=y^{\prime },$ in the figure above. Interesting results emerge when we take the dot product of this with either ${\hat{e}}_k$ or ${\hat{e}}_k^{\prime },$ because for unit vectors, $\underset{\_}{A}\cdot \underset{\_}{B}=AB\cos \theta ,$ leads to:

$\begin{array}{cc}{\hat{e}}_1\cdot {\hat{e}}_1^{\prime }=\cos \theta & {\hat{e}}_1\cdot {\hat{e}}_2^{\prime }=-\sin \theta \\ {\hat{e}}_2\cdot {\hat{e}}_1^{\prime }=\sin \theta & {\hat{e}}_2\cdot {\hat{e}}_2^{\prime }=\cos \theta \end{array}$

$\begin{array}{cc}{\hat{e}}_1\cdot {\hat{e}}_1^{\prime }=\cos \theta & {\hat{e}}_1\cdot {\hat{e}}_2^{\prime }=-\sin \theta \\ {\hat{e}}_2\cdot {\hat{e}}_1^{\prime }=\sin \theta & {\hat{e}}_2\cdot {\hat{e}}_2^{\prime }=\cos \theta \end{array}$

If find writing this as, ${\underset{\_}{x}}^{\prime }=\underset{\_}{\underset{\_}{R}}\cdot \underset{\_}{x},$ to be simultaneously vague and useful. To understand why, we "derive" the rotation tensor $\underset{\_}{\underset{\_}{R}}$ for the two dimensional graph shown above. Let

The great virtue of the unit vector is that it's dot product of any vector with a unit vector reveals a component of that vector. The tensor subscript notation permits us to "vaguely" take the dot product with each unit vector in the rotated reference frame, i.e., with ${\hat{e}}_k^{\prime }$:

(depending on whether $j=1,2$ or $j=1,2,3$.) The underline convention will facilitate the introduction of tensors:

start

$\begin{array}{ccc}\underset{\_}{a}\cdot {\hat{e}}_k^{\prime }& =& \sum _j{a}_j{\hat{e}}_j\cdot {\hat{e}}_k^{\prime }\\ a_k^{\prime }& =& \sum _j{a}_j{\hat{e}}_j\cdot {\hat{e}}_k^{\prime }\\ & =& \sum _j({\hat{e}}_k^{\prime }\cdot {\hat{e}}_j)a_j\end{array}$

$\begin{array}{ccc}\underset{\_}{a}\cdot {\hat{e}}_k^{\prime }& =& \sum _j{a}_j{\hat{e}}_j\cdot {\hat{e}}_k^{\prime }\\ a_k^{\prime }& =& \sum _j{a}_j{\hat{e}}_j\cdot {\hat{e}}_k^{\prime }\\ a_{\begin{array}{cc}k& \\ \downarrow & \\ j\end{array}}^{\prime }& =& \sum _{\begin{array}{cc}j& \\ \downarrow & \\ k\end{array}}({\hat{e}}_{\begin{array}{cc}k& \\ \downarrow & \\ j\end{array}}^{\prime }\cdot {\hat{e}}_{\begin{array}{cc}j& \\ \downarrow & \\ k\end{array}})a_{\begin{array}{cc}j& \\ \downarrow & \\ k\end{array}}\end{array}$