Relating Basis Vectors in Rotated Reference Frames

Question

Consider two three-dimensional reference frames $F$ and $F'$, related by a rotation matrix $R$. In general, $R^{-1} = R^T.$ The components of a vector $\vec{a} = \sum_{i} a_i \hat{x}_i = \sum_{i} a_i' \hat{x}_i'$ are related by $a_i' = \sum_{j} R_{ij} a_j.$ I would wish to show that $\hat{x}_i' = \sum_j R_{ji} \hat{x}_j$, for which I am having some difficulty with the indices. Here is my work so far:

• I know that $a_i = \sum_{j} R_{ij}^{-1} a_j' = \sum_{j} R_{ji} a_j'.$
• Now, $$\vec{a} = \sum_{i} a_i \hat{x}_i = \sum_{i} a_i' \hat{x}_i' \Rightarrow \sum_i \left(\sum_j R_{ji} a_j'\right) \hat{x}_i = \sum_{i} a_i' \hat{x}_i' \Rightarrow \sum_j \sum_i R_{ij} a_i'\hat{x}_j = \sum_i a_i' \hat{x}_i \Rightarrow ??$$

Somehow, by comparison of the left and right-hand sides, $\hat{x}_i' = \sum_j R_{ji}\hat{x}_j$ should naturally "fall out", but the indices and the double-sums are rendering it difficult for me to see how this is so.

Answer 1

You said

• I know that $a_i = \sum_{j} R_{ij}^{-1} a_j' = \sum_{j} R_{ji} a_j'.$

Then, consider, for example, the vector $\hat{x}_1'$ with coordinates $a_1'=1, a_2'=0,\ldots$. Then its coordinates in the other basis are, according to you, $$ a_i=\sum_j R_{ji} a_j'=R_{1i} $$ and therefore $$ \hat{x}_1'=R_{11} \hat{x}_1+\cdots=\sum_j R_{1j} \hat{x}_j $$

But the index 1 has nothing special so in general, and according to you sentence: $$ \hat{x}_i'=\sum_j R_{ij} \hat{x}_j $$

This is not what you want, so if I am right (I don't find any errors in my approach) you have a mistake in your text. If I am wrong let me know.

Answer 2

Write in matrix multiplication form \begin{align} \vec{a}=a^\top \hat{x}={a'}^\top\hat{x}' \end{align} and insert between those matrices an identitiy matrix $I=R^\top R\,.$ This gives $$ a^\top R^\top R\,\hat{x}={a'}^\top\hat{x}'\,. $$ When ${a'}^\top=(1,0,\dots,0)$ then $a$ is the first row of $R$ and then $$ (R\,\hat{x})_1=\hat{x}'_1\,. $$ Repeat this with ${a'}^\top=(0,1,0,\dots,0)$ and so on to get finally $R\,\hat{x}=\hat{x}'\,.$

• Note that for basis vectors and components we have the transformation laws $R\,\hat{x}=\hat{x}'$ and $R\,a=a'$ in which the first one is different from yours.

• This seems confusing because we are told that components are contravariant and basis vectors covariant. The point is however that components are typically written as row vectors which makes their transformation contravariant $$ a'^\top=a^\top R^\top $$ because $R^\top=R^{-1}$ is used here.