- #1

shinobi20

- 270

- 20

- TL;DR Summary
- I got confused with the rotation matrices for the basis and the components, i.e., why they are written that way. Eventually, from linear algebra, I discussed how I think it should be done. I would like to confirm if my discussion is correct or maybe I missed something.

I'm reading Group Theory by A. Zee , specifically, chapter I.3 on rotations. He used the passive transformation in analyzing a point ##P## in space. There are two observers, one labeled with unprimed coordinates and the other with primed coordinates. From the figure below, he deduced the relation between the two observer coordinates,

\begin{equation}\tag{1}

x' = x \cos\theta + y \sin\theta, \qquad y' = -x \sin\theta + y \cos\theta

\end{equation}

This can be written in vector notation as ##\; \vec{r}' = R(\theta) \vec{r} \;## where,

\begin{equation}\tag{2}

\vec{r} = \begin{bmatrix} x \\ y \end{bmatrix}, \quad \vec{r}' = \begin{bmatrix} x' \\ y' \end{bmatrix}, \quad R(\theta) = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}

\end{equation}

I'll briefly discuss what I know from linear algebra. Given a vector ##\vec{v}##, we can always express it in different bases. Here, the vector ##\vec{v}## is the arrow from the origin to the point ##P##. I'll use an unprimed basis ##\{\hat{x},\hat{y}\}## and primed basis ##\{\hat{x}',\hat{y}'\}##. We can rotate the basis ##\{\hat{x},\hat{y}\}## using the rotation operator ##R(\theta)## so that it results to the new basis ##\{\hat{x}',\hat{y}'\}##. In equations,

\begin{equation}\tag{3}

\begin{bmatrix} \hat{x}' \\ \hat{y}' \end{bmatrix} = R(\theta) \begin{bmatrix} \hat{x} \\ \hat{y} \end{bmatrix} = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} \hat{x} \\ \hat{y} \end{bmatrix}

\end{equation}

This can be seen in the figure below where the unprimed basis is colored in blue while the primed basis is colored in red,

We can write,

\begin{equation}\tag{4}

\vec{v} = x' \hat{x}' + y' \hat{y}' = \begin{bmatrix} x' & y' \end{bmatrix} \begin{bmatrix} \hat{x}' \\ \hat{y}' \end{bmatrix} = \begin{bmatrix} x' & y' \end{bmatrix} R(\theta) \begin{bmatrix} \hat{x} \\ \hat{y} \end{bmatrix}

\end{equation}

which means,

\begin{equation}\tag{5}

\begin{bmatrix} x & y \end{bmatrix} = \begin{bmatrix} x' & y' \end{bmatrix} R(\theta)

\end{equation}

taking the transpose we get,

\begin{equation}\tag{6}

\begin{bmatrix} x \\ y \end{bmatrix} = R(\theta)^T \begin{bmatrix} x' \\ y' \end{bmatrix}

\end{equation}

This tells us that there exist an operator ##\tilde{R}(\theta)## such that,

\begin{equation}\tag{7}

\begin{bmatrix} x' \\ y' \end{bmatrix} = \tilde{R}(\theta)^T \begin{bmatrix} x \\ y \end{bmatrix}

\end{equation}

so that,

\begin{equation}\tag{8}

\begin{bmatrix} x \\ y \end{bmatrix} = R(\theta)^T \begin{bmatrix} x' \\ y' \end{bmatrix} = R(\theta)^T \tilde{R}(\theta)^T \begin{bmatrix} x \\ y \end{bmatrix} = (\tilde{R}(\theta) R(\theta))^T \begin{bmatrix} x \\ y \end{bmatrix}

\end{equation}

This means ##\tilde{R}(\theta) = R(\theta)^{-1} = R(\theta)^T## (we are working on orthogonal transformations so inverse is just transpose). This implies that,

\begin{equation}\tag{9}

\begin{bmatrix} x' \\ y' \end{bmatrix} = R(\theta) \begin{bmatrix} x \\ y \end{bmatrix}

\end{equation}

My original confusion while reading is that given the basis transforms through ##R(\theta)##, we know that the coordinates should transform inversely so as to preserve the vector ##\vec{v}##, i.e.,

\begin{equation}\tag{10}

\begin{bmatrix} \hat{x}' \\ \hat{y}' \end{bmatrix} = R(\theta) \begin{bmatrix} \hat{x} \\ \hat{y} \end{bmatrix} = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} \hat{x} \\ \hat{y} \end{bmatrix} \qquad \leftrightarrow \qquad \begin{bmatrix} x' \\ y' \end{bmatrix} = R(\theta)^T \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}

\end{equation}

This led me to be confused with Zee's rotation matrix in eq.##(2)## stemming from the equation for the coordinate transformation in eq.##(1)## since the rotation matrix is different from what I expected which is ##R(\theta)^T##.

Eventually, I think I know the reason. It has something to do with the row-column dot product to write the vector. From the linear algebra discussion above, to preserve the vector, and considering eq.##(3)## and eq.##(9)##,

\begin{equation}\tag{11}

\vec{v} = x' \hat{x}' + y' \hat{y}' = \begin{bmatrix} x' & y' \end{bmatrix} \begin{bmatrix} \hat{x}' \\ \hat{y}' \end{bmatrix} = \begin{bmatrix} x & y \end{bmatrix} R(\theta)^T R(\theta) \begin{bmatrix} \hat{x} \\ \hat{y} \end{bmatrix} = \begin{bmatrix} x & y \end{bmatrix} \begin{bmatrix} \hat{x} \\ \hat{y} \end{bmatrix} = x \hat{x} + y \hat{y}

\end{equation}

So, this is what we mean by "preserve" the vector. The naive notion that the coordinates transform inversely compared to the basis as is written in eq.##(10)##, i.e., both are written in a column-wise manner, is incorrect. The key here is the expression ##\begin{bmatrix} x' & y' \end{bmatrix} = \begin{bmatrix} x & y \end{bmatrix} R(\theta)^T##. In order words, what we mean by preserve is that the basis is written in column while the components is in a row. However, we usually express this as a column vector so that instead of the right-hand side equation in eq.##(10)##, we have

\begin{equation}\tag{12}

\begin{bmatrix} x' \\ y' \end{bmatrix} = R(\theta) \begin{bmatrix} x \\ y \end{bmatrix}

\end{equation}

which matches Zee's rotation matrix. This column version of the rotation of coordinates may lead one to think that both the basis and coordinates transform through ##R(\theta)##, but in fact it is doesn't, the coordinates transform through ##R(\theta)^T##, and the column version is just a convenient way of expressing the rotation of coordinates. I think this should be correct (obviously from the linear algebra) but can anyone comment if my realization is correct?

\begin{equation}\tag{1}

x' = x \cos\theta + y \sin\theta, \qquad y' = -x \sin\theta + y \cos\theta

\end{equation}

This can be written in vector notation as ##\; \vec{r}' = R(\theta) \vec{r} \;## where,

\begin{equation}\tag{2}

\vec{r} = \begin{bmatrix} x \\ y \end{bmatrix}, \quad \vec{r}' = \begin{bmatrix} x' \\ y' \end{bmatrix}, \quad R(\theta) = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}

\end{equation}

I'll briefly discuss what I know from linear algebra. Given a vector ##\vec{v}##, we can always express it in different bases. Here, the vector ##\vec{v}## is the arrow from the origin to the point ##P##. I'll use an unprimed basis ##\{\hat{x},\hat{y}\}## and primed basis ##\{\hat{x}',\hat{y}'\}##. We can rotate the basis ##\{\hat{x},\hat{y}\}## using the rotation operator ##R(\theta)## so that it results to the new basis ##\{\hat{x}',\hat{y}'\}##. In equations,

\begin{equation}\tag{3}

\begin{bmatrix} \hat{x}' \\ \hat{y}' \end{bmatrix} = R(\theta) \begin{bmatrix} \hat{x} \\ \hat{y} \end{bmatrix} = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} \hat{x} \\ \hat{y} \end{bmatrix}

\end{equation}

This can be seen in the figure below where the unprimed basis is colored in blue while the primed basis is colored in red,

We can write,

\begin{equation}\tag{4}

\vec{v} = x' \hat{x}' + y' \hat{y}' = \begin{bmatrix} x' & y' \end{bmatrix} \begin{bmatrix} \hat{x}' \\ \hat{y}' \end{bmatrix} = \begin{bmatrix} x' & y' \end{bmatrix} R(\theta) \begin{bmatrix} \hat{x} \\ \hat{y} \end{bmatrix}

\end{equation}

which means,

\begin{equation}\tag{5}

\begin{bmatrix} x & y \end{bmatrix} = \begin{bmatrix} x' & y' \end{bmatrix} R(\theta)

\end{equation}

taking the transpose we get,

\begin{equation}\tag{6}

\begin{bmatrix} x \\ y \end{bmatrix} = R(\theta)^T \begin{bmatrix} x' \\ y' \end{bmatrix}

\end{equation}

This tells us that there exist an operator ##\tilde{R}(\theta)## such that,

\begin{equation}\tag{7}

\begin{bmatrix} x' \\ y' \end{bmatrix} = \tilde{R}(\theta)^T \begin{bmatrix} x \\ y \end{bmatrix}

\end{equation}

so that,

\begin{equation}\tag{8}

\begin{bmatrix} x \\ y \end{bmatrix} = R(\theta)^T \begin{bmatrix} x' \\ y' \end{bmatrix} = R(\theta)^T \tilde{R}(\theta)^T \begin{bmatrix} x \\ y \end{bmatrix} = (\tilde{R}(\theta) R(\theta))^T \begin{bmatrix} x \\ y \end{bmatrix}

\end{equation}

This means ##\tilde{R}(\theta) = R(\theta)^{-1} = R(\theta)^T## (we are working on orthogonal transformations so inverse is just transpose). This implies that,

\begin{equation}\tag{9}

\begin{bmatrix} x' \\ y' \end{bmatrix} = R(\theta) \begin{bmatrix} x \\ y \end{bmatrix}

\end{equation}

My original confusion while reading is that given the basis transforms through ##R(\theta)##, we know that the coordinates should transform inversely so as to preserve the vector ##\vec{v}##, i.e.,

\begin{equation}\tag{10}

\begin{bmatrix} \hat{x}' \\ \hat{y}' \end{bmatrix} = R(\theta) \begin{bmatrix} \hat{x} \\ \hat{y} \end{bmatrix} = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} \hat{x} \\ \hat{y} \end{bmatrix} \qquad \leftrightarrow \qquad \begin{bmatrix} x' \\ y' \end{bmatrix} = R(\theta)^T \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}

\end{equation}

This led me to be confused with Zee's rotation matrix in eq.##(2)## stemming from the equation for the coordinate transformation in eq.##(1)## since the rotation matrix is different from what I expected which is ##R(\theta)^T##.

Eventually, I think I know the reason. It has something to do with the row-column dot product to write the vector. From the linear algebra discussion above, to preserve the vector, and considering eq.##(3)## and eq.##(9)##,

\begin{equation}\tag{11}

\vec{v} = x' \hat{x}' + y' \hat{y}' = \begin{bmatrix} x' & y' \end{bmatrix} \begin{bmatrix} \hat{x}' \\ \hat{y}' \end{bmatrix} = \begin{bmatrix} x & y \end{bmatrix} R(\theta)^T R(\theta) \begin{bmatrix} \hat{x} \\ \hat{y} \end{bmatrix} = \begin{bmatrix} x & y \end{bmatrix} \begin{bmatrix} \hat{x} \\ \hat{y} \end{bmatrix} = x \hat{x} + y \hat{y}

\end{equation}

So, this is what we mean by "preserve" the vector. The naive notion that the coordinates transform inversely compared to the basis as is written in eq.##(10)##, i.e., both are written in a column-wise manner, is incorrect. The key here is the expression ##\begin{bmatrix} x' & y' \end{bmatrix} = \begin{bmatrix} x & y \end{bmatrix} R(\theta)^T##. In order words, what we mean by preserve is that the basis is written in column while the components is in a row. However, we usually express this as a column vector so that instead of the right-hand side equation in eq.##(10)##, we have

\begin{equation}\tag{12}

\begin{bmatrix} x' \\ y' \end{bmatrix} = R(\theta) \begin{bmatrix} x \\ y \end{bmatrix}

\end{equation}

which matches Zee's rotation matrix. This column version of the rotation of coordinates may lead one to think that both the basis and coordinates transform through ##R(\theta)##, but in fact it is doesn't, the coordinates transform through ##R(\theta)^T##, and the column version is just a convenient way of expressing the rotation of coordinates. I think this should be correct (obviously from the linear algebra) but can anyone comment if my realization is correct?