1

So I am doing some exercises in Susan Lea's book Mathematics for Physicists, currently question 7 chapter 1:

  1. Find the matrix that represents the transformation obtained by (a) rotating about the x-axis by $45^\circ$ counterclockwise and then (b) rotating about the y'-axis by $30^\circ$ clockwise.

This is the relevant part of the question. The first part I get the correct rotation matrix. For the second part I get this rotation matrix:

$\left[\begin{array}{2}cos(-30)&cos(90)&cos(60)\\cos(90)&cos(0)&cos(90)\\cos(-120)&cos(90)&cos(-30)\end{array}\right]$

which is equivalent to :

$\left[\begin{array}{2}\frac{\sqrt3}{2}&0&\frac{1}{2}\\0&1&0\\-\frac{1}{2}&0&\frac{\sqrt3}{2}\end{array}\right]$

Whereas the book's solution is:

$\left[\begin{array}{2}\frac{\sqrt3}{2}&0&-\frac{1}{2}\\0&1&0\\ \frac{1}{2}&0&\frac{\sqrt3}{2}\end{array}\right]$

I believe there may be something wrong in my understanding. I am using the formula for a rotation matrix:

$\mathbf A _{ij} = cos(\theta_{ij})$

where $\theta_{ij}$ is the angle between the ith new axis and jth axis in the original system.

Can someone explain where I went wrong if I did.

QuantumPanda
  • 191
  • I've never seen that formula for the elements of the rotation matrix before, is it from that book or is it found somewhere else?? – Daddy Kropotkin Aug 12 '18 at 22:25
  • It is from the book listed as well as in Thornton and Marion's Classical Dynamics of particles and systems 4th edition. – QuantumPanda Aug 12 '18 at 22:48
  • 2
    Hi. You are rotating clockwise on y-axis so you rotate it as though the direction of the y-axis is the direction of the front face of clock. So look at initial angle of axis $x$ to final axis $z$ in element $A_{13}$. Its $90^0$ (initial xz angle) + the rotation $30^0$ clockwise = $120^0$. Same as element $A_{31}$, initial z to final x, now $60^0(90^0-30^0)$. – philip_0008 Aug 12 '18 at 22:53
  • @N.Steinle - It's correct. It's why an alternative name for a rotation/transformation matrix is the direction cosine matrix, and the elements of such a matrix are called direction cosines. One has to take great care, however, in considering which is which. – David Hammen Aug 13 '18 at 04:32
  • @QuantumPanda - You cannot and must not use the concepts from one textbook in answering questions from another. There are several arbitrary choices when it comes to rotation / transformation matrices, and between any two textbooks, the odds are strong that Murphy's Law applies. – David Hammen Aug 13 '18 at 05:56
  • @DavidHammen I am only mentioning the other textbook as it reaffirms the concept in similar notation. To be clear I am using formulas from the Susan Lea textbook to solve its respective problems. – QuantumPanda Aug 13 '18 at 14:53
  • How is this a physics question? right now (v.1) it is a pure math problem. – AccidentalFourierTransform Aug 13 '18 at 22:41

1 Answers1

1

Your matrix is the right one. The matrix given by Susan Lea is incorrect.

The matrix for the transformation of coordinates from an initial cartesian right-handed system of coordinates $\;Ox_1x_2x_3\;$ to another cartesian right-handed system of coordinates $\;Ox'_1x'_2x'_3\;$ rotated with respect to the first around a direction $\;\mathbf{n}=\left(n_1,n_2,n_3\right),\Vert\mathbf{n}\Vert=1, \;$ through an angle $\;\theta\;$ (positive if counterclockwise with respect to $\;\mathbf{n}$) is(1)

\begin{equation} \mathbb{A}\left(\mathbf{n}, \theta\right)= \begin{bmatrix} \cos\theta+(1-\cos\theta)n_1^2&(1-\cos\theta)n_1n_2+\sin\theta n_3&(1-\cos\theta)n_1n_3-\sin\theta n_2 \vphantom{\dfrac12}\\ (1-\cos\theta)n_2n_1-\sin\theta n_3&\cos\theta+(1-\cos\theta)n_2^2&(1-\cos\theta)n_2n_3+\sin\theta n_1\vphantom{\dfrac12}\\ (1-\cos\theta)n_3n_1+\sin\theta n_2&(1-\cos\theta)n_3n_2-\sin\theta n_1&\cos\theta+(1-\cos\theta)n_3^2\vphantom{\dfrac12} \end{bmatrix} \tag{01}\label{eq01} \end{equation}

So, the transformation matrices around axes $\;x_1,x_2,x_3\;$ through angles $\;\theta_1,\theta_2,\theta_3\;$ respectively are \begin{align} \mathbb{A}\left(x_1, \theta_1\right)& = \begin{bmatrix} \hphantom{-}1 & \hphantom{-}0 & 0 \vphantom{\dfrac12}\\ \hphantom{-}0 & \hphantom{-}\cos\theta_1 & \sin\theta_1\vphantom{\dfrac12}\\ \hphantom{-}0 & -\sin\theta_1 & \cos\theta_1\vphantom{\dfrac12} \end{bmatrix} \tag{02.1}\label{eq02.1}\\ \mathbb{A}\left(x_2, \theta_2\right)& = \begin{bmatrix} \cos\theta_2 & \hphantom{-}0 & -\sin\theta_2\vphantom{\dfrac12}\\ 0 & \hphantom{-}1 & \hphantom{-}0 \vphantom{\dfrac12}\\ \sin\theta_2 & \hphantom{-}0 & \hphantom{-}\cos\theta_2\vphantom{\dfrac12} \end{bmatrix} \tag{02.2}\label{eq02.2}\\ \mathbb{A}\left(x_3, \theta_3\right)& = \begin{bmatrix} \hphantom{-}\cos\theta_3 & \sin\theta_3 & 0 \hphantom{-} \vphantom{\dfrac12}\\ -\sin\theta_3 & \cos\theta_3 & 0 \hphantom{-}\vphantom{\dfrac12}\\ \hphantom{-}0 & \hphantom{-}0 & 1\hphantom{-} \vphantom{\dfrac12}\\ \end{bmatrix} \tag{02.3}\label{eq02.3} \end{align}

After a careful look in above equations we note that in the cases of rotation around $\;x_1,x_3\;$ [equations \eqref{eq02.1},\eqref{eq02.3}] the term $\;\boldsymbol{+}\sin\theta_\jmath\;$ is up-right while the term $\;\boldsymbol{-}\sin\theta_\jmath\;$ is bottom-left. But in the case of $\;x_2$, equation \eqref{eq02.2}, we have the inverse : the term $\;\boldsymbol{+}\sin\theta_3\;$ is bottom-left while the term $\;\boldsymbol{-}\sin\theta_3\;$ is up-right. In this difference lies the error in Susan Lea, see Figure below (page extracted from $^{\prime}$Student Solutions Manual for Mathematics for Physicists$^{\prime}$ by Susan Lea)

enter image description here

The corrected matrix $\;\mathbb{A}_2\;$ must be \begin{equation} \mathbb{A}_2= \begin{pmatrix} \cos\left(-30^\circ\right) & \hphantom{-}0 & -\sin\left(-30^\circ\right)\vphantom{\dfrac12}\\ 0 & \hphantom{-}1 & \hphantom{-}0 \vphantom{\dfrac12}\\ \sin\left(-30^\circ\right) & \hphantom{-}0 & \hphantom{-}\cos\left(-30^\circ\right)\vphantom{\dfrac12} \end{pmatrix} = \begin{pmatrix} \hphantom{-}\frac{\sqrt{3}}{2} & \hphantom{-}0 & \hphantom{-}\frac{1}{2}\vphantom{\dfrac12}\\ \hphantom{-}0 & \hphantom{-}1 & \hphantom{-}0 \vphantom{\dfrac12}\\ -\frac{1}{2} & \hphantom{-}0 & \hphantom{-}\frac{\sqrt{3}}{2}\vphantom{\dfrac12} \end{pmatrix} \tag{03}\label{eq03} \end{equation}

(1) For the matrix of equation \eqref{eq01} see equation (08) in my answer there : Rotation of a vector. But since this matrix of equation (08) therein concerns the rotation of a vector (active view) we replace $\;\theta\;$ by $\;\boldsymbol{-}\theta\;$ to get the matrix \eqref{eq01} herein in order to have the transformation of coordinates (passive view).

Frobenius
  • 15,613