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I was compiling some solutions to Griffith's E&M, and I can't get my solution to square with the others I have found online. The question asks:

1.9 Find the transformation matrix R that describes a rotation by $120^\circ$ about an axis from the origin through the point $(1, 1, 1)$. The rotation is clockwise as you look down the axis toward the origin.

The 'standard' solution is to look down the rotation axis $(1,1,1)$ to get the following view: enter image description here

Now, we can easily note that $$ \begin{align*} x' &= z \\ y' &= x \\ z' &=y \end{align*} $$

where the primes indicate the image under $R$, the matrix we are trying to write down. This suggests to me that our rotation matrix should be: $$ R = \begin{pmatrix} 0&1&0\\ 0&0&1 \\ 1&0&0 \end{pmatrix} $$ A brief check indeed shows us that: $$ x' = R(\hat{x}) = \begin{pmatrix} 0&1&0\\ 0&0&1 \\ 1&0&0 \end{pmatrix} \begin{pmatrix} 1\\ 0 \\ 0 \\ \end{pmatrix} = \begin{pmatrix} 0\\ 0 \\ 1 \end{pmatrix} = z $$ as suggested by the above image.

However, the official instructor's manual (as well as other solutions online), cite the rotation matrix as the transpose of the above: $$ R_s \stackrel{?}{=} \begin{pmatrix} 0&0&1\\ 1&0&0\\ 0&1&0 \end{pmatrix} $$

What I think might be happening is this: The solution manual is strangely working in the dual space and saying that $$ \begin{pmatrix} A_x'\\ A_y'\\ A_z'\\ \end{pmatrix} = \begin{pmatrix} 0&0&1\\ 1&0&0\\ 0&1&0 \end{pmatrix} \begin{pmatrix} A_x\\ A_y\\ A_z \end{pmatrix} $$ Where $A_x = (1,0,0)$ for example. This indeed shows us that $$ A_x' = A_z \rightarrow R_s\left((1,0,0) \right) = (0,0,1) $$ or in Dirac notation (which might be a bit clearer), the manual finds: $$ \langle x|R_s = \langle z | $$ when in reality we want $$ R|x\rangle = |z\rangle $$

So, the manual should make the final step of taking the Hermitian conjugate (in this case just the transpose) to write $$ (R_s)^T|x\rangle = |z\rangle $$

I definitely feel like I am overthinking this, but I don't see any other reason the solution manual (and others) would write down the matrix they have. Can anyone clarify what is going on here, and why the manual might have gotten the solution it did?

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  • $\begingroup$ Active vs passive issue perhaps? $\endgroup$ – jacob1729 Jun 12 at 16:02
  • $\begingroup$ @jacob1729 Perhaps, but it is not immediately clear to me how so. $\endgroup$ – gabe Jun 12 at 16:51
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    $\begingroup$ If considering the basis vectors to transform as $\hat{e}\mapsto R\hat{e}$ then the components $v_i = \hat{e}_i \cdot \vec{v} = \hat{e}_i^T \vec{v}$ transform via $v_i \mapsto R^T v_i$. $\endgroup$ – jacob1729 Jun 12 at 18:05
  • $\begingroup$ And, noting, $R^{T}=R^{-1}$, it's clearly active vs passive. $\endgroup$ – Cinaed Simson Jun 14 at 20:13
  • $\begingroup$ @jacob1729 After more consideration, I agree with your interpretation. If you want to type up an answer explaining this, I will happily give you this bounty. $\endgroup$ – gabe Jun 16 at 18:00
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I believe what's going on is that the solution is interpreting the question as a 'passive' transformation whilst you are interpreting it as an 'active transformation'. That is, the question is assuming we want to change basis by rotating our basis vectors by $120^\circ$ about the $(1,1,1)$ axis.

For orthonormal coordinate systems the components of a vector can be extracted by dotting with the basis vectors:

$$\vec{v} = v^i \hat{e}_i \iff v^i = \hat{e}_i \cdot \vec{v}$$

(this only holds for orthonormal bases, but fortunately rotations are specifically the transformations that keep the basis orthonormal so that's alright). If we rotate the basis vectors by a rotation $R$ then we have:

$$v'^i = \hat{e}'_i\cdot \vec{v} = (R\hat{e}_i)\cdot \vec{v}$$

$$v'^i = (R\hat{e}_i)^T \vec{v} = R^Tv^i$$

which means that if you think of the basis vectors as rotating, and the coordinates transforming such that vectors stay constant, then the components transform using $R^T$. The general rule is that if the basis vectors transform via $R$ the coordinates must transform via $R^{-1}$ (so that the combination $\vec{v}=v^i\hat{e}_i$ stays constant) and for a rotation $R^{-1}$ and $R^T$ coincide.


(Aside on passive transformations:) the reason you might want to do this, despite active transformations being probably simpler to visualise is because this corresponds to analysing what happens if we don't change the physical system, but merely change the coordinates we use to describe it. A simple example might be that we usually use three numbers in the order $(x,y,z)$ to mean a vectors projection along the $x,y,z$ axes, but we could obviously have chosen to list those projections in other orders such as $(y,x,z)$. There are multiple ways of describing this change in convention: we could say something like "the new basis is such that the first new basis vector is the second old basis vector and vice versa". This is an accurate description of what we've done. The other way to describe it is to say "we've reflected the vector in the plane $x=y$". This is a bit awkward since we didn't mean to reflect any vectors, just to change our coordinate system (although, in the end we'll get the same results).

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  • $\begingroup$ Upon looking back at the exposition in Griffiths regarding rotation, he indeed rotates the coordinate system as opposed to the vectors (passive v. active as you point out). Thanks for clarifying this. $\endgroup$ – gabe Jun 16 at 20:24
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The rotational matrix $R$ about the axis $\vec{{d}}$ with clockwise rotational angle $\varphi$ about this axis is:

$$R=I_3+\sin(\varphi)\,\tilde{{d}}+(1-\cos(\varphi))\,\tilde{{d}}\,\tilde{{d}}$$

Where:

$\vec{d}$ is the rotational axis with $\vec{d}\cdot \vec{d}=1$ and

$\tilde{d}=\begin{bmatrix} 0 & -d_z & d_y \\ d_z & 0 & -d_x \\ -d_y & d_x & 0 \\ \end{bmatrix}$

The transformation matrix $R$ transformed vector from body fixed coordinate system to space coordinate system

Example:

$\vec{d}=\frac{1}{\sqrt{3}}\,\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}$ and $\varphi=2/3\,\pi$

you get

$$R=\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{bmatrix}$$

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