Alright, let's walk through how to derive this transformation matrix step by step. I'll break it down into clear, actionable parts so you can follow along easily.

Step 1: Map the spherical coordinate unit vector to Cartesian coordinates #

First, let's write the unit vector $\hat{z}'$ (this will be the z-axis of our new coordinate system $A'$) in standard Cartesian terms using the given spherical parameters $\theta$ (tilt from the z-axis) and $\phi$ (angle of the xy-plane projection from the x-axis):
$$\hat{z}' = \left( \sin\theta\cos\phi,\ \sin\theta\sin\phi,\ \cos\theta \right)$$
This makes sense: the z-component is $\cos\theta$ (since $\theta$ is the angle from the z-axis), the xy-plane projection has length $\sin\theta$, and we split that into x/y components using $\cos\phi$ and $\sin\phi$.

Step 2: Construct the orthonormal basis for $A'$ #

We need three mutually perpendicular unit vectors for $A'$: $\hat{x}'$, $\hat{y}'$, $\hat{z}'$. We already have $\hat{z}'$, so let's build the other two (which lie in plane $S$, perpendicular to $\hat{z}'$):

1. Define $\hat{x}'$:
   We want $\hat{x}'$ to lie in plane $S$, and we're told the x'y' axes have a 0 rotation angle around $\hat{z}'$. A standard way to do this is to create a vector that's perpendicular to $\hat{z}'$ and aligns with the "xz-plane direction" when $\phi=0$. We can construct this by first making a vector perpendicular to $\hat{z}'$ in the xz-plane (when $\phi=0$) and then rotating it around the z-axis by $\phi$. The result is:
   $$\hat{x}' = \left( \cos\theta\cos\phi,\ \cos\theta\sin\phi,\ -\sin\theta \right)$$
   You can verify this is perpendicular to $\hat{z}'$ by taking their dot product—it will equal 0—and that it's a unit vector (its magnitude is 1).

2. Define $\hat{y}'$:
   For a right-handed coordinate system, we can get $\hat{y}'$ by taking the cross product of $\hat{z}'$ and $\hat{x}'$ (this guarantees it's perpendicular to both $\hat{z}'$ and $\hat{x}'$):
   $$\hat{y}' = \hat{z}' \times \hat{x}'$$
   Calculating this cross product gives:
   $$\hat{y}' = \left( -\sin\phi,\ \cos\phi,\ 0 \right)$$
   Again, check the dot products with $\hat{z}'$ and $\hat{x}'$—they'll be 0—and confirm it's a unit vector.

Step 3: Build the transformation matrix $R(\theta, \phi)$ #

The problem states that $(x,y,z) = R(x',y',z')$. This means each coordinate in the original system $A$ is a linear combination of the coordinates in $A'$, where the coefficients are the components of $A'$'s basis vectors in $A$.

In matrix terms, the columns of $R$ are exactly the basis vectors $\hat{x}'$, $\hat{y}'$, $\hat{z}'$ we just found. Putting it all together:

$$R(\theta, \phi) = \begin{pmatrix}
\cos\theta\cos\phi & -\sin\phi & \sin\theta\cos\phi \
\cos\theta\sin\phi & \cos\phi & \sin\theta\sin\phi \
-\sin\theta & 0 & \cos\theta
\end{pmatrix}$$

Quick Verification #

• This is an orthogonal matrix, meaning its transpose is its inverse ($R^T R = I$). This makes sense for coordinate transformations, as we don't want to scale or skew the coordinate system.
• If $\theta=0$ (the new z-axis aligns with the original z-axis), the matrix simplifies to a standard z-axis rotation by $\phi$, which matches our intuition.
• If $\phi=0$, the matrix simplifies to a rotation around the y-axis by $\theta$, which also checks out.

内容的提问来源于stack exchange，提问作者yaniv