# Spherical coordinates and rotations of axes

Let $$P_o$$ denote the position with the coordinates $$(1,0,0)$$ in the Descartes coordinate system $$(x,y,z)$$.

The point $$P_o$$ is rotated about the z-axis so that the line $$OP$$ turns directly towards the positive y-axis through an angle $$\phi$$. The position of the point after this rotation is denoted by $$P_1$$

$$P_1$$ is then rotated about the line in the x-y plane perpendicular to $$OP_1$$ so that the line $$OP$$ turns directly towards the positive z-axis through an angle $$\lambda$$, giving $$P_2$$. Find the coordinates of $$P_2$$

Initially I approached this question using the spherical coordinates: $$x=r\sin(\theta)\cos(\phi), y=r\sin(\theta)\sin(\phi), z=r\cos(\theta) \tag{1}$$ where $$\theta$$ is the polar angle and $$\phi$$ the azimuthal angle.

Setting $$\theta=\frac{\pi}{2}-\lambda$$ and $$\phi=\phi$$ gave me the correct the answer: $$P_2=(\cos(\phi)\cos(\lambda),\sin(\phi)\cos(\lambda),\sin(\lambda))\tag{2}$$

However, I then tried an alternative method by rotating the coordinate axes and an incorrect answer was obtained:

I rotated the coordinate axes $$(x,y,z)$$ by the angle $$\phi$$ anticlockwise about the z-axis. Denoting the new coordinate axes by $$(\bar{x},\bar{y},\bar{z})$$, we have \begin{align} x&=\bar{x}\cos(\phi)-\bar{y}\sin(\phi)\\ y&=\bar{x}\sin(\phi)+\bar{y}\cos(\phi)\\ z&=\bar{z}\end{align} \tag{3}

since $$\left(\begin{matrix} x\\y\\z \end{matrix} \right)=\left(\begin{matrix} \cos{(\phi)}&-\sin(\phi)&0 \\ \sin(\phi)&\cos(\phi)&0\\0&0&1\end{matrix}\right) \left(\begin{matrix} \bar{x}\\ \bar{y} \\ \bar{z} \end{matrix}\right) \tag{4}$$

Now in the $$(\bar{x},\bar{y},\bar{z})$$ coordinate system, $$P_1$$ has the coordinates $$(1,0,0)$$.

Rotating $$P_1$$ through the angle $$\lambda$$ anticlockwise about $$\bar{y}$$ gives $$P_2$$

$$\left(\begin{matrix} \cos(\lambda)&0&\sin(\lambda)\\0&1&0\\-\sin{(\lambda)}&0&\cos(\lambda) \end{matrix}\right) \left( \begin{matrix}1\\0\\0 \end{matrix}\right)=\left(\begin{matrix}\cos(\lambda)\\0\\-\sin(\lambda) \end{matrix}\right)=\left(\begin{matrix}\bar{x}\\\bar{y}\\\bar{z}\end{matrix}\right) \tag{5}$$

Solving for $$x,y$$ and $$z$$ via $$(3)$$ in the original coordinate system yields

$$P_2=(\cos(\phi)\cos(\lambda),\sin(\phi)\cos(\lambda),-\sin(\lambda))\tag{6}$$

which is not the correct the answer and the problem seems to originate from the $$z$$ component which has an extra minus sign in front of it.

What conceptual errors are present in my working?

The angle in spherical coordinates is measured clockwise from the positive $$z$$ axis. Whereas the rotation angle is measured anti-clockwise about the rotation axis. So to take a vector to $$\theta$$ in the spherical coordinates, we must rotate clockwise about the $$\bar y$$ axis. This means your $$\lambda$$ must be negative in the second case. • Doesn't the second rotation matrix $(5)$ (i.e. the conventional rotation matrix about y R(y)) already take into account that the y-axis points into the page when looking at the Z-X plane (i.e. a clockwise rotation about y)? Feb 27 '20 at 21:03