# Angular velocity in Fick Spherical coordinates

On wikipedia and mathworld it is explained how angular velocity is calculated in spherical coordinates. The coordinate system used there is however not what I am using. In my field, coordinates follow the Fick gimbal. The transforms for this system are: \begin{align} x &= \rho\cos{\phi}\sin{\theta}\\ y &= \rho\sin{\phi}\\ z &= \rho\cos{\phi}\cos{\theta} \end{align} and the inverses: \begin{align} \rho &= \sqrt{x^2+y^2+z^2}\\ \theta &= \tan^{-1} \frac{x}{z}\\ \phi &= \tan^{-1} \frac{y}{\sqrt{x^2+z^2}}\\ \end{align}

That is, $$\theta$$ is the azimuthal angle/rotation in the $$x$$-$$z$$ plane, and $$\phi$$ the elevation/rotation out of the $$x$$-$$z$$ plane.

Now, i would like to determine the expression for angular velocity of a point in this coordinate system. I know from secondary sources that the answer is $$v=\sqrt{\dot\theta^2\cos^2\phi+\phi^2}$$ but i have trouble arriving at it. I have done the following:

I found the following two texts, which explain how to determine the angular derivatives based on the unit vectors of the spherical system, and this answer to another question for how to derive the unit vectors.

Following the steps of the latter link to find $$\boldsymbol{\hat\rho}$$, $$\boldsymbol{\hat\theta}$$, and $$\boldsymbol{\hat{\phi}}$$, we take the derivative of the spherical coordinate representation $$(\rho\cos{\phi}\sin{\theta}, \rho\sin{\phi}, \rho\cos{\phi}\cos{\theta})$$ with respect to $$\rho$$, $$\theta$$, and $$\phi$$, respectively, and then normalize each one. This yields: $$\begin{bmatrix}\boldsymbol{\hat\rho} \\ \boldsymbol{\hat\theta} \\ \boldsymbol{\hat\phi} \end{bmatrix} = \begin{bmatrix} \cos\phi\sin\theta & \sin\phi & \cos\phi\cos\theta \\ \cos\phi\cos\theta & 0 & -\cos\phi\sin\theta \\ -\sin\phi\sin\theta & \cos\phi & -\cos\theta\sin\phi \end{bmatrix} \begin{bmatrix} \mathbf{\hat x} \\ \mathbf{\hat y} \\ \mathbf{\hat z} \end{bmatrix}$$ If i understand correctly, each row is a unit vector here.

Then, following here, \begin{align} \boldsymbol{v}&=\frac{d}{dt}\boldsymbol{\rho}\\ &=\boldsymbol{\hat\rho}\cdot\frac{d\rho}{dt}+\rho\cdot\frac{d\boldsymbol{\hat\rho}}{dt}\\ &=\boldsymbol{\hat\rho}\cdot\frac{d\rho}{dt}+\rho\cdot\biggl(\frac{\delta\boldsymbol{\hat\rho}}{\delta \rho}\cdot\frac{d\rho}{dt}+\frac{\delta\boldsymbol{\hat\rho}}{\delta \theta}\cdot\frac{d\theta}{dt}+\frac{\delta\boldsymbol{\hat\rho}}{\delta \phi}\cdot\frac{d\phi}{dt}\biggr) \end{align}

To solve that equation, we need the partial derivatives of $$\hat{\boldsymbol{\rho}}$$ with respect to $$\rho$$, $$\theta$$, and $$\phi$$, which are: \begin{align} \frac{\delta\boldsymbol{\hat\rho}}{\delta\rho} &= 0\\ \frac{\delta\boldsymbol{\hat\rho}}{\delta\theta} &= \cos\phi\cos\theta - \cos\phi\sin\theta\\ &= \boldsymbol{\hat\theta}\\ \frac{\delta\boldsymbol{\hat\rho}}{\delta\phi} &= - \sin\phi\sin\theta + \cos\phi - \cos\theta\sin\phi \\ &= \boldsymbol{\hat\phi} \end{align}

which would make: $$\boldsymbol{v}=\hat{\boldsymbol{\rho}}\cdot\frac{d\rho}{dt}+\rho\cdot\biggl(\boldsymbol{\hat\theta}\cdot\frac{d\theta}{dt}+\boldsymbol{\hat\phi}\cdot\frac{d\phi}{dt}\biggr)$$ and thereby $$v=\sqrt{\dot\theta^2+\dot\phi^2},$$ which isn't right!

Where am i going wrong?

\begin{align} \frac{\delta\boldsymbol{\hat\rho}}{\delta\rho} &= 0\\ \frac{\delta\boldsymbol{\hat\rho}}{\delta\theta} &= \sqrt{(\cos\phi\cos\theta)^2+( - \cos\phi\sin\theta)^2}\boldsymbol{\hat\theta} \\ &= \cos\phi\boldsymbol{\hat\theta}\\ \frac{\delta\boldsymbol{\hat\rho}}{\delta\phi} &= \sqrt{(- \sin\phi\sin\theta)^2 + \cos^2\phi +(- \cos\theta\sin\phi)^2 }\boldsymbol{\hat\phi}\\ &= \boldsymbol{\hat\phi} \end{align}
$$v=\sqrt{\dot\theta^2\cos^2\phi+\dot\phi^2},$$
• Ha, wikipedia to the rescue. I see that to calculate the coefficients for determining how $\hat\rho$ changes with each of the spherical coordinates, we need to take the vector magnitude of the partial diffs. Thanks! – Diederick C. Niehorster Sep 26 at 22:49