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?


1 Answer 1


You need to perform the vector summation in the calculation below

$$ \begin{align} \frac{\delta\boldsymbol{\hat\rho}}{\delta\rho} &= 0\\ \frac{\delta\boldsymbol{\hat\rho}}{\delta\theta} &= \sqrt{(\cos\phi\cos\theta)^2+( - \cos\phi\sin\theta)^2}\boldsymbol{\hat\theta} \\ &= \cos\phi\boldsymbol{\hat\theta}\\ \frac{\delta\boldsymbol{\hat\rho}}{\delta\phi} &= \sqrt{(- \sin\phi\sin\theta)^2 + \cos^2\phi +(- \cos\theta\sin\phi)^2 }\boldsymbol{\hat\phi}\\ &= \boldsymbol{\hat\phi} \end{align} $$

Then, you will get

$$ v=\sqrt{\dot\theta^2\cos^2\phi+\dot\phi^2}, $$

  • $\begingroup$ Hmm, that does the trick. But, why? Are you able to elucidate on/point me to the step i missed? It seems quite different from the two documents i linked to my (admittedly) untrained eye $\endgroup$ Commented Sep 26, 2019 at 22:07
  • $\begingroup$ 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! $\endgroup$ Commented Sep 26, 2019 at 22:49
  • 1
    $\begingroup$ Correct, which is why you need to take the vector sum at the end. $\endgroup$
    – Quanto
    Commented Sep 26, 2019 at 23:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .