Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Line element for Spherical Polar unit vectors using Geometric Algebra.

Posted by peeterjoot on October 9, 2009

[Click here for a PDF incorporating this post with nicer formatting]

The line element for the particle moving on a spherical surface can be calculated by calculating the derivative of the spherical polar unit vector \hat{\mathbf{r}}

\begin{aligned}\frac{d\hat{\mathbf{r}}}{dt} = \frac{\partial {\hat{\mathbf{r}}}}{\partial {\phi}} \frac{d\phi}{dt} +\frac{\partial {\hat{\mathbf{r}}}}{\partial {\theta}} \frac{d\theta}{dt}\end{aligned} \quad\quad\quad(17)

than taking the magnitude of this vector. We can start either in coordinate form

\begin{aligned}\hat{\mathbf{r}} &= \mathbf{e}_3 \cos\theta + \mathbf{e}_1 \sin\theta \cos\phi + \mathbf{e}_2 \sin\theta \sin\phi\end{aligned} \quad\quad\quad(18)

or, instead do it the fun way, first grouping this into a complex exponential form. Writing i = \mathbf{e}_1 \mathbf{e}_2, the first factorization is

\begin{aligned}\hat{\mathbf{r}} &= \mathbf{e}_3 \cos\theta + \mathbf{e}_1 \sin\theta e^{i\phi} \end{aligned} \quad\quad\quad(19)

The unit vector \boldsymbol{\rho} = \mathbf{e}_1 e^{i\phi} lies in the x,y plane perpendicular to \mathbf{e}_3, so we can form the unit bivector \mathbf{e}_3\boldsymbol{\rho} and further factor the unit vector terms into a \cos + i \sin form

\begin{aligned}\hat{\mathbf{r}} &= \mathbf{e}_3 \cos\theta + \mathbf{e}_1 \sin\theta e^{i\phi} \\ &= \mathbf{e}_3 (\cos\theta + \mathbf{e}_3 \boldsymbol{\rho} \sin\theta) \\ \end{aligned}

This allows the spherical polar unit vector to be expressed in complex exponential form (really a vector-quaternion product)

\begin{aligned}\hat{\mathbf{r}} = \mathbf{e}_3 e^{\mathbf{e}_3 \boldsymbol{\rho} \theta} = e^{-\mathbf{e}_3 \boldsymbol{\rho} \theta} \mathbf{e}_3\end{aligned} \quad\quad\quad(20)

Now, calcuating the unit vector velocity, we get

\begin{aligned}\frac{d\hat{\mathbf{r}}}{dt} &= \mathbf{e}_3 \mathbf{e}_3 \boldsymbol{\rho} e^{\mathbf{e}_3 \boldsymbol{\rho} \theta} \dot{\theta} + \mathbf{e}_1 \mathbf{e}_1 \mathbf{e}_2 \sin\theta e^{i\phi} \dot{\phi} \\ &= \boldsymbol{\rho} e^{\mathbf{e}_3 \boldsymbol{\rho} \theta} \left(\dot{\theta} + e^{-\mathbf{e}_3 \boldsymbol{\rho} \theta} \boldsymbol{\rho} \sin\theta e^{-i\phi} \mathbf{e}_2 \dot{\phi}\right) \\ &= \left( \dot{\theta} + \mathbf{e}_2 \sin\theta e^{i\phi} \dot{\phi} \boldsymbol{\rho} e^{\mathbf{e}_3 \boldsymbol{\rho} \theta} \right) e^{-\mathbf{e}_3 \boldsymbol{\rho} \theta} \boldsymbol{\rho}\end{aligned}

The last two lines above factor the \boldsymbol{\rho} vector and the e^{\mathbf{e}_3 \boldsymbol{\rho} \theta} quaternion to the left and to the right in preparation for squaring this to calculate the magnitude.

\begin{aligned}\left( \frac{d\hat{\mathbf{r}}}{dt} \right)^2&=\left\langle{{ \left( \frac{d\hat{\mathbf{r}}}{dt} \right)^2 }}\right\rangle \\ &=\left\langle{{ \left( \dot{\theta} + \mathbf{e}_2 \sin\theta e^{i\phi} \dot{\phi} \boldsymbol{\rho} e^{\mathbf{e}_3 \boldsymbol{\rho} \theta} \right) \left(\dot{\theta} + e^{-\mathbf{e}_3 \boldsymbol{\rho} \theta} \boldsymbol{\rho} \sin\theta e^{-i\phi} \mathbf{e}_2 \dot{\phi}\right) }}\right\rangle \\ &=\dot{\theta}^2 + \sin^2\theta \dot{\phi}^2+ \sin\theta \dot{\phi} \dot{\theta}\left\langle{{ \mathbf{e}_2 e^{i\phi} \boldsymbol{\rho} e^{\mathbf{e}_3 \rho \theta}+e^{-\mathbf{e}_3 \rho \theta} \boldsymbol{\rho} e^{-i\phi} \mathbf{e}_2}}\right\rangle \\ \end{aligned}

This last term (\in \text{span} \{\rho\mathbf{e}_1, \rho\mathbf{e}_2, \mathbf{e}_1\mathbf{e}_3, \mathbf{e}_2\mathbf{e}_3\}) has only grade two components, so the scalar part is zero. We are left with the line element

\begin{aligned}\left(\frac{d (r\hat{\mathbf{r}})}{dt}\right)^2 = r^2 \left( \dot{\theta}^2 + \sin^2\theta \dot{\phi}^2 \right)\end{aligned} \quad\quad\quad(21)

In retrospect, at least once one sees the answer, it seems obvious. Keeping \theta constant the length increment moving in the plane is ds = \sin\theta d\phi, and keeping \phi constant, we have ds = d\theta. Since these are perpendicular directions we can add the lengths using the Pythagorean theorem.

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