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## spherical polar line element (continued).

Posted by peeterjoot on October 25, 2009

[Click here for a PDF of this sequence of posts with nicer formatting]

# Line element.

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. This factorization was done above, but starting over allows this to be done a bit more effectively for this particular problem. As above, 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, calculating 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.

## Line element using an angle and unit bivector parameterization.

Parameterizing using scalar angles is not the only approach that we can take to calculate the line element on the unit sphere. Proceding directly with a alternate polar representation, utilizing a unit bivector $j$, and scalar angle $\theta$ is

\begin{aligned}\mathbf{x} = r \mathbf{e}_3 e^{j\theta}\end{aligned} \quad\quad\quad(22)

For this product to be a vector $j$ must contain $\mathbf{e}_3$ as a factor ($j = \mathbf{e}_3 \wedge m$ for some vector m.) Setting $r = 1$ for now, the deriviate of $\mathbf{x}$ is

\begin{aligned}\dot{\mathbf{x}} &= \mathbf{e}_3 \frac{d}{dt} \left( \cos\theta + j \sin\theta \right) \\ &= \mathbf{e}_3 \dot{\theta} \left( -\sin\theta + j \cos\theta \right) + \mathbf{e}_3 \frac{d j}{dt} \sin\theta \\ &= \mathbf{e}_3 \dot{\theta} j \left( j \sin\theta + \cos\theta \right) + \mathbf{e}_3 \frac{d j}{dt} \sin\theta \\ \end{aligned}

This is

\begin{aligned}\dot{\mathbf{x}} = \mathbf{e}_3 \left( \frac{d\theta}{dt} j e^{j\theta} + \frac{d j}{dt} \sin\theta \right)\end{aligned} \quad\quad\quad(23)

Alternately, we can take derivatives of $\mathbf{x} = r e^{-j\theta} \mathbf{e}_3$, for

Or

\begin{aligned}\dot{\mathbf{x}} = -\left( \frac{d\theta}{dt} j e^{-j\theta} + \frac{d j}{dt} \sin\theta \right) \mathbf{e}_3\end{aligned} \quad\quad\quad(24)

Together with 23, the line element for position change on the unit sphere is then

\begin{aligned}\dot{\mathbf{x}}^2 &= \left\langle{{-\left( \frac{d\theta}{dt} j e^{-j\theta} + \frac{d j}{dt} \sin\theta \right) \mathbf{e}_3 \mathbf{e}_3 \left( \frac{d\theta}{dt} j e^{j\theta} + \frac{d j}{dt} \sin\theta \right) }}\right\rangle \\ &= \left\langle{{-\left( \frac{d\theta}{dt} j e^{-j\theta} + \frac{d j}{dt} \sin\theta \right) \left( \frac{d\theta}{dt} j e^{j\theta} + \frac{d j}{dt} \sin\theta \right) }}\right\rangle \\ &=\left(\frac{d\theta}{dt}\right)^2 - \left(\frac{d j}{dt}\right)^2 \sin^2\theta - \frac{d\theta}{dt} \sin\theta \left\langle{{ \frac{dj}{dt} j e^{j\theta} + j e^{-j\theta} \frac{dj}{dt} }}\right\rangle \\ \end{aligned}

Starting with cyclic reordering of the last term, we get zero

\begin{aligned}\left\langle{{ \frac{dj}{dt} j e^{j\theta} + j e^{-j\theta} \frac{dj}{dt} }}\right\rangle &=\left\langle{{ \frac{dj}{dt} j \left( e^{j\theta} + e^{-j\theta} \right) }}\right\rangle \\ &=\left\langle{{ \frac{dj}{dt} j 2 j \sin\theta }}\right\rangle \\ &=- 2 \sin\theta \frac{d}{dt} \underbrace{\left\langle{{ j }}\right\rangle}_{=0} \\ \end{aligned}

The line element (for constant $r$) is therefore

\begin{aligned}\dot{\mathbf{x}}^2 =r^2 \left( \dot{\theta}^2 - \left(\frac{dj}{dt}\right)^2 \sin^2 \theta \right)\end{aligned} \quad\quad\quad(25)

This is essentially the same result as we got starting with an explicit $r, \theta, \phi$. Repeating for comparision that was

\begin{aligned}\dot{\mathbf{x}}^2 = r^2 \left( \dot{\theta}^2 + \sin^2\theta \dot{\phi}^2 \right)\end{aligned} \quad\quad\quad(26)

The bivector that we have used this time encodes the orientation of the plane of rotation from the polar axis down to the position on the sphere corresponds to the angle $\phi$ in the
scalar parameterization. The negation in sign is expected due to the negative bivector square.

Also comparing to previous results it is notable that we can explicitly express this bivector in terms of the scalar angle if desired as

\begin{aligned}\boldsymbol{\rho} &= \mathbf{e}_1 e^{\mathbf{e}_1 \mathbf{e}_2 \phi} = \mathbf{e}_1 \cos\phi + \mathbf{e}_2 \sin\phi \\ j &= \mathbf{e}_3 \wedge \boldsymbol{\rho} = \mathbf{e}_3 \boldsymbol{\rho}\end{aligned} \quad\quad\quad(27)

The inverse mapping, expressing the scalar angle using the bivector representation is also possible, but not unique. The principle angle for that inverse mapping is

\begin{aligned}\phi &= -\mathbf{e}_1 \mathbf{e}_2 \ln(\mathbf{e}_1 \mathbf{e}_3 j) \end{aligned} \quad\quad\quad(29)

## Allowing the magnitude to vary.

Writing a vector in polar form

\begin{aligned}\mathbf{x} = r \hat{\mathbf{r}}\end{aligned} \quad\quad\quad(30)

and also allowing $r$ to vary, we have

\begin{aligned}\left(\frac{d\mathbf{x}}{dt}\right)^2 &= \left( \frac{dr}{dt} \hat{\mathbf{r}} + r \frac{d\hat{\mathbf{r}}}{dt} \right)^2 \\ &= \left( \frac{dr}{dt} \right)^2 + r^2 \left( \frac{d\hat{\mathbf{r}}}{dt} \right)^2 + 2 r \frac{dr}{dt} \hat{\mathbf{r}} \dot{c} \frac{d\hat{\mathbf{r}}}{dt} \end{aligned}

The squared unit vector derivative was previously calculated to be

\begin{aligned}\left(\frac{d \hat{\mathbf{r}}}{dt}\right)^2 = \dot{\theta}^2 + \sin^2\theta \dot{\phi}^2 \end{aligned} \quad\quad\quad(31)

Picturing the geometry is enough to know that $\dot{\hat{\mathbf{r}}} \dot{c} \hat{\mathbf{r}} = 0$ since $\dot{\hat{\mathbf{r}}}$ is always tangential to the sphere. It should also be possible to algebraically show this, but without going through the effort we at least now know the general line element

\begin{aligned}\dot{\mathbf{x}}^2 = {\dot{r}}^2 + r^2 \left( \dot{\theta}^2 + \sin^2\theta \dot{\phi}^2 \right)\end{aligned} \quad\quad\quad(32)