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Spherical basis vectors in terms of exponential rotation operators

July 24, 2009 7 Comments

The spherical basis vectors $\mathbf e_r$ , $\mathbf e_\theta$ , and $\mathbf e_\phi$ may be expressed as

$(1a)\qquad\mathbf e_r= e^{-i\mathbf e_3\phi/2} e^{-i\mathbf e_2\theta/2}\mathbf e_3 e^{i\mathbf e_2\theta/2}e^{i\mathbf e_3\phi/2}$ ,
$(1b)\qquad\mathbf e_\theta= e^{-i\mathbf e_3\phi/2} e^{-i\mathbf e_2\theta/2}\mathbf e_1 e^{i\mathbf e_2\theta/2}e^{i\mathbf e_3\phi/2}$ ,
$(1c)\qquad\mathbf e_\phi= e^{-i\mathbf e_3\phi/2} e^{-i\mathbf e_2\theta/2}\mathbf e_2 e^{i\mathbf e_2\theta/2}e^{i\mathbf e_3\phi/2}$ ,

Equations (1a) to (1c) state that $\mathbf e_r$ , $\mathbf e_\theta$ , $\mathbf e_\phi$ are the unit vectors $\mathbf e_3$ , $\mathbf e_1$ , and $\mathbf e_2$ rotated counterclockwise about the the vector $\mathbf e_2$ by an angle $\theta$ and then rotated counterclockwise about the vector $\mathbf e_3$ by an angle $\phi$ .

Using the exponential relations

$(2a)\qquad \mathbf e_1 e^{i\mathbf e_2\theta/2}=e^{-i\mathbf e_2\theta/2}\mathbf e_1$ ,
$(2b)\qquad \mathbf e_2 e^{i\mathbf e_2\theta/2}=e^{i\mathbf e_2\theta/2}\mathbf e_2$ ,
$(2c)\qquad \mathbf e_3 e^{i\mathbf e_2\theta/2}=e^{-i\mathbf e_2\theta/2}\mathbf e_3$ ,

Eqs. (1a) to (1c) simplifies to

$(3a)\qquad \mathbf e_r=e^{-i\mathbf e_3\phi/2}\mathbf e_3 e^{i\mathbf e_2\theta}e^{i\mathbf e_3\phi/2}$ ,
$(3b)\qquad \mathbf e_\theta=e^{-i\mathbf e_3\phi/2}\mathbf e_1 e^{i\mathbf e_2\theta}e^{i\mathbf e_3\phi/2}$ ,
$(3c)\qquad \mathbf e_\phi=e^{-i\mathbf e_3\phi/2}\mathbf e_2 e^{i\mathbf e_3\phi/2}$ .

Furthermore, employing the Euler identities

$(4a)\qquad\mathbf e_1e^{i\mathbf e_2\theta}=\mathbf e_1\cos\theta-\mathbf e_3\sin\theta$ ,
$(4b)\qquad\mathbf e_3e^{i\mathbf e_2\theta}=\mathbf e_3\cos\theta+\mathbf e_1\sin\theta$ ,

together with the exponential realations

$(5a)\qquad\mathbf e_1 e^{i\mathbf e_3\phi/2}=e^{-i\mathbf e_3\phi/2}\mathbf e_1$ ,
$(5b)\qquad\mathbf e_2 e^{i\mathbf e_3\phi/2}=e^{-i\mathbf e_3\phi/2}\mathbf e_2$ ,
$(5c)\qquad\mathbf e_3 e^{i\mathbf e_3\phi/2}=e^{i\mathbf e_3\phi/2}\mathbf e_3$ ,

Eqs. (3a) to (3c) becomes

$(6a)\qquad\mathbf e_r=\mathbf e_3\cos\theta+\mathbf e_1 e^{i\mathbf e_3\phi}\sin\theta$ ,
$(6b)\qquad\mathbf e_\theta=\mathbf e_1 e^{i\mathbf e_3\phi}\cos\theta-\mathbf e_3\sin\theta$ ,

$(6c)\qquad\mathbf e_\phi=\mathbf e_2 e^{i\mathbf e_3\phi}$

Finally, using the Euler identities

$(7a)\qquad \mathbf e_1 e^{i\mathbf e_3\phi}=\mathbf e_1\cos\phi + \mathbf e_2\sin\phi$ ,
$(7b)\qquad\mathbf e_2 e^{i\mathbf e_3\phi}=\mathbf e_2\cos\phi-\mathbf e_1\sin\phi$ ,

Eqs. (6a) to (6c) reduces to

$(8a)\qquad\mathbf e_r=\mathbf e_3\cos\theta+\mathbf e_1\sin\theta\cos\phi + \mathbf e_2\sin\theta\sin\phi$ ,
$(8b)\qquad\mathbf e_\theta=\mathbf e_1\cos\theta\cos\phi + \mathbf e_2\cos\theta\sin\phi-\mathbf e_3\sin\theta$ ,

$(8c)\qquad\mathbf e_\phi=\mathbf e_2\cos\phi-\mathbf e_1\sin\phi$ .

This corresponds to the standard expressions for the spherical basis vectors in rectangular coordinates (see Wikipedia).

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Filed under Uncategorized Tagged with Euler identity, exponential functions, geometric algebra, rotation operators, spherical coordinates

About Quirino M. Sugon Jr
Theoretical Physicist in Manila Observatory

7 Responses to Spherical basis vectors in terms of exponential rotation operators

peeterjoot says:

November 11, 2009 at 1:50 pm

(6a), (6b) can be reduced nicely to a form similar to (6c). Introduce a unit bivector $j$ for the plane of rotation from $\mathbf{e}_3$ through the point at which your basis vectors are sitting on the unit sphere. That is

$\begin{aligned}j = \mathbf{e}_3 \wedge (\mathbf{e}_1 e^{i \mathbf{e}_3 \phi})\end{aligned}$

Now you can factor (6a) like so

$\begin{aligned}\mathbf e_r &= \mathbf e_3\cos\theta+\mathbf e_1 e^{i\mathbf e_3\phi}\sin\theta \\ &= \mathbf e_3 (\cos\theta+ \mathbf{e}_3 \mathbf e_1 e^{i\mathbf e_3\phi}\sin\theta) \\ &= \mathbf e_3 (\cos\theta + j \sin\theta) \\ &= \mathbf e_3 e^{j \theta}\end{aligned}$

Like (6c) you have something very much like a complex polar representation, where a quaternion exponential rotates your vector from its initial position to the final position by straight multiplication.

(6b) is similar, and I get

$\begin{aligned}\mathbf e_\theta &= \mathbf e_1 e^{i\mathbf e_3\phi}\cos\theta-\mathbf e_3\sin\theta \\ &= \mathbf e_1 e^{i\mathbf e_3\phi}( \cos\theta - e^{-i\mathbf e_3\phi} \mathbf e_1\mathbf e_3\sin\theta) \\ &= \mathbf e_1 e^{i\mathbf e_3\phi}( \cos\theta + j \sin\theta) \\ &= \mathbf{e}_3 j e^{j\theta} \\ \end{aligned}$

Since $j$ contains $\mathbf{e}_3$ as a factor, the product $\mathbf{e}_3 j$ is a vector, and you have the same form in the end, vector times exponential producing a new vector.
Quirino M. Sugon Jr says:

November 12, 2009 at 1:46 am

Thanks, Peeter. That’s a neat trick. The vector $\mathbf e_3 j = \mathbf e_1 e^{j\theta}$ . This is the one rotated counterclockwise in the plane $j$ by an angle $\theta$ . Thanks also for the \begin{aligned} command.
peeterjoot says:

November 12, 2009 at 2:05 am

That vector isn’t $\mathbf{e}_1 e^{j \theta}$ , but is $\mathbf{e}_3 j = \mathbf{e}_1 e^{i \mathbf{e}_3 \phi}$ . This is $\mathbf{e}_1$ , rotated in the x,y plane rotated towards the $\mathbf{e}_2$ axis by $\phi$ (what is probably usually designated by $\mathbf{\rho}$ in cylindrical polar coordinates, the outwards facing radial unit vector in such a coordinate system).

re. aligned. I have cobbled together the following perl script for converting standalone latex to wordpress format:

http://sites.google.com/site/peeterjoot/math2009/tex2blog

you may be interested in it. It has some bugs but does the job fairly well. There’s also a python script that is similar:

http://lucatrevisan.wordpress.com/latex-to-wordpress/

(I use mine since it has citation and \ref support for equation #s, and also supports my personal newcommand’s … it wasn’t obvious to me how to modify the above python script to support multiple argument expression replacement.)
Quirino M. Sugon Jr says:

November 13, 2009 at 8:12 am

Thanks, Peeter. Another typo mistake. I’ll try the links.
Christiaan Mantz says:

May 7, 2011 at 10:27 am

Nice site! I would like to point out a typo first (1b) er => etheta

In (4a) you use a relationship for the basis vectors: e1e2 = i e3.
This relationship is true in 3 dimensions. One can choose Pauli matrices then to show that this is true.
Is it also true in four dimensions? What representation would you use then?
Quirino M. Sugon Jr says:

May 28, 2011 at 3:12 am

Christiaan,

If we think of the algebra Cl_{3,0} generated by {e_1, e_2, e_3} as a subset of the algebra Cl_{4,0} generated by {e_1, e_2, e_3, e_4}, then e_1 e_2 = i e_3, where i = e_1 e_2 e_3, is still valid in Cl_{4,0}. If are thinking of the pseudoscalar I = e_1 e_2 e_3 e_4, then e_1 e_2 \neq I e_3, because I e_3 = -e_1 e_2 e_4.
Christiaan Mantz says:

May 29, 2011 at 3:56 pm

In the mean while I discovered the answer. Replace in eq 1 the basis vectors in the exponentials by the wedge products of the two other basis vectors. So replace ie_3 by e_1 e_2 this is now valid in arbitrary dimension since the wedge product of two basis vectors are the generators of a liegroup in arbitrary dimensions and the basis vector are only generators in 3 dimensions

Thanks! take care

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Spherical basis vectors in terms of exponential rotation operators

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