# Geometric algebra: longitude and latitude rotor ordering?

1. May 11, 2008

### Peeter

[SOLVED] geometric algebra: longitude and latitude rotor ordering?

Was playing around with what is probably traditionally a spherical trig type problem using geometric algebra (locate satellite position using angle measurements from two well separated points). Origin of the problem was just me looking at my Feynman Lectures introduction where there is a diagram illustrating how trianglulation could be used to locate "Sputnik" and thought I'd try such a calculation, but in a way that I thought was more realistic.

I'm pretty sure that I solved the problem, but have a question about a small detail that I glossed over. I'll describe a bit of the context as background.

Part of my solution requires the measured unit vector be rotated back to a reference frame (ie: measure direction cosines in a local frame with frame dir vectors north facing, east facing, and up facing). If I fix a reference frame at (0,0) ( equator/gren. intersection, e1 = east, e2 = north, e3 = up at that point), I can rotate to any given longitude/latitude (and thus inverse rotate my measured direction vector to the satellite location) using the following rotor:

$$R = R_{g} R_{e} = \exp(-e_3 \wedge e_1 \theta/2)\exp(-e_3 \wedge e_1 \alpha/2)$$

where $\theta$ is the east directed angle measurement, and $\alpha$ is the angle to the north from the equator. This gives a combined rotor equation of:

$$x' = R x R^\dagger$$

Now, looking at a globe, it seems clear to me that these "perpendicular" (abusing the word) rotations could be applied in either order, but their rotors definitely don't commute, so I assume that together the non-commutive bits of the rotors "cancel out".

Any idea how to demonstrate that the end effect of applying these rotors in either order is the same?

ie: for R_g and R_e above:

$$x' = R_g R_e x R_e^\dagger R_g^\dagger = R_e R_g x R_g^\dagger R_e^\dagger$$

In general order of rotations should matter, but here (ie: for the rotors above) I don't think it will. I started doing a brute force expansion of sine and cosine terms, but decided I don't want to do that without a symbolic calculator.

2. May 13, 2008

### Peeter

I figured this out (examining the difference of the two rotation varients applied to an arbitrary vector). These longitude and latitude rotations only commute when applied to an outwards facing vector (ie: a point on the sphere represented by a vector). This is what intuition tells you, but that breaks down if you are trying to rotate a frame positioned at that point on the sphere, or any direction vector in that frame that has a component in the east/west or north/south directions.

I suspect that nobody but me is interested in the math involved to prove this so I won't post it unless asked.