Vector Reflection in Spherical Coordinates Proof

Question

Say I have a vector a = (1, $\theta_1, \phi_1$), and another vector b = (1, $\theta_2, \phi_2$). I want to come up with a formula for reflecting a through b in spherical coordinates which in Cartesian coordinates is defined as: $$c = (2bb^T-I)a = 2(b \cdot a) b - a $$

Intuitively/geometrically it seems like the resultant vector must be (1, $2\theta_2 - \theta_1, 2\phi_2 - \phi_1$) since b should be the mid-point in the rotation that gets one from a to c (a good mental picture would be to imagine all 3 points on a disc slicing the unit sphere).

The reason I suspect c = (1, $2\theta_2 - \theta_1, 2\phi_2 - \phi_1$) is because we can go from a to b by 2 rotations $\theta_2 - \theta_1$, and $\phi_2 - \phi_1$ along the longitude and the latitude; hence repeating the 2 rotations again should result in c. However, I don't know how to prove this. I tried the brute force way of converting to Cartesian and then applying the transformation but I end up with a huge mess of sines and cosines. Is my suspicion even right? If so, how do I prove it? Here's the formula for the dot product in spherical coordinates if it helps:

$$(r_1, \theta_1, \phi_1)\cdot(r_2, \theta_2, \phi_2) = r_1r_2 (\sin\phi_1\sin\phi_2\cos(\theta_1-\theta_2)+\cos\phi_1\cos\phi_2)$$

Answer 1

"However, I don't know how to prove this. I tried the brute force way of converting to Cartesian and then applying the transformation but I end up with a huge mess of sines and cosines. "

Unfortunately I think that is the best you can hope for. The latitude and longitude lines on a sphere are poorly adapted to handling general isometries of the sphere.

BTW. Think carefully about which type of "reflection " you wish to model. In 3D space you can (i) reflect about a plane in space or (ii) reflect through the origin using the map $A\to -A$ (called the antipodal map or reflection through the origin. All other isometries of the sphere in 3D space are actually rotations about some axis. The plane perpendicular to that axis carves a great circle on the sphere.

If you want to " (iii) reflect $A$ about $B$" by traveling on a great circle arc from $A$ through $B$ to the other equidistant point $A'$ on that great circle, that is actually rotation of the sphere. It can however be expressed as a composition of two reflections: first (ii) then (i)

Here are the details.

The reflection formula for what I referred to as case (i) above (reflecting across the plane whose normal vector is $B$) is written in Cartesian vector form as (i) $A\to A- 2(A\cdot B) B$. As a reality check, note that every $A$ that lies in the plane perpendicular to $B$ will stay fixed by this formula (i), and $A$ gets reversed.

If you first perform (ii) and then perform (i) you get the rotation map that sends $A\to - A\to -A + 2 (A\cdot B) B =f(A)$

For example, if $B$ is the North Pole, then each point $A$ on the equator is mapped to $f(A)=-A$ on the opposite side of equator. But if $A$ lies in the northern hemisphere (so $A\cdot B>0$) then $f(A)\cdot B= A\cdot B >0$ so $f(A)$ is also in that northern hemisphere.

Thus on further reflection I realize that we were talking in circles (pun!): your original posted formula is correct, but the descriptive term "reflection" is a bit ambiguous.

Anyway, your best hope is to use Cartesian coordinates to compute $f(A)$.