# Defining a rotation vector as a function of theta

I'm reading a paper and came across the following definition of a rotation vector.

$P_r = 2sin\frac{\theta}{2}\begin{bmatrix} n_1 & n_2 & n_3 \end{bmatrix}^T, 0 \leq \theta \leq \pi$

and "$R$ is a simple function of $P_r$ without any trigonometric functions"

$R = (1 -\frac{|P_r|^2}{2})I + \frac{1}{2}(P_rP_r^T+\alpha \cdot Skew(P_r))$

$\alpha = \sqrt{4 - |P_r|^2}$

I'm not sure where these two equations come from. I am, however familiar with the matrix formulation of the Rodrigues formula for rotation.

$R = I + Skew(n)sin(\theta) + Skew(n)^2(1-cos(\theta))$

where $n$ is the axis of rotation and $\theta$ is the angle of rotation.

I think the two equations are somehow related, but I don't know a whole lot about how the first one was derived.

• See Rotation matrix from axis and angle at Wikipedia, and particularly the Exponential map section (Lie algebra), then the Exponential map section in the Rotation group SO(3) article. I'm not a mathematician myself, so perhaps someone with better grasp of the correct terminology could distill the most salient points here? – Nominal Animal Jul 2 '18 at 19:56
• @NominalAnimal thanks, this certainly gives a hint towards the right direction. – Carpetfizz Jul 2 '18 at 20:08
• Presumably $n=(n_1,n_2,n_3)^T$ is a unit vector? – amd Jul 2 '18 at 21:39
• @amd yes that’s right – Carpetfizz Jul 2 '18 at 21:41
• I’m a bit amused by the “without trigonometric functions” assertion since the definition of $P_r$ involves... a trigonometric function. They’re there, but in disguise. – amd Aug 15 '18 at 23:38

## 1 Answer

We can get the Rodrigues formula:

$v' = (\cos \theta) v + (\sin \theta) n \times v + (1 - \cos \theta) n ( n \cdot v)$

Or in matrix form:

$v' = ((\cos \theta) I + (\sin \theta) Skew(n) + (1 - \cos \theta) n n^T) v$

From the equation that you have. First notice that:

$\| P_r \| = 2 \sin \theta /2$

$\| P_r \|^2 = 2(1 - \cos \theta)$

So the first term is just:

$(1 - \| P_r \|^2/2) I = (\cos \theta) I$

Using the same identity, the term $P_r P_r^T$ is just the same as $2(1 - \cos \theta) n n^T$

The third term $\alpha Skew (P_r)$ is just $(2 \sin \theta) Skew(n)$ since:

$\alpha Skew (P_r)= \alpha (2 \sin \theta/2) Skew (n)$

$\alpha = \sqrt{4 - 4\sin^2(\theta/2)} = 2 \sqrt{1 - \sin^2(\theta/2)} = 2 \cos \theta/2$

Using the identity $\sin \theta = 2 \cos(\theta/2) \sin(\theta/2)$:

$\alpha Skew (P_r) = 2 \sin \theta Skew (n)$

Joining the three pieces we finally get:

$R = (\cos \theta) I + (\sin \theta) Skew(n) + (1 - \cos \theta) n n^T$

Which is Rodrigues formula

• Perfect. Thanks! – Carpetfizz Jul 3 '18 at 14:21