Rotation Matrix Time Derivatives

Problem Setup

For a point $x_0 \in\mathbb R^4$ , the rotation is a transformation defined as

x(\Delta t) = R(\Delta t) x_0

where $R(\Delta t) \in SO(3)$ .

Then, through the transformation through time $t$ will create a trajectory. We come up with the velocity of $x$ as

\dot x (\Delta t) = \dot R(\Delta t) x_0

So that the velocity is the time derivative of the rotation matrix.

Angular Velocity

Without loss of generality, assume $x$ is rotated along the rotation with the axis of rotation $\vec a\in\mathbb R^3$ . Therefore, $x$ is travel in a circle.
Let $\dot\theta$ be the change in angle.
Let $v: \mathbb R\rightarrow \mathbb R^3$ be the velocity of $x$ , decompose $v(t) = a(t)d(t)$ where $a$ is the magnitude and $d$ is the direction.

Since we are rotating around $\vec a$ , i.e. the plane that contains the circle is orthogonal to $\vec a$ . Hence $d\perp \vec a$ .
In addition, a rotation is orthogonal matrix so that $v\perp x$ , since $\vec a,v,x$ are mutually orthogonal, we can uniquely determine $d$ from

d = \frac{\vec a \times x}{\|x\|}

Then, consider the magnitude $a$ , let $y = x + \Delta tv$ . The angle formed by $y$ and $x$ is $\dot\theta \Delta t$ . So that we can have

a\Delta t = \|x\|\tan(\dot\theta) = \|x\|\frac{\sin(\dot\theta\Delta t)}{\cos(\dot\theta\Delta t)}

Therefore, we can have

\lim_{\Delta t\rightarrow 0} \|x\|\frac{\sin(\dot\theta\Delta t)}{\cos(\dot\theta\Delta t)} = \|x\|\frac{\dot\theta\Delta t}{1} = \|x\|\dot\theta\Delta t

so that

a = \|x\|\dot\theta

and then

v = ad=\dot\theta\|x\|\frac{\vec a\times x}{\|x\|} = (\dot\theta \vec a) \times x = \omega \times x

Therefore, we obtain the angular velocity $\omega$ , which includes the velocity of angle and the rotation direction.

Rotation Matrix

Since cross product can be written into cross matrix form as matrix multiplication, $v=\omega\times x $ can then be understood as

\frac{dx}{dt} = [\omega]_\times x

which is a linear ODE, and has analytical solution

x(t) = \exp([\omega]_\times t) x

where $\exp(M)$ is the matrix exponential.

Matrix Exponential

For an invertible matrix $A\in\mathbb R^{n\times n}$ . the matrix exponential $\exp(A)$ is given as

\exp(A) = V\begin{bmatrix}e^{\lambda_1}&\cdots&0\\\vdots&\ddots&\vdots\\0&\cdots&e^{\lambda_n}\end{bmatrix}V^{-1}

Where $A=V\Lambda A^{-1}$ is the Eigen decomposition. However, as we did the decomposition, it is not very efficient.

For our problem, since $[\omega]_{\times}$ is a cross product matrix, hence $3\times 3$ skew-symmetric, we can use Rodrigues' Rotation Formula. First, we can break $\omega t$ into the axis of rotation and angle of rotation, as $\vec a$ and $\theta$

\omega t = \frac{\omega}{\|\omega\|}{\|\omega\|t} = \vec a \theta

then Rodrigues' Rotation Formula gives

R(t) = I + \sin(\theta)[\vec a]_\times + (1-\cos(\theta)) {[\vec a]_\times}^2

Relationship between R and w

Note that we have $\dot x(t) = \dot R(t)x_0$ and the equation above gives $\dot x(t) = [\omega]_\times x$ , therefore, we have found the relation that

\dot R = [\omega]_\times

General Equation

Consider the explicit equation with a fixed $t_0$

x(t_0+\Delta t) = \Delta R(\Delta t) R(t_0)x_0

and its time derivative is

v = \frac{dx}{d\Delta t} = \Delta \dot R(\Delta t) R(t_0)x_0

Note that from the derivations above, since $R(t_0)x_0$ is a fixed point, $v$ is just the time derivative of rotation at time $t_0$ so that

v = [\omega]_\times Rx_0

Another form of this equation is

\begin{align*} v &= \omega \times Rx_0\\ &= -(Rx_0)\times \omega\\ &= R{[x_0]_\times}^TR^T \omega \end{align*}

then this form gives that $v$ is linear in $\omega$ .

Source code