In the theory of three-dimensional rotation, Rodrigues' rotation formula, named after Olinde Rodrigues, is an efficient algorithm for rotating a vector in space, given an axis and angle of rotation. By extension, this can be used to transform all three basis vectors to compute a rotation matrix in, the group of all rotation matrices, from an axis–angle representation. In terms of Lie theory, the Rodrigues' formula provides an algorithm to compute the exponential map from the Lie algebra to its Lie group .
This formula is variously credited to Leonhard Euler, Olinde Rodrigues, or a combination of the two. A detailed historical analysis in 1989 concluded that the formula should be attributed to Euler, and recommended calling it "Euler's finite rotation formula."[1] This proposal has received notable support,[2] but some others have viewed the formula as just one of many variations of the Euler–Rodrigues formula, thereby crediting both.[3]
If is a vector in and is a unit vector describing an axis of rotation about which rotates by an angle according to the right hand rule, the Rodrigues formula for the rotated vector is
The intuition of the above formula is that the first term scales the vector down, while the second skews it (via vector addition) toward the new rotational position. The third term re-adds the height (relative to
bf{k}
An alternative statement is to write the axis vector as a cross product of any two nonzero vectors and which define the plane of rotation, and the sense of the angle is measured away from and towards . Letting denote the angle between these vectors, the two angles and are not necessarily equal, but they are measured in the same sense. Then the unit axis vector can be written
k=
a x b | |
|a x b| |
=
a x b | |
|a||b|\sin\alpha |
.
This form may be more useful when two vectors defining a plane are involved. An example in physics is the Thomas precession which includes the rotation given by Rodrigues' formula, in terms of two non-collinear boost velocities, and the axis of rotation is perpendicular to their plane.
Let be a unit vector defining a rotation axis, and let be any vector to rotate about by angle (right hand rule, anticlockwise in the figure), producing the rotated vector
vrot
Using the dot and cross products, the vector can be decomposed into components parallel and perpendicular to the axis,
v=v\parallel+v\perp,
where the component parallel to is called the vector projection of on,
v\parallel=(v ⋅ k)k
and the component perpendicular to is called the vector rejection of from :
v\perp=v-v\parallel=v-(k ⋅ v)k=-k x (k x v)
where the last equality follows from the vector triple product formula: . Finally, the vector
k x v\perp=k x v
v\perp
k
R3
Under the rotation, the component
v\parallel
v\parallelrot=v\parallel;
while the perpendicular component will retain its magnitude but rotate its direction in the perpendicular plane spanned by
v\perp
k x v
v\perprot=\cos(\theta)v\perp+\sin(\theta)k x v\perp=\cos(\theta)v\perp+\sin(\theta)k x v,
in analogy with the planar polar coordinates in the Cartesian basis, :
r=r\cos(\theta)ex+r\sin(\theta)ey.
Now the full rotated vector is:
vrot=v\parallelrot+v\perprot=v\parallel+\cos(\theta)v\perp+\sin(\theta)k x v.
v\perp=v-v\|
v\|=v-v\perp
The linear transformation on
v\isinR3
v\mapstok x v
\begin{bmatrix}(k x v)x\ (k x v)y\ (k x v)z\end{bmatrix}=\begin{bmatrix}kyvz-kzvy\ kzvx-kxvz\ kxvy-kyvx\end{bmatrix}=\left[\begin{array}{rrr} 0 &-kz&ky\\ kz&0 &-kx\\ -ky&kx&0 \end{array}\right] \begin{bmatrix}vx\ vy\ vz\end{bmatrix}.
That is, the matrix of this linear transformation (with respect to standard coordinates) is the cross-product matrix:
K=\left[\begin{array}{rrr} 0 &-kz&ky\\ kz&0 &-kx\\ -ky&kx&0 \end{array}\right].
k x v=Kv, k x (k x v)=K(Kv)=K2v.
vrot=v+(\sin\theta)Kv+(1-\cos\theta)K2v.
vrot=Rv
ak{so}(3)
ak{so}(3)
In terms of the matrix exponential,
R=\exp(\thetaK).
To see that the last identity holds, one notes that
R(\theta)R(\phi)=R(\theta+\phi), R(0)=I,
For an alternative derivation based on this exponential relationship, see exponential map from
ak{so}(3)
ak{so}(3)
The Hodge dual of the rotation
R
R*=-\sin(\theta)k
\begin{align} \sin(\theta)&=\sigma\left|R*\right|\\[3pt] k&=-
\sigmaR* | |
\left|R*\right| |
\end{align}
\sigma=\pm1
I
K2
K*=-k