Rodrigues' rotation formula

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 other words, the Rodrigues' formula provides an algorithm to compute the exponential map from , the Lie algebra of , to without actually computing the full matrix exponential.

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." This proposal has received notable support, but some others have viewed the formula as just one of many variations of the Euler–Rodrigues formula, thereby crediting both.

Statement

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 $\textbf$) that was lost by the first term.

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

:$\mathbf = \frac = \frac\,.$

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.

Derivation

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).

Using the dot and cross products, the vector can be decomposed into components parallel and perpendicular to the axis ,

:$\mathbf = \mathbf_\parallel + \mathbf_\perp \,,$

where the component parallel to is

:$\mathbf_\parallel = (\mathbf \cdot \mathbf) \mathbf$

called the vector projection of on , and the component perpendicular to is

:$\mathbf_ = \mathbf - \mathbf_ = \mathbf - (\mathbf \cdot \mathbf) \mathbf = - \mathbf\times(\mathbf\times\mathbf)$

called the vector rejection of from .

The vector can be viewed as a copy of rotated anticlockwise by 90° about , so their magnitudes are equal but directions are perpendicular. Likewise the vector a copy of rotated anticlockwise through about , so that and are equal in magnitude but in opposite directions (i.e. they are negatives of each other, hence the minus sign). Expanding the vector triple product establishes the connection between the parallel and perpendicular components, for reference the formula is given any three vectors , , .

The component parallel to the axis will not change magnitude nor direction under the rotation,

:$\mathbf_ = \mathbf_\parallel \,,$

only the perpendicular component will change direction but retain its magnitude, according to

:$\begin \left, \mathbf_\ &= \left, \mathbf_\perp\ \,, \\ \mathbf_ &= \cos(\theta) \mathbf_\perp + \sin(\theta) \mathbf\times\mathbf_\perp \,, \end$

and since and are parallel, their cross product is zero , so that

:$\mathbf \times \mathbf_\perp = \mathbf \times \left(\mathbf - \mathbf_\right) = \mathbf \times \mathbf - \mathbf \times \mathbf_ = \mathbf \times \mathbf$

and it follows

:$\mathbf_ = \cos(\theta) \mathbf_\perp + \sin(\theta) \mathbf\times\mathbf \,.$

This rotation is correct since the vectors and have the same length, and is rotated anticlockwise through about . An appropriate scaling of and using the trigonometric functions sine and cosine gives the rotated perpendicular component. The form of the rotated component is similar to the radial vector in 2D planar polar coordinates in the Cartesian basis

:$\mathbf = r\cos(\theta) \mathbf_x + r\sin(\theta) \mathbf_y \,,$

where , are unit vectors in their indicated directions.

Now the full rotated vector is

:$\mathbf_ = \mathbf_ + \mathbf_ \,,$

By substituting the definitions of and in the equation results in

:$\begin \mathbf_ & = \mathbf_\parallel + \cos(\theta) \, \mathbf_\perp + \sin(\theta) \, \mathbf\times\mathbf \\ & = \mathbf_\parallel + \cos(\theta) \left(\mathbf - \mathbf_\parallel\right) + \sin(\theta) \, \mathbf\times\mathbf \\ & = \cos(\theta) \, \mathbf + (1 - \cos\theta)\mathbf_\parallel + \sin(\theta) \, \mathbf\times\mathbf \\ & = \cos(\theta) \, \mathbf + (1 - \cos\theta)(\mathbf \cdot \mathbf)\mathbf + \sin(\theta) \, \mathbf\times\mathbf \end$

Matrix notation

Representing and as column matrices, the cross product can be expressed as a matrix product

:$\begin (\mathbf\times\mathbf)_x \\ (\mathbf\times\mathbf)_y \\ (\mathbf\times\mathbf)_z \end = \begin k_y v_z - k_z v_y \\ k_z v_x - k_x v_z \\ k_x v_y - k_y v_x \end = \begin 0 & -k_z & k_y \\ k_z & 0 & -k_x \\ -k_y & k_x & 0 \end \begin v_x \\ v_y \\ v_z \end \,.$

By , denote the " cross-product matrix" for the unit vector ,
: \mathbf=
\left begin
0 & -k_z & k_y \\
k_z & 0 & -k_x \\
-k_y & k_x & 0
\end\right,.

That is to say,
: $\mathbf\mathbf = \mathbf\times\mathbf$
for any vector . (In fact, is the unique matrix with this property. It has eigenvalues 0 and ).

It follows that iterating the cross product is equivalent to multiplying by the cross-product matrix on the left; specifically:
: $\mathbf(\mathbf\mathbf) = \mathbf^2\mathbf = \mathbf\times(\mathbf\times\mathbf) \,.$

The previous rotation formula in matrix language is therefore

:$\begin \mathbf_\mathrm & = \mathbf \cos\theta + (\mathbf \times \mathbf)\sin\theta + \mathbf ~(\mathbf \cdot \mathbf) (1 - \cos\theta) \\ & = \mathbf \cos\theta + (\mathbf \times \mathbf)\sin\theta + (\mathbf - \mathbf_) (1 - \cos\theta) \\ & = \mathbf \cos\theta + (\mathbf \times \mathbf)\sin\theta + (\mathbf + \mathbf\times(\mathbf\times\mathbf))(1 - \cos\theta) \\ & = \mathbf \cos\theta + (\mathbf \times \mathbf)\sin\theta + (\mathbf + \mathbf(\mathbf\mathbf)) (1 - \cos\theta) \\ & = \mathbf (\cos\theta + 1 - \cos\theta) + (\mathbf \times \mathbf)\sin\theta + \mathbf(\mathbf\mathbf) (1 - \cos\theta) \end$

So we have:

:$\mathbf_ = \mathbf + (\sin\theta) \mathbf\mathbf + (1 - \cos\theta)\mathbf^2\mathbf \,,\quad \, \mathbf\, _2 = 1\,.$
Note the coefficient of the leading term is ''now'' 1, in this notation: see the Lie-Group discussion below.

Factorizing the allows the compact expression
:$\mathbf_\mathrm = \mathbf\mathbf$

where

is the rotation matrix through an angle counterclockwise about the axis , and the identity matrix. This matrix is an element of the rotation group of , and is an element of the Lie algebra $\mathfrak(3)$ generating that Lie group (note that is skew-symmetric, which characterizes $\mathfrak(3)$).

In terms of the matrix exponential,
:$\mathbf = \exp (\theta\mathbf)\,.$

To see that the last identity holds, one notes that
:$\mathbf(\theta) \mathbf(\phi) = \mathbf (\theta+\phi), \quad \mathbf(0) = \mathbf\,,$
characteristic of a one-parameter subgroup, i.e. exponential, and that the formulas match for infinitesimal .

For an alternative derivation based on this exponential relationship, see exponential map from $\mathfrak(3)$ to . For the inverse mapping, see log map from to $\mathfrak(3)$.

The Hodge dual of the rotation $\mathbf$ is just $\mathbf^* = -\sin(\theta)\mathbf$ which enables the extraction of both the axis of rotation and the sine of the angle of the rotation from the rotation matrix itself, with the usual ambiguity,
:\begin
\sin(\theta) &= \sigma \left, \mathbf^*\ \\ pt \mathbf &= -\frac
\end
where $\sigma = \pm 1$. The above simple expression results from the fact that the Hodge duals of $\mathbf$ and $\mathbf^2$ are zero, and $\mathbf^* = -\mathbf$.

When applying the Rodrigues' formula, however, the usual ambiguity could be removed with an extended form of the formula.

See also

* Axis angle
* Rotation (mathematics)
* SO(3) and SO(4)
* Euler–Rodrigues formula

References

*Leonhard Euler, "Problema algebraicum ob affectiones prorsus singulares memorabile", ''Commentatio 407 Indicis Enestoemiani, Novi Comm. Acad. Sci. Petropolitanae'' 15 (1770), 75–106.
*Olinde Rodrigues, "Des lois géométriques qui régissent les déplacements d'un système solide dans l'espace, et de la variation des coordonnées provenant de ces déplacements considérés indépendants des causes qui peuvent les produire", ''Journal de Mathématiques Pures et Appliquées'' 5 (1840), 380–440
online

*Friedberg, Richard (2022). " Rodrigues, Olinde: "Des lois géométriques qui régissent les déplacements d'un systéme solide...", translation and commentary". ''arXiv:2211.07787''.
*Don Koks, (2006) ''Explorations in Mathematical Physics'', Springer Science+Business Media,LLC. . Ch.4, pps 147 et seq. ''A Roundabout Route to Geometric Algebra''
*

External links

* Johan E. Mebius
Derivation of the Euler-Rodrigues formula for three-dimensional rotations from the general formula for four-dimensional rotations.
''arXiv General Mathematics'' 2007.
* For another descriptive example see: http://chrishecker.com/Rigid_Body_Dynamics#Physics_Articles, Chris Hecker, physics section, part 4. "The Third Dimension" – on page 3, section ``Axis and Angle'', http://chrishecker.com/images/b/bb/Gdmphys4.pdf

{{DEFAULTSORT:Rodrigues' Rotation Formula
Rotation in three dimensions
Euclidean geometry
Orientation (geometry)

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