If v is a vector in ℝ3 and k is a unit vector describing an axis of rotation about which v rotates by an angle θ according to the right hand rule, the Rodrigues formula is
An alternative statement is to write the axis vector as a cross producta × b of any two nonzero vectors a and b which define the plane of rotation, and the sense of the angle θ is measured away from a and towards b. Letting α denote the angle between these vectors, the two angles θ and α are not necessarily equal, but they are measured in the same sense. Then the unit axis vector can be written
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.
The vector k×v can be viewed as a copy of v⊥ rotated anticlockwise by 90° about k, so their magnitudes are equal but directions are perpendicular. Likewise the vector k×(k×v) a copy of v⊥ rotated anticlockwise through 180° about k, so that k×(k×v) and v⊥ 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 a×(b×c) = (a·c)b − (a·b)c given any three vectors a, b, c.
The component parallel to the axis will not change magnitude nor direction under the rotation,
only the perpendicular component will change direction but retain its magnitude, according to
and since k and v∥ are parallel, their cross product is zero k×v∥ = 0, so that
and it follows
This rotation is correct since the vectors v⊥ and k×v have the same length, and k×v is v⊥ rotated anticlockwise through 90° about k. An appropriate scaling of v⊥ and k×v using the trigonometric functionssine and cosine gives the rotated perpendicular component. The form of the rotated component is similar to the radial vector in 2d plane polar coordinatesr, θ in the Cartesian basis
where ex, ey are unit vectors in their indicated directions.
Now the full rotated vector is
so eliminating the parallel component gives
while the elimination of the perpendicular component gives
The factors (1 − cosθ) and sinθ can be seen geometrically in the diagram above.
for any vector v. (In fact, K is the unique matrix with this property. It has eigenvalues 0 and ±i).
Iterating the cross product on the left is equivalent to multiplying by the cross product matrix on the left, in particular
The previous rotation formula in matrix language is therefore
and factorizing the v allows the compact expression
is the rotation matrix through an angle θ anticlockwise about the axis k, and I the 3×3 identity matrix. This matrix R is an element of the rotation group SO(3) of ℝ3, and K is an element of the Lie algebraso(3) generating that Lie group (note that K is skew-symmetric, which characterizes so(3)). In terms of the matrix exponential,
To see that the last identity holds, one notes that
characteristic of a one-parameter subgroup, i.e. exponential, and that the formulas match for infinitesimal θ.
Leonhard Euler, "Problema algebraicum ob affectiones prorsus singulares memorabile", Commentatio 407 Indicis Enestoemiani, Novi Comm. Acad. Sci. Petropolitanae15 (1770), 75–106.
Olinde Rodrigues, "Des lois géometriques qui regissent les déplacements d' un systéme solide dans l' espace, et de la variation des coordonnées provenant de ces déplacement considérées indépendent des causes qui peuvent les produire", J. Math. Pures Appl.5 (1840), 380–440.
Don Koks, (2006) Explorations in Mathematical Physics, Springer Science+Business Media,LLC. ISBN 0-387-30943-8. Ch.4, pps 147 et seq. A Roundabout Route to Geometric Algebra'