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Euler pole.

In kinematics, Euler's rotation theorem states that, in three-dimensional space, any displacement of a rigid body such that a point on the rigid body remains fixed, is equivalent to a rotation about a fixed axis through that point. The theorem is named after Leonhard Euler, who proved this in 1775 by an elementary geometric argument. The axis of rotation is known as an Euler pole.

In mathematical terms, this is a statement that, in 3D space, any two Cartesian coordinate systems with a common origin are related by a rotation about some fixed axis. This also means that the product of two rotation matrices is again a rotation matrix. A (non-identity) rotation matrix has a real eigenvalue which is equal to unity. The eigenvector corresponding to this eigenvalue is the axis of rotation connecting the two systems.

[edit] Euler's theorem (1776)

Euler states the theorem as follows:^[1]

Theorema. Quomodocunque sphaera circa centrum suum conuertatur, semper assignari potest diameter, cuius directio in situ translato conueniat cum situ initiali.

or (in free translation):

When a sphere is moved around its centre it is always possible to find a diameter whose direction in the displaced position is the same as in the initial position.

To prove this, Euler considers a great circle on the sphere and the great circle to which it is transported by the movement. These two circles intersect in two (opposite) points of which one, say A, is chosen. This point lies on the initial circle and thus is transported to a point a on the second circle. On the other hand, A lies also on the translated circle, and thus corresponds to a point α on the initial circle. Now Euler considers the symmetry plane of the angle αAa (which passes through the centre C of the sphere) and the symmetry plane of the arc Aa (which also passes through C). These two planes intersect in a diameter whose endpoint O on the sphere remains fixed under the movement because the triangle OαA is transported onto the triangle OAa (since αA is mapped on Aa and the triangles have the same angles).

This also shows that the rotation of the sphere can be seen as two consecutive reflections about the two planes described above. Points in a mirror plane are invariant under reflection, and hence the points on their intersection (a line: the axis of rotation) are invariant under both the reflections, and hence under the rotation.

[edit] Matrix proof

An algebraic proof starts from the fact that a rotation is a linear map in one-to-one correspondence withi a 3×3 rotation matrix R, i.e, a matrix for which

$\mathbf{R}^\mathrm{T}\mathbf{R} = \mathbf{R}\mathbf{R}^\mathrm{T} = \mathbf{E},$

where E is the 3×3 identity matrix and superscript T indicates the transposed matrix. Clearly a rotation matrix has determinant ±1, for invoking some properties of determinants, one can prove

$1=\det(\mathbf{E})=\det(\mathbf{R}^\mathrm{T}\mathbf{R}) = \det(\mathbf{R}^\mathrm{T})\det(\mathbf{R}) = \det(\mathbf{R})^2 \quad\Longrightarrow \quad \det(\mathbf{R}) = \pm 1.$

The matrix with positive determinant is a proper rotation and with a negative determinant an improper rotation (is equal to a reflection times a proper rotation).

It will now be shown that a rotation matrix R has at least one invariant vector n, i.e., R n = n. Note that this is equivalent to stating that the vector n is an eigenvector of the matrix R with eigenvalue λ = 1.

A proper rotation matrix R has at least one unit eigenvalue. Using the two relations:

$\det(-\mathbf{R}) = (-1)^3 \det(\mathbf{R}) = - \det(\mathbf{R}) \quad\hbox{and}\quad\det(\mathbf{R}^{-1} ) = 1,$

we find

$\begin{align} \det(\mathbf{R} - \mathbf{E}) =& \det\big((\mathbf{R} - \mathbf{E})^{\mathrm{T}}\big) =\det\big((\mathbf{R}^{\mathrm{T}} - \mathbf{E})\big) = \det\big((\mathbf{R}^{-1} - \mathbf{E})\big) = \det\big(-\mathbf{R}^{-1} (\mathbf{R} - \mathbf{E}) \big) \\ =& - \det(\mathbf{R}^{-1} ) \; \det(\mathbf{R} - \mathbf{E}) = - \det(\mathbf{R} - \mathbf{E})\quad \Longrightarrow\quad \det(\mathbf{R} - \mathbf{E}) = 0 \end{align}$

From this follows that λ = 1 is a root (solution) of the secular equation, that is,

$\det(\mathbf{R} - \lambda \mathbf{E}) = 0\quad \hbox{for}\quad \lambda=1.$

In other words, the matrix R − E is singular and has a non-zero kernel, that is, there is at least one non-zero vector, say n, for which

$(\mathbf{R} - \mathbf{E}) \mathbf{n} = \mathbf{0} \quad \Longleftrightarrow \quad \mathbf{R}\mathbf{n} = \mathbf{n}$

The line μn for real μ is invariant under R, i.e, μn is a rotation axis. This proves Euler's theorem.

[edit] Equivalence of an orthogonal matrix to a rotation matrix

A proper orthogonal matrix is equivalent to

$\mathbf{R} \sim \begin{pmatrix} \cos\phi & -\sin\phi & 0 \\ \sin\phi & \cos\phi & 0 \\ 0 & 0 & 1\\ \end{pmatrix}, \qquad 0\le \phi \le 2\pi.$

If R has more than one invariant vector then φ = 0 and R = E. Any vector is an invariant vector of E.

[edit] Excursion into matrix theory

In order to prove the previous equation some facts from matrix theory must be recalled.

An m×m matrix A has m orthogonal eigenvectors if and only if A is normal, that is, if A^†A = AA^†. ^[2] This result is equivalent to stating that normal matrices can be brought to diagonal form by a unitary similarity transformation:

$\mathbf{A}\mathbf{U} = \mathbf{U}\; \mathrm{diag}(\alpha_1,\ldots,\alpha_m)\quad \Longleftrightarrow\quad \mathbf{U}^\dagger \mathbf{A}\mathbf{U} = \operatorname{diag}(\alpha_1,\ldots,\alpha_m),$

and U is unitary, that is,

$\mathbf{U}^\dagger = \mathbf{U}^{-1}.$

The eigenvalues α₁, ..., α_m are roots of the secular equation. If the matrix A happens to be unitary (and note that unitary matrices are normal), then

$\left(\mathbf{U}^\dagger\mathbf{A} \mathbf{U}\right)^\dagger = \mathrm{diag}(\alpha^*_1,\ldots,\alpha^*_m) = \mathbf{U}^\dagger\mathbf{A}^{-1} \mathbf{U} = \mathrm{diag}(1/\alpha_1,\ldots,1/\alpha_m)$

and it follows that the eigenvalues of a unitary matrix are on the unit circle in the complex plane:

$\alpha^*_k = 1/\alpha_k \quad\Longleftrightarrow \alpha^*_k\alpha_k = |\alpha_k|^2 = 1,\qquad k=1,\ldots,m.$

Also an orthogonal (real unitary) matrix has eigenvalues on the unit circle in the complex plane. Moreover, since its secular equation (an mth order polynomial in λ) has real coefficients, it follows that its roots appear in complex conjugate pairs, that is, if α is a root then so is α^∗.

After recollection of these general facts from matrix theory, we return to the rotation matrix R. It follows from its realness and orthogonality that

$\mathbf{R} \mathbf{U} = \mathbf{U} \begin{pmatrix} e^{i\phi} & 0 & 0 \\ 0 & e^{-i\phi} & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$

with the third column of the 3×3 matrix U equal to the invariant vector n. Writing u₁ and u₂ for the first two columns of U, this equation gives

$\mathbf{R}\mathbf{u}_1 = e^{i\phi}\, \mathbf{u}_1 \quad\hbox{and}\quad \mathbf{R}\mathbf{u}_2 = e^{-i\phi}\, \mathbf{u}_2$

If u₁ has eigenvalue 1, then φ= 0 and u₂ has also eigenvalue 1, which implies that in that case R = E.

Finally, the matrix equation is transformed by means of a unitary matrix,

$\mathbf{R} \mathbf{U} \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{i}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & \frac{-i}{\sqrt{2}} & 0 \\ 0 & 0 & 1\\ \end{pmatrix} = \mathbf{U} \underbrace{ \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{i}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & \frac{-i}{\sqrt{2}} & 0 \\ 0 & 0 & 1\\ \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{-i}{\sqrt{2}} & \frac{i}{\sqrt{2}} & 0 \\ 0 & 0 & 1\\ \end{pmatrix} }_{=\;\mathbf{E}} \begin{pmatrix} e^{i\phi} & 0 & 0 \\ 0 & e^{-i\phi} & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{i}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & \frac{-i}{\sqrt{2}} & 0 \\ 0 & 0 & 1\\ \end{pmatrix}$

which gives

$\mathbf{U'}^\dagger \mathbf{R} \mathbf{U'} = \begin{pmatrix} \cos\phi & -\sin\phi & 0 \\ \sin\phi & \cos\phi & 0 \\ 0 & 0 & 1\\ \end{pmatrix} \quad\hbox{with}\quad \mathbf{U'} = \mathbf{U} \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{i}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & \frac{-i}{\sqrt{2}} & 0 \\ 0 & 0 & 1\\ \end{pmatrix} .$

The columns of U′ are orthonormal. The third column is still n, the other two columns are perpendicular to n. This result implies that any orthogonal matrix R is equivalent to a rotation over an angle φ around an axis n.

[edit] Equivalence classes

It is of interest to remark that the trace (sum of diagonal elements) of the real rotation matrix given above is 1 + 2cosφ. Since a trace is invariant under an orthogonal matrix transformation:

$\mathrm{Tr}[\mathbf{A} \mathbf{R} \mathbf{A}^\mathrm{T}] = \mathrm{Tr}[ \mathbf{R} \mathbf{A}^\mathrm{T}\mathbf{A}] = \mathrm{Tr}[\mathbf{R}]\quad\hbox{with}\quad \mathbf{A}^\mathrm{T} = \mathbf{A}^{-1},$

it follows that all matrices that are equivalent to R by an orthogonal matrix transformation have the same trace. The matrix transformation is clearly an equivalence relation, that is, all equivalent matrices form an equivalence class. In fact, all proper rotation 3×3 rotation matrices form a group, usually denoted by SO(3) (the special orthogonal group in 3 dimensions) and all matrices with the same trace form an equivalence class in this group. Elements of such an equivalence class share their rotation angle, but all rotations are around different axes. If n is a eigenvector of R with eigenvalue 1, then An is an eigenvector of ARA^T, also with eigenvalue 1. Unless A = E, n and An are different.

[edit] Applications

[edit] Generators of rotations

Main articles: Rotation matrix and Rotation group

Suppose we specify an axis of rotation by a unit vector [x, y, z] , and suppose we have an infinitely small rotation of angle Δθ about that axis. To first order the rotation matrix ΔR is represented as:

$\Delta R = \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} + \begin{bmatrix} 0 & z&-y\\ -z& 0& x\\ y &-x& 0 \end{bmatrix}\,\Delta \theta = \mathbf{I}+\mathbf{A}\,\Delta \theta.$

A finite rotation through angle θ about this axis may be seen as a succession of small rotations about the same axis. Approximating Δθ as θ/N where N is a large number, a rotation of θ about the axis may be represented as:

$R =\left(\mathbf{1}+\frac{\mathbf{A}\theta}{N}\right)^N \approx e^{\mathbf{A}\theta}.$

It can be seen that Euler's theorem essentially states that all rotations may be represented in this form. The product $\mathbf{A}\theta$ is the "generator" of the particular rotation. Analysis is often easier in terms of these generators, rather than the full rotation matrix. Analysis in terms of the generators is known as the Lie algebra of the rotation group.

[edit] Quaternions

Main articles: Rotation operator (vector space) and Quaternions and spatial rotation

It follows from Euler's theorem that the relative orientation of any pair of coordinate systems may be specified by a set of four numbers. Three of these numbers are the direction cosines that orient the eigenvector. The fourth is the angle about the eigenvector that separates the two sets of coordinates. Such a set of four numbers is called a quaternion.

While the quaternion as described above, does not involve complex numbers, if quaternions are used to describe two successive rotations, they must be combined using the non-commutative quaternion algebra derived by William Rowan Hamilton through the use of imaginary numbers.

Rotation calculation via quaternions has come to replace the use of direction cosines in Aerospace applications through their reduction of the required calculations, and their ability to minimize round-off errors. Also, in computer graphics the ability to perform spherical interpolation between quaternions with relative ease is of value.

[edit] See also

Euler angles
Euler-Rodrigues parameters
Rotation representation
Rotation operator (vector space)
Screw theory, based on the similar fact that all rigid-body motions can be described as a so-called "screw motion".

[edit] Notes

^ Novi Commentarii academiae scientiarum Petropolitanae 20, 1776, pp. 189-207 (E478)
^ The dagger symbol † stands for complex conjugation followed by transposition. For real matrices complex conjugation does nothing and daggering a real matrix is the same as transposing it.

[edit] References

This article incorporates material from the Citizendium article "Euler's theorem (rotation)", which is licensed under the Creative Commons Attribution/Share-Alike 3.0 Unported License but not under the GFDL. Euler's theorem and its proof are contained in paragraphs 24-26 of the appendix (Additamentum. pp. 201-203)

L. Eulero (Leonhard Euler)
Formulae generales pro translatione quacunque corporum rigidorum (General formulas for the translation of arbitrary rigid bodies),

presented to the St. Petersburg Academy on October 9, 1775, and first published in

Novi Commentarii academiae scientiarum Petropolitanae 20, 1776, pp. 189-207 (E478)

and was reprinted in

Theoria motus corporum rigidorum, ed. nova, 1790, pp. 449-460 (E478a)

and later in his collected works

Opera Omnia, Series 2, Volume 9, pp. 84 - 98.

[edit] External links

Euler's original treatise in The Euler Archive: entry on E478, first publication 1976 (pdf)

Euler's original text (in Latin) and English translation (by Johan Sten) (pdf)

[0] Novi Commentarii academiae scientiarum Petropolitanae 20, 1776, pp. 189-207 (E478)

[1] The dagger symbol † stands for complex conjugation followed by transposition. For real matrices complex conjugation does nothing and daggering a real matrix is the same as transposing it.

[1]

[2]

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