Euler angles

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Figure 1. Euler angles. From left to right: initial configuration, after rotation over angle α, after rotation over angle β, and after rotation over angle γ.

In physics, mathematics, and engineering, Euler angles are three rotation angles, commonly denoted by 0 ≤ α ≤ 2π, 0 ≤ β ≤ π, and 0 ≤ γ ≤ 2π. Any rotation of a 3-dimensional object can be performed by three consecutive rotations over the three Euler angles.

Different conventions are in use: a rotation can be active (the object is rotated, the system of axes is fixed in space), or passive (the object is fixed in space, the axes are rotated).

The choice of rotation axes may vary; a very common active convention is the z′-y′-z convention. Attach a system of Cartesian coordinate axes to the body that is to be rotated (the coordinate frame is fixed to the body and is rotated simultaneously with it); in the figure the body-fixed frame is shown in red and labeled by lowercase letters. First rotate around z, then around the new body-fixed y-axis, y′, and finally around z′. Another convention often used is the z-x′-z′ convention, where instead of over the new y-axis the second rotation is over the new x-axis. Also the z-y-x convention is used (and will be discussed below).

The right-hand screw rule is practically always followed: the rotation axis is a directed line and a positive rotation is as a cork screw driven into the positive direction of the axis.

Euler angles are used in many different branches of physics and engineering. The present article is written from the point of view of molecular physics, where the objects to be rotated are molecules.

The angles are named after the 18th century mathematician Leonhard Euler who introduced in 1765 two of the three for an axially symmetric body where the third angle, γ, does not play a role.

Geometric discussion

In Figure 1 the space-fixed (laboratory) axes are labeled by capital X, Y, and Z and are shown in black. The body to be rotated is not shown, but a system of axes fixed to it is shown in red. One may use any convenient orthonormal frame as a body-fixed frame. Often the body-fixed axes are principal axes, that means that they are eigenvectors of the inertia tensor of the body. Also symmetry axes, when present, may be used. When the body has symmetry axes, the principal axes often coincide with these.

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Figure 2. Rotation of r to r′. On the left around z-axis over α (φ increases), on the right around y-axis over β (φ decreases). Both rotation axes point toward the reader.

The z-y′-z′ convention will be followed. Initially, the two frames coincide, and the path to a final arbitrary orientation of the body—and its frame—is depicted on Figure 1. The first rotation is around the z-axis, which coincides with the Z-axis. The x- and y-axis move in a plane perpendicular to the z-axis over an angle α. The second rotation is in a plane through the origin perpendicular to the y′-axis. The angle is β. The present convention has the great practical advantage that the z′-axis has the usual spherical polar coordinates α ≡ φ (longitude angle) and β ≡ θ (colatitude angle) with respect to the space-fixed frame. (For instance, due to this fact the complex conjugate of the m = 0 column of the Wigner D-matrix consists of spherical harmonics). The final rotation is in a plane perpendicular to the z′-axis over an angle γ. From geometric considerations follows that any orientation of the body-fixed frame in space may be obtained.

Write $\mathbf {R} (\varphi ,{\hat {n}})$ for the rotation matrix that describes a rotation around the unit vector ${\hat {n}}$ over an angle $\varphi$. Clearly the three consecutive Euler rotations correspond to rotations around the unit vectors along the body-fixed axes z, y′, and z′ over angles α β, and γ, respectively. Because a matrix acts on a column vector to its right, the order in the matrix product is as in the leftmost term in the following equation. It will be shown that the corresponding matrix product can be written in reverse order (but around fixed, unprimed, axes z, y, z), that is,

${\begin{aligned}\mathbf {R} (\gamma ,{\hat {e}}_{z'})\mathbf {R} (\beta ,{\hat {e}}_{y'})\mathbf {R} (\alpha ,{\hat {e}}_{z})&=\mathbf {R} (\alpha ,{\hat {e}}_{z})\mathbf {R} (\beta ,{\hat {e}}_{y})\mathbf {R} (\gamma ,{\hat {e}}_{z})\\&={\begin{pmatrix}\cos \alpha &-\sin \alpha &0\\\sin \alpha &\cos \alpha &0\\0&0&1\\\end{pmatrix}}{\begin{pmatrix}\cos \beta &0&\sin \beta \\0&1&0\\-\sin \beta &0&\cos \beta \\\end{pmatrix}}{\begin{pmatrix}\cos \gamma &-\sin \gamma &0\\\sin \gamma &\cos \gamma &0\\0&0&1\\\end{pmatrix}}\\&={\begin{pmatrix}\cos \alpha \cos \beta \cos \gamma -\sin \alpha \cos \gamma \;&\;-\cos \alpha \cos \beta \sin \gamma -\sin \alpha \cos \gamma \;&\;\cos \alpha \sin \beta \\\sin \alpha \cos \beta \cos \gamma +\cos \alpha \sin \gamma \;&\;-\sin \alpha \cos \beta \sin \gamma +\cos \alpha \cos \gamma \;&\;\sin \alpha \sin \beta \\-\sin \beta \cos \gamma \;&\;\sin \beta \sin \gamma \;&\;\cos \beta \end{pmatrix}}\end{aligned}}$

Note that the third column contains indeed the spherical polar coordinates of the unit vector along the z′-axis.

Before proving the first equality in the above equation (reversal of order), we derive the matrix for a rotation around the z-axis, see the left drawing in Figure 2. The rotated vector has components

${\begin{pmatrix}\cos(\alpha +\phi )\\\sin(\alpha +\phi )\\0\end{pmatrix}}={\begin{pmatrix}\cos \alpha \cos \phi -\sin \alpha \sin \phi \\\sin \alpha \cos \phi +\cos \alpha \sin \phi \\0\end{pmatrix}}={\begin{pmatrix}\cos \alpha &-\sin \alpha &0\\\sin \alpha &\cos \alpha &0\\0&0&1\\\end{pmatrix}}{\begin{pmatrix}\cos \phi \\\sin \phi \\0\\\end{pmatrix}}\equiv \mathbf {R} (\alpha ,{\hat {e}}_{z}){\begin{pmatrix}\cos \phi \\\sin \phi \\0\end{pmatrix}}.$

We used here the relations well-known from trigonometry for the sine and cosine of a sum angle. The derivation of the matrix for a rotation around the y-axis proceeds along the same lines. Note, however, that the angle of a vector with the x-axis decreases by a rotation around the positive y-axis (see right-hand drawing in Figure 2).

To prove the first equality (reversal of the order in the angles), a property of rotation matrices is used. A rotation (orthogonal 3×3) matrix A, transforming a rotation axis, gives rise to the following similarity equation,

$\mathbf {R} (\varphi ,\mathbf {A} {\hat {n}})=\mathbf {A} \mathbf {R} (\varphi ,{\hat {n}})\mathbf {A} ^{\mathrm {T} },$

where the superscript T indicates the transpose of the matrix. For rotation matrices the transposed matrix is equal to the inverse of the matrix. From this similarity relation follows that

$\mathbf {R} (\gamma ,{\hat {e}}_{z'})=\mathbf {R} (\beta ,{\hat {e}}_{y'})\mathbf {R} (\gamma ,{\hat {e}}_{z})\mathbf {R} (\beta ,{\hat {e}}_{y'})^{\mathrm {T} },$

so that

$\mathbf {R} (\gamma ,{\hat {e}}_{z'})\mathbf {R} (\beta ,{\hat {e}}_{y'})\mathbf {R} (\alpha ,{\hat {e}}_{z})=\mathbf {R} (\beta ,{\hat {e}}_{y'})\mathbf {R} (\gamma ,{\hat {e}}_{z})\mathbf {R} (\beta ,{\hat {e}}_{y'})^{\mathrm {T} }\;\mathbf {R} (\beta ,{\hat {e}}_{y'})\mathbf {R} (\alpha ,{\hat {e}}_{z})=\mathbf {R} (\beta ,{\hat {e}}_{y'})\mathbf {R} (\gamma ,{\hat {e}}_{z})\mathbf {R} (\alpha ,{\hat {e}}_{z}).$

Also

$\mathbf {R} (\beta ,{\hat {e}}_{y'})=\mathbf {R} (\alpha ,{\hat {e}}_{z})\mathbf {R} (\beta ,{\hat {e}}_{y})\mathbf {R} (\alpha ,{\hat {e}}_{z})^{\mathrm {T} },$

so that

$\mathbf {R} (\beta ,{\hat {e}}_{y'})\mathbf {R} (\gamma ,{\hat {e}}_{z})\mathbf {R} (\alpha ,{\hat {e}}_{z})=\mathbf {R} (\alpha ,{\hat {e}}_{z})\mathbf {R} (\beta ,{\hat {e}}_{y})\mathbf {R} (\alpha ,{\hat {e}}_{z})^{\mathrm {T} }\;\mathbf {R} (\gamma ,{\hat {e}}_{z})\mathbf {R} (\alpha ,{\hat {e}}_{z})=\mathbf {R} (\alpha ,{\hat {e}}_{z})\mathbf {R} (\beta ,{\hat {e}}_{y})\mathbf {R} (\gamma ,{\hat {e}}_{z}),$

where it is used that rotations around the same axis commute, that is,

$\mathbf {R} (\alpha ,{\hat {e}}_{z})^{\mathrm {T} }\mathbf {R} (\gamma ,{\hat {e}}_{z})\mathbf {R} (\alpha ,{\hat {e}}_{z})=\mathbf {R} (\gamma ,{\hat {e}}_{z})\mathbf {R} (\alpha ,{\hat {e}}_{z})^{\mathrm {T} }\mathbf {R} (\alpha ,{\hat {e}}_{z})=\mathbf {R} (\gamma ,{\hat {e}}_{z})$

and the required result is proved.

Algebraic treatment

In the proof that any rotation can be written as three consecutive rotations, an appeal was made to the geometric insight of the reader. The same result can be proved more rigorously by algebraic means. To that end the notation is somewhat shortened:

$\mathbf {R} _{z}(\omega )\equiv \mathbf {R} (\omega ,{\hat {e}}_{z})\quad {\hbox{and}}\quad \mathbf {R} _{y}(\omega )\equiv \mathbf {R} (\omega ,{\hat {e}}_{y}).$

Theorem

A proper rotation matrix R can be factorized thus

$\mathbf {R} =\mathbf {R} _{z}(\omega _{3})\;\mathbf {R} _{y}(\omega _{2})\;\mathbf {R} _{x}(\omega _{1})$

which is referred to as the Euler z-y-x parametrization, or also as

$\mathbf {R} =\mathbf {R} _{z}(\alpha )\;\mathbf {R} _{y}(\beta )\;\mathbf {R} _{z}(\gamma )\quad$

the Euler z-y-z parametrization.

Proof

First the Euler z-y-x-parametrization will be proved by an algorithm for the factorization of a given matrix R ≡ (r₁, r₂, r₃). Second the z-y-z parametrization will be proved; this parametrization is—as shown above—equivalent to the z′-y′-z parametrization with angles in reverse order. A Fortran subroutine based on the algorithm is given on the code page.

To prove the z-y-x parametrization we consider the matrix product

$\mathbf {R} _{z}(\omega _{3})\,\mathbf {R} _{y}(\omega _{2})={\begin{pmatrix}\cos \omega _{3}\cos \omega _{2}&-\sin \omega _{3}&\cos \omega _{3}\sin \omega _{2}\\\sin \omega _{3}\cos \omega _{2}&\cos \omega _{3}&\sin \omega _{3}\sin \omega _{2}\\-\sin \omega _{2}&0&\cos \omega _{2}\\\end{pmatrix}}\equiv (\mathbf {a} _{1},\mathbf {a} _{2},\mathbf {a} _{3}).$

The columns of the matrix product are for ease of reference designated by a₁, a₂, and a₃. Note that the multiplication by

$\mathbf {R} _{x}(\omega _{1})\equiv {\begin{pmatrix}1&0&0\\0&\cos \omega _{1}&-\sin \omega _{1}\\0&\sin \omega _{1}&\cos \omega _{1}\end{pmatrix}}$

on the right does not affect the first column, so that a₁ = r₁ (the first column of R). Solve $\omega _{2}\;$ and $\omega _{3}\;$ from the first column of R (which is known),

$\mathbf {a} _{1}={\begin{pmatrix}\cos \omega _{3}\;\cos \omega _{2}\\\sin \omega _{3}\;\cos \omega _{2}\\-\sin \omega _{2}\\\end{pmatrix}}={\begin{pmatrix}R_{11}\\R_{21}\\R_{31}\\\end{pmatrix}}\equiv \mathbf {r} _{1}.$

This is possible. First solve $\omega _{2}\;$ for $-\pi /2\leq \omega _{2}\leq \pi /2$ from

$\sin \omega _{2}=-R_{31}.\,$

Then solve $\omega _{3}\;$ for $0\leq \omega _{3}\leq 2\pi$ from the two equations:

$\cos \omega _{3}={R_{11} \over \cos \omega _{2}},\qquad \sin \omega _{3}={R_{21} \over \cos \omega _{2}}.$

The angles $\omega _{2}\;$ and $\omega _{3}\;$ determine fully the vectors a₂ and a₃.

Since a₁, a₂ and a₃ are the columns of a proper rotation matrix they form an orthonormal right-handed system. The plane spanned by a₂ and a₃ is orthogonal to $\mathbf {a} _{1}\equiv \mathbf {r} _{1}$ and hence the plane contains $\mathbf {r} _{2}$ and $\mathbf {r} _{3}$. Thus the latter two vectors are a linear combination of the first two,

$(\mathbf {r} _{2},\mathbf {r} _{3})=(\mathbf {a} _{2},\mathbf {a} _{3}){\begin{pmatrix}\cos \omega _{1}&-\sin \omega _{1}\\\sin \omega _{1}&\cos \omega _{1}\\\end{pmatrix}}.$

Since $\mathbf {r} _{2},\;\mathbf {a} _{2},\;\mathbf {a} _{3}$ are known orthonormal vectors, we can compute

${\begin{aligned}\mathbf {a} _{2}\cdot \mathbf {r} _{2}=&\cos \omega _{1}\\\mathbf {a} _{3}\cdot \mathbf {r} _{2}=&\sin \omega _{1}.\end{aligned}}$

These equations give $\omega _{1}\;$ with $0\leq \omega _{1}\leq 2\pi$.

The angle ω₁ gives the matrix $\mathbf {R} _{x}(\omega _{1})$ with

${\begin{aligned}\mathbf {R} \equiv (\mathbf {r} _{1},\mathbf {r} _{2},\mathbf {r} _{3})=(\mathbf {r} _{1},\mathbf {a} _{2},\mathbf {a} _{3})\mathbf {R} _{x}(\omega _{1})=(\mathbf {a} _{1},\mathbf {a} _{2},\mathbf {a} _{3})\mathbf {R} _{x}(\omega _{1})=\mathbf {R} _{z}(\omega _{3})\,\mathbf {R} _{y}(\omega _{2})\,\mathbf {R} _{x}(\omega _{1}).\end{aligned}}$

This gives the required z-y-x factorization of the arbitrary proper orthogonal matrix R.

The proof of the Euler z-y-z parametrization is obtained by a small modification of the previous proof. We start by retrieving the spherical polar coordinates $\omega _{2}\;$ and $\omega _{3}\;$ of the unit vector $\mathbf {r} _{3}=\mathbf {a} _{3}$, the third column [the rightmost multiplication by R_z(ω₁) does not affect r₃]. Then consider

$(\mathbf {r} _{1},\;\mathbf {r} _{2})=(\mathbf {a} _{1},\;\mathbf {a} _{2}){\begin{pmatrix}\cos \omega _{1}&-\sin \omega _{1}\\\sin \omega _{1}&\cos \omega _{1}\\\end{pmatrix}}$

or, $\mathbf {a} _{1}\cdot \mathbf {r} _{1}=\cos \omega _{1}\;,\quad \mathbf {a} _{2}\cdot \mathbf {r} _{1}=\sin \omega _{1}.$ The equation for R can be written as

$(\mathbf {r} _{1},\mathbf {r} _{2},\mathbf {r} _{3})=(\mathbf {a} _{1},\mathbf {a} _{2},\mathbf {r} _{3})\,\mathbf {R} _{z}(\omega _{1})=\mathbf {R} _{z}(\omega _{3})\,\mathbf {R} _{y}(\omega _{2})\,\mathbf {R} _{z}(\omega _{1})\;,$

which proves the Euler z-y-z parametrization. Clearly, this factorization is equal to the one given in the previous section, with

$\omega _{3}\equiv \alpha ,\quad \omega _{2}\equiv \beta ,\quad \omega _{1}\equiv \gamma .$