The use of the quaternions for describing the Earth’s rotation

Abstract

We derive the quaternions expressing the rotation transformation between celestial and terrestrial reference systems as functions of the parameters giving the position of the Celestial Intermediate Pole (CIP) in those two systems, and of the rotation angle. A first version is associated with the common Earth Orientation Parameters (EOP) recommended by the International Earth Rotation and Reference System Service (IERS) and IAU; a second version uses the direction cosines of the CIP in the International Terrestrial Reference System (ITRS) instead of the conventional pole coordinates. Whereas matrix and quaternion methods are numerically comparable, the quaternion formulation offers a very elegant and concise analytical representation, that does not exist for Earth rotation matrix and that can be easily programmed. In our view, this quaternion representation reinforces the scope of the nowadays parameters adopted for describing Earth rotation.

Appendices

Appendices

Definition of the quaternions and properties

1.1 Definition

The quaternions were discovered by the Irish Mathematician William Rowan Hamilton in the early 1840’s (Hamilton 1866). The problem was to find a multiplication law $\otimes$ for n-uplets of real number $X=[x_1,x_2,\ldots ,x_n]$ such that the modulus of a product $X \otimes X'$ is equal to the product of the modulus $|X| |X'|$:

$$\begin{aligned} \vert X \vert \vert X' \vert = \vert X \otimes X' \vert \end{aligned}$$

(A1)

the modulus being defined by $|X|=\sqrt{X \overline{X}}$ where $\overline{X}=[x_1,-x_2,-x_3,\ldots ]$ is the conjugate.

Numbers obeying such a law constitute a generalization of complex numbers. Hamilton had already searched for a solution to this problem for the tri-uplet, but unsuccessfully. Considering then a fourth dimension, he obtained a multiplication law, which satisfies Eq. (A1). Considering two quadruplets $q_1=[t_1,x_1,y_1,z_1]$ and $q_2=[t_2,x_2,y_2,z_2]$, their multiplication is given by

$$\begin{aligned} q_{1} \otimes q_{2} = \left| \begin{array}{c} t_1 t_2 - x_1 x_2 - y_1 y_2 - z_1 z_2\\ t_1 x_2 + x_1 t_2 + y_1 z_2 - z_1 y_2\\ t_1 y_2 - x_1 z_2 + y_1 t_2 + z_1 x_2\\ t_1 z_2 + x_1 y_2 - y_1 x_2 + z_1 t_2 \end{array} \right. \end{aligned}$$

(A2)

Endowed with this internal composition, these quadruplets of real numbers $q=[t,x,y,z]$ are called quaternions and have a structure of group.

The addition of two quaternions is defined by

$$\begin{aligned} q_{1} + q_{2}=[t_1+t_2,x_1+x_2,y_1+y_2,z_1+z_2]. \end{aligned}$$

(A3)

It can be easily shown that the multiplication law is associative and distributive with respect to the addition. However, it is not commutative. Let be $i=[0,1,0,0]$, $j=[0,0,1,0]$, $k=[0,0,0,1]$, any quaternion [t, x, y, z] can be expressed by

$$\begin{aligned} q= t+ i x + j y + k z. \end{aligned}$$

(A4)

Any quaternion of the form (t, 0, 0, 0) is reduced to the real number t. From the multiplication law (A2) i, j and k check the properties

$$\begin{aligned} \begin{array}{l} i \otimes j= -j \otimes i = k,\; j \otimes k= -k \otimes j = i,\\ \quad \; k \otimes i= -i \otimes k = j \\ i^2 = i\otimes i = j^2 = k^2= i j k = [-1,0,0,0] \end{array}. \end{aligned}$$

(A5)

These relations determine fully the multiplication of two quaternions.

1.2 Scalar and vectorial parts

We can consider (x, y, z) as the components of a vector ${V}$ in a direct orthonormal basis, so that every quaternion can be expressed by

$$\begin{aligned} q=[t,{V}], \end{aligned}$$

(A6)

where ${V}$ is the vectorial part of the quaternion and t is its scalar part. As the vectorial part is referred to a direct orthonormal basis, (A2) can be rewritten

$$\begin{aligned} \begin{aligned} q_1 \otimes q_2 =&[t_1,{V}_1] \otimes [t_2,{V}_2] = q_3 = [t_3,{V}_3] \\ \textrm{with}&\;\; t_3 = t_1 t_2 - {V}_1 \cdot {V}_2 \;\; \textrm{and} \;\;\\&\quad {V}_3 = t_2 {V}_1 + t_1 {V}_2 + {V}_1\wedge {V}_2, \end{aligned} \end{aligned}$$

(A7)

where ${V}_1 \cdot {V}_2$ is the scalar product of the vectorial parts and ${V}_1\wedge {V}_2$ their vectorial product.

The non-commutativity of the product of two quaternions comes from the non-commutativity of the vectorial product ${V}_1\wedge {V}_2$.

1.3 Symmetric and antisymmetric parts

According to (A7), the symmetric part of the product of two quaternions is

$$\begin{aligned}{} & {} <q_1 \otimes q_2> ={\displaystyle \frac{1}{2}} (q_1 \otimes q_2 + q_2 \otimes q_1) \nonumber \\{} & {} \quad =[t_1 t_2 - {V}_1.{V}_2, t_2 {V}_1 + t_1 {V}_2 ], \end{aligned}$$

(A8)

and the antisymmetric part

$$\begin{aligned}{}[q_1 \otimes q_2] ={\displaystyle \frac{1}{2}} (q_1 \otimes q_2 - q_2 \otimes q_1) =[0,{V}_1\wedge {V}_2]. \end{aligned}$$

(A9)

1.4 Conjugate and modulus

The conjugate of a quaternion $q=[t,{V}]$ is defined by $\overline{q}=[t,-{V}]$. Then, it can be easily shown that the conjugate of a product of two quaternions is the inverse product of the conjugates:

$$\begin{aligned} \overline{q_1 \otimes q_2} =\overline{q_2} \otimes \overline{q_1}. \end{aligned}$$

(A10)

The norm or the modulus of a quaternion q is

$$\begin{aligned} \vert q \vert = \sqrt{q \otimes \overline{q}}=\sqrt{\overline{q} \otimes q}=\sqrt{t^2+x^2+y^2+z^2}. \end{aligned}$$

(A11)

1.5 Inverse

From A11, any non-zero quaternion q has an inverse, written $q^{-1}$, such as $q \otimes q^{-1} = q^{-1} \otimes q = (1,0,0,0)$ and given by

$$\begin{aligned} q^{-1} = {\displaystyle \frac{\overline{q}}{\vert q \vert ^2}}. \end{aligned}$$

(A12)

1.6 An useful lemme

Let $q=(t,{\Gamma })$ be any quaternion and $(0,{V})$ a pure vectorial quaternion, the following expressions can be easily derived:

$$\begin{aligned}{} & {} q \otimes (0,{V}) \otimes \overline{q}\nonumber \\{} & {} \quad = \left[ 0, 2 \,({\Gamma } \cdot {V}) {\Gamma } + (t^2-{\Gamma }^2){V} + 2 t \,{\Gamma } \wedge {V} \right] , \end{aligned}$$

(A13)

and

$$\begin{aligned}{} & {} \overline{q} \otimes [0,{V}] \otimes q\nonumber \\{} & {} \quad =\left[ 0, 2 \,({\Gamma } \cdot {V}) {\Gamma } + (t^2-{\Gamma }^2){V} - 2 t \,{\Gamma } \wedge {V} \right] . \end{aligned}$$

(A14)

Representation of a rotation by a unit quaternion

1.1 Quaternion of rotation

Any rotation can be fully defined by the rotation angle $\phi$ and the direction of its rotation axis, characterized by its direction cosines $(\alpha ,\beta ,\gamma )=\hat{u}$ in a direct orthonormal basis. From a given vector ${r}$ and the vector $\hat{u}$, we can build another orthonormal direct basis $(\hat{u},\hat{v},\hat{w})$ such as

$$\begin{aligned} \begin{array}{c} \hat{w} = {\displaystyle \frac{\hat{u}\wedge {r}}{\Vert \hat{u}\wedge {r} \Vert }}\\ \hat{v} = \hat{w} \wedge \hat{u}= {\displaystyle \frac{{r}-(\hat{u}\cdot {r})\hat{u}}{\Vert \hat{u}\wedge {r} \Vert }} \end{array}. \end{aligned}$$

(B1)

Let us apply the rotation of angle $\phi$ and direction vector $\hat{u})$ on the vector ${r}$, which becomes ${r}\,'$. The projection of ${r}\,'$ on the rotation axis $\hat{u}$ is unchanged and equal to ${r}\cdot \hat{u}$. Its second component along $\hat{v}$, namely $r \cdot \hat{v}$, is rotated by the angle $\phi$ in the plane ($\hat{v},\hat{w}$). Its third component, along $\hat{w}$ is equal to zero, for $\hat{w}$ is perpendicular to t${r}$. It results

$$\begin{aligned} {r}\,'= ( {r} \cdot \hat{u}) \,\hat{u}+ ( {r} \cdot \hat{v} ) \, \cos \phi \, \hat{v} + ({r}\cdot \hat{v}) \, \sin \phi \, \hat{w}. \end{aligned}$$

(B2)

By using the last expression of (B1), we get ${\hat{v}} \cdot \hat{v} = 1 = r \cdot \hat{v} / \Vert \hat{u}\wedge {r} \Vert$, that is $r \cdot \hat{v} = \Vert \hat{u}\wedge {r} \Vert$, allowing to derive $\hat{v} = [ {r}-(\hat{u}\cdot {r})\hat{u} ] / r \cdot \hat{v}$ and $\hat{w} = \hat{u} \wedge r / r \cdot \hat{v}$, and finally

$$\begin{aligned} {r}\,'= & {} ( 1-\cos \phi )(\hat{u} \cdot {r}) \hat{u} \nonumber \\{} & {} +\,\cos \phi {r} +\sin \phi \, \hat{u}\wedge {r}. \end{aligned}$$

(B3)

On the other hand, applying the transformation (A13) to the quaternion vector $[0,{r}\,']$ and quaternion $q =(t,{\Gamma })$, yields

$$\begin{aligned}{} & {} q \otimes [0,{r}\,] \otimes \overline{q}= [0,r\,'] \nonumber \\{} & {} \quad = \left[ 0, 2 \,({\Gamma } \cdot {r}) {\Gamma } + (t^2-{\Gamma }^2)\,{r}+ 2 t \,{\Gamma } \wedge {r} \right] . \end{aligned}$$

(B4)

So, the quaternion $q=(t,\Gamma )$ corresponds to the rotation transformation (B3) if it obeys the system

$$\begin{aligned} \begin{array}{ll} 2 ({\Gamma } \cdot {r}) {\Gamma } = (1-\cos \phi )( \hat{u} \cdot {r}) \,\hat{u} \\ t^2 -{\Gamma }^2 = \cos \phi \\ 2 t \,{\Gamma } \wedge {r}= \sin \phi \, \hat{u} \wedge {r} \end{array}, \end{aligned}$$

(B5)

with solutions

$$\begin{aligned} q= (t, \Gamma ) = \pm \left[ \cos \left( {\displaystyle \frac{\phi }{2}}\right) ,\sin \left( {\displaystyle \frac{\phi }{2}}\right) \hat{u} \right] . \end{aligned}$$

(B6)

By definition, the quaternion of rotation $q_{(\phi ,\hat{u})}$ associated with the rotation $(\phi ,\hat{u})$ is

$$\begin{aligned} q_{(\phi ,\hat{u})}=\left[ \cos \left( {\displaystyle \frac{\phi }{2}}\right) ,\sin \left( {\displaystyle \frac{\phi }{2}}\right) \hat{u} \right] . \end{aligned}$$

(B7)

The rotation of same axis but of opposite angle is given by the conjugate quaternion $\overline{q}_{(\phi ,\hat{u})}$.

Both these quaternions allows to obtain the transform vector $r\,'$ according to

$$\begin{aligned}{}[0,{r}\,']= q_{(\phi ,\hat{u})} \otimes [0,{r}\,] \otimes \overline{q}_{(\phi ,\hat{u})}. \end{aligned}$$

(B8)

1.2 Coordinate transformation

Let (x, y, z) be the coordinates of a given vector $r$ in the Cartesian frame Oxyz endowed with the orthonormal basis $(\hat{i},\hat{j},\hat{k})$. By applying the quaternion of rotation $q_{(\phi ,\hat{u})}$ on $(\hat{i},\hat{j},\hat{k})$, this one is transformed into the basis $(\hat{i}',\hat{j}',\hat{k}')$. Let $(x',y',z')$ be the coordinate of $r$ in this new basis. Then, two equal vectorial quaternions give $r$:

$$\begin{aligned} x' [0,\hat{i}'] + y' [0,\hat{j}']+ z'[0,\hat{k}'] = x[0, \hat{i}] + y [0,\hat{j}]+z [0,\hat{k}].\nonumber \\ \end{aligned}$$

(B9)

According to the transformation law (B8), this relation becomes

$$\begin{aligned} \begin{aligned}&x' q_{(\phi ,\hat{u})} \otimes [0,\hat{i}] \otimes \overline{q}_{(\phi ,\hat{u})} + y' q_{(\phi ,\hat{u})} \otimes [0,\hat{j}] \otimes \overline{q}_{(\phi ,\hat{u})} \\&\qquad +\, z' q_{(\phi ,\hat{u})} \otimes [0,\hat{k}] \otimes \overline{q}_{(\phi ,\hat{u})} \\&\quad = x [0,\hat{i}] + y [0,\hat{j}] + z [0,\hat{k}]. \end{aligned}\nonumber \\ \end{aligned}$$

(B10)

Then

$$\begin{aligned} \begin{aligned}&\overline{q}_{(\phi ,\hat{u})} \otimes q_{(\phi ,\hat{u})} \left( x' [0,\hat{i}] + y' [0,\hat{j}] + z' [0,\hat{k}] \right) \\&\overline{q}_{(\phi ,\hat{u})} \otimes q_{(\phi ,\hat{u})} = \overline{q}_{(\phi ,\hat{u})} \otimes \left( x [0, \hat{i}] + y [0,\hat{j}]+z [0,\hat{k}] \right) \\&\qquad \qquad \qquad \qquad \quad \otimes q_{(\phi ,\hat{u})}. \end{aligned}\nonumber \\ \end{aligned}$$

(B11)

As $\overline{q}_{(\phi ,\hat{u})} \otimes q_{(\phi ,\hat{u})} = \overline{q}_{(\phi ,\hat{u})} \otimes q_{(\phi ,\hat{u})} = \vert q_{(\phi ,\hat{u})} \vert ^2 = 1$, we obtain

$$\begin{aligned}{} & {} x' [0,\hat{i}] + y' [0,\hat{j}] +z' [0,\hat{k}] \nonumber \\{} & {} \quad = \overline{q}_{(\phi ,\hat{u})} \otimes \left( x [0, \hat{i}] + y [0,\hat{j}] + z [0,\hat{k}] \right) \otimes q_{(\phi ,\hat{u})}.\nonumber \\ \end{aligned}$$

(B12)

or

$$\begin{aligned}{}[0,x',y',z'] = \overline{q}_{(\phi ,\hat{u})} \otimes [0,x,y,z] \otimes q_{(\phi ,\hat{u})}. \end{aligned}$$

(B13)

Noting $\underline{x} = (x,y,z)$ and $\underline{x'}=(x',y',z')$, the former relation can be expressed by

$$\begin{aligned}{} & {} [0,\underline{x}']= \overline{q}_{(\phi ,\hat{u})} \otimes [0,\underline{x}] \otimes q_{(\phi ,\hat{u})}. \end{aligned}$$

(B14)

This is the passive transformation in the sense that it only transforms the vector coordinates and not the vector itself, in contrast to the active transformation (B8).

1.3 Equivalent matrix for coordinate transformation

Equation (B3) allows to calculate the components of the new basis in the old one, i.e., the transformation matrix P from the basis (${\hat{i}},{\hat{j}},{\hat{k}}$) to the basis (${\hat{i}}',{\hat{j}}',{\hat{k}}'$):

$$\begin{aligned} P= & {} (1-\cos \phi )\left[ \begin{array}{ccc} \alpha ^2 &{}\quad \beta \alpha &{}\quad \gamma \alpha \\ \alpha \beta &{}\quad \beta ^2 &{} \quad \gamma \beta \\ \alpha \gamma &{}\quad \beta \gamma &{}\quad \gamma ^2 \end{array}\right] +\cos \phi I\nonumber \\{} & {} + \sin \phi \left[ \begin{array}{ccc} 0 &{} \quad -\gamma &{}\quad \beta \\ \gamma &{} \quad 0 &{} \quad -\alpha \\ -\beta &{}\quad \alpha &{} \quad 0 \end{array}\right] , \end{aligned}$$

(B15)

where $\alpha$, $\beta$ and $\gamma$ are the direction cosines of $\hat{u}$.

Considering the components $t=\cos \phi /2$, $a=\alpha \sin \phi /2$, $b = \beta \sin \phi /2$, $c= \gamma \sin \phi /2$ of the quaternion $q_{(\phi ,\hat{u})}=[t,a,b,c]$, P takes the form

$$\begin{aligned} P=\left[ \begin{array}{ccc} t^2+a^2-b^2-c^2 &{}\quad 2(ab-ct) &{} \quad 2(ac+bt) \\ 2(ab+ct) &{}\quad t^2-a^2+b^2-c^2 &{} \quad 2(bc-at) \\ 2(ac-bt) &{} \quad 2(bc+at) &{}\quad t^2-a^2-b^2+c^2 \end{array}\right] .\nonumber \\ \end{aligned}$$

(B16)

Then, the coordinate transformation can be written as

$$\begin{aligned} \underline{x'}= P^{T} \, \underline{x} = R \,\underline{x}, \end{aligned}$$

(B17)

where the coordinate transformation using the matrix R is equivalent to that using the quaternion q given by (B14).

1.4 Quaternion representing the product of two rotations

Let $q_1$ and $q_2$ be the quaternions pertaining to the rotations $R_1$ and $R_2$, respectively (with secant axes). From (B8), the active sequence of rotations $R_2 R_1$ (first $R_1$ then $R_2$) applied to the vector ${r}$ gives

$$\begin{aligned}{}[0,{r}\,']= & {} (q_2 \otimes q_1) \otimes [0,{r}\,] \otimes (\overline{q}_1 \otimes \overline{q}_2) \nonumber \\= & {} (q_2 \otimes q_1) \otimes [0,{r}\,] \otimes \overline{q_2 \otimes q_1}. \end{aligned}$$

(B18)

According to (B14), the passive sequence transforming the coordinates $\underline{x}$ of $r$ to $\underline{x}'$ is

$$\begin{aligned}{}[0,\underline{x}']= & {} (\overline{q}_2 \otimes \overline{q}_1) \otimes [0,\underline{x}] \otimes (q_1 \otimes q_2) \nonumber \\= & {} \overline{q_1 \otimes q_2} \otimes [0,\underline{x}] \otimes (q_1 \otimes q_2). \end{aligned}$$

(B19)

So, the product $R_{2} R_{1}$ is associated with the quaternion $q_2 \otimes q_1$ in the case of the active transformation and to $q_1 \otimes q_2$ in the case of the passive transformation.

Matrix-based calculation of $s'-s''$

According to (23), we derive

$$\begin{aligned}{} & {} \cos (s'-s'')=1 - 2{\displaystyle \frac{2}{1+z_p}}\sin ^2{\displaystyle \frac{x}{2}}\nonumber \\{} & {} \quad \sin ^2{\displaystyle \frac{y}{2}} ={\displaystyle \frac{\cos x+\cos y}{1+z_p}}. \end{aligned}$$

(C1)

This difference actually comes from a residual rotation between the 2-rotation sequence $R_2(x) R_1(y)$ and the 3-rotation sequence $W_p = R_3(-\lambda _p) R_2(\theta _p) R_3(\lambda _p)$. To see this, let us apply those rotation sequences on the unit vector $\hat{x}$ of the ITRS basis ($\hat{x}$, $\hat{y}$, $\hat{z}$). The resulting vector in the ITRS basis is the first column of the transpose of the corresponding transformation coordinate matrix.

So, for the transformation $R_2(x) R_1(y)$, we have to consider the matrix

$$\begin{aligned} R_1(-y) R_2(-x) =\left[ \begin{array}{ccc} \cos x &{}\quad 0 &{} \quad \sin x \\ \sin x \sin y &{} \quad \cos y &{}\quad -\cos x \sin y \\ -\sin x \cos y &{}\quad \sin y &{}\quad \cos x \cos y \end{array} \right] \nonumber \\ \end{aligned}$$

(C2)

of which the first column gives the vector

$$\begin{aligned} \hat{x}_1 = \cos x \,\hat{x} + \sin x \sin y \,\hat{y} -\sin x \cos y \,\hat{z}. \end{aligned}$$

(C3)

For the 3 rotation sequence $W_p$, it should be noticed that the transformation coordinate $W_p^T R_3(s'')$ from the TIRS to the ITRS is similar to $PN(X,Y) R_3(s)$ going from the CIRS to the GCRS. So $W_p^T$ can be easily written by replacing X with $x_p$ and Y with $y_p$ in the conventional precession–nutation matrix PN(X, Y) given by (2), taking as well $a=1/(1+z_p)=1/(1+\cos x \cos y)$:

$$\begin{aligned}{} & {} W_p^T = R_3(\lambda _p) R_2(-\theta _p)R_3(-\lambda _p) \nonumber \\{} & {} \quad =\left[ \begin{array}{ccc} 1-a x_p^2 &{} -a x_p y_p &{} x_p \\ -a x_p y_p &{} 1-a y_p^2 &{} y_p \\ -x_p &{} -y_p &{} 1-a (x_p^2+y_p^2) \end{array} \right] . \end{aligned}$$

(C4)

The corresponding transformed vector is

$$\begin{aligned} \hat{x}_2 = (1-ax_p^2) \,\hat{x} - a x_p y_p \,\hat{y} - x_p \,\hat{z} \end{aligned}$$

(C5)

that is, from (18),

$$\begin{aligned}{} & {} \hat{x}_2 = \left( 1-{\displaystyle \frac{\sin ^2 x}{1+\cos x \cos y}}\right) \nonumber \\{} & {} \quad \,\hat{x} + {\displaystyle \frac{\sin x \cos x \sin y}{1+\cos x \cos y}} \,\hat{y} -\sin x \,\hat{z}. \end{aligned}$$

(C6)

Then the angle between the vectors $\hat{x}_1$ and $\hat{x}_2$ is precisely the difference $s''-s'$ with the cosine

$$\begin{aligned} \begin{array}{ll} \cos (s''-s') &{}={x_1}\cdot {x_2} \\ &{} ={\displaystyle \frac{1}{1+z_p}}(\cos x+\cos ^2 x\cos y \\ &{}\quad -\cos x\sin ^2 x+\sin ^2x\cos x\sin ^2 y \\ &{}\quad +\sin ^2 x\cos y +\sin ^2 x\cos x\cos ^2y)\\ &{} ={\displaystyle \frac{\cos x+\cos y}{1+z_p}}, \end{array} \end{aligned}$$

(C7)

which is exactly the expression (C1) that we found by comparing quaternion expressions.