Euler angles

We have seen how we can solve Euler's equations to determine the properties of a rotating body in the co-rotating body frame. Let us now investigate how we can determine the same properties in the inertial fixed frame.

The fixed frame and the body frame share the same origin. Hence, we can transform from one to the other by means of an appropriate rotation of our coordinate axes. In general, if we restrict ourselves to rotations about one of the Cartesian axes, three successive rotations are required to transform the fixed frame into the body frame. There are, in fact, many different ways to combined three successive rotations in order to achieve this goal. In the following, we shall describe the most widely used method, which is due to Euler.

We start in the fixed frame, which has coordinates $x$, $y$, $z$, and unit vectors ${\bf e}_x$, ${\bf e}_y$, ${\bf e}_z$. Our first rotation is counterclockwise (if we look down the axis) through an angle $\phi$ about the $z$-axis. The new frame has coordinates $x''$, $y''$, $z''$, and unit vectors ${\bf e}_{x''}$, ${\bf e}_{y''}$, ${\bf e}_{z''}$. According to Section A.6, the transformation of coordinates can be represented as follows:

$$\left(\begin{array}{c}x''\\ y''\\ z''\end{array}\right)= \left(\b... ...\ 0,&0,&1 \end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right).$$ (8.45)

The angular velocity vector associated with $\phi$ has the magnitude $\skew{5}\dot{\phi}$, and is directed along ${\bf e}_z$ (i.e., along the axis of rotation). Hence, we can write

$$\mbox{\boldmath$\omega$} _\phi = \skew{5}\dot{\phi}\,{\bf e}_z.$$ (8.46)

Clearly, $\skew{5}\dot{\phi}$ is the precession rate about the $z$-axis, as seen in the fixed frame.

The second rotation is counterclockwise (if we look down the axis) through an angle $\theta$ about the $x''$-axis. The new frame has coordinates $x'''$, $y'''$, $z'''$, and unit vectors ${\bf e}_{x'''}$, ${\bf e}_{y'''}$, ${\bf e}_{z'''}$. By analogy with Equation (8.45), the transformation of coordinates can be represented as follows:

$$\left(\begin{array}{c}x'''\\ y'''\\ z'''\end{array}\right)= \left... ...heta \end{array}\right)\left(\begin{array}{c}x''\\ y''\\ z''\end{array}\right).$$ (8.47)

The angular velocity vector associated with $\theta$ has the magnitude $\skew{5}\dot{\theta}$, and is directed along ${\bf e}_{x''}$ (i.e., along the axis of rotation). Hence, we can write

$$\mbox{\boldmath$\omega$} _\theta = \skew{5}\dot{\theta}\,{\bf e}_{x''}.$$ (8.48)

The third rotation is counterclockwise (if we look down the axis) through an angle $\psi$ about the $z'''$-axis. The new frame is the body frame, which has coordinates $x'$, $y'$, $z'$, and unit vectors ${\bf e}_{x'}$, ${\bf e}_{y'}$, ${\bf e}_{z'}$. The transformation of coordinates can be represented as follows:

$$\left(\begin{array}{c}x'\\ y'\\ z'\end{array}\right)= \left(\begi... ...1 \end{array}\right)\left(\begin{array}{c}x'''\\ y'''\\ z'''\end{array}\right).$$ (8.49)

The angular velocity vector associated with $\psi$ has the magnitude $\skew{5}\dot{\psi}$, and is directed along ${\bf e}_{z''}$ (i.e., along the axis of rotation). Note that ${\bf e}_{z'''}={\bf e}_{z'}$, because the third rotation is about ${\bf e}_{z'''}$. Hence, we can write

$$\mbox{\boldmath$\omega$} _\psi= \skew{5}\dot{\psi}\,{\bf e}_{z'}.$$ (8.50)

Clearly, $\skew{5}\dot{\psi}$ is minus the precession rate about the $z'$-axis, as seen in the body frame.

The full transformation between the fixed frame and the body frame is rather complicated. However, the following results can easily be verified:

$${\bf e}_z = \sin\psi\,\sin\theta\,{\bf e}_{x'} + \cos\psi\,\sin\theta\,{\bf e}_{y'} + \cos\theta\,{\bf e}_{z'},$$ (8.51)

$${\bf e}_{x''} = \cos\psi\,{\bf e}_{x'} -\sin\psi\,{\bf e}_{y'}.$$ (8.52)

It follows from Equation (8.51) that ${\bf e}_z\!\cdot\!{\bf e}_{z'} =\cos\theta$. In other words, $\theta$ is the angle of inclination between the $z$- and $z'$-axes. Finally, because the total angular velocity can be written

$$\mbox{\boldmath$\omega$} = \mbox{\boldmath$\omega$} _\phi + \mbox{\boldmath$\omega$} _\theta+ \mbox{\boldmath$\omega$} _\psi,$$ (8.53)

Equations (8.46), (8.48), and (8.50)–(8.52) yield

$$\omega_{x'} = \sin\psi\,\sin\theta\,\,\skew{5}\dot{\phi} +\cos\psi\,\,\skew{5}\dot{\theta},$$ (8.54)

$$\omega_{y'} = \cos\psi\,\sin\theta\,\,\skew{5}\dot{\phi}-\sin\psi\,\,\skew{5}\dot{\theta},$$ (8.55)

$$\omega_{z'} = \cos\theta\,\,\skew{5}\dot{\phi} +\skew{5}\dot{\psi}.$$ (8.56)

The angles $\phi$, $\theta$, and $\psi$ are termed Euler angles. Each has a clear physical interpretation: $\phi$ is the angle of precession about the $z$-axis in the fixed frame, $\psi$ is minus the angle of precession about the $z'$-axis in the body frame, and $\theta$ is the angle of inclination between the $z$- and $z'$- axes. Moreover, we can express the components of the angular velocity vector $\omega$ in the body frame entirely in terms of the Eulerian angles, and their time derivatives. [See Equations (8.54)–(8.56).]

Consider a freely rotating body that is rotationally symmetric about one axis (the $z'$-axis). In the absence of an external torque, the angular momentum vector ${\bf L}$ is a constant of the motion. [See Equation (8.3).] Let ${\bf L}$ point along the $z$-axis. In the previous section, we saw that the angular momentum vector subtends a constant angle $\theta$ with the axis of symmetry; that is, with the $z'$-axis. Hence, the time derivative of the Eulerian angle $\theta$ is zero. We also saw that the angular momentum vector, the axis of symmetry, and the angular velocity vector are coplanar. Consider an instant in time at which all of these vectors lie in the $y'$-$z'$ plane. This implies that $\omega_{x'}=0$. According to the previous section, the angular velocity vector subtends a constant angle $\alpha$ with the symmetry axis. It follows that $\omega_{y'}=\omega\,\sin\alpha$ and $\omega_{z'} = \omega\,\cos\alpha$. Equation (8.54) gives $\psi=0$. Hence, Equation (8.55) yields

$$\omega\,\sin\alpha = \sin\theta\,\skew{5}\dot{\phi}.$$ (8.57)

This can be combined with Equation (8.44) to give

$$\skew{5}\dot{\phi} = \omega\left[1 + \left(\frac{{\cal I}_\parallel^{\,2}}{{\cal I}_\perp^{\,2}}-1\right)\cos^2\alpha\right]^{1/2}.$$ (8.58)

Finally, Equation (8.56), together with Equations (8.44) and (8.57), yields

$$\skew{5}\dot{\psi} = \omega\,\cos\alpha-\cos\theta\,\skew{5}\dot{... ...t)= \omega\,\cos\alpha\left(1-\frac{{\cal I}_\parallel}{{\cal I}_\perp}\right).$$ (8.59)

A comparison of this equation with Equation (8.40) gives

$$\skew{5}\dot{\psi}=-{\mit\Omega}.$$ (8.60)

Thus, as expected, $\skew{5}\dot{\psi}$ is minus the precession rate (of the angular momentum and angular velocity vectors) in the body frame. On the other hand, $\skew{5}\dot{\phi}$ is the precession rate (of the angular velocity vector and the symmetry axis) in the fixed frame. Note that $\skew{5}\dot{\phi}$ and ${\mit \Omega }$ are quite dissimilar. For instance, ${\mit \Omega }$ is negative for elongated bodies ( ${\cal I}_\parallel<{\cal I}_\perp$) whereas $\skew{5}\dot{\phi}$ is positive definite. It follows that the precession is always in the same sense as $L_z$ in the fixed frame, whereas the precession in the body frame is in the opposite sense to $L_{z'}$ for elongated bodies. We found, in the previous section, that for a flattened body the angular momentum vector lies between the angular velocity vector and the symmetry axis. This means that, in the fixed frame, the angular velocity vector and the symmetry axis lie on opposite sides of the fixed angular momentum vector, about which they precess. See Figure 8.2. On the other hand, for an elongated body we found that the angular velocity vector lies between the angular momentum vector and the symmetry axis. This means that, in the fixed frame, the angular velocity vector and the symmetry axis lie on the same side of the fixed angular momentum vector, about which they precess. See Figure 8.2. (Recall that the angular momentum vector, the angular velocity vector, and the symmetry axis are coplanar.)

Figure: 8.2 A freely rotating object that is elongated along its axis of symmetry, ${\bf e}_{z'}$ (left), and a freely rotating object that is flattened along its axis of symmetry (right). The ${\bf L}$ vector is fixed.