Eulerian 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. 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 (looking 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 Equations (A.1277)-(A.1279), the transformation of coordinates can be represented as follows:

$$\left(\begin{array}{c}x''\\ y''\\ z''\end{array}\right)= \le... ...rray}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right).$$ (528)

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

$$\mbox{\boldmath$\omega$}_\phi = \dot{\phi}\,{\bf e}_z.$$ (529)

Clearly, $\dot{\phi}$ is the precession rate about the ${\bf e}_z$ axis, as seen in the fixed frame.

The second rotation is counterclockwise (looking 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 (528), the transformation of coordinates can be represented as follows:

$$\left(\begin{array}{c}x'''\\ y'''\\ z'''\end{array}\right)= ... ...right)\left(\begin{array}{c}x''\\ y''\\ z''\end{array}\right).$$ (530)

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

$$\mbox{\boldmath$\omega$}_\theta = \dot{\theta}\,{\bf e}_{x''}.$$ (531)

The third rotation is counterclockwise (looking 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:

$$\left(\begin{array}{c}x'\\ y'\\ z'\end{array}\right)= \left(... ...ht)\left(\begin{array}{c}x'''\\ y'''\\ z'''\end{array}\right).$$ (532)

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

$$\mbox{\boldmath$\omega$}_\psi= \dot{\psi}\,{\bf e}_{z'}.$$ (533)

Clearly, $\dot{\psi}$ is minus the precession rate about the ${\bf e}_{z'}$ 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:

$${\bf e}_z = \sin\psi\,\sin\theta\,{\bf e}_{x'} + \cos\psi\,\sin\theta\,{\bf e}_{y'} + \cos\theta\,{\bf e}_{z'},$$ (534)

$${\bf e}_{x''} = \cos\psi\,{\bf e}_{x'} -\sin\psi\,{\bf e}_{y'}.$$ (535)

It follows from Equation (534) 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, since the total angular velocity can be written

$$\mbox{\boldmath$\omega$} = \mbox{\boldmath$\omega$}_\phi +\mbox{\boldmath$\omega$}_\theta+\mbox{\boldmath$\omega$}_\psi,$$ (536)

Equations (529), (531), and (533)-(535) yield

$$\omega_{x'} = \sin\psi\,\sin\theta\,\dot{\phi} +\cos\psi\,\dot{\theta},$$ (537)

$$\omega_{y'} = \cos\psi\,\sin\theta\,\dot{\phi}-\sin\psi\,\dot{\theta},$$ (538)

$$\omega_{z'} = \cos\theta\,\dot{\phi} +\dot{\psi}.$$ (539)

The angles $\phi$, $\theta$, and $\psi$ are termed Eulerian angles. Each has a clear physical interpretation: $\phi$ is the angle of precession about the ${\bf e}_z$ axis in the fixed frame, $\psi$ is minus the angle of precession about the ${\bf e}_{z'}$ axis in the body frame, and $\theta$ is the angle of inclination between the ${\bf e}_z$ and ${\bf e}_{z'}$ 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 (537)-(539)].

Consider a rigid body which is constrained to rotate about a fixed axis with the constant angular velocity $\omega$. Let the fixed angular velocity vector point along the $z$-axis. In the previous section, we saw that the angular momentum and the torque were both steady in the body frame. Since there is no precession of quantities in the body frame, it follows that the Eulerian angle $\psi$ is constant. Furthermore, since the angular velocity vector is fixed in the body frame, as well as the fixed frame [as can be seen by applying Equation (502) to $\omega$ instead of ${\bf L}$], it must subtend a constant angle with the ${\bf e}_{z'}$ axis. Hence, the Eulerian angle $\theta$ is also constant. It follows from Equations (537)-(539) that

$$\omega_{x'} = \sin\psi\,\sin\theta\,\dot{\phi},$$ (540)

$$\omega_{y'} = \cos\psi\,\sin\theta\,\dot{\phi},$$ (541)

$$\omega_{z'} = \cos\theta\,\dot{\phi},$$ (542)

which implies that $\omega\equiv (\omega_{x'}^{\,2}+\omega_{y'}^{\,2}+\omega_{z'}^{\,2})^{1/2}=\dot{\phi}$. In other words, the precession rate, $\dot{\phi}$, in the fixed frame is equal to $\omega$. Hence, in the fixed frame, the constant torque and angular momentum vectors found in the body frame precess about the angular velocity vector (i.e., about the $z$-axis) at the rate $\omega$. As discussed in the previous section, for the special case where the angular velocity vector is parallel to one of the principal axes of the body, the angular momentum vector is parallel to the angular velocity vector, and the torque is zero. Thus, in this case, there is no precession in the fixed frame.

Consider a rotating device such as a flywheel or a propeller. If the device is statically balanced then its center of mass lies on the axis of rotation. This is desirable since, otherwise, gravity, which effectively acts at the center of mass, exerts a varying torque about the axis of rotation as the device rotates, giving rise to unsteady rotation. If the device is dynamically balanced then the axis of rotation is also a principal axis, so that, as the device rotates its angular momentum vector, ${\bf L}$, remains parallel to the axis of rotation. This is desirable since, otherwise, the angular momentum vector is not parallel to the axis of rotation, and, therefore, precesses around it. Since $d{\bf L}/dt$ is equal to the torque, a precessing torque must also be applied to the device (at right-angles to both the axis and ${\bf L}$). The result is a reaction on the bearings which can give rise to violent vibration and wobbling, even when the device is statically balanced.

Consider a freely rotating body which 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 (456)]. 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: i.e., 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 (537) yields $\psi=0$. Hence, Equation (538) yields

$$\omega\,\sin\alpha = \sin\theta\,\dot{\phi}.$$ (543)

This can be combined with Equation (525) to give

$$\dot{\phi} = \omega\left[1 + \left(\frac{I_\parallel^{\,2}}{I_\perp^{\,2}}-1\right)\cos^2\alpha\right]^{1/2}.$$ (544)

Finally, Equations (539), together with (525) and (543), yields

$$\dot{\psi} = \omega\,\cos\alpha-\cos\theta\,\dot{\phi} = \om... ... \omega\,\cos\alpha\left(1-\frac{I_\parallel}{I_\perp}\right).$$ (545)

A comparison of the above equation with Equation (521) gives

$$\dot{\psi}=-{\mit\Omega}.$$ (546)

Thus, as expected, $\dot{\psi}$ is minus the precession rate (of the angular momentum and angular velocity vectors) in the body frame. On the other hand, $\dot{\phi}$ is the precession rate (of the angular velocity vector and the symmetry axis) in the fixed frame. Note that $\dot{\phi}$ and ${\mit\Omega}$ are quite dissimilar. For instance, ${\mit\Omega}$ is negative for elongated bodies ( $I_\parallel<I_\perp$) whereas $\dot{\phi}$ 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. 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. (Recall that the angular momentum vector, the angular velocity vector, and the symmetry axis, are coplanar.)

As an example, consider the free rotation of a thin disk. It is easily demonstrated (from the perpendicular axis theorem) that

$$I_\parallel = 2\,I_\perp$$ (547)

for such a disk. Hence, from Equation (521), the precession rate in the body frame is

$${\mit\Omega} = \omega\,\cos\alpha.$$ (548)

According to Equation (544), the precession rate in the fixed frame is

$$\dot{\phi} = \omega\left[1+ 3\,\cos^2\alpha\right]^{1/2}.$$ (549)

In the limit in which $\alpha$ is small (i.e., in which the angular velocity vector is almost parallel to the symmetry axis), we obtain

$${\mit\Omega} \simeq \omega,$$ (550)

$$\dot{\phi} \simeq 2\,\omega.$$ (551)

Thus, the symmetry axis precesses in the fixed frame at approximately twice the angular speed of rotation. This precession is manifest as a wobbling motion.

It is known that the axis of rotation of the Earth is very slightly inclined to its symmetry axis (which passes through the two geographic poles). The angle $\alpha$ is approximately $0.2$ seconds of an arc (which corresponds to a distance of about $6\,{\rm m}$ on the Earth's surface). It is also known that the ratio of the terrestrial moments of inertia is about $I_\parallel/I_\perp=1.00327$, as determined from the Earth's oblateness--see Section 12.7. Hence, from (521), the precession rate of the angular velocity vector about the symmetry axis, as viewed in a geostationary reference frame, is

$${\mit\Omega} = 0.00327\,\omega,$$ (552)

giving a precession period of

$$T'= \frac{2\pi}{{\mit\Omega}} = 305\,\,{\rm days}.$$ (553)

(Of course, $2\pi/\omega = 1$ day.) The observed period of precession is about 440 days. The disagreement between theory and observation is attributed to the fact that the Earth is not perfectly rigid. The Earth's symmetry axis subtends an angle $\theta\simeq \alpha = 0.2''$ [see (525)] with its angular momentum vector, but lies on the opposite side of this vector to the angular velocity vector. This implies that, as viewed from space, the Earth's angular velocity vector is almost parallel to its fixed angular momentum vector, whereas its symmetry axis subtends an angle of $0.2''$ with both vectors, and precesses about them. The (theoretical) precession rate of the Earth's symmetry axis, as seen from space, is given by Equation (544):

$$\dot{\phi} = 1.00327\,\omega.$$ (554)

The associated precession period is

$$T = \frac{2\pi}{\dot{\phi}} = 0.997\,\,{\rm days}.$$ (555)

The free precession of the Earth's symmetry axis in space, which is known as the Chandler wobble, since it was discovered by the American astronomer Seth Chandler in 1891, is superimposed on a much slower forced precession, with a period of about 26,000 years, caused by the small gravitational torque exerted on the Earth by the Sun and the Moon, as a consequence of the Earth's slight oblateness--see Section 12.10.