Rotating reference frames

Suppose that a given object has position vector ${\bf r}$ in some inertial (i.e., non-rotating) reference frame. Let us observe the motion of this object in a non-inertial reference frame that rotates with constant angular velocity $\Omega$ about an axis passing through the origin of the inertial frame. Suppose, first of all, that our object appears stationary in the rotating reference frame. Hence, in the non-rotating frame, the object's position vector ${\bf r}$ will appear to precess about the origin with angular velocity $\Omega$. It follows from Section A.7 that, in the non-rotating reference frame,

$$\frac{d{\bf r}}{dt} = \mbox{\boldmath$\Omega$} \times{\bf r}.$$ (6.1)

Suppose, now, that our object appears to move in the rotating reference frame with instantaneous velocity ${\bf v}'$. It is fairly obvious that the appropriate generalization of the preceding equation is simply

$$\frac{d{\bf r}}{dt} = {\bf v}'+ \mbox{\boldmath$\Omega$} \times{\bf r}.$$ (6.2)

Let $d/dt$ and $d/dt'$ denote apparent time derivatives in the non-rotating and rotating frames of reference, respectively. Because an object that is stationary in the rotating reference frame appears to move in the non-rotating frame, it is clear that $d/dt\neq d/dt'$. Writing the apparent velocity, ${\bf v}'$, of our object in the rotating reference frame as $d{\bf r}/dt'$, Equation (6.2) takes the form

$$\frac{d{\bf r}}{dt} = \frac{d{\bf r}}{dt'}+ \mbox{\boldmath$\Omega$} \times{\bf r},$$ (6.3)

or

$$\frac{d}{dt} =\frac{d}{dt'}+ \mbox{\boldmath$\Omega$} \,\times,$$ (6.4)

because ${\bf r}$ is a general position vector. Equation (6.4) expresses the relationship between apparent time derivatives in the non-rotating and rotating reference frames.

Operating on the general position vector ${\bf r}$ with the time derivative in Equation (6.4), we get

$${\bf v} = {\bf v}' + \mbox{\boldmath$\Omega$} \times{\bf r}.$$ (6.5)

This equation relates the apparent velocity, ${\bf v} = d{\bf r}/dt$, of an object with position vector ${\bf r}$ in the non-rotating reference frame to its apparent velocity, ${\bf v}' = d{\bf r}/dt'$, in the rotating reference frame.

Operating twice on the position vector ${\bf r}$ with the time derivative in Equation (6.4), we obtain

$${\bf a} = \left(\frac{d}{dt'}+ \mbox{\boldmath$\Omega$}\,\times\right)\left({\bf v}' + \mbox{\boldmath$\Omega$}\times{\bf r}\right),$$ (6.6)

or

$${\bf a} = {\bf a}' + \mbox{\boldmath$\Omega$} \times( \mbox{\boldmath$\Omega$} \times{\bf r}) + 2\, \mbox{\boldmath$\Omega$} \times{\bf v}'.$$ (6.7)

This equation relates the apparent acceleration, ${\bf a}= d^{\,2}{\bf r}/dt^{\,2}$, of an object with position vector ${\bf r}$ in the non-rotating reference frame to its apparent acceleration, ${\bf a}'= d^{\,2}{\bf r}/dt'^{\,2}$, in the rotating reference frame.

Applying Newton's second law of motion in the inertial (i.e., non-rotating) reference frame, we obtain

$$m\,{\bf a} = {\bf f}.$$ (6.8)

Here, $m$ is the mass of our object, and ${\bf f}$ is the (non-fictitious) force acting on it. These quantities are the same in both reference frames. Making use of Equation (6.7), the apparent equation of motion of our object in the rotating reference frame takes the form

$$m\,{\bf a}' = {\bf f} - m\, \mbox{\boldmath$\Omega$} \times( \mbox{\boldmath$\Omega$} \times{\bf r}) -2\,m\, \mbox{\boldmath$\Omega$} \times{\bf v}'.$$ (6.9)

The last two terms in this equation are so-called fictitious forces; such forces are always needed to account for motion observed in non-inertial reference frames. Fictitious forces can always be distinguished from non-fictitious forces in Newtonian mechanics because the former have no associated reactions. Let us now investigate the two fictitious forces appearing in Equation (6.9).