## Non-inertial frames of reference

Posted by Diego Assencio on 2014.10.17 under Physics (Mechanics)

Some physical systems can be studied much more easily when we consider their equations of motion on non-inertial frames of reference. As an example, studying ocean tides is easier if we use a non-inertial frame with origin at the center of the Earth; this frame is non-inertial since the Earth is under the action of gravitational forces caused by other bodies in the solar system such as the Sun and the Moon.

Consider a fixed inertial frame of reference $S$ with origin at a point $O$ and a non-inertial frame of reference $S'$ whose origin $O'$ is, at time $t$, positioned at ${\bf R}(t)$ as measured in $S$ (see figure 1). Our goal is to study the motion of a particle of mass $m$ on both $S$ and $S'$ and then find how the positions, velocities and accelerations of the particle as measured on each of these frames are related. After that, we will study the so-called fictitious forces which arise on non-inertial frames of reference. The axes of the fixed frame $S$ are labeled as $x_i$ while the axes of $S'$ are labeled as $x_i'$.

 Fig. 1: A fixed inertial frame of reference $S$ and a non-inertial frame of reference $S'$. At time $t$, the position of the origin $O'$ of $S'$ is positioned at ${\bf R}(t)$ as measured in the frame $S$.

From figure 1, we see that: $$\boxed{ {\bf x}(t) = {\bf R}(t) + {\bf x}'(t) } \label{post_51416695075e0ba12a5bb95029450fdb_x_x_prime}$$ This equation relates the coordinates ${\bf x}(t)$ of the particle on $S$ to its coordinates ${\bf x}'(t)$ on $S'$.

With respect to the fixed frame $S$, the motion of the frame $S'$ over an infinitesimal amount of time $dt$ can be broken into two parts: the translation of its origin and an infinitesimal rotation of its axes around an axis which passes through the origin of $S'$. To make this clearer, consider the case in which the origins of $S$ and $S'$ always coincide, i.e., ${\bf R}(t) = {\bf 0}$ for all $t$. Over an infinitesimal amount of time $dt$, the axes of $S'$ can do nothing but to rotate an infinitesimal angle $d\theta$ around some axis which passes through its origin. This axis, which is called the instantaneous axis of rotation of $S'$, can vary over time. This means the infinitesimal rotations can be done around different axes at different times. By letting the origin $O'$ of $S'$ move as well, we see that the full motion of $S'$ can be fully described by the translation of its origin plus the rotation of its axes.

Consider now how both $S$ and $S'$ see changes on ${\bf x}'(t)$ over an infinitesimal amount of time $dt$ (from now on, the dependency on $t$ will be omitted unless confusion would otherwise arise). If $S'$ sees a change $(d{\bf x}')_{S'}$, and since the axes of $S'$ might undergo an infinitesimal rotation during the period of time $dt$, then $S$ sees a change of $(d{\bf x}')_{S'}$ plus this small rotation: $$(d{\bf x}')_S = (d{\bf x}')_{S'} + d{\pmb\theta} \times {\bf x}' \label{post_51416695075e0ba12a5bb95029450fdb_dx_prime_S_S_prime}$$ where above $d{\pmb\theta} = d{\pmb\theta}(t)$ is a vector which is parallel to the rotation axis of $S'$ at time $t$ and has magnitude $d\theta$. Dividing both sides of the equation above by $dt$, we obtain: $$\left(\frac{d{\bf x}'}{dt}\right)_S = \left(\frac{d{\bf x}'}{dt}\right)_{S'} + \frac{d{\pmb\theta}}{dt} \times {\bf x}' = \left(\frac{d{\bf x}'}{dt}\right)_{S'} + {\pmb\omega} \times {\bf x}' \label{post_51416695075e0ba12a5bb95029450fdb_v_prime_S_S_prime}$$ where ${\pmb\omega} = {\pmb\omega}(t)$ is the angular velocity of the axes of $S'$ as measured in $S$ at time $t$. Notice that the derivation of \eqref{post_51416695075e0ba12a5bb95029450fdb_dx_prime_S_S_prime} does not explicitly use the fact that ${\bf x}'$ is the position of some particle as measured in $S'$. This means a change $(d{\bf q})_{S'}$ on any vector ${\bf q}$ as measured in $S'$ during $dt$ is seen by $S$ as the same change $(d{\bf q})_{S'}$ plus an additional term which comes from the infinitesimal rotation of $S'$ during $dt$: $$(d{\bf q})_S = (d{\bf q})_{S'} + d{\pmb\theta} \times {\bf q} \label{post_51416695075e0ba12a5bb95029450fdb_dq_prime_S_S_prime}$$ Dividing both sides by $dt$, we obtain: $$\displaystyle\left(\frac{d{\bf q}}{dt}\right)_S = \left(\frac{d{\bf q}}{dt}\right)_{S'} + \left(\frac{d{\pmb\theta}}{dt}\right)_S \times {\bf q} = \left(\frac{d{\bf q}}{dt}\right)_{S'} + {\pmb\omega} \times {\bf q} \label{post_51416695075e0ba12a5bb95029450fdb_dq_dt_S_S_prime}$$ Using equation \eqref{post_51416695075e0ba12a5bb95029450fdb_v_prime_S_S_prime}, we can relate the velocities of the particle measured in the frames $S$ and $S'$. Consider the perspective from frame $S$. From equation \eqref{post_51416695075e0ba12a5bb95029450fdb_x_x_prime}, we have that: $$\displaystyle \left(\frac{d{\bf x}}{dt}\right)_S = \left(\frac{d{\bf R}}{dt}\right)_S + \left(\frac{d{\bf x}'}{dt}\right)_S \label{post_51416695075e0ba12a5bb95029450fdb_v_step1}$$ But using equation \eqref{post_51416695075e0ba12a5bb95029450fdb_v_prime_S_S_prime} on \eqref{post_51416695075e0ba12a5bb95029450fdb_v_step1}, we obtain: $$\displaystyle \left(\frac{d{\bf x}}{dt}\right)_S = \left(\frac{d{\bf R}}{dt}\right)_S + \left(\frac{d{\bf x}'}{dt}\right)_{S'} + {\pmb\omega} \times {\bf x}'$$ A better way to write the equation above is: $$\boxed{ {\bf v}_S = \dot{\bf R} + {\bf v}_{S'} + {\pmb\omega}\times{\bf x}' } \label{post_51416695075e0ba12a5bb95029450fdb_v_final}$$ where:

 ${\bf v}_S$ is the velocity of the particle as measured in the inertial frame $S$ $\dot{\bf R}$ is the velocity of the origin of $S'$ with respect to $S$ ${\bf v}_{S'}$ is the velocity of the particle as measured in the non-inertial frame $S'$ ${\pmb\omega}$ is the (instantaneous) angular velocity of the axes of $S'$ ${\bf x}'$ is the position of the particle as measured in $S'$

We can now relate the accelerations of the particle as measured in the frames $S$ and $S'$. Consider again the perspective from frame $S$. From equation \eqref{post_51416695075e0ba12a5bb95029450fdb_v_final}, we have that: $$\displaystyle\left(\frac{d{\bf v}_S}{dt}\right)_S = \left(\frac{d\dot{\bf R}}{dt}\right)_S + \left(\frac{d{\bf v}_{S'}}{dt}\right)_S + \left(\frac{d{\pmb\omega}}{dt}\right)_S \times{\bf x}' + {\pmb\omega} \times \left(\frac{d{\bf x}'}{dt}\right)_S \label{post_51416695075e0ba12a5bb95029450fdb_a_step}$$ Using equation \eqref{post_51416695075e0ba12a5bb95029450fdb_dq_dt_S_S_prime} with ${\bf q}$ replaced by ${\bf v}_{S'}$ and equation \eqref{post_51416695075e0ba12a5bb95029450fdb_v_prime_S_S_prime} on equation \eqref{post_51416695075e0ba12a5bb95029450fdb_a_step}, we get: $$\begin{eqnarray} \displaystyle\left(\frac{d{\bf v}_S}{dt}\right)_S = \left(\frac{d\dot{\bf R}}{dt}\right)_S + \left(\frac{d{\bf v}_{S'}}{dt}\right)_{S'} &+& {\pmb\omega}\times{\bf v}_{S'} + \left(\frac{d{\pmb\omega}}{dt}\right)_S \times{\bf x}' \nonumber\\[5pt] &+& {\pmb\omega} \times \left[\left(\frac{d{\bf x}'}{dt}\right)_{S'} + {\pmb\omega}\times{\bf x}'\right] \nonumber\\[5pt] \label{post_51416695075e0ba12a5bb95029450fdb_a_step2} \end{eqnarray}$$ Using our definition above for ${\bf v}_{S'}$: $${\bf v}_{S'} = \left(\frac{d{\bf x}'}{dt}\right)_{S'}$$ equation \eqref{post_51416695075e0ba12a5bb95029450fdb_a_step2} can finally be written in a simpler form: $$\boxed{ {\bf a}_S = \ddot{\bf R} + {\bf a}_{S'} + \dot{\pmb\omega} \times{\bf x}' + {\pmb\omega} \times ({\pmb\omega}\times{\bf x}') + 2{\pmb\omega}\times{\bf v}_{S'} } \label{post_51416695075e0ba12a5bb95029450fdb_a_final}$$ where:

 ${\bf a}_S$ is the acceleration of the particle as measured in the inertial frame $S$ $\ddot{\bf R}$ is the acceleration of the origin of $S'$ with respect to $S$ $\dot{\pmb\omega}$ is the (instantaneous) angular acceleration of the axes of $S'$ ${\bf a}_{S'}$ is the acceleration of the particle as measured in the non-inertial frame $S'$

Equation \eqref{post_51416695075e0ba12a5bb95029450fdb_a_final} is very important since it allows us to compute effective forces on non-inertial frames of reference. Before we do that, notice that Newton's second law of motion states that, in an inertial frame of reference $S$, the net force ${\bf F}_S$ on an object with constant mass $m$ satisfies: $${\bf F}_S = m{\bf a}_S \label{post_51416695075e0ba12a5bb95029450fdb_newton_2nd_law}$$ Using equations \eqref{post_51416695075e0ba12a5bb95029450fdb_a_final} and \eqref{post_51416695075e0ba12a5bb95029450fdb_newton_2nd_law}, we can then define an effective force acting on the mass $m$ as measured in $S'$: $$\boxed{ {\bf F}^{\textrm{eff}}_{S'} = m{\bf a}_{S'} = {\bf F}_S - m\ddot{\bf R} - m\dot{\pmb\omega} \times{\bf x}' - m{\pmb\omega} \times ({\pmb\omega}\times{\bf x}') - 2m{\pmb\omega}\times{\bf v}_{S'} } \label{post_51416695075e0ba12a5bb95029450fdb_Feff}$$ According to Newton's laws, the real force is the one experienced in the inertial frame $S$, so on equation \eqref{post_51416695075e0ba12a5bb95029450fdb_Feff} all terms on the right-hand side except for ${\bf F}_S$ are fictitious forces which arise from the fact that we wrote an equation in the form ${\bf F}_{S'} = m{\bf a}_{S'}$ on the non-inertial frame of reference $S'$ even though Newton's second law states that ${\bf F} = m{\bf a}$ only on inertial frames of reference!

To finalize our study, let's analyze each one of the fictitious forces on equation \eqref{post_51416695075e0ba12a5bb95029450fdb_Feff}:

 $-m\ddot{\bf R}$ arises from the fact that $S'$ can be accelerating with respect to $S$ $-m\dot{\pmb\omega} \times{\bf x}'$ arises from the fact that the axes of $S'$ can be rotating non-uniformly when seen from $S$, i.e., this term can only be nonzero if the angular velocity ${\pmb\omega}$ changes over time $-m{\pmb\omega} \times ({\pmb\omega}\times{\bf x}')$ this term is often called the centrifugal force because it always points outwards from the axis of rotation; its magnitude is $m\omega^2\rho$, where $\rho$ is the distance between the particle and the axis of rotation (see figure 2) $- 2m{\pmb\omega}\times{\bf v}_{S'}$ this term is called the Coriolis force and arises from the motion of the particle as measured in $S'$; this force always points to the right of the direction along which the particle travels if ${\pmb\omega}$ is taken as the "upwards" direction (see figure 3)
 Fig. 2: The centrifugal force. From the figure, one can see that this force always points outwards from the axis of rotation. The magnitude of the centrifugal force is $m\omega^2\|{\bf x'}\|\sin\phi = m\omega^2\rho$, where $\rho = \|{\bf x}'\|\sin\phi$ is the distance between the axis of rotation and the particle.
 Fig. 3: The Coriolis force. The velocity of the particle on the frame $S'$ is ${\bf v}_{S'} = {\bf v}^{\parallel}_{S'} + {\bf v}_{S'}^{\perp}$, where ${\bf v}^{\parallel}_{S'}$ and ${\bf v}^{\perp}_{S'}$ are the components of ${\bf v}_{S'}$ which are parallel and orthogonal to ${\pmb\omega}$ respectively. Since ${\pmb\omega}\times{\bf v}^{\parallel}_{S'} = {\bf 0}$, we see that the Coriolis force is equal to $-2m{\pmb\omega}\times{\bf v}^{\perp}_{S'}$, which always points to the right of ${\bf v}_{S'}$ if the system is seen from the perspective in which ${\pmb\omega}$ points to the observer.

### References

 [1] S. Thornton, J. Marion, Classical Dynamics of Particles and Systems, Thomson Brooks/Cole; 5th edition (2003) [2] H. M. Nussenzveig, Curso de Física Básica - Vol. 1 (Mecânica), Edgard Blücher; 4th edition (2002)