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) |

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