Newton's laws of motion and special relativity are said to not hold for non-inertial frame of references. Are there other laws that govern non-inertial frames of reference? If so, what are they? If not, can we just use the usual laws like we normally do provided that we include fictious forces that arise in such reference frames?
1 Answers
Start from the laws for an inertial observer and use relative kinematics. Leaving some references:
- Relative kinematics for points
- Rotations. Section about successive rotations provides the relation used below, for relative kinematics of orientation, angular velocity and acceleration
- Equations of motion for rigid bodies
Relative kinematics: position and momentum
More explicitly, inserting the expression of acceleration from relative kinematics of a point $P$ w.r.t. an inertial observer $O$ as a function of the motion of a generic observer $Q$ and the kinematics w.r.t. $Q$,
$$\begin{aligned} (P - O) & = ( Q - O ) + ( P - Q ) && & \text{position} \\ \vec{v}^O_{P/O} & =\vec{v}^O_{Q/O} + \vec{v}^Q_{P/Q} + \vec{\omega}_{Q/O} \times ( P - Q ) && & \text{velocity} \\ \vec{a}^O_{P/O} & = \vec{a}^O_{Q/O} + \vec{a}^{Q}_{P/Q} + 2 \vec{\omega}_{Q/P} \times \vec{v}^Q_{P/Q} + \vec{\alpha}_{Q/O} \times (P - Q) + \vec{\omega}_{Q/P} \times [ \, \vec{\omega}_{Q/P} \times (P - Q) \, ] && & \text{acceleration} \end{aligned}$$
into momentum equation w.r.t. the inertial reference frame
$$m \vec{a}_{P/O}^O = \vec{F} \ ,$$
you get the momentum equation of point system $P$ w.r.t. non inertial reference frame,
$$\begin{aligned} m \vec{a}_{P/Q}^Q & = \vec{F} + \\ & - \vec{a}^O_{Q/O} - 2 \vec{\omega}_{Q/P} \times \vec{v}^Q_{P/Q} - \vec{\alpha}_{Q/O} \times (P - Q) - \vec{\omega}_{Q/P} \times [ \, \vec{\omega}_{Q/P} \times (P - Q) \, ] \ , \end{aligned}$$
being the terms after $\vec{F}$, depending on the kinematics of $Q$ the so called apparent forces. The contributions depending only on the velocity and angular velocity of the observer $Q$ are
- Coriolis force $- 2 \vec{\omega}_{Q/P} \times \vec{v}^Q_{P/Q}$
- centrifugal force $-\vec{\omega}_{Q/P} \times [ \, \vec{\omega}_{Q/P} \times (P - Q) \, ]$
while the other two contributions depend on the acceleration and angular acceleration of the non inertial observer.
Relative kinematics: orientation and angular momentum
For the the balance of angular momentum, we need the relative kinematics of orientation, angular velocity and angular acceleration. While orientation is described by rotation tensor and relative orientation is obtained through composition of rotation (dot product of tensors), relative kinematics of angular velocity and acceleration results in summation, see here. The orientation of a system $P$ w.r.t. a reference frame material with the observer $O$ and a reference frame material with the non inertial observer $Q$ reads
$$\begin{aligned} \mathbb{R}^{O \rightarrow P} & = \mathbb{R}^{Q \rightarrow P} \cdot \mathbb{R}^{O \rightarrow Q} \\ \vec{\omega}_{P/O} & = \vec{\omega}_{P/Q} + \vec{\omega}_{Q/O} \\ \vec{\alpha}_{P/O} & = \vec{\alpha}_{P/Q} + \vec{\alpha}_{Q/O} \end{aligned}$$
Introducing these relations in the expression of the angular momentum equation you like the most, see here, you get the desired expression of angular momentum equation for a non inertial reference frame. Just as an example, introducing the relative kinematics relations into the angular momentum equation for a rigid body using the center of mass as the reference point,
$$ \mathbb{I}_G \cdot \vec{\alpha}_{/O} + \vec{\omega}_{/O} \times \mathbb{I}_G \cdot \vec{\omega}_{/O} = \vec{M}_G^e \, $$
you get
$$ \mathbb{I}_G \cdot \vec{\alpha}_{/Q} + \vec{\omega}_{/Q} \times \mathbb{I}_G \cdot \vec{\omega}_{/Q} = \vec{M}_G^e - \mathbb{I}_G \cdot \vec{\alpha}_{Q/O} - \vec{\omega}_{/Q} \times \mathbb{I}_G \cdot \vec{\omega}_{Q/O} - \vec{\omega}_{Q/O} \times \mathbb{I}_G \cdot \vec{\omega}_{/Q} - \vec{\omega}_{Q/O} \times \mathbb{I}_G \cdot \vec{\omega}_{Q/O}\, $$
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