Section 2.1 Referentials
We have defined in
Section 1.1 a referential
\(\cE\) as a three-dimensional space, typically associated with the rigid extension of a physical body. A referential is equipped with a clock allowing the measurements of the time
\(t\) of occurrence of (simultaneous) events at various locations of
\(\cE\text{.}\) Now consider another referential
\(\cF\) in motion relative to
\(\cE\text{.}\) In each referential, an observer is equipped with a clock to record the occurrence of events. We assume that the clocks of
\(\cE\) and
\(\cF\) have been synchronized.
In classical mechanics, we make the fundamental assumption that two events which occur simultaneously in \(\cE\) must occur simultaneously in \(\cF\text{.:}\) this implies that the clocks of \(\cE\) and \(\cF\) indicate identical times of occurrence of a specific event. Similarly, we assume that the distances measured in \(\cE\) and \(\cF\) between two simultaneous events are identical. In particular, the length measured between any two points of a referential (or rigid body) remains constant relative to any other referential.
The main task we face in mechanics is the determination the time rate of change \(d\bV /dt\) of vector functions \(\bV (t)\text{.}\) However, vector functions, as opposed to scalar functions, have different time-derivatives in different referentials in relative motion: hence it is appropriate to denote \(d\bV /dt\) relative to referential \(\cE\) as
\begin{equation*}
\left( {d \bV \over dt} \right)_{\cE}
\end{equation*}
in order to specify the dependence upon referential.
We will always do so whenever necessary. 1
In practice, the calculation of the time-derivative of
\(\bV\) relative to
\(\cE\) can be done by resolving
\(\bV\) on a basis
\((\be_1 , \be_2 , \be_3 )\) of unit vectors fixed in
\(\cE\text{:}\) \(\bV= V_1 \be_1 + V_2 \be_2 + V_3 \be_3\text{.}\) If
\(\bV\) is a function of time, its components should also be functions of time. Then since
\((\be_1 , \be_2 , \be_3 )\) are constant in
\(\cE\text{,}\) \(d\bV /dt\) is given by
2
\begin{equation}
\left( \frac{d \bV}{dt}\right)_{\cE} =
\frac{d}{dt} ( V_1 \be_1 + V_2 \be_2 + V_3 \be_3) =
\dot{V}_1 \be_1 + \dot{V}_2 \be_2 + \dot{V}_3 \be_3 \tag{2.1.1}
\end{equation}
This approach is adequate for most applications in particle kinematics, but it will become quickly laborious in rigid body kinematics. We shall learn in
Chapter 3 how the concept of angular velocity will allow us to perform time derivatives of vector functions.
Let us mention properties of the time-differentiation w.r.t. to a given referential
\begin{equation}
\frac{d}{dt} ( \bV + \bW ) = \frac{d\bV}{dt} + \frac{d\bV}{dt} \tag{2.1.2}
\end{equation}
\begin{equation}
\frac{d}{dt} ( \lambda \bV ) = \dot{\lambda} \bV + \lambda \frac{d\bV}{dt}\tag{2.1.3}
\end{equation}
\begin{equation}
\frac{d}{dt} ( \bV \cdot \bW ) = \frac{d\bV}{dt} \cdot \bW + \bV \cdot \frac{d\bW}{dt} \label{scaprod}\tag{2.1.4}
\end{equation}
\begin{equation}
\frac{d}{dt} ( \bV \times \bW ) = \frac{d\bV}{dt} \times \bW + \bV \times \frac{d\bW}{dt}\tag{2.1.5}
\end{equation}
for two arbitrary vectors \(\bV\) and \(\bW\text{.}\) Here \(\bV \cdot \bW\) and \(\bV\times \bW\) are the scalar product and vector product between \(\bV\) and \(\bW\text{,}\) respectively.
As an application of these properties, here are two useful theorems:
Theorem 2.1.2.
A vector \(\bV (t)\) remains of constant magnitude if and only if \(\bV\) remains perpendicular to its derivative \(d\bV /dt\) (relative to any referential \(\cE\)).
Proof.
The time-derivative of the scalar \(|\bV|^2 = \bV \cdot \bV\) vanishes since vector \(\bV\) has a constant magnitude. This gives
\begin{equation*}
{d\bV \over dt} \cdot \bV + \bV \cdot {d\bV \over dt} = 0
\end{equation*}
Hence \(\bV \cdot (d\bV / dt) = 0\text{:}\) \(\bV\) and \(d \bV / dt\) are orthogonal.
Theorem 2.1.3.
A non-zero vector \(\bV (t)\) remains of constant direction in a referential \(\cE\) if and only if \(\bV\) remains collinear to its time-derivative \(d\bV /dt\) in \(\cE\text{.}\)
Proof.
To prove this property, introduce the unit vector \(\bv = \mu \bV\) with \(\mu = 1/ |\bV|\) (\(\bV\) is not zero) which is collinear to \(\bV\text{.}\) If \(\bV (t)\) remains of constant direction, then \(\bv (t)\) is constant vector. We then take the cross-product of both sides of \(d \bv /dt = \dot{\mu} \bV + \mu (d\bV / dt) = \bze\) with vector \(\bV\) to find
\begin{equation}
\bV \times {d\bV \over dt} = 0 \tag{2.1.6}
\end{equation}
which shows that
\(\bV\) and
\(d\bV /dt\) are collinear. Conversely, assume that
(2.1.6) holds, then it is easy to show that
\(\bv \times d\bv / dt = 0\text{.}\) Hence
\(\bv\) is collinear to
\(d\bv /dt\text{.}\) Furthermore, from theorem
Theorem 2.1.2,
\(\bv\) is also orthogonal to
\(d\bv /dt\text{,}\) since
\(\bv\) is a unit vector. Then necessarily
\(d\bv /dt = \bze\text{,}\) that is,
\(\bv\) is a constant vector in
\(\cE\text{:}\) \(\bV\) remains of constant direction.
