Relative Motion

Consider two reference frames: a fixed (inertial) frame S and a moving frame S'. A point A is observed from both. Three vectors describe its position:

• $\vec{r}_A$ — absolute position of A as measured in S (from O to A).
• $\vec{r}_{O'}$ — position of the origin of S' as seen from S. It tells us where S' is located relative to S. Its time derivative gives the transport velocity.
• $\vec{r}_{A/S'}$ — relative position of A as measured in S'. This is the position of A relative to O' — the coordinates that an observer in S' assigns to A. Physically: if you were riding on S', this is where you would say "A is relative to me".

The fundamental vector relation is: $$\vec{r}_A = \vec{r}_{O'} + \vec{r}_{A/S'}$$ Interpretation: to find where A is in the fixed frame S, add the position of S' (relative to S) to the position of A as seen from S'. It is a vector sum: if S' is displaced by $\vec{r}_{O'}$ and inside S' the point A is at $\vec{r}_{A/S'}$, then in S the position of A is the diagonal of the parallelogram formed by the two vectors. Notation: the subscript "A/S'" is read "of A relative to S'". The slash "/" separates the object (A) from the reference frame (S') in which the measurement is taken.

If S' translates relative to S (no rotation), differentiating the position relation gives the velocity composition law for pure translation: $$\vec{v}_A = \vec{v}_{A/S'} + \vec{v}_{S'/S}$$ where:

• $\vec{v}_A$ — absolute velocity of A measured in S (derivative of $\vec{r}_A$).
• $\vec{v}_{A/S'}$ — relative velocity of A measured by an observer in S' (derivative of $\vec{r}_{A/S'}$).
• $\vec{v}_{S'/S}$ — transport velocity (velocity of S' relative to S), the derivative of $\vec{r}_{O'}$. Independent of A.

In compact form: $$\vec{v}_{abs} = \vec{v}_{rel} + \vec{v}_{transport}$$ Important: this relation holds only for pure translation. If S' rotates, an additional term $\vec{\omega} \times \vec{r}_{A/S'}$ appears (see §4). Physical example: if you walk inside a moving train, your velocity relative to the ground is the sum of the train's velocity plus your walking velocity relative to the train.

Differentiating the velocity relation gives the acceleration composition for pure translation: $$\vec{a}_A = \vec{a}_{A/S'} + \vec{a}_{S'/S}$$

• $\vec{a}_A$ — absolute acceleration (measured in S).
• $\vec{a}_{A/S'}$ — relative acceleration (measured in S').
• $\vec{a}_{S'/S}$ — transport acceleration (of S' relative to S).

Case 1 — S' inertial (ULM): if $\vec{a}_{S'/S} = 0$, then $\vec{a}_A = \vec{a}_{A/S'}$. Accelerations measured in the two frames coincide. This leads to the Galilean Principle of Relativity: the laws of mechanics are identical in all inertial frames. No mechanical experiment can tell if the frame is at rest or in uniform motion. Case 2 — S' accelerating: if $\vec{a}_{S'/S} \neq 0$, Newton's second law does not hold in its simple form $\vec{F} = m\vec{a}_{A/S'}$ within S'. To restore it, we must introduce a fictitious (pseudo) force: $$\vec{F}_{fict} = -m\,\vec{a}_{S'/S}$$ This is not a real force (no physical agent exerts it), but it is necessary to describe motion in the accelerating frame as if it were inertial. Example: in a lift accelerating upward, you feel heavier (the floor exerts a larger normal force). An external observer (S) sees $N - mg = ma$; inside S', the fictitious force $-m\vec{a}_{S'/S}$ adds to the real weight.

When S' rotates with angular velocity $\vec{\omega}$ relative to S, the unit vectors $\hat{e}_i'$ of S' are themselves rotating: their time derivative is not zero but $\dot{\hat{e}}_i' = \vec{\omega} \times \hat{e}_i'$. The relative derivative theorem (Poisson's formula) relates the time derivative of any vector $\vec{A}$ as seen from S versus from S': $$\left(\frac{d\vec{A}}{dt}\right)_S = \left(\frac{d\vec{A}}{dt}\right)_{S'} + \vec{\omega}\times\vec{A}$$ Interpretation:

• $\left(d\vec{A}/dt\right)_S$ — change of $\vec{A}$ seen from the fixed frame S.
• $\left(d\vec{A}/dt\right)_{S'}$ — change of $\vec{A}$ seen from the rotating frame S'.
• $\vec{\omega}\times\vec{A}$ — correction term due to the rotation of S'. If $\vec{\omega} = 0$ (no rotation), the two derivatives coincide.

Why this term appears: in S', $\vec{A} = \sum A_i' \hat{e}_i'$. Differentiating gives $\sum \dot{A}_i' \hat{e}_i' + \sum A_i' \dot{\hat{e}}_i' = (d\vec{A}/dt)_{S'} + \vec{\omega}\times\vec{A}$. The second term comes from the rotation of the basis vectors. This theorem is the key tool for deriving velocity and acceleration in rotating frames (see §5).

Applying the relative derivative theorem to $\vec{r}_A$ gives the velocity in a rotating-and-translating frame S': $$\vec{v}_A = \vec{v}_{A/S'} + \vec{\omega}\times\vec{r}_{A/S'} + \vec{v}_{O'/S}$$ where $\vec{\omega}\times\vec{r}_{A/S'}$ is the rotational transport term: even if A is stationary in S', it rotates with S'. Differentiating again gives the acceleration. When multiplied by mass, the extra terms become fictitious forces: 1. Centrifugal force $\vec{F}_{cf} = m\omega^2 r_\perp \hat{r}_\perp$

• Points radially outward from the rotation axis.
• Depends only on position ($r_\perp$ = distance from axis), not on relative velocity.
• Causes the equatorial bulge and reduces effective $g$ at the equator by ~0.3%.

2. Coriolis force $\vec{F}_{Cor} = -2m\,\vec{\omega}\times\vec{v}_{A/S'}$

• Depends on relative velocity $\vec{v}_{A/S'}$: a body stationary in S' experiences no Coriolis force.
• Always perpendicular to both $\vec{\omega}$ and $\vec{v}_{A/S'}$ → deflects motion without doing work.
• On Earth: deflects moving objects to the right in the northern hemisphere, to the left in the southern. Explains cyclone rotation and trade wind patterns.

3. Linear transport force (if S' also accelerates linearly): $\vec{F}_{trans} = -m\vec{a}_{O'/S}$. For the special case of S' rotating uniformly about a fixed axis ($\vec{v}_{O'/S}=0$, $\vec{a}_{O'/S}=0$): $$\vec{a}_A = \vec{a}_{A/S'} + 2\vec{\omega}\times\vec{v}_{A/S'} + \vec{\omega}\times(\vec{\omega}\times\vec{r}_{A/S'})$$ where $2\vec{\omega}\times\vec{v}_{A/S'}$ is the Coriolis acceleration and $\vec{\omega}\times(\vec{\omega}\times\vec{r}_{A/S'}) = -\omega^2 r_\perp\hat{r}_\perp$ is the centrifugal acceleration. Both are fictitious — they vanish when viewed from an inertial frame.