Relative Motion: Vector Equations

1. The Importance of Reference Frames

Before we begin deriving equations, it's crucial to be precise about our language. In physics, any measurement of motion is meaningless without context.

To summarize the key ideas:

• A coordinate system is a mathematical construct for assigning numbers to points in space (e.g., Cartesian axes $\hat{i},\hat{j},\hat{k}$). You can change your coordinate system (e.g., from Cartesian to polar) without changing the physics.
• A reference frame is a physical object or environment relative to which motion is measured. An observer on a train is in a different reference frame from an observer on the ground. To relate their observations, we need the laws of physics.

For our derivations, we will consider two frames:

• Frame S: A "fixed" or "absolute" frame. In Newtonian mechanics, this is an inertial frame (one that is not accelerating). Let its origin be O.
• Frame S': A "moving" or "relative" frame. Its origin, O', can be moving, and its axes can be rotating relative to S.

2. Relative Position

Let's consider a particle P. Its position can be described by vectors from the origins of both frames.

Let's define our vectors:

• ${\vec{r}}_P$: The position vector of particle P as measured in the fixed frame S (vector from O to P).
• ${\vec{r}}_{S^{\prime }}$: The position vector of the origin of the moving frame, O', as measured in the fixed frame S (vector from O to O').
• ${\vec{r}}_{P/S^{\prime }}$: The position vector of particle P as measured in the moving frame S' (vector from O' to P).

From the principle of vector addition, it is clear that:
${\vec{r}}_P={\vec{r}}_{S^{\prime }}+{\vec{r}}_{P/S^{\prime }}$
This simple equation is the foundation for everything that follows. It connects the position vectors in the two frames.

3. The General Velocity Transformation

To find the relationship between velocities, we must differentiate the position equation with respect to time. The derivative must be taken by an observer in the fixed frame S, as we want to find ${\vec{v}}_P=d{\vec{r}}_P/dt$.

$\frac{d{\vec{r}}_P}{dt}=\frac{d{\vec{r}}_{S^{\prime }}}{dt}+\frac{d{\vec{r}}_{P/S^{\prime }}}{dt}$
Let's analyze each term:

• $\frac{d{\vec{r}}_P}{dt}={\vec{v}}_P$, the absolute velocity of P in frame S.
• $\frac{d{\vec{r}}_{S^{\prime }}}{dt}={\vec{v}}_{S^{\prime }}$, the velocity of the origin of the moving frame S' in frame S.

The third term, $\frac{d{\vec{r}}_{P/S^{\prime }}}{dt}$, is the most subtle. We are taking the time derivative in the fixed frame S of a vector, ${\vec{r}}_{P/S^{\prime }}$, which is most naturally described by its components in the moving frame S'. If frame S' is rotating with an angular velocity $\vec{\Omega }$ relative to S, we must account for the rotation of the basis vectors of S'.

This is precisely where the Transport Theorem (or Coriolis Equation), which you've seen in the context of rotating rigid bodies, becomes essential. For any vector $\vec{A}$, its time derivative in frame S is related to its time derivative in frame S' by:
${\left(\frac{d\vec{A}}{dt}\right)}_S={\left(\frac{d\vec{A}}{dt}\right)}_{S^{\prime }}+\vec{\Omega }\times \vec{A}$

Applying this theorem to the vector ${\vec{r}}_{P/S^{\prime }}$:
$\frac{d{\vec{r}}_{P/S^{\prime }}}{dt}={\left(\frac{d{\vec{r}}_{P/S^{\prime }}}{dt}\right)}_{S^{\prime }}+\vec{\Omega }\times {\vec{r}}_{P/S^{\prime }}$

The term ${\left(\frac{d{\vec{r}}_{P/S^{\prime }}}{dt}\right)}_{S^{\prime }}$ is the time derivative of the relative position vector as measured within the moving frame S'. This is, by definition, the relative velocity of P in S', which we will denote as ${\vec{v}}_{P/S^{\prime }}$.

Substituting this back into our main velocity equation gives the general velocity transformation:
${\vec{v}}_P={\vec{v}}_{S^{\prime }}+{\vec{v}}_{P/S^{\prime }}+\vec{\Omega }\times {\vec{r}}_{P/S^{\prime }}$

This equation states that the absolute velocity of P is the sum of three components:

1. ${\vec{v}}_{S^{\prime }}$: The velocity of the moving frame's origin ("drag" velocity).
2. ${\vec{v}}_{P/S^{\prime }}$: The velocity of P relative to the moving frame.
3. $\vec{\Omega }\times {\vec{r}}_{P/S^{\prime }}$: The velocity arising from the rotation of the moving frame.

4. The General Acceleration Transformation

The final step is to derive the transformation for acceleration by differentiating the general velocity equation with respect to time in frame S. This is a complex derivation that requires careful use of the product rule for cross products and repeated application of the Transport Theorem.

After performing the full differentiation as shown in the resource, we arrive at the general acceleration transformation:
${\vec{a}}_P={\vec{a}}_{S^{\prime }}+{\vec{a}}_{P/S^{\prime }}+{\vec{\alpha }}_{S^{\prime }}\times {\vec{r}}_{P/S^{\prime }}+\vec{\Omega }\times (\vec{\Omega }\times {\vec{r}}_{P/S^{\prime }})+2(\vec{\Omega }\times {\vec{v}}_{P/S^{\prime }})$

Let's dissect this formidable equation, as understanding each term's physical meaning is crucial.

• ${\vec{a}}_P$: The absolute acceleration of P in the fixed frame S.
• ${\vec{a}}_{S^{\prime }}$: The acceleration of the origin of the moving frame S'.
• ${\vec{a}}_{P/S^{\prime }}$: The relative acceleration of P as measured by an observer in the moving frame S'.
• ${\vec{\alpha }}_{S^{\prime }}\times {\vec{r}}_{P/S^{\prime }}$: The tangential acceleration component, arising from the angular acceleration (${\vec{\alpha }}_{S^{\prime }}=d\vec{\Omega }/dt$) of the frame S'.
• $\vec{\Omega }\times (\vec{\Omega }\times {\vec{r}}_{P/S^{\prime }})$: The centripetal acceleration component, arising from the rotation of frame S' at angular velocity $\vec{\Omega }$.
• $2(\vec{\Omega }\times {\vec{v}}_{P/S^{\prime }})$: The Coriolis acceleration. This is perhaps the most fascinating term. It arises only when a particle has a velocity relative to a rotating frame. It is responsible for phenomena like the circulation of hurricanes and the observed deflection of long-range projectiles on Earth.

5. Special Case: Inertial Frames

The power of this general framework is that it simplifies correctly for less complex scenarios. Consider the case of two frames moving at a constant velocity with respect to each other (the Galilean transformation).

In this case, the moving frame S' is not rotating and its origin is not accelerating.

• $\vec{\Omega }=\vec{0}$
• ${\vec{\alpha }}_{S^{\prime }}=\vec{0}$
• ${\vec{a}}_{S^{\prime }}=\vec{0}$

Conclusion

Today, we have rigorously derived the complete kinematic relationships for observing motion from an arbitrary reference frame. This is a significant theoretical milestone in our study of kinematics.

Key Takeaways:

• The position vector transformation is a simple vector sum: ${\vec{r}}_P={\vec{r}}_{S^{\prime }}+{\vec{r}}_{P/S^{\prime }}$.
• The general velocity and acceleration transformations are derived by successive time differentiations, requiring the use of the Transport Theorem to handle the rotating basis vectors of the moving frame.
• The general velocity is: ${\vec{v}}_P={\vec{v}}_{S^{\prime }}+{\vec{v}}_{P/S^{\prime }}+\vec{\Omega }\times {\vec{r}}_{P/S^{\prime }}$.
• The general acceleration contains five distinct physical terms: ${\vec{a}}_P={\vec{a}}_{S^{\prime }}+{\vec{a}}_{P/S^{\prime }}+{\vec{\alpha }}_{S^{\prime }}\times {\vec{r}}_{P/S^{\prime }}+\vec{\Omega }\times (\vec{\Omega }\times {\vec{r}}_{P/S^{\prime }})+2(\vec{\Omega }\times {\vec{v}}_{P/S^{\prime }})$.
• Understanding the physical origin of each term (e.g., Centripetal, Coriolis) is as important as the mathematical derivation itself.

Preview of the Next Lesson:
With this powerful theoretical machinery now established, we can begin to apply it. In the next lesson, we will focus on solving practical problems. We will start with 1D relative motion scenarios, such as chase problems, and see how the general vector equations simplify and provide a systematic way to find solutions.