Problem 19 - Classical Mechanics

Question

The co-ordinates \((x, y, z)\) of a particle with respect to a uniformly rotating frame may be related to those with respect to a fixed inertial frame, \(\left(x^{*}, y^{*}, z^{*}\right)\), by the transformation $$ \left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{ccc} \cos \omega t & \sin \omega t & 0 \\ -\sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} x^{*} \\ y^{*} \\ z^{*} \end{array}\right] $$ (Here, we use matrix notation: this stands for three separate equations, \(x=\cos \omega t \cdot x^{*}+\sin \omega t \cdot y^{*}\) etc.) Write down the inverse relation giving \(\left(x^{*}, y^{*}, z^{*}\right)\) in terms of \((x, y, z)\). By differentiating with respect to \(t\), rederive the relation (5.15) between \(\mathrm{d}^{2} \boldsymbol{r} / \mathrm{d} t^{2}\) and \(\ddot{\boldsymbol{r}}\). [Hint: Note that \(\ddot{\boldsymbol{r}}=(\ddot{x}, \ddot{y}, \ddot{z})\), while \(\mathrm{d}^{2} \boldsymbol{r} / \mathrm{d} t^{2}\) is the vector obtained by applying the above transformation \(\left.\operatorname{to}\left(\ddot{x}^{*}, \ddot{y}^{*}, \ddot{z}^{*}\right) .\right]\)

Step-by-Step Solution

Answer

Question: Find the inverse relation for the given transformation matrix and rederive the relation between \(\frac{d^2 \textbf{r}}{dt^2}\) and \(\ddot{\textbf{r}}\). Answer: The inverse relation can be found by taking the transpose of the given orthogonal matrix: $$ \left[\begin{array}{c} x^* \\ y^* \\ z^* \end{array}\right]=\left[\begin{array}{ccc} \cos \omega t & -\sin \omega t & 0 \\ \sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right] $$ After differentiating the matrix elements and applying the transformation, the rederived relation between \(\frac{d^2 \textbf{r}}{dt^2}\) and \(\ddot{\textbf{r}}\) is: $$ \frac{d^2 \textbf{r}}{dt^2} = \left[\begin{array}{ccc} \cos \omega t & -\sin \omega t & 0 \\ \sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right] \ddot{\textbf{r}} $$

Step 1: Find the Inverse Relation

To find the inverse relation, we need to find the inverse of the transformation matrix. The given transformation matrix is an orthogonal matrix, so its inverse is equal to its transpose. Therefore, the inverse relation can be obtained by transposing the given matrix: $$ \left[\begin{array}{c} x^* \\ y^* \\ z^* \end{array}\right]=\left[\begin{array}{ccc} \cos \omega t & -\sin \omega t & 0 \\ \sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right] $$

Step 2: Differentiate the Inverse Relation

Now, we differentiate the inverse relation with respect to time \(t\): $$ \frac{d}{dt}\left[\begin{array}{c} x^* \\ y^* \\ z^* \end{array}\right]=\frac{d}{dt}\left[\begin{array}{ccc} \cos \omega t & -\sin \omega t & 0 \\ \sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right] + \left[\begin{array}{ccc} \cos \omega t & -\sin \omega t & 0 \\ \sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right]\frac{d}{dt}\left[\begin{array}{c} x \\ y \\ z \end{array}\right] $$

Step 3: Differentiate the Matrix Elements

To find \(\frac{\text{d}^2 \textbf{r}}{\text{d}t^2}\) we need to differentiate both the elements of the inverse matrix and the coordinate vector \((x, y, z)\): $$ \frac{d^2}{dt^2}\left[\begin{array}{c} x^* \\ y^* \\ z^* \end{array}\right]=\frac{d^2}{dt^2}\left[\begin{array}{ccc} \cos \omega t & -\sin \omega t & 0 \\ \sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right] + 2\frac{d}{dt}\left[\begin{array}{ccc} \cos \omega t & -\sin \omega t & 0 \\ \sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right]\frac{d}{dt}\left[\begin{array}{c} x \\ y \\ z \end{array}\right] + \left[\begin{array}{ccc} \cos \omega t & -\sin \omega t & 0 \\ \sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right]\frac{d^2}{dt^2}\left[\begin{array}{c} x \\ y \\ z \end{array}\right] $$

Step 4: Derive the Relation between \(\frac{\text{d}^2 \textbf{r}}{\text{d}t^2}\) and \(\ddot{\textbf{r}}\)

After differentiating the matrix elements, we need to apply the transformation to \((\ddot{x^*}, \ddot{y^*}, \ddot{z^*})\) and rederive the relation: $$ \frac{d^2 \textbf{r}}{dt^2} = \left[\begin{array}{ccc} \cos \omega t & -\sin \omega t & 0 \\ \sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} \ddot{x^*} \\ \ddot{y^*} \\ \ddot{z^*} \end{array}\right] $$ Now, we apply the transformation to \((\ddot{x^*}, \ddot{y^*}, \ddot{z^*})\) and substitute \(\ddot{\textbf{r}} = (\ddot{x}, \ddot{y}, \ddot{z})\): $$ \frac{d^2 \textbf{r}}{dt^2} = \left[\begin{array}{ccc} \cos \omega t & -\sin \omega t & 0 \\ \sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right] \ddot{\textbf{r}} $$ This is the rederived relation between \(\frac{d^2 \textbf{r}}{dt^2}\) and \(\ddot{\textbf{r}}\), as we were asked to find.

Key Concepts

Coordinate Transformation

Coordinate transformation is crucial when dealing with different reference frames.
It allows us to switch the viewpoint of measured positions, velocities, and accelerations between moving reference frames and fixed ones. In our exercise, we are dealing with two types of frames: a rotating frame and an inertial frame.
For a coordinate transformation between a non-inertial (rotating) frame and an inertial frame, a matrix is used to express the relationship between the two sets of coordinates. This involves using trigonometric functions of time \( (\cos \omega t, \sin \omega t) \) applied to the components of the coordinate vectors.

The importance of this transformation lies in understanding the physical laws in different frames. Each frame may interpret the same physical event differently. It's like translating a sentence from one language to another. The information should remain the same, but the representation looks different. This is achieved through the transformation matrix, which for the rotating and inertial frame of this exercise, was found through orthogonal matrix properties.

Inertial Frame

An inertial frame is a reference frame in which bodies not subjected to any force move in straight lines with constant velocities.
It is a critical concept in physics, acting as a standard to explore dynamics and mechanics. In this exercise, the inertial frame is the fixed frame with coordinates \((x^*, y^*, z^*)\). It holds the laws of physics in their simplest forms.

In contrast to a rotating frame, inertial frames are those where Newton's laws of motion apply without the need for additional, fictitious forces. Developing a proper understanding of how things appear in an inertial frame can help analyze how they will appear in more complex, non-inertial frames like a rotating one. Understanding these frames simplifies the calculation process, making complex problems less intimidating and more approachable.
Distinguishing between these frames is vital when solving physics problems, especially those involving rotations. When dealing with forces and accelerations, interpreting these in an inertial frame can give a clearer picture of the dynamics involved.

Time Derivative

The concept of the time derivative is about understanding how quantities change over time.
In physics, it's essential as it provides insights into the rates at which things shift and evolve. Differentiating a quantity with respect to time will give you its rate of change — its "derivative."

In our exercise context, differentiating the coordinates with respect to time gives velocities, and further differentiating those gives accelerations. We focus here primarily on acceleration — the derivative of velocity with respect to time. Using time derivatives with a coordinate transformation lets us understand how these rates of change differ between frames. For example, by applying the transformation to the accelerations, as seen in the matrix operations, you can convert how accelerations appear in the rotating frame back into the inertial frame.
Mathematically, differentiating such transformations involves product rules and chain rules of calculus, which can get intricate. But breaking it down, the time derivative operation offers a powerful look into motion and its deeper mechanics when parsed between frames. This enables us to apply Newton's laws effectively across various references.