Vectors and the Laws of Physics — Question 2

Question 2

A right-angle frame change is a sign test

A fixed vector has components $(-4,7)$ m. Two observers use axes rotated respectively $90^\circ$ and $180^\circ$ counterclockwise from the original axes. For perpendicular axes rotated counterclockwise by $\phi$ about the same origin, a fixed vector transforms as $A_{x\prime}=A_x\cos\phi+A_y\sin\phi$ and $A_{y\prime}=-A_x\sin\phi+A_y\cos\phi$.

1. Find its components for each observer.

2. An algorithm uses $(A_{x\prime},A_{y\prime})=(-A_y,A_x)$ for a $90^\circ$ axis rotation. Diagnose the error.

3. Apply the correct $90^\circ$ change four times and interpret the result.

Question 2 - Solution

A right-angle frame change is a sign test

Here $Q_\phi$ transforms components when axes turn counterclockwise by $\phi$; $R_\phi=Q_{-\phi}$ rotates the arrow in fixed axes. An unsubscripted $Q$ denotes the same component transformation, and $I$ is the identity.

(a) Use the axis directions directly.

$$\boxed{\vec A'_{90}=(A_y,-A_x)=(7,4)\,\mathrm m,\quad\vec A'_{180}=(-A_x,-A_y)=(4,-7)\,\mathrm m}$$ At $90^\circ$, positive $x\prime$ is old positive $y$ and positive $y\prime$ is old negative $x$. Thus the pair is $(7,4)$ m. At $180^\circ$, both axes reverse, giving $(4,-7)$ m. Both magnitudes remain $\sqrt{65}$ m.

(b) Distinguish two opposite operations.

$$Q_{90}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}y\\-x\end{pmatrix},\qquad Q_{-90}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}-y\\x\end{pmatrix}$$ The formula $(-A_y,A_x)$ rotates the physical arrow counterclockwise in fixed axes, or rotates the axes clockwise for a fixed arrow. It is the wrong sense for the stated counterclockwise axis change. The correct passive mapping is $(A_y,-A_x)$.

(c) Check a full turn.

$$(-4,7)\to(7,4)\to(4,-7)\to(-7,-4)\to(-4,7),\quad Q_{90}^4=I$$ The successive pairs are $(-4,7)\to(7,4)\to(4,-7)\to(-7,-4)\to(-4,7)$ m. Four quarter turns return the basis to its initial orientation and therefore restore the original components. The arrow itself has remained fixed throughout.