Vectors and the Laws of Physics — Question 5

Question 5

Which combinations are independent of axis orientation?

A fixed planar vector has components $(3,4)$ m. Consider $P=A_x^2+A_y^2$, $Q=A_x+A_y$, and $T=A_x^2-A_y^2$. Rotate the axes counterclockwise by $90^\circ$, so the new components are $(4,-3)$ m.

1. Evaluate P, Q and T in both frames with their units.

2. Which expression is guaranteed to be invariant under every planar rotation? Justify beyond this one example.

3. Explain why a component being a signed number does not make it a rotationally invariant physical scalar.

Question 5 - Solution

Which combinations are independent of axis orientation?

(a) Calculate before classifying.

$$(P,Q,T)=(25\,\mathrm{m^2},7\,\mathrm m,-7\,\mathrm{m^2}),\quad(P',Q',T')=(25\,\mathrm{m^2},1\,\mathrm m,7\,\mathrm{m^2})$$ Originally $P=25\ \mathrm{m^2}$, $Q=7$ m, and $T=-7\ \mathrm{m^2}$. After the turn they are $P\prime=25\ \mathrm{m^2}$, $Q\prime=1$ m, and $T\prime=7\ \mathrm{m^2}$. Q and T fail even this simple rotation test.

(b) Use the orthogonality of the transformation.

$$\boxed{P'=(A_xc+A_ys)^2+(-A_xs+A_yc)^2=(c^2+s^2)(A_x^2+A_y^2)=P}$$ Write $c=\cos\phi$ and $s=\sin\phi$. For a general angle, substituting $A_{x\prime}=A_x\cos\phi+A_y\sin\phi$ and $A_{y\prime}=-A_x\sin\phi+A_y\cos\phi$ into P cancels the mixed terms. Since $\sin^2\phi+\cos^2\phi=1$, $P\prime=A_x^2+A_y^2$. P is the squared physical magnitude.

(c) Specify the meaning of scalar.

$$A_x'=A_x\cos\phi+A_y\sin\phi\ne A_x\text{ in general},\qquad A'=A$$ A component is a scalar coefficient in a chosen basis, but generally changes when that basis rotates. A rotationally invariant physical scalar, such as the vector magnitude, has the same value for all such orientations. The word scalar can refer to these different roles; the transformation behavior resolves the distinction.