Transverse and Longitudinal Waves — Question 6

Question 6

A circular local orbit can still be purely transverse

A wave travels along $z$. At a fixed material label, its displacement is $\mathbf u=A\hat{\mathbf i}\cos\Phi+A\hat{\mathbf j}\sin\Phi$, with phase advancing uniformly in time.

1. Describe the local path and classify the wave displacement.

2. What changes if the second amplitude becomes $A/2$? Does the wave acquire a longitudinal component?

Question 6 - Solution

A circular local orbit can still be purely transverse

Method and mathematical working

Eliminate the common phase to identify the particle path. Independently test the component along the propagation axis; the shape of the path does not replace that test.$$\begin{aligned}\left(\frac{u_x}{A}\right)^2+\left(\frac{u_y}{B}\right)^2&=1,\qquad u_z=0.\\[4pt]u_x&=A\cos\Phi,\quad u_y=A\sin\Phi,\quad u_z=0,\\u_x^2+u_y^2&=A^2;\qquad B=A/2\Rightarrow\frac{u_x^2}{A^2}+\frac{u_y^2}{(A/2)^2}=1.\end{aligned}$$

(a) Calculation and conclusion

$u_x^2+u_y^2=A^2$, so the local path is a circle in the plane perpendicular to propagation. Since $u_z=0$, the displacement is purely transverse, despite having two changing components.

(b) Interpretation and limits

The path becomes an ellipse, $u_x^2/A^2+u_y^2/(A/2)^2=1$. It remains transverse because both components are perpendicular to $z$. Circular or elliptical local motion does not imply displacement along propagation.

Physical check

Changing $B/A$ changes the ellipse shape, but never introduces a $z$ component. The circular limit is $B=A$ and the linearly polarized limit is $B\to0$.