Question

Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px , py and pz . Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.

Answer 1

lx = ypz – zpy
ly = zpx – xpz
lz = xpy –ypx
Linear momentum of the particle, p→ =$\vec p =p_x\^i+p_y\^j+p_z\^k$
Position vector of the particle, $\vec r=x\^i+y\^j+z\^k$
Angular momentum, $\vec{l} =\vec{r} ×\vec{p}$

$= (x \^i + y \^j + z \^k) ×(p_x\^i+p_y\,\^J+p_z\^k)$
$\begin{bmatrix} \^i& \^j& \^k\\\ x& y& z\\\ p_x& p_y& p_z\end{bmatrix}$

$l_x\^i+\^j+l_z\^k=\^i(yp_z-zp_y)-\^j(xp_z-zp_x)+\^k(xp_y-zp_x)$

Comparing the coefficients of$\^i,\^j$, and $\^k$, we get :
\left.\begin{matrix} l_x& =yp_z& =zp_y\\\ l_y& =xp_z& -zp_x\\\ l_z& =xp_y& yp_x\end{matrix}\right\\} $…..(i)$

The particle moves in the x-y plane. Hence, the z-component of the position vector and linear momentum vector becomes zero, i.e.,
z = pz = 0
Thus, equation (i) reduces to :

\left.\begin{matrix} l_x=0& & \\\ l_Y=0& & \\\ l_z=xp_y-yp_x& & \end{matrix}\right\\}

Therefore, when the particle is confined to move in the x - y plane, the direction of angular momentum is along the z-direction.