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 'p_z`. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.
Solution 1
lx = ypz – zpy
ly = zpx – xpz
lz = xpy –ypx
Linear momentum of the particle,`vecp = p_x hati + p_y hatj + p_z hatk`
Position vector of the particle, `vecr = xhati + yhatj + zhatk`
Angular momentum, `hatl = hatr xx hatp`
=`(xhati + yhatj + zhatk) xx (p_x hati + p_y hatj + p_z hatk)`
`=|(hati,hatj,hatk),(x,y,z), (p_x, p_y,p_z)|`
`l_xhati + l_yhatj + l_z hatk = hati (yp_z - zp_y) - hatj(xp_z - zp_x) + hatk (xp_y - zp_x)`
Comparing the coefficients of `hati, hatj, hatk` we get:
`((l_x = yp_z - zp_y),(l_y = xp_z -zp_x),(l_z = xp_y - yp_x))}...(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:
`((l_x=0),(l_y=0),(l_z=xp_y -yp_x))} `
Therefore, when the particle is confined to move in the x-y plane, the direction of angular momentum is along the z-direction.
Solution 2
We know that angular momentum `vecl` of a particle having position vector `vecr` and momentum `vecp` is given by
`vecl = vecr xx vecp`
But vecr = [xveci + yhatj + zveck], where x, y,z are the component of `vecr and vecp = [p_xveci + p_yhatj + p_zhatk]`
`:. vecl = vecr xx vecp = [x hati + yhatj + zhatk] xx [p_xhati + p_y hatj + p_z hatk]`
or `(l_xhati + l_yhatj + l_zhatk) `= `|(hati ,hatj ,hatk),(x,y,z), (p_x,p_y,p_z)|`
`=(yp_z - zp_y)hati + (zp_x - xp_z)hatj + (xp_y - yp_x) hatk`
From this relation we conclude that
`l_x = yp_z - zp_y, l_y = zp_x - xp_z, l_z = xp_y- yp_x`
if the given particle moves only in the x - y plane then z = 0 and 'p_z = 0' and hence
'vecl = (xp_y - yp_x)hatk` which is onl;y the z- component of hatl
it means that for a particle moving only in the x-y plane, the angular momentum has only the z-component
