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.
Let p be the momentum of a particle which is defined as,
p = p x i ^ + p y j ^ + p z k ^ (1)
Here, p x is the angular momentum in x-axis, p y is the angular momentum in y-axis and p z is the angular momentum in z-axis.
Let r be the position vector of the particle which is defined as,
r = x i ^ + y j ^ + z k ^ (2)
Here, x is the position vector along x-axis, y is the position vector along y-axis and z is the position vector along z-axis.
Let l be the angular momentum of the particle which is defined as,
l = l x i ^ + l y j ^ + l z k ^
Here, l x is the angular momentum along x-axis, l y is the angular momentum along y-axis and l z is the angular momentum along z-axis.
The relation of l with p and r is,
l = r × p
From equation (1) and (2),
l = ( x i ^ + y j ^ + z k ^ ) × ( p x i ^ + p y j ^ + p z k ^ )
The above vector product can be solved as,
l = | i ^ j ^ k ^ x y z p x p y p z | l x i ^ + l y j ^ + l z k ^ = i ^ ( y p z − z p y ) − j ^ ( x p z − z p x ) + k ^ ( x p y − y p x )
Hence, from the above expressions,
l x = ( y p z − z p y ) l y = − ( x p z − z p x ) l z = ( x p y − y p x ) (3)
As the particle moves only in x-y plane, hence z component of both the positions and linear momentum vectors will be 0 .
Substitute z = 0 and p z = 0 in equation (3),
l x = 0 l y = 0 l z = ( x p y − y p x )
Thus, the above expression shows that if the particle moves in x-y plane, then the angular momentum has only z component.
Let
Here,
Let
Here,
Let
Here,
The relation of
From equation (1) and (2),
The above vector product can be solved as,
Hence, from the above expressions,
As the particle moves only in x-y plane, hence
Substitute
Thus, the above expression shows that if the particle moves in x-y plane, then the angular momentum has only
