If the angular momentum of a particle is conserved and it is also 0, then is it true that the particle moves along a line? If so, how can we derive the equation for the trajectory from both the above fact and knowledge of initial position?
(This question was inspired by the fact that if angular momentum is conserved and not 0 then a particle always moves on the same plane, but nothing is said if it is 0).
As always, any hint or comment is highly appreciated!
Best Answer
Yes, if the angular momentum of a particle is conserved and it is zero, the particle must move along a straight line.
Indeed, by using the triple product identity, we have $$ {\bf r} \times ( {\bf r} \times {\bf v} ) = ( {\bf r} \cdot {\bf v} ) {\bf r} - r^2 {\bf v}. $$ Therefore, if the angular momentum is zero, we must have at every time $t$ $$ ( {\bf r} \cdot {\bf v} ) {\bf r} = r^2 {\bf v} $$ which means that velocity always has the direction of the initial position vector, i.e., the motion is always along a line.