Angular Momentum of a Particle

Linear momentum p = mv belongs to the particle alone. Angular momentum does not: it measures how the particle's motion carries it around one chosen point O, so every calculation opens with a declaration, about O. Draw the position vector r from O to the particle and define L = r×p = m r×v. Everything that follows is fixed by that choice. r is in metres, p in kg m s⁻¹, and L in kg m² s⁻¹, which is also J s, the unit of Planck's constant. A particle at rest has L = 0 about every point. A moving particle has L = 0 only about points on its own line of motion, where r is parallel or antiparallel to p. About every other point the value depends on where O was put, which is why the angular momentum of the ball, with no point named, is an incomplete sentence.

The magnitude of a cross product is |L| = r p sin φ, with φ the angle between r and p. Group it two ways. As r⊥ p, it is the momentum times the lever arm r⊥ = r sin φ, the perpendicular distance from O to the line along which p points. As r p⊥, it is the distance times the component of momentum across r, the only part of the motion that swings r round as seen from O; the component along r just changes the distance. Take a 0.50 kg ball 2.0 m from O, moving at 6.0 m s⁻¹ at φ = 30° to r. Then p = 3.0 kg m s⁻¹, the lever arm is 2.0 m × sin 30° = 1.0 m, and |L| = 3.0 kg m² s⁻¹; the sideways reading gives 2.0 m × 1.5 kg m s⁻¹, the same. At φ = 0° or 180° the ball heads straight away from or straight at O, and L = 0. At φ = 90°, r and p are perpendicular and |L| = r p, the largest value possible at that distance.

L is perpendicular to both r and p, so it is perpendicular to the plane they span. To choose between the two perpendiculars, point the fingers of your right hand along r, curl them towards p through the smaller angle, and your thumb gives L. Order matters: p×r = −L. In components, L = (y p_z − z p_y, z pₓ − x p_z, x p_y − y pₓ), where x, y, z are the components of r in a right-handed frame centred on O. For motion in the xy-plane with O in that plane, z = 0 and p_z = 0, so only L_z = x p_y − y pₓ survives and the vector reduces to a signed number. Positive means counterclockwise about O seen from +z, drawn ⊙ out of the page; negative means clockwise, drawn ⊗ into it. A particle at r = (2.0, 1.0) m with p = (−3.0, 4.0) kg m s⁻¹ has L_z = 2.0 × 4.0 − 1.0 × (−3.0) = 11 kg m² s⁻¹: counterclockwise, out of the page.

A free particle moves along a straight line at constant velocity, and nothing about it turns. It still carries angular momentum about any point off that line, and the lever-arm reading shows the value cannot change. As the particle travels, r lengthens or shortens and φ swings from near 180° through 90° towards 0°, but r sin φ is always the perpendicular distance d from O to the line, a fixed length. So |L| = m v d at every instant, and the direction, set by which side of O the line passes, never flips. Put d = 1.5 m and p = 4.0 kg m s⁻¹. With the particle 2.0 m before the foot of the perpendicular, r = 2.5 m and sin φ = 0.60; 3.0 m past it, r = 3.35 m and sin φ = 0.447. Both products give 1.5 m, so |L| = 6.0 kg m² s⁻¹ throughout. This is how a ball thrown past the hinge of a door can set the door turning: it arrives already carrying angular momentum about the hinge.

Shift the origin from O to O′, with R pointing from O to O′. Every position vector becomes r′ = r − R, so L′ = (r − R)×p = L − R×p. The correction R×p is zero only when R is parallel to p: sliding the origin along the line of motion changes nothing, while moving it a distance Δ perpendicular to that line changes L_z by ±pΔ and reverses its sign once O′ crosses the line. Return to the particle at r = (2.0, 1.0) m with p = (−3.0, 4.0) kg m s⁻¹, where L_z = 11 kg m² s⁻¹. About O′ at R = (1.0, −1.0) m, (R×p)_z = 1.0 × 4.0 − (−1.0)(−3.0) = 1.0 kg m² s⁻¹, so L′_z = 10 kg m² s⁻¹, and computing directly with r′ = (1.0, 2.0) m gives the same. About (0.5, 3.0) m, a point on the line of motion, L′_z = 11 − 11 = 0. For several particles the correction is R×P with P their total momentum, so a system with P = 0 has the same L about every point.

For motion on a circle of radius r, measured from the centre, r is radial and v is tangential, so φ = 90° at every instant and |L| = m v r. With v = r ω this becomes L = m r² ω, and the right-hand rule points L along the axis of the circle, up for counterclockwise motion seen from above. A 420 kg satellite in a circular orbit of radius 6.93 × 10⁶ m travels at 7.58 × 10³ m s⁻¹, so about Earth's centre L = 420 × 7.58 × 10³ × 6.93 × 10⁶ = 2.21 × 10¹³ kg m² s⁻¹, constant in size and in direction. That constancy is a statement about the centre. About a point on the orbit itself, the same motion gives an L that is zero as the satellite passes that point and reaches 2mvr half an orbit later. Why the centre is the natural origin for an orbit, with gravity having no lever arm about it, is the subject of the orbital angular momentum topic.