Furthermore if the potential is central, U ( x , y , z ) = U ( r ) and so the force is along the radial direction. In this case, the cross product of Newton's second law of motion with the particle's position vector r must vanish since the cross product of two parallel vectors is zero: r × × m r ¨ ¨ = d d t ( r × × m r ˙ ˙ ) − − ( r ˙ ˙ × × m r ˙ ˙ ) = r × × F = 0 but r ˙ ˙ × × m r ˙ ˙ = 0 (cross product of parallel vectors), so d d t ( r × × m r ˙ ˙ ) = 0 The term in parentheses on the left hand side is a constant, label this with L the angular momentum, In particular, in polar coordinates, L = m r v φ φ = r 2 φ φ ˙ ˙ or φ φ ˙ ˙ = L m r 2 Further, r ˙ ˙ 2 = v r 2 + v φ φ 2 = r ˙ ˙ 2 + ( r φ φ ˙ ˙ ) 2 , so the energy equation may be simplified with the angular momentum as E t o t = 1 2 m r ˙ ˙ 2 + 1 2 m r 2 φ φ ˙ ˙ 2 + U ( r ) = 1 2 m r ˙ ˙ 2 + L 2 2 m r 2 + U ( r ) = 1 2 m r ˙ ˙ 2 + U e f f ( r ) This indicates that the angular momentum contributes an effective potential energy U e f f ( r ) = U ( r ) + L 2 2 m r 2 Solve this equation for r ˙ ˙ r ˙ ˙ = d r d t = 2 m E t o t − − U e f f ( r ) which may be converted to the derivative of r with respect to the azimuthal angle φ φ as d r d φ φ = d r / d t d φ φ / d t = r ˙ ˙ φ φ ˙ ˙ = m r 2 L 2 m E t o t − − U e f f ( r ) This is a separable first order differential equation. Integrating and yields the formula φ φ = φ φ 0 + L 2 m ∫ ∫ r d r r 2 E t o t − − U e f f ( r )