Editing Torque (section)

==== Proof of the equivalence of definitions ====
The definition of angular momentum for a single point particle is:
<math display="block">\mathbf{L} = \mathbf{r} \times \mathbf{p}</math>
where '''p''' is the particle's [[linear momentum]] and '''r''' is the position vector from the origin. The time-derivative of this is:

<math display="block">\frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \mathbf{r} \times \frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t} + \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} \times \mathbf{p}.</math>

This result can easily be proven by splitting the vectors into components and applying the [[product rule]]. But because the rate of change of linear momentum is force <math display="inline">\mathbf{F}</math> and the rate of change of position is velocity <math display="inline">\mathbf{v}</math>,

<math display="block">\frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \mathbf{r} \times \mathbf{F} + \mathbf{v} \times \mathbf{p} </math>

The cross product of momentum <math>\mathbf{p}</math> with its associated velocity <math>\mathbf{v}</math> is zero because velocity and momentum are parallel, so the second term vanishes. Therefore, torque on a particle is ''equal'' to the [[Derivative#Notation for differentiation|first derivative]] of its angular momentum with respect to time. If multiple forces are applied, according [[Newton's second law]] it follows that<math display="block">\frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} =  \boldsymbol{\tau}_{\mathrm{net}}.</math>

This is a general proof for point particles, but it can be generalized to a system of point particles by applying the above proof to each of the point particles and then summing over all the point particles. Similarly, the proof can be generalized to a continuous mass by applying the above proof to each point within the mass, and then [[Integral calculus|integrating]] over the entire mass.