I am reading An Introduction to Mechanics by Kleppner and Kolenkow (2014). On page 241 is the definition of the angular momentum:

Here is the formal definition of the angular momentum $\vec{L}$ of a particle that has momentum $\vec{p}$ and is at position $\vec{r}$ with respect to a given coordinate system: $$\vec{L}=\vec{r} \times \vec{p}$$

In the book there is no explanation why this formula should be true. From this equation the formulas for torque and moment of inertia are derived.

My question is: Why is the formula above correct? Why isn't the formula for angular momentum something completely different, like $\vec{L}=\sqrt{(\vec{r} \times \vec{p})2\pi M}$? Is the formula for angular momentum just a mere definition and if not, how to derive it? How did people come across that particular formula?

  • $\begingroup$ Hi @user50224: I removed your last question (v1) which was primarily opinion-based. $\endgroup$ – Qmechanic Sep 21 '14 at 15:12
  • $\begingroup$ Putting aside the history of it, angular momentum is defined to be $r\times p$. Why this particular form and not the one you propose? Primarily because it is a conserved quantity. $\endgroup$ – lemon Sep 21 '14 at 15:29
  • $\begingroup$ So, it is only an arbitrary definition? $\endgroup$ – user50224 Sep 21 '14 at 15:32
  • 2
    $\begingroup$ @user50224 that depends what you mean by 'arbitrary'. You could redefine it as $L=2r\times p$, for instance, and as long as you adjust all of the other maths accordingly then you'll get the same answers. But it's the simplest expression containing both $r$ and $p$ that is conserved - I'm reluctant to call that 'arbitrary'. $\endgroup$ – lemon Sep 21 '14 at 15:34
  • $\begingroup$ But why should the quantity "angular momentum" contain both $r$ and $p$? $\endgroup$ – user50224 Sep 21 '14 at 15:37

You know Newton's second law:

$$\vec F = m \vec a~.$$

Now you multiply both sides by "$\vec r \times~$", then you get

$$\vec r \times \vec F = \vec r \times m \vec a = \frac{d}{dt} (\vec r \times \vec p)~.\tag{1}$$

(If you carry out the time derivative, the first term from the product rule is $\vec v\times\vec v$, which of course is $0$.)

This is very similar to the other form of Newton's second law:

$$\vec F = \frac{d}{dt} \vec p~,$$

which can be read as: a force gives us the rate of change of momentum.

Likewise, equation (1) reads: a force applied about an axis produces a rate of change of the momentum about the same axis. This means we identify the left hand side of (1) with the torque $\vec \tau$ and the the quantity in the brackets of the right hand side with the angular momentum $\vec L=\vec r \times m\vec v$, so eqn. (1) can be written as

$$\vec \tau=\frac{d}{dt}\vec L~.$$

So in a sense it's in comparison with Newton's second law that you define this quantity as angular momentum.


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