Why is angular momentum defined as $\mathbf r \times \mathbf p$ and not as $\mathbf p \times \mathbf r$?

Question

Why is angular momentum defined as $ \vec{r} \times \vec p $ and not as $ \vec p \times \vec r$ ?

Answer 1

Because you need to choose a convention: it doesn't matter which one, but you need to choose one of the two signs, or you're just unable to move. Ultimately, this is a choice of handedness: in the usual convention, if you use your right-hand fingers to curl with the rotational motion, your thumb points along $\vec L$. There's no specific reason to choose the right hand, and you could equally well use the left hand, but we need to choose one convention so we all speak the same language.

Answer 2

The alternative you propose would multiply all angular momenta by $-1$. There's no physical reason we can't do that.

One advantage to the usual definition is that $\mathbf{r},\,\mathbf{p},\,\mathbf{L}$ form a right-handed coordinate system for radial forces, since then $\mathbf{L}$ is conserved and the motion is planar, viz.$$\mathbf{r}=r\hat{\mathbf{r}},\,\mathbf{p}=m\dot{r}\hat{\mathbf{r}}+r\dot{\theta}\hat{\boldsymbol{\theta}},\,\mathbf{L}=mr^2\dot{\theta}\hat{\mathbf{k}}.$$

In general $\dot{\mathbf{L}}=\mathbf{r}\times\mathbf{F}+\mathbf{v}\times\mathbf{p}$; the last term vanishes since it's a cross-product of parallel vectors, so $\dot{\mathbf{L}}=\mathbf{r}\times\mathbf{F}$ is the usual definition of a moment. Note that if we adopted your proposed alternative convention it would also make sense to "reverse" the moment definition, so that the total moment still gives the rate of change of angular momentum.