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

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    $\begingroup$ A matter of convention. Since this vector product is antisymmetric the two definitions differ only by an overall minus sign. $\endgroup$ – Alpha001 Feb 20 '17 at 8:25
  • $\begingroup$ Because most of us are right handed. So, we're going to define angular momentum such that if a particle moves in the direction of your curved right hand fingers, the angular momentum will point in the direction of your thumb. $\endgroup$ – Count Iblis Feb 20 '17 at 8:29
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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.

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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.

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    $\begingroup$ ... but why choose $\mathbf L$ such that $(\mathbf r,\mathbf p,\mathbf L)$ is a right-handed triad instead of a left-handed one? That's just hiding the core of the question underneath even more terminology. $\endgroup$ – Emilio Pisanty Feb 20 '17 at 8:40
  • $\begingroup$ @EmilioPisanty Because we want our polar coordinates to also be right-handed. Right-handed bases of 3D vector spaces are typically preferred in physics, but again it's just a convention. $\endgroup$ – J.G. Feb 20 '17 at 8:42
  • $\begingroup$ Precisely: because convention. There's no other reason. Insisting on "stuff being right-handed" doesn't work, either: you could equally well demand that $(\mathbf p,\mathbf r,\mathbf L)$ be a right-handed triad, and it would sound just as reasonable. $\endgroup$ – Emilio Pisanty Feb 20 '17 at 8:51
  • $\begingroup$ @EmilioPisanty You could, but the reason we care more about the $\left(\mathbf{r},\,\mathbf{p},\,\mathbf{L}\right)$ order is that we start with position, then we differentiate it, then we take a cross product whose conservation constrains to a plane the first two vectors, whose order is chosen so each of the three vectors introduces one new polar coordinate. It would be an odd convention to take a basis vector parallel to $\mathbf{p}$ for $\dot{r}\ne 0$. $\endgroup$ – J.G. Feb 20 '17 at 9:44
  • $\begingroup$ 1. The OP is not proposing an alternative, just asking why one option is used rather than another. 2. I suspect the math is over the OP's head. May suit others though. $\endgroup$ – sammy gerbil Feb 20 '17 at 18:52

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