# Why is the angular momentum defined as the cross product of linear momentum and the radius? [duplicate]

I first searched why is Torque defined as a cross product and a lot of people answered that it is defined as a cross product because angular momentum is defined that way and it is only logical that Torque and angular momentum have the same direction.

I know the cross product is used for the direction of angular momentum but why does the direction of angular momentum have to be in the direction of axis of rotation and even if it has to be in the direction of axis of rotation then why does it have to be a cross product and not some other mathematical product.

also why is the magnitude of angular momentum defined as $$||p||*||r||*\sin{\theta}$$

• The last point about the magnitude just comes from the definition of the cross product. Mar 30, 2020 at 11:51

Direction
As for the direction: in three dimensions any other axis would have a projection on the rotation plane, which would pose a problem, since for rotational motion all the directions in this plane should be equivalent. In my opinion, this is largely a matter of convention, although such a definition has lots of mathematical conveniences (e.g., we can add angular momenta).

Magnitude
The magnitude has to do with the part of the force that causes the rotation: it is the projection on the direction perpendicular to the radius, whereas the projection along the radius pulls on the axis.

As I explained in this answer angular momentum is the moment of momentum, just as torque is the moment of force and velocity is the moment of rotation.

And the conservation of angular momentum is really about preserving the geometry of momentum. Momentum has a magnitude and direction and a special line in space called the axis of percussion.

The conservation law for angular momentum (coupled with the conservation law for linear momentum) just states that not only the magnitude and direction of momentum is conserved but also the line in space where moment acts through is also conserved. So not only which direction is momentum point, by where is space it exists.

The idea is that the cross product $$\text{(position)} \times \text{(vector)}$$ only uses the perpendicular components of position for the calculation. The same applies for other quantities as well: $$\matrix{ \text{quantity} & \text{derivation} & \text{description} \\ \hline \text{linear velocity} & \boldsymbol{v} = \boldsymbol{r} \times \boldsymbol{\omega} & \text{moment of rotation} \\ \text{angular momentum} & \boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p} & \text{moment of momentum} \\ \text{torque} & \boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F} & \text{moment of force} \\ } \tag{3}$$ As a mnemonic you can remember that $$\boldsymbol{r} \times \rightarrow \text{(moment of)}$$

In summary, for "moment-of" type of quantities, the cross product provides the perpendicular distance to the special line.

This idea is expanded upon on this post about the nature of cross products in physics.

The direction of this moment (like that of torque) is always perpendicular its special axis, such that you can use the right-hand rule to find its direction.

But the magnitude and direction of the moment vector tells you where in space the axis is. The direction of the percussion axis is only defined from the direction of linear momentum.

Here is a decomposition of geometry of momentum given a coordinate system on the center of mass, the momentum vector $$\boldsymbol{p}$$ and angular momentum about the center of mass $$\boldsymbol{L}$$

Quantity Formula
Mangitude $$p = \| \boldsymbol{p} \|$$
Direction $$\boldsymbol{\hat{e}} = \boldsymbol{p}/p$$
Position $$\boldsymbol{r} = \frac{ \boldsymbol{p} \times \boldsymbol{L}}{p^2}$$
Pitch $$h = \frac{\boldsymbol{p}\cdot\boldsymbol{L}}{p^2}$$