What's the definition of angular momentum on a two-body system?

Question

If some body it's rotating about the origin $O$ with constant speed, we can define the angular momentum as

$$\vec{L}=\vec{r}\times\vec{p}\implies\frac{d\vec{L}}{dt}=0$$ Being $\vec{r}$ the position vector and $\vec{p}$ the momentum vector.

And it feels natural. Then I wonder if I select another origin that isn't the Center of rotation then $\vec{L}$ will change or not.

$$\vec{r}=\vec{r}_C+\vec{R} \implies \vec{L}=(\vec{r}_C+\vec{R})\times\vec{p}\implies\frac{d\vec{L}}{dt}=\vec{r}_C\times\frac{d\vec{p}}{dt}\neq0$$

Being $\vec{r}$ the position vector, $\vec{r}_C$ the vector from the origin to the center of rotation and $\vec{R}$ the vector from the center of rotation to the body.

With that result, if I measure the angular momentum outside of the center of rotation, it'll change through time, and that's physically strange. Obviously, if we can choose an origin, it's convenient to use the Center of rotation.

But, what happen if we have a system of two or more bodies rotating (like the image)

What choice of Origin it's the best? Or is it necessary to measure the Angular momentum of each body relative to his Center of rotation?

Answer 1

The angular momentum $\vec{L}$ and torque $\vec{\tau}$ are both dependent on the choice of the origin. But the relation between them is not dependent on the choice of coordinate. Angular momentum is not neccesary a constant in time, its rate of change is equal to the torque.

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

Let see how this relation is preserved in change origin:

In frame A:

$$ \vec{L} = \vec{r} \times \vec{p};\\ \vec{\tau} = \vec{r} \times \vec{F};\\ $$

The rate change ot angular momentum:

$$ \frac{d\vec{L}}{dt} = \frac{d}{dt} \left( \vec{r} \times \vec{p} \right) = \vec{v}\times\vec{p} +\vec{r}\times \frac{d\vec{p}}{dt} = \vec{r} \times \vec{F} = \vec{\tau}. $$

Say that we move the origin to $\vec{R}$ (a constant vector), such that $\vec{r} \to \vec{r}' = \vec{r} + \vec{R}$. Both $\vec{L}$ and $\vec{\tau}$ are changed in accordance with the movement of the origin:

In frame B:

$$ \vec{L}_B = \left(\vec{r} + \vec{R} \right) \times \vec{p} \ne \vec{L};\\ \vec{\tau}_B = \left( \vec{r} + \vec{R} \right) \times \vec{F} \ne \vec{\tau};\\ $$

But the rate change of angular momentum relation to the torque is the same:

$$ \frac{d\vec{L}_B}{dt} = \frac{d}{dt} \left\{ (\vec{r} + \vec{R}) \times \vec{p} \right\} \\ = \vec{v}\times\vec{p} + (\vec{r} +\vec{R} )\times \frac{d\vec{p}}{dt} = (\vec{r} +\vec{R} ) \times \vec{F} = \vec{\tau}_B. $$

Since $\vec{R}$ is a constant vector, $\frac{d\vec{R}}{dt} = 0$.

Fro many bodies, $\vec{L}$ and $\vec{\tau}$ are both a vector sum of each individual angular momentum or torque. And the relation between rate change of total angular momentum and total torque is the same.

Answer 2

So you have two objects each with momentum $ \boldsymbol{p}_1 = m_1 \boldsymbol{v}_1 $ and $\boldsymbol{p}_2 = m_2 \boldsymbol{v}_2$. Angular momentum is defined from the body positions as $\boldsymbol{L}_1 = \boldsymbol{r}_1 \times \boldsymbol{p}_1$ and $\boldsymbol{L}_2 = \boldsymbol{r}_2 \times \boldsymbol{p}_2 $, all defined from a common origin.

Together you get the total momentum as the sum of the individual parts

$$ \begin{aligned} \boldsymbol{p} & = \boldsymbol{p}_1 + \boldsymbol{p}_2 & \boldsymbol{L} & = \boldsymbol{L}_1 + \boldsymbol{L}_2 \end{aligned}$$

Now it is unclear what the kinematics of the two bodies are, which leads to further simplifications. For example if the two bodies are in orbit around each other, then they both rotate about a common bary-center defined by $$\boldsymbol{r}_C = \frac{ m_1 \boldsymbol{r}_1 + m_2 \boldsymbol{r}_2 }{m_1 +m_2} $$

Now decompose each position as $\boldsymbol{r}_1 = \boldsymbol{r}_C + \boldsymbol{d}_1$ and $\boldsymbol{r}_2 = \boldsymbol{r}_C + \boldsymbol{d}_2$ and note that for the barycenter to be correct you must have $m_1 \boldsymbol{d}_1 + m_2 \boldsymbol{d}_2 = 0$.

If they orbit each other, then they share a common rotational velocity $\boldsymbol{\omega}$ about the barycenter, and the kinematics are as follows:

$$ \begin{aligned} \boldsymbol{v}_1 &= \boldsymbol{v}_C + \boldsymbol{\omega} \times \boldsymbol{d}_1 & \boldsymbol{v}_2 &= \boldsymbol{v}_C + \boldsymbol{\omega} \times \boldsymbol{d}_2 \end{aligned}$$

Now take the total momentum

$$\require{cancel} \begin{aligned} \boldsymbol{p} & = m_1 \boldsymbol{v}_1 + m_2 \boldsymbol{v}_2 \\ & = m_1 (\boldsymbol{v}_C + \boldsymbol{\omega} \times \boldsymbol{d}_1) + m_2 ( \boldsymbol{v}_C + \boldsymbol{\omega} \times \boldsymbol{d}_2) \\ & = (m_1 + m_2) \boldsymbol{v}_C + \boldsymbol{\omega} \times \cancel{(m_1 \boldsymbol{d}_1 + m_2 \boldsymbol{d}_2 )} \\ & = (m_1 + m_2)\,\boldsymbol{v}_C \end{aligned} $$

And the total angular momentum about the barycenter

$$ \begin{aligned} \boldsymbol{L}_C & = \boldsymbol{d}_1 \times \boldsymbol{p}_1 + \boldsymbol{d}_2 \times \boldsymbol{p}_2 \\ & = \boldsymbol{d}_1 \times m_1 (\boldsymbol{v}_C + \boldsymbol{\omega} \times \boldsymbol{d}_1) + \boldsymbol{d}_2 \times m_2 (\boldsymbol{v}_C + \boldsymbol{\omega} \times \boldsymbol{d}_2) \\ & = \cancel{ ( m_1 \boldsymbol{d}_1 + m_2 \boldsymbol{d}_2 )}\times \boldsymbol{v}_C + \boldsymbol{d}_1 \times ( \boldsymbol{\omega}\times \boldsymbol{d}_1) + \boldsymbol{d}_2 \times ( \boldsymbol{\omega}\times \boldsymbol{d}_2) \\ & = \mathbf{I}_1 \boldsymbol{\omega} + \mathbf{I}_2 \boldsymbol{\omega} = ( \mathbf{I}_1 + \mathbf{I}_2 ) \boldsymbol{\omega} \end{aligned}$$

where $\mathbf{I}_i$ is a 3×3 mass moment of inertia tensor derived from some mathematical trickery by factoring out $\boldsymbol{\omega}$ from $\boldsymbol{d}_i \times ( \boldsymbol{\omega}\times \boldsymbol{d}_i)$.

Now to transfer the angular momentum to the origin you do the following

$$ \boldsymbol{L} = \boldsymbol{L}_C + \boldsymbol{r}_C \times \boldsymbol{p} $$

What is conserved here is $\boldsymbol{L}_C$ not just in magnitude, but in direction also and also $\boldsymbol{p}$ since no external forces are present. As a result $\boldsymbol{L}$ is conserved also. A condition being that $\boldsymbol{r}_C \times \boldsymbol{v}_C = 0 $.

For each body $\boldsymbol{L}_i = \boldsymbol{d}_i \times m_i ( \boldsymbol{\omega} \times \boldsymbol{d}_i )$ is conserved if $\boldsymbol{\omega}\cdot \boldsymbol{d}_i=0$.