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)

Two bodies rotating about two different axis and Centers

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?

  • $\begingroup$ Are you aware of writing angular momentum in combined translation and rotation? Something along the lines "translation of center of mass and rotation about center of mass". $\endgroup$
    – Notwen
    Feb 24, 2021 at 0:58
  • $\begingroup$ The important thing about angular momentum, as well as linear momentum, is that it is conserved. So this is how you find your momenta - look for quantities that are conserved. One way to do it is by writing down the Lagrangian for the two particles. Note however that a particle will not spin around an axis for no reason, so to make the system closed you will either need to consider spinning disks, or at least particles spinning around their centre, or introduce some potential that forces circular orbits. $\endgroup$
    – Cryo
    Feb 24, 2021 at 1:14
  • $\begingroup$ It's just a "mathematical" question, I mean, I want to know the most useful definition to use on abstract situations $\endgroup$
    – johnbear02
    Feb 24, 2021 at 1:45
  • $\begingroup$ And, as the definition of "momentum of a system of particles", it is just the sum of the momentum of all the particles, regardless of the choice of origin. How would it be with the angular momentum $\endgroup$
    – johnbear02
    Feb 24, 2021 at 1:56
  • $\begingroup$ To be more precise: if one is rotating, $\vec{v}$ cannot be constant. If you refer to its modulus, then $L$ do not have to be constant. That is the case for a central force system for example. On the other side: consider a much more simple scenario: an object moving towards me in rectilinear motion, do not have $L$ in my reference frame, but in other one that is not along the line between me and the object, it does has it. So: angular momentum is indeed relative. $\endgroup$
    – fich
    Feb 24, 2021 at 3:13

2 Answers 2


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.

  • $\begingroup$ $\frac{d\vec{L}_A}{dt} = \vec{\tau}_A$ is not generally true if the summation point A isn't fixed or co-moving with the center of mass. The full relationship is $$\frac{d\vec{L}_A}{dt} = \vec{\tau}_A - \vec{v}_A \times \vec{p} $$ see physics.stackexchange.com/a/599062/392 $\endgroup$ Feb 24, 2021 at 13:22
  • 1
    $\begingroup$ @JohnAlexiou: That doesn't entirely make sense to me. What if $\vec{v}_A$ is a constant, but non-zero? Does that mean that $\vec{\tau}_A \neq d\vec{L}_A/dt$ relative to that choice of origin? Wouldn't that mean that $\vec{\tau}_A = d\vec{L}_A/dt$ in some inertial frames, but not others? $\endgroup$ Feb 24, 2021 at 21:49
  • $\begingroup$ @JohnAlexiou That corresponding to the my case of $\frac{d\vec{R}}{dt} \ne 0$., which I had exclused It is due to improper measuring the velocity. If frame A is moving w.r.t the COM frame C, that $\vec{p}$ is no longer the momentum measured in frame A. That statement is very missleading. You should not transform angular momentum like that. Because if frame A and frame C have relative motion, then $\vec{p}$ cannot be the momentum for both frame A and fram C. $L_C = L_A + (r_c - r_A) \times p$ is not correct, when A has relative motion with C. $p_a \ne p_c$, $\endgroup$
    – ytlu
    Feb 24, 2021 at 22:03
  • $\begingroup$ @JohnAlexiou Your derivation and conclusion were deadly wrong. I suggest that you should withdraw that series of posts. $\endgroup$
    – ytlu
    Feb 24, 2021 at 23:21
  • $\begingroup$ @ytlu - Deadly? here is not the place to discuss this, but I think there is a misunderstanding regarding measuring momentum from different locations. It is possible there is mistake in my calc or that it only applies on special circumstances. I would like to discuss under my original post and not here. Maybe here is the correct place to discuss. $\endgroup$ Feb 25, 2021 at 13:22

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 $\boldsymbol{\omega}\cdot \boldsymbol{d}_i=0$.


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