# Angular momentum in non-inertial frame of reference (König's theorem)

König's theorem states that:
$$\vec{L} =\vec {r}_{CoM}\times \sum \limits _{i}m_{i}{\vec {v}}_{CoM}+{\vec {L}}'=\vec {L}_{CoM}+\vec {L}'$$
where $$\vec{L}'$$ is the angular momentum with respect to the center of mass in a non-inertial frame of reference integral with the CoM.
For instance, we could take a rigid body rotating around its center of mass with velocity $$\vec{\omega}$$ and we could use as non-inertial frame of reference to calculate $$\vec{L}'$$ a frame of reference centered in the center of mass and rotating with the same angular velocity $$\vec{\omega}$$. In this case, we can notice that each particle of the rigid body will have zero velocity in our non-inertial frame of reference; therefore, $$\vec{L}'$$ will be 0. Hence, $$\vec{L} =\vec {L}_{CoM}$$. If we consider, instead, as non-inertial frame of reference a frame of reference integral with the CoM, but without rotation, is easy to demonstrate that $$\vec{L}'$$ is not zero and $$\vec{L} =\vec {L}_{CoM}+\vec {L}'$$.
It is clear that something is wrong with my considerations. May you explain me where I am wrong?

## 2 Answers

In König's theorem, you aren't free to choose any non-inertial frame for $$\vec{L}'$$.

$$\vec{L}'$$ is defined in what's called the center of mass frame: its origin is the center of mass of the system (as you already stated), but its axes must be identical to those of an inertial frame. In other words, this frame is in a pure translation with respect to the inertial frame.

So you can't compute $$\vec{L}'$$ in a frame where all points of the system are at rest, unless of course the system isn't rotating at all in the inertial frame.

Edit: that last paragraph is true for a rigid body. For a generic system of moving points, $$\vec{L}'$$ could be zero without every point being at rest, but it doesn't change the fact that $$\vec{L}'$$ is defined in the center-of-mass frame defined above.

• Thank you very much for clearing my doubts! Jun 1 at 18:55

The expression you provide is the angular momentum for a system of particles (not necessarily a rigid body) segregated into two terms: (1) the angular momentum of the CM plus (2) the angular momentum about the center of mass (CM).

See "König's theorem (kinetics)" on Wikipedia, or Goldstein, Classical Mechanics, or Symon, Mechanics, for the details. Also, see Internal/Rotational angular momentum on this exchange.

$$\vec L^{'}$$ is the angular momentum with respect to the CM and is never zero for a rigid body rotating about the CM. So, your statement "we could use as non-inertial frame of reference to calculate $$\vec L^{'}$$ a frame of reference centered in the center of mass and rotating with the same angular velocity $$\vec \omega$$" is not appropriate since $$\vec L^{'}$$ is evaluated with respect to the CM.

Perhaps I am missing something in your question?

• To calculate an Angular momentum you need to define a pole and a system of reference with respect to which you calculate the velocities. For L' I chose cm as a pole and as system of reference a system of reference integral with CoM and rotating with its same angular velocity. However, as @Miyase corrected me, the system of reference is supposed by Koenig's theorem to not rotate. Jun 1 at 19:44
• Yes, the answer by @Miyase is the better answer. One point, if we are dealing with a system of particles that is not a rigid body, individual particles can move independently such that the total angular momentum with respect to the center of mass is zero; for example, two particles of equal mass moving away from each other with equal speed. Jun 1 at 20:57
• You're right, I assumed a rigid body. I added another paragraph at the end of my answer. Thanks. Jun 1 at 21:07