Internal/Rotational angular momentum

Question

I have some difficulties to understand the relation between the internal and the rotational angular momentum of a rigid body which is also known as König's theorem, so what physical intuition lies behind this equation? (I googled it but I can't find anything, so I will also appreciate any references if possible.)

edit:

$$H(O/Rg) = OG \times p + H^\star$$ (as vector quantities)

$p$ is the moment of momentum , $G$ is the center of mass , $Rg$ is a gallilean reference of frame , $H$ is the angular momentum with respect to a point $O$ fixed in that reference , and $H^\star$ is the angular momentum in the CM reference .

Answer 1

König's theorem is essentially a statement of conservation of angular momentum. If you consider the angular momenta of a system of particles, they had better add up to the same value no matter what frame you consider, provided that you are computing the angular momentum about the same fixed point in space. In a fixed frame, a system of particles has some angular momentum, which you have denoted $H(O/Rg)$. In a frame of reference co-moving with the center of mass of that system of particles, the total angular momentum appears to be $H^*$. König's theorem states that the connection between these two frames that resolves the apparent discrepancy in observed angular momentum is to adjust for the angular momentum of the center of mass itself, as viewed from the Gallilean refrence frame. This angular momentum is $\mathbf{r}_{CM} \times \mathbf{p}_{CM}$ where $\mathbf{r}_{CM}$ is the position of the center of mass relative to the origin $O$ and $\mathbf{p}_{CM}$ is the momentum of the center of mass of that system of particles, again relative to $O$.

Answer 2

Let a system consist of particles with positions $\mathbf x_i$ as measured in some inertial frame, and let $\mathbf x'$ denote the position of the center of mass of the system, then if we define the center of mass positions by $$ \mathbf x'_i = \mathbf x_i -\mathbf x' $$ then we have $$ \mathbf x = \mathbf x' + \mathbf x_i' $$ And the angular momentum of the system is \begin{align} \mathbf L &= \sum_i \mathbf x_i\times(m_i \dot {\mathbf x}_i) \\ &= \sum_i m_i(\mathbf x'+\mathbf x_i')\times (\dot {\mathbf x}' + \dot{\mathbf x}_i') \\ &= \mathbf x'\times (M\dot{\mathbf x}')+\mathbf x'\times\left(\sum_i m_i\dot{\mathbf x}_i'\right)+ \mathbf x'\times\left(\sum_i m_i{\mathbf x}_i'\right)\times \dot{\mathbf x}' +\sum_i \mathbf x_i'\times(m_i \dot{\mathbf x}_i') \\ \end{align} where $$ M = \sum_i m_i $$ is the total mass. The second and third terms vanish because $$ \sum_i m_i \mathbf x_i' = \sum_im_i(\mathbf x_i - \mathbf x') = \sum_i m_i \mathbf x_i - M \mathbf x' = M\mathbf x' - M\mathbf x' = \mathbf 0 $$ So we finally get $$ \boxed{\mathbf L = \mathbf L' + \mathbf L_{\mathrm{cm}}} $$ where $\mathbf L'$ denotes the angular momentum of the center of mass and $\mathbf L_{\mathrm{cm}}$ denotes the angular momentum of the system about the center of mass; $$ \mathbf L' = \mathbf x'\times (M\dot{\mathbf x}'), \qquad \mathbf L_{\mathrm{cm}} = \sum_i \mathbf x_i'\times(m_i \dot{\mathbf x}_i') $$