# Independence of angular momentum with respect to origin for rotation about center of mass

Why is angular momentum for rotation about the center of mass independent of origin of the coordinate system?

• Because you are calculating it according to a definite origin, i.e. in the center on mass frame! – Ali Jul 21 '13 at 14:28

$$\vec L = \sum_i \vec {r_i} \times \vec{p_i}$$

Now changing coordinates to CM, i.e. $\vec{r_i} \rightarrow \vec{R_{CM}}+\vec{\mathfrak{r}_i}$, where $\vec{\mathfrak{r}_i}$ is the new coordinates of the particle according to the CM frame.

$$\vec{L}=\sum_i \left(\vec{R_{CM}}+\vec{\mathfrak{r}_i} \right)\times \vec{p_i}=\vec{R_{CM}} \times \sum_i \vec{p_i}+\sum_i \vec{\mathfrak{r}_i}\times\vec{p_i}$$

Now the first term is zero, because $\sum_i \vec{p_i}$ is zero in the CM frame; and the latter is just the angular momentum in the CM. Ergo:

$$\vec{L}=\vec{L_{CM}}$$

I think this is what OP meant in the question.

• +1, it would be more readable if you used $r'_{i}$ for the new coordinates. – Physiks lover Jul 21 '13 at 17:27
• @Physikslover I changed my notation, how does it look now? – Ali Jul 22 '13 at 14:29
• Looks nice - very arty ;) – Physiks lover Jul 22 '13 at 14:52