# 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
Commented Jul 21, 2013 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. Commented Jul 21, 2013 at 17:27
• @Physikslover I changed my notation, how does it look now?
– Ali
Commented Jul 22, 2013 at 14:29
• Looks nice - very arty ;) Commented Jul 22, 2013 at 14:52