# Angular momentum of a system about the center of mass

Let $$\mathbf{R}$$ be the center of mass of a system of particles. Then the angular momentum of the system is \begin{align} \mathbf{L} &= \sum \mathbf{r}_i \times \mathbf{p}_i\\\\ &= \sum \left(\mathbf{R} + \mathbf{r}_i^\prime\right) \times \mathbf{p}_i\\\\ &= \mathbf{R} \times \sum \mathbf{p}_i + \sum \mathbf{r}_i^\prime \times \mathbf{p}_i\\\\ &= \mathbf{R} \times \mathbf{P} + \sum \mathbf{r}_i^\prime \times \mathbf{p}_i, \end{align}

yet Goldstein claims that $$\mathbf{L} = \mathbf{R} \times \mathbf{P} + \sum \mathbf{r}_i^\prime \times \mathbf{p}_i^\prime,\tag{1.28}$$ which is identical to what I derived with the exception that $$\mathbf{p}_i$$ gets replaced with $$\mathbf{p}_i^\prime.$$ How does one reconcile these similar yet different expressions for $$\mathbf{L}$$?

• You did not define p in your expression, m*v, relative to some coordinate system. v = Vcm + v', then you have more terms in your derivation. – ggcg Nov 12 '18 at 11:24
• To add to what @ggcg is saying, $p_i=p_{cm}+p'_i$ – Aaron Stevens Nov 12 '18 at 11:48

Note that $$\boldsymbol p'_i:=\boldsymbol p_i-m_i\dot{\boldsymbol R}$$. Therefore, $$\sum_i \boldsymbol{r}_i^\prime \times \boldsymbol{p}_i-\sum_i \boldsymbol{r}_i^\prime \times \boldsymbol{p}_i^\prime\equiv\sum_im_i\boldsymbol r'_i\times\dot{\boldsymbol R}$$ which is easily seen to vanish, inasmuch as $$\sum_im_i\boldsymbol r'_i\equiv0$$ by definition.

Thus, the expression in the OP and the one in Goldstein are identical.

AccidentalFourierTransform's answer is correct and complete. I would add that there really are two sensible ways of defining "the angular momentum about a moving point". It can mean the instantaneous angular momentum about that point in the lab frame, or the instantaneous angular momentum about that point in the point's frame. Your result is written in the first way, and Goldstein's is written in the second. Both notions are useful in different contexts.

Of course, the two coincide when the point is the center of mass, so there's nothing to worry about.

The answer by AccidentalFourierTransform is incorrect in that Goldstein doesn't define $$\sum_i m_i {\bf r}'_i = 0$$. Between 1.27 (which states: $$\bf{r_i}=\bf{R}+\bf{r'_i}$$) and 1.28 there is an argument (he doesn't put everything in symbols!),

[T]he factor $$\sum m_i \bf r'_i \bf$$, which, it will be recognized, defines the radius vector of the center of mass in the very coordinate system whose origin is the center of mass and is therefore a null vector.

but it isn't by definition that the sum vanishes, as some say. The straight forward proof of it is just a smidgen of algebra along with the definition of center of mass a few might have glossed over. Using double index as a implicit sum: $$m_i {\bf r}_i = M \bf R$$

Now, multiply through 1.27 by $$m_i$$, and use the definition of $$\bf R$$: $$m_i {\bf r}_i = m_i {\bf R} + m_i {\bf r}'_i$$ There, that's why $$m_i {\bf r}'_i = 0$$.