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Proof related to angular momentum
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[QUOTE="Saitama, post: 4535560, member: 331656"] Angular momentum for a particle is defined as $$\vec{L}=\vec{r}\times \vec{p}$$ where ##\vec{r}## is the radius vector of the particle from some fixed point (origin) and ##\vec{p}## is the momentum of particle. For a system of particle, $$\vec{L}=\sum_i \vec{r_i} \times \vec{p_i}$$ where i denotes the ith particle. Shifting the origin, the new position vector for the ith particle is $$\vec{r_i'}=\vec{r_i}-\vec{R}$$ Hence, $$\vec{L'}=\sum_i \vec{r_i'}\times \vec{p_i}=\sum_i(\vec{r_i}-\vec{R})\times p_i$$ $$\Rightarrow \vec{L'}=\vec{L}-\sum_i \vec{R}\times \vec{p_i}$$ The second term is zero, hence proved. Is this the correct way? Did I use the right words? Thank you TSny! :) [/QUOTE]
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Proof related to angular momentum
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