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This question already has an answer here:

Working in 3-dimensions, if we are given a Lagrangian containing $N$ particles. Say, through Noether's theorem, we know that the sum of the linear momentum of all $N$ particles in each direction are constant. We also know that the sum of the angular momentum in each direction are constant.

My question is, if I know that linear momentum is conserved, can I actually automatically say that angular momentum is also conserved?

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marked as duplicate by sammy gerbil, stafusa, Kyle Kanos, Sebastian Riese, ZeroTheHero Jun 5 '18 at 21:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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No you cannot. Consider the two-particle Lagrangian $$L = \frac{1}{2}m(\dot{\vec{x}}_1^2+\dot{\vec{x}}_2^2) - \vec{a}\cdot(\vec{x}_2-\vec{x}_1)$$ for some arbitrary vector $\vec{a}$. This is fully translation invariant, and so Noether's theorem implies that momentum is conserved. However, it is not invariant under rotations because of the preferred direction $\vec{a}$, so angular momentum is not conserved.

Note that this Lagrangian is symmetric under rotations around $\vec{a}$ so it does conserve the angular momentum around that axis. We can make it conserve no components of angular momentum by adding another term like $(\vec{b}\cdot(\vec{x}_2-\vec{x}_1))^2$ where $\vec{b}$ is not parallel to $\vec{a}$.

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