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I'm having a fairly hard time understanding the intuition behind Noether's derivation of the conservation of angular momentum from the rotational invariance of the Lagrangian, though I do understand it mathematically, I just don't get it intuitively.

I'm saying this because it appears so clearly derivable from the original conservation of energy statement, but I know they're quite different, qualitatively. What am I missing?

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up vote 4 down vote accepted

As a simple example, let $\phi$ be the azimuthal angle and let the angular momentum $\vec{L}$ point in the $z$-direction. (This is just business as usual.)

If the lagrangian $L(q_{i},\dot{q_{i}})$ is invariant under rotations about $z$ (in other words, changes in $\phi$) we can say that $$\frac{\partial L}{\partial \phi}=0$$

By the euler-lagrange equations we conclude that $$\frac{d}{dt}\frac{\partial L}{\partial \dot{\phi}}=0 \rightarrow \frac{\partial L}{\partial \dot{\phi}} = \text{const.}$$

As a further illustration for your edification (it rhymes) recall that the parts of the lagrangian that depend on the coordinates correspond to potentials, and that the derivatives of these potentials correspond to forces.

If your lagragian depends on $\phi$, you'll find that you have a force pointing the $\hat{\phi}$-direction, applying a torque to the system under consideration.

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If it's helpful for your intuition, consider this counterexample. You are probably sitting right now in a noninertial reference frame, with a preferred direction ("down"). Is angular momentum conserved in your reference frame? No! If you spin a gyroscope or a bicycle wheel, only if the angular momentum vector points vertically is its direction constant; if the angular momentum makes an angle to the vertical, you get precession. Since the symmetry of space is broken, you have nonconservation of angular momentum. If you could restore that symmetry (by, say, operating in free-fall) you'd see much better conservation of angular momentum.

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