Sufficient proof that the angular momentum has a constant norm.

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(I'm posting this here instead of at physics.* because of the type of questions/answers I've found here compared to there)

So, in a problem I'm working on they ask me to prove that the angular momentum $\vec{L}$ is such that $||\vec{L}||$ is constant. I have that $$\vec{L}= \{aw_1,0,bw_3\}\quad;\quad \dot{\vec{L}}=\{0,c,0\},$$ so I figured out that since $\vec{L}\cdot\dot{\vec{L}}=0$ I can safely say that the momentum doesn't grow "in it's own direction" so the norm has to remain constant over time. However, in the solutions manual they do this:

$$ \frac{d}{dt}||\vec{L}||=\frac{d}{dt}(\vec{L}\cdot\vec{L})=2\,\vec{L}\cdot\dot{\vec{L}}=0. $$

Are both solutions as valid as the other? Or did I miss some nuance with the way I "proved" it?

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What you write is a very good way of intuitively seeing why $\|\vec L\|$ is constant. There's no sharp boundary between “a very good way of intuitively seeing why” and “a rigorous proof”. Every proof, unless it's completely formalized, which in practice it almost never is, consists of statements that are sufficiently obvious to the competent reader that they don't need to be further justified. What's sufficiently obvious to the competent reader may depend on the reader. So unless you completely formalize a proof, you're always making a judgement about what you can expect your peers to accept without further proof.

However, history and experience show that what seems obvious may turn out to be wrong. The more formal and rigorous you make a proof, the more precaution you take against your intuition turning out to be wrong after all. That doesn't mean you should always be $100\%$ formal and rigorous (which is practically impossible anyway); it's just something to keep in mind.

In your case, the two proofs are on quite different points along the spectrum between intuition and formalism. The argument that the moment “doesn't grow in its own direction” is certainly a useful image to have in mind when thinking about these things, but I suspect that many people wouldn't regard it as a proof. It's the sort of thing that feels very plausible but might turn out to be mistaken after all. It essentially hides a distinction between first-order and second-order changes (since of course a vector does grow if you add multiples of an orthogonal vector to it, but only to second order), and that distinction might not be correctly reflected in our intuition. The proof using derivatives is more formal and will probably feel more reliable to many people.