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Nov 14 at 23:35 vote accept Danyel
Nov 12 at 16:24 answer added Cosmas Zachos timeline score: 1
Nov 12 at 12:03 comment added Cosmas Zachos Try this
Nov 12 at 6:58 comment added ZeroTheHero You’re trying to understand angular momentum using an incorrect pictorial representation that is not clearly explained. No wonder you are lost (as anyone would be!). It’s like trying to get to a precise destination using the wrong map and vague instructions.
Nov 12 at 5:34 history edited Qmechanic CC BY-SA 4.0
added 2 characters in body; edited tags
Nov 12 at 5:19 comment added Danyel You mean that $J_\perp^2$ in the drawing is one to one to that of the actual vector?
Nov 12 at 5:08 comment added naturallyInconsistent No, they are saying that if you scaled things so that $J^2=j(j+1)$ then you will get the correct behaviour that $J_x^2+J_y^2=j(j+1)-m^2$
Nov 12 at 4:39 history asked Danyel CC BY-SA 4.0