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Apr 19, 2024 at 3:05 vote accept CBBAM
Jan 7, 2023 at 9:39 comment added Quillo Good general question, it may be useful to list here some related questions for future users: physics.stackexchange.com/q/457591/226902 physics.stackexchange.com/q/455387/226902 physics.stackexchange.com/q/525092/226902 physics.stackexchange.com/q/110360/226902 physics.stackexchange.com/q/374900/226902
Jan 7, 2023 at 9:20 comment added CBBAM @JeanbaptisteRoux By a non-degenerate inner product I assume you mean $(u,v) = 0$ for all $v$ implies $u = 0$?
Jan 7, 2023 at 9:17 answer added Qmechanic timeline score: 3
Jan 7, 2023 at 9:11 comment added Jeanbaptiste Roux I don't have Folland's book at hand, so I will not post an answer as I don't have the context. The idea is to use the fact that the inner product in Euclidean space is non-degenerate, as opposed to the Lorentzian one. This means we can safely use the "ball parametrization" of the space, where the radius variable we integrate on is zero only at the origin.
Jan 7, 2023 at 8:55 history edited Qmechanic
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Jan 7, 2023 at 8:50 history asked CBBAM CC BY-SA 4.0