Can someone please explain the Right Hand Rule (used in Physics for magnetism), in basic calculus? As we have simply been given a 'vague general rule to use', and no other vector-calculus has actually been taught to us, I have trouble understanding the more "fancy" mathmatical explanations out there, but find the simple "use your hand like this" rules too vague. Hope someone can help!
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The RHR is direct result of the cross product learned in a standard intro linear algebra course. Wikipedia's page on it is pretty solid tbh. There are also many great youtube videos on the topic. – SescoMath Sep 12 '18 at 09:39
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Would it be possible to express without basic linear algebra? – Nuclear Sep 12 '18 at 12:44
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RHR comes from the definition of cross product. Because when you define a base to analyse your system, you want the set of vectors to be orthogonal direct (convention + it's convenient for EM). In a direct orthogonal base $(e_x, e_y, e_z)$, $e_x \wedge e_y = e_z$. However, in a indirect orthogonal base, $e_x \wedge e_y = - e_z$. Which is very unconvenient because it would add a bunch of minus signs in EM. Also it is a convention, even for domains of physics that don't use cross product.
PackSciences
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@Nuclear : EM means electromagnetic; so everything related to rot operator is designed from cross product. – PackSciences Sep 12 '18 at 17:06
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@Nuclear : Orthogonal means "they are not a sum in the canonical base". And the canonical base is defined with a direct base, so it can be confusing that there is some kind of circular reasoning. Basically orthogonal means that in a canonical base (direct), the vectors of the new base aren't a sum (but they can be multiplied by a scalar/number) of those canonical base vectors. – PackSciences Sep 12 '18 at 17:09
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@Nuclear As an example, I create a Portal 1 mod called Portal: Canonical Base of $R^3$ in which I put the game upside down. The name is a reference to the fact that we AREN'T using the canonical base (my game titles are always the opposite), but using the base $e_1 = x$, $e_2 = y$ and $e_3 = - z$ not $e_3 = z$ – PackSciences Sep 12 '18 at 17:10