Tuesday, 12 April 2016

vectors - How to define pseudovector mathematically?


The textbook describes pseudovector like this:



Let $a,b$ be vectors and $c=a\times b$, $P$ be the parity operator. Then $P(a)=-a,P(b)=-b$ by definition. But $P(c)=c$ since both $a$ and $b$ reverse direction. So $c$ is a pseudovector.



I can't understand this. For I know from my math course that $c$ is perfectly a regular vector. So $P(c)$ should be equal to $-c$ by definition. In the example above, however, the text seems to have assumed $P(c)=P(a)\times P(b)$, which is nonsense if $c$ is a self-governed entity. What I can see is that $c$ depends on $a$ and $b$ and any transformation on $c$ must be computed after $a$ and $b$ have changed accordingly. Thus I tend to define $c:=<\times,(a,b)>$ and $P(c)$ should actually be written as $\tilde{P}(c)$ where $\tilde{P}()=P(f(a),f(b))$. As a result, I think a pseudovector is actually a function of vectors equipped with preallocated arguments and every transformation on a vector has a counterpart on a pseudovector like $\tilde{P}$.



Am I right?




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