# Vectors on Different Coordinate Systems? [closed]

Consider the product of vectors coordinated relative to a given coordinate frame, defined by $$\vec{a}\square\vec{b}=((a_{1},b_{1})\square(a_{2},b_{2})):=(a_{1}b_{1},a_{2}b_{2})$$ Explain why that in the context of vectors this is no good.

My work:

I think that it is no good because if you change the coordinate system, you may change the vector. (Like a vector in Cartesian coordinates would be a different vector in polar coordinates [Not a 100% sure about that, but seems to make sense]).

If anyone could expand as to why this is true, it would be greatly appreciated. Also please note, that I am fairly new to vectors, so if the advanced physics jargon could be kept to a minimum it would be appreciated as well.

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Vectors are the same in any coordinate system. Although its components may change, if you change the coordinate system, the vector is still the same. –  Anuar Jun 4 '13 at 17:12
Your operation could be also seen as a multiplication between tensorial products of two vectors, that is : $(a_1 \otimes a_2) * (b_1 \otimes b_2) = (a_1 b_1\otimes a_2 b_2)$ –  Trimok Jun 4 '13 at 18:16
The only reason why this is worse than, say, the cross product, is that the definition of this operation requires you to choose a preferred coordinate frame. –  Peter Kravchuk Jun 4 '13 at 18:16
Note that in real-life situations you can have preferred coordinate frames. E.g. if you are working with crystal. Still, however, relevant operations should respect the remaining discrete symmetry. –  Peter Kravchuk Jun 4 '13 at 18:25

## closed as too localized by Emilio Pisanty, Brandon Enright, Waffle's Crazy Peanut, David Z♦Jun 6 '13 at 3:56

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So for example, suppose we write the rotated version of $\vec{a}$ and $\vec{b}$ as $R(\vec{a})$ and $R(\vec{b})$. Then we might reasonably demand that your product is invariant: $R(\vec{a}) \square R(\vec{b}) = R(\vec{a} \square \vec{b})$. Now we have to check whether this is true.
If we rotate by an angle $\theta$, we can write $R(\vec{a}) = (a_1\, \cos\theta - a_2\, \sin\theta, a_1\, \sin\theta + a_2\, \cos\theta)$, and similarly for $R(\vec{b})$. With your definition, we have $$\begin{split} R(\vec{a}) \square R(\vec{b}) = \big( (a_1\, \cos\theta - a_2\, \sin\theta)\, (b_1\, \cos\theta - b_2\, \sin\theta), \\ (a_1\, \sin\theta + a_2\, \cos\theta)\, (b_1\, \sin\theta + b_2\, \cos\theta) \big)~. \end{split}$$ But we also have $$R(\vec{a} \square \vec{b})= \big( (a_1\, b_1\, \cos\theta - a_2\, b_2\, \sin\theta), (a_1\, b_1\, \sin\theta + a_2\, b_2\, \cos\theta) \big)~,$$ which is not the same. (Your job is to show why these things are true, and show that they are different.)
So your product is not invariant under rotations. And different coordinate systems might just express these vectors as rotations of one another. So the meaning of your product depends on the particular coordinate system you are using. That's usually not a good idea. You can check that $R(\vec{a}) \cdot R(\vec{b}) = R(\vec{a} \cdot \vec{b}) = \vec{a} \cdot \vec{b}$ and $R(\vec{a}) \times R(\vec{b}) = R(\vec{a} \times \vec{b})$ for the usual dot and cross products (the latter of which requires three dimensions). That's why they are so useful.