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I have seen the answer for related topics, and it makes sense to me for the trivial contranvariant expression for a vector, $$\pmb{v} = v^i\hat{e}_i\tag{1}$$ and it is said that if the base $\hat{e}_i$ double its length, the component $v^i$ should change contraryly to halve its length to make the vector invariant. However, the formal definition of contravariant vector by a transformation should be, $$x^{\alpha} \rightarrow x'^{\alpha} = \frac{\partial x'^{\alpha}}{\partial x^{\beta}}x^{\beta}\tag{2}$$ $$A^{\alpha} \rightarrow A'^{\alpha} = \frac{\partial x'^{\alpha}}{\partial x^{\beta}}A^{\beta}\tag{3}$$ It seems to me that, in formal definition, the component in contravariant manner would change corresponding to the change of coordinates since $x'^{\alpha}$ is in numerator in (3), isn't the component changing conversely to the change in coordinates? Please help me figure it out!

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You're almost there, you just have the reasoning reversed. Suppose that $x'^\alpha = 2 x^\alpha$. Then all of the contravariant components double, just as you said. But a vector is left invariant because all the basis vectors become half as long. Formally, that's because the basis vectors carry a lower index. Intuitively, it's because a basis vector represents the distance you need to go to increase the coordinate by $1$. If you double the coordinates, then you only need to go half as far, so the basis vector is halved.

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