Trouble understanding contravariant transformations for vectors

  • #1
whisperzone
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TL;DR Summary
Don't understand equation of contravariant transformation for vectors
Hey, so I've been studying some math on my own and I'm really confused by this one bit. I understand what contravariant components of a vector are, but I don't understand the ways in which they transform under a change of coordinate system.

For instance, let's say we have two coordinate systems ##(x^1, x^2, ..., x^n)## and ##(\overline{x}^1, \overline{x}^2, ..., \overline{x}^n)## and a coordinate transformation relating the two systems: ##\overline{x}^j = \overline{x}^j (x^1, x^2, ..., x^n)##. Everything I've read says that

$$\overline{a}^j = \sum_{k = 1}^{n} \frac{\partial \overline{x}^j}{\partial x^k} a^k, \text{ where } j = 1, 2, ... n$$

but intuitively this makes no sense to me at all. For instance, consider a coordinate transformation ##(x^1, x^2, ..., x^n) \rightarrow (2x^1, 2x^2, ..., 2x^n)##. Then the partial derivative of ##\overline{x}^j## with respect to ##x^k## should be 2, right (or am I missing something here)? And if the partial derivative is 2, is that really a contravariant transformation, since whatever you're applying the transformation to would also get multiplied by 2?

Apologies if anything was too unclear, I've just been struggling with this for a while.
 

Answers and Replies

  • #2
Orodruin
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Contravariant refers to transforming the opposite way of the tangent basis vectors, which are defined as
$$
\vec E_i = \frac{\partial \vec x}{\partial x^i}.
$$
This means that
$$
\vec E_i ' = \frac{\partial \vec x}{\partial x^j} \frac{\partial x^j}{\partial x'^i}.
$$
In the case of your transformation, you therefore have ##\vec E_i' = \vec E_i / 2## so there is a factor of 1/2 in the transformation of the basis vectors. The components therefore (which are multiplied by 2) transform in the opposite fashion, i.e., contravariantly.
 

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