How to apply tensor transformation rule

Suppose I have a Cartesian Coordinate system (x,y) and a polar coordinate system (##r, \theta##). The position vector (3,4) and (5, ##\arctan \frac{4}{3}##) are the same except the representation. The position vector is a tensor, how does the position vector follow the tensor transformation rule? Surely I cannot write ##x = r \frac{\partial x}{\partial r} + \theta \frac{\partial x}{\partial \theta}##

It's clear for a function ##f(x(r, \theta),y(r, \theta))##, its derivative ##\frac{\partial f}{\partial r}## which is the gradient vector follows the transformation rule.

Does the transformation rule apply to a position vector?

guv said:

Surely I cannot write ##x = r \frac{\partial x}{\partial r} + \theta \frac{\partial x}{\partial \theta}##

You can obviously write it (you just did), but it would be very wrong in general.

guv said:

Does the transformation rule apply to a position vector?

Yes! Try it in polar coordinates on the Euclidean plane!

$$
x = r \cos(\theta), \quad y = r\sin(\theta)
$$

Thanks, I know how ##x = r \cos \theta , y = r \sin \theta## works. What makes me wonder is why you can't use the tensor transformation rule on the position as I initially wrote.

You can if you do it correctly.

Would you mind showing how that works, I am very curious to see how. Thanks!

For example, the position vector in polar coordinates is ##X = r\partial_r##. In other words, the only non-zero component is ##X^r##. Hence
$$
X^x = X^r \frac{\partial x}{\partial r}
= r \cos(\theta) = x
$$
Similarly for the y-component.

Don't you need to include both ##r## and ##\theta##? i.e. exactly what I wrote initially? Why is ##r## non-zero but ##\theta## is zero? ##\theta## is not necessarily zero? Sorry I am not getting it.

guv said:

Don't you need to include both ##r## and ##\theta##? i.e. exactly what I wrote initially? Why is ##r## non-zero but ##\theta## is zero? ##\theta## is not necessarily zero? Sorry I am not getting it.

I did include the ##\theta## component (it is zero).

The position vector in polar coordinates does not have ##r## and ##\theta## as its components. It only has a radial component with value ##r##. Whatever point you pick, its position vector is fully in the radial direction.

If it makes you feel better we can always write
$$
X^x = X^r \frac{\partial x}{\partial r} +
\underbrace{X^\theta}_{= 0}\frac{\partial x}{\partial \theta}
= r \cos(\theta) = x
$$

Silly me. I get it now. Thanks!