How to apply tensor transformation rule

[guv]

TL;DR Summary

how is the tensor transformation rule applied on a position vector? $$v^\alpha = v^{*\beta} \frac{\partial u^\alpha}{\partial u^{* \beta}}$$

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?

 

[Orodruin]

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.

Does the transformation rule apply to a position vector?

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

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

 

[guv]

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.

 

[Orodruin]

You can if you do it correctly.

 

[guv]

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

 

[Orodruin]

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.

 

[guv]

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.

 

[Orodruin]

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.

 

[Orodruin]

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

 

[guv]

Silly me. I get it now. Thanks!