Coord. Transf.: V'μ from (dy/dx)*Vν

[TimeRip496]

V′^μ=((∂y^μ)/(∂x^ν))*V^ν

This is a contravariant vector transformation. (Guys I am really sorry for making the formula above looks so incomprehensible as I still new to this.)

For the y in the partial derivative, is y a function in terms of x? In that sense, is it formula that maps x to y? Is it something like how much units of x correspond to each unit of y?

 

[Nugatory]

Yes, each $y^i$ is a function of the $x^i$, and vice versa.

A common and easy example is Cartesian and polar coordinates applied to the two-dimensional Euclidean plane. Using your notation, and with the $y^i$ as the Cartesian coordinates, the $x^i$ as the polar coordinates, and $i$ takes on the values 0 and 1 because this is a two-dimensional space:

$y^0=x^0cos(x^1)$; you're more used to seeing $x=rcos\theta$
$y^1=x^0sin(x^1)$; you're more used to seeing $y=rsin\theta$

If you have the components of a vector in polar coordinates $V^\mu$, the formula you posted will give you the components of the same vector in Cartesian coordinates $V'^\mu$ (using the Einstein summation convention, of course).

 

[TimeRip496]

Yes, each $y^i$ is a function of the $x^i$, and vice versa.

A common and easy example is Cartesian and polar coordinates applied to the two-dimensional Euclidean plane. Using your notation, and with the $y^i$ as the Cartesian coordinates, the $x^i$ as the polar coordinates, and $i$ takes on the values 0 and 1 because this is a two-dimensional space:

$y^0=x^0cos(x^1)$; you're more used to seeing $x=rcos\theta$
$y^1=x^0sin(x^1)$; you're more used to seeing $y=rsin\theta$

If you have the components of a vector in polar coordinates $V^\mu$, the formula you posted will give you the components of the same vector in Cartesian coordinates $V'^\mu$ (using the Einstein summation convention, of course).

Thanks! This really help a lot.