# Contravariant vectors and Transformation Equations

1. Jan 21, 2012

### ashwinnarayan

1. The problem statement, all variables and given/known data

This is me doing some independent study on Tensors because I eventually hope to understand General Relativity.

My question is about the following equation which describe hoe the components of a displacement vector transform when there is a change in the coordinate system.

$d{y}^1 = \frac{∂y^1}{∂x^1} dx^1 + \frac{∂y^1}{∂x^2} dx^2$

To understand with and example, I drew out a normal 2D Cartesian Coordinate system with the axes $x^1$ and$x^2$ and drew a new coordinate system with axes $y^1$ and $y^2$. This new coordinate system was just the x system rotated anticlockwise by 30 degrees. I drew a displacement vector which went from (1,1) to (2,2) essentially a displacement vector of $(dx^1,dx^2)$

Since I already know linear algebra fairly well, I used the inverse transformation of a clockwise rotation of 30 degrees
$\left( \begin{array}{ccc} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{-1}{2} & \frac{\sqrt{3}}{2} \\ \end{array} \right)$
to find the values of $y^1$ and $y^2$ and got
$dy^1 =\frac{ \sqrt{3}}{2} dx^2 + \frac{1}{2} dx^2$

Now try as I might I cannot seem to be able to figure out how the heck $\frac{∂y^1}{∂x^1} = \frac{\sqrt{3}}{2}$ and $\frac{∂y^1}{∂x^2} = \frac{1}{2}$

Can someone help me out?

2. Relevant equations

3. The attempt at a solution

Last edited: Jan 21, 2012
2. Jan 21, 2012

### genericusrnme

1. If you're doing tensors this book (it's free) is the single best book I've read on introducing the subject
http://www.math.odu.edu/~jhh/counter2.html

2. if the equation you wrote for $dy^1$ is correct then $\frac{\partial y^1}{\partial x^1} = \frac{1}{2}$ and $\frac{\partial y^1}{\partial x^2} = \frac{\sqrt{3}}{2}$
I think you may have mixed you indices up, try writing out the transformation equations in the form of
$y^1 = f_1(x^1)+g_1(x^2)$
$y^2 = f_2(x^1)+g_2(x^2)$

3. Jan 21, 2012

### ashwinnarayan

Oh! I'm really sorry! It should have been

$dy^1 =\frac{ \sqrt{3}}{2} dx^1 + \frac{1}{2} dx^2$

And that book is really good. Thanks!

About trying to write the equation in that format I think the problem maybe that I'm not writing it down correctly.

In my coordinate system transformation (which I've attached) am I right in saying that

$y^1 = \frac{1}{\sqrt{3}}x^1$ because $y^1$ is essentially a straight line with a gradient of $\frac{1}{\sqrt{3}}$?

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4. Jan 21, 2012

### genericusrnme

Yeah, if $dy^1 = \frac{\sqrt{3}}{2} dx^1 + \frac{1}{2}dx^2$ then your partial derivatives are correct since the partial derivative with respect to one variable is just what you get when you set the 'd'*other variables to zero

For your transformation equation, think about what happens to your position in the y coordinates as you increase x1 then when you increase x2, each time you change your x1 coordinate you are changing your position in both of the y coordinates, if you understand what I mean.

if you move from x1=0 to x1=3, you've increase your y1 position but you've also decreased your y2 position since you are not below the y1 axis

5. Jan 21, 2012

### ashwinnarayan

Ah! I think I get it!!

I've been considering the cartesian equation instead of the vector equation!

So $y^1 = \frac{\sqrt{3}}{2} x^1 + \frac{1}{2} x^2$ !

Then I can get $\frac{∂y^1}{∂x^1} = \frac{\sqrt{3}}{2}$ and $\frac{∂y^1}{∂x^2} = \frac{1}{2}$

Yes, I get it now. Thanks a lot for helping me!