Contravariant vectors and Transformation Equations

[ashwinnarayan]

Homework Statement

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.

[itex]d{y}^1 = \frac{∂y^1}{∂x^1} dx^1 + \frac{∂y^1}{∂x^2} dx^2[/itex]

To understand with and example, I drew out a normal 2D Cartesian Coordinate system with the axes [itex]x^1[/itex] and[itex] x^2 [/itex] and drew a new coordinate system with axes [itex]y^1[/itex] and [itex]y^2[/itex]. 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 [itex](dx^1,dx^2)[/itex]

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

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

Can someone help me out?

Homework Equations

The Attempt at a Solution

 

[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 [itex]dy^1[/itex] is correct then [itex]\frac{\partial y^1}{\partial x^1} = \frac{1}{2}[/itex] and [itex] \frac{\partial y^1}{\partial x^2} = \frac{\sqrt{3}}{2}[/itex]
I think you may have mixed you indices up, try writing out the transformation equations in the form of
[itex]y^1 = f_1(x^1)+g_1(x^2)[/itex]
[itex]y^2 = f_2(x^1)+g_2(x^2)[/itex]
it'll help you out a lot

 

[ashwinnarayan]

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

[itex]dy^1 =\frac{ \sqrt{3}}{2} dx^1 + \frac{1}{2} dx^2[/itex]


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

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

 

[genericusrnme]

Yeah, if [itex]dy^1 = \frac{\sqrt{3}}{2} dx^1 + \frac{1}{2}dx^2[/itex] 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

 

[ashwinnarayan]

Ah! I think I get it!

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

So [itex] y^1 = \frac{\sqrt{3}}{2} x^1 + \frac{1}{2} x^2 [/itex] !

Then I can get [itex]\frac{∂y^1}{∂x^1} = \frac{\sqrt{3}}{2}[/itex] and [itex]\frac{∂y^1}{∂x^2} = \frac{1}{2}[/itex]

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