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Homework Help: Contravariant vectors and Transformation Equations

  1. Jan 21, 2012 #1
    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.

    [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?

    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Jan 21, 2012
  2. jcsd
  3. Jan 21, 2012 #2
    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
     
  4. Jan 21, 2012 #3
    Oh! I'm really sorry! :frown: 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!:biggrin:

    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]?
     

    Attached Files:

  5. Jan 21, 2012 #4
    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
     
  6. Jan 21, 2012 #5
    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!
     
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