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GR notation, newbie question

  1. Sep 6, 2008 #1
    I'm trying to learn some GR from Carrol's textbook, but I'm a little lost there. For example, this simple problem:

    In Euclidean 3-space, let p be the point with coordinates (x,y,z) = (1,0,-1). Consider the curve passing through p:

    [tex]x^i(\lambda) = (\lambda, (\lambda-1)^2, -\lambda)[/tex]

    Calculate the components of tangent vectors to these curves at p in the coordinate basis [tex]\{\partial_x, \partial_y, \partial_z\}[/tex].

    The attempt at a solution

    The components of tangent vectors are given by

    [tex]V^i = \frac{dx^i}{d\lambda}[/tex]

    It is of course in the basis of x,y,z. But I don't understand what does the basis [tex]\{\partial_x, \partial_y, \partial_z\}[/tex] mean. The notation is new to me, but I think that

    [tex]\partial_x \equiv \frac{\partial}{\partial x}[/tex]

    so how can this be used as a basis? If you just take these derivatives at each component of the curve, you always get (1,1,1), right?
     
  2. jcsd
  3. Sep 6, 2008 #2

    Dick

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    In GR, and differential geometry in general, you often define a vector as a directional derivative. So the vector ai+bj+ck (where i,j,k are the usual unit vectors) is written as a*d/dx+b*d/dy+c*d/dz. Since the derivative along the direction i is d/dx etc.
     
  4. Sep 7, 2008 #3
    Alright, so this is a trivial problem then, right?
     
  5. Sep 7, 2008 #4

    Dick

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    Sure. Just like what you did in Calc I but written in GR type notation.
     
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