1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Contravariant components to covariant

  1. Jul 2, 2010 #1
    Hi, everyone

    I was playing with the coordinate transformations and metric tensors to get a feeling of how it all behaves, and got stuck with some basic problem I am hoping you can help me with.

    So, I have defined a coordinate system (s,t), with the s axis going along the x axis in the cartesian coordinates, and t axis going along the y=x line:
    s = x-y
    t = y*sqrt(2)

    with inverse transformation:

    x = s + t/sqrt(2)
    y = t/sqrt(2)


    If I am differentiating correctly, the metric tensor in these coordinates looks like:
    1 (2+sqrt(2))/2
    (2+sqrt(2))/2 1

    g11 = g22 = 1,
    g21=g12 = (2+sqrt(2))/2

    Now, I pick a point (3,1) in cartesian coordinates, and transform it to my new frame, and get the contravariant coordinates as (2, sqrt(2)).
    So far so good. What I am trying to do is find out what its covariant coordinates are going to be. I think, that covariant coordinates are supposed to be the lengths of orthogonal projections of the vector on the respective axes. From basic geometry, I get (3, 2*sqrt(2)).

    The problem is that when I try to multiply my metric tensor by the contravariant vector, I get a different answer - (3+sqrt(2), 2+2*sqrt(2))
    Clearly, there is something I am doing wrong here, but I can't figure out what it is :( Can somebody please help me spot the problem?

    Thanks a lot for your help!
     
  2. jcsd
  3. Jul 3, 2010 #2
    Your metric is wrong. The mixed components (g12 and g21) should be 1/sqrt(2).
    Because you did not show how you got your metric tensor, I can't say where you went wrong, but if your check your index dropping with the correct metric, you'll see that it fits.

    Hope, it helps ...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook