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Induced metrics

  1. Dec 9, 2013 #1
    So, by accident, while deriving the induced metric for a sphere in 3 dimensions I realized that the transpose of the jacobi matrix multiplied by the jacobi matrix (considering it as 3 row/column vectors)will work out the induced metric. Why is it that i≠j ends up being superfluous. One would have X=a 2-sphere in parameterized coordinates, and then g_ij= <X_;i,X_;j>. Thus one would compute <X_;1,X_;2> and the same for 2,3. Is this because the embedded manifold is an immersion, or is there something else? Thanks for any elucidation and best.
     
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  3. Dec 9, 2013 #2

    Ben Niehoff

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    Try a different parametrization and see what happens:

    [tex]\begin{align}x &= \sin \theta \cos (\phi + \theta), \\ y &= \sin \theta \sin (\phi + \theta), \\ z &= \cos \theta.\end{align}[/tex]
     
  4. Dec 11, 2013 #3
    Does that, in fact, parameterize a sphere? It's not obvious to me. I included an r term in the coordinates since I want it in general, not the only the unit case. The induced metric was:
    $$
    \left(
    \begin{array}{ccc}
    1 & 0 & 0 \\
    0 & r^2 sin^2\theta & r^2 sin^2\theta \\
    0 & r^2 sin^2\theta & r^2 sin^2\theta
    \end{array}
    \right)
    $$

    I haven't had time to write it out again, so I could have made a mistake, but this looks bad to me. The matrix is singular, and thus it can't be a metric, correct? Assuming this is correct, I don't see how the Jacobian will account for i ≠ j. I'll work through it again. Thanks
     
  5. Dec 11, 2013 #4

    Ben Niehoff

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    The term in the center of your matrix is wrong.
     
  6. Dec 12, 2013 #5
    Okay, so I went through it again:

    let X be the 2 sphere with your suggested parameterization

    $$
    X_{;\theta} = \left(
    \begin{array}{c} x^1 = rcos\theta cos \left(\phi + \theta \right) - rsin \theta sin \left(\phi + \theta \right) \\
    x^2 = rcos \theta sin \left( \phi + \theta \right) + rsin\theta cos \left(\phi + \theta \right) \\
    x^3 = -rsin\theta
    \end{array}
    \right)
    $$

    When I dot that with itself I got the same answer again. Mathematica, with a bit of tinkering, gave me

    $$ r^2 (Cos(\phi)^2 + Cos(2 \theta + \phi)^2 + sin(\theta)^2) $$

    What should g_22 be?
     
  7. Dec 12, 2013 #6

    Ben Niehoff

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    You should end up with

    [tex]r^2 (1 + \sin^2 \theta)[/tex]

    Check your algebra again. You shouldn't really need Mathematica, it's pretty easy to do.
     
  8. Dec 13, 2013 #7
    Yeah, I combined the trig terms into a 0 instead of a 1, so that's it. I see why it works for all cases now, in hindsight it's obvious. Thanks!
     
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