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Orthogonal Lines and their line element

  1. Jul 3, 2012 #1
    In one of the early chapters of Gravity by Hartle, he is developing the line element on a sphere in preparation for developing the concept of a spacetime interval. Whilst finishing up the proof Hartle sort of implicitly says that if two lines are orthogonal the line element connecting two points on the two lines can be given by a pythagorean-like equation (the distance)2=(position on one line)2+(position on the other)2.

    Is the more general question of a line element being the sum of the squares of the two orthogonal lines true?

    Sorry if this question is a bit messy.
  2. jcsd
  3. Jul 3, 2012 #2


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    In general, if you have a "surface" or "space" with a coordinate system with coordinates [itex]x^1[/itex], [itex]x^2[/itex], ..., [itex]x^n[/itex], then the distance is given by a quadratic form [itex]\sum a_{ij}dx^idx^j[/itex].

    Is that what you are talking about?
    In particular, on the surface of a sphere of radius, R, we can identify each point by using spherical coordinates with the radial coordinate, [itex]\rho[/itex], fixed as the constant, R.

    [itex]x= Rcos(\theta)sin(\phi)[/itex] so that [itex]dx= -Rsin(\theta)sin(\phi)d\theta+ Rcos(\theta)cos(\phi)d\phi[/itex]
    [itex]y= Rsin(\theta)sin(\phi)[/itex] so that [itex]dy= Rcos(\theta)sin(\phi)d\theta+ Rsin(\theta)cos(\phi)d\phi[/itex]
    [itex]z= R cos(\phi)[/itex] so that [itex]dz= -R sin(\phi)d\phi[/itex]

    And now,
    [itex]dx^2= R^2cos^2(\theta)sin^2(\phi)d\theta^2- 2R^2sin(\theta)cos(theta)sin(\phi)cos(\phi)d\theta d\phi+ R^2cos^2(\theta)cos^2(\phi)d\phi^2[/itex]
    [itex]dy^2= R^2sin^2(\theta)sin^2(\phi)d\theta^2+ 2R^2 sin(\theta)cos(\theta)sin(\phi)cos(\phi)d\theta d\phi+ R^2sin^2(\theta)cos^2(\phi)d\phi^2[/itex]
    so that [itex]dx^2+ dy^2= R^2sin^2(\phi)d\theta^2+ R^2cos^2(\phi)d\phi^2[/itex]

    And [itex]dz^2= R^2 sin^2(\phi) d\phi^2[/itex] so that [itex]dx^2+ dy^2+ dz^2= R^2sin^2(\phi)d\theta^2+ R^2d\phi^2[/itex]

    That can be written as [itex]\sum g_{ij}dx^idx^j[/itex] or, as a matrix product, in this particular coordinate system,
    [tex]\begin{pmatrix}d\theta & d\phi\end{pmatrix}\begin{pmatrix}R^2sin^2(\phi) & 0 \\ 0 & R^2\end{pmatrix}\begin{pmatrix}d\theta \\ d\phi\end{pmatrix}[/tex]

    There is no "[itex]d\theta d\phi[/itex]" term, and the matrix is diagonal, precisely because the constant [itex]\theta[/itex] and constant [itex]\phi[/itex] curves are always orthogonal.
    Last edited by a moderator: Jul 3, 2012
  4. Jul 3, 2012 #3
    I think I followed that, and I do think that is what I was talking about.

    Thank you.
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