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Non-Orthogonality of Frames Del/Delx^i ; i=1, n

  1. Dec 30, 2007 #1

    WWGD

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    Hi, everyone:

    In an effort to show that at any point p in a Riemannian mfld. M
    there is an orthonormal basis --relatively straightforward--a new
    question came up:

    Why aren't the coordinate vector fields always orthonormal?.
    I know these are orthonormal when M is locally isometric to
    IR^n, but cannot see how?.

    We can prove the existence of the orthonormal frames using
    Gram-Schmidt. I tried applying Gram-Schmidt to the coord.
    V.Fields, see if the projections cancelled out, but this is
    not working.

    Any Ideas?.
    Thanks.
     
  2. jcsd
  3. Dec 30, 2007 #2
    Well, although it is possible to construct ortonormal coordinates, this does not mean that every coordinate system is ortonormal.

    The essential property of coordinate vector fields is that they commute. Whether a coordinate system is ortonormal depends on the metric.
     
  4. Dec 30, 2007 #3

    WWGD

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    Right, but wouldn't the metric be the Riemannian metric?. I assume the issue is
    that we cannot always assign a Riemannian metric such that the coord. V.Fields
    would be orthogonal, unless the Riemannian mfld. is locally flat. I hope this is not
    a dumb comment, I am fuzzy on this area.
     
  5. Dec 31, 2007 #4
    I am not sure I understand your question. Are you wondering when it is possible to extend an ortonormal frame at some point to an ortonormal local coordinate system?

    To get such a coordinate system one would like to parallel transport the frame to each point in some neighbourhood and thus hope to get a set of coordinate vector fields. For that to work out, the manifold must, just as you mention, be flat.

    But note that although ortonormal coordinate systems are not possible on curved manifolds, ortogonal are. Usual azimuth/zenith coordinates on the sphere is an example of this.
     
  6. Dec 31, 2007 #5

    WWGD

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    Thanks again, OOT

    But, if we have orthogonality, can't we just normalize each vector to have length
    1 , to get orthonormality?.

    I was working under this setup:

    The Riemannian metric determines orthogonality, in that

    Xi_p orthogonal Xj_p , (i.e, both are based at the same tangent space), if

    g_p<Xi_p,Xj_p>=0 , where g_p is the 2-tensor field at p .


    If I apply Gram-Schmidt to a basis, I get an ortho. basis. If I start with

    an ortho basis and apply Gr.-Schmidt , I end up with the same basis.


    I was then trying to see that the coord. V.Fields are not always Orthonormal

    by applying Gram-Schmidt (on a fixed tangent space to p ) to the coord. V.Fields

    { Del/Delx^i } at p , and seeing why we would not get {Del/Delx^i} back after

    G-Schmidt -- but I don't see why we don't :


    First step: Given {Del/Delx^i } at p, turn it into an orthonormal base:{Y1,...Yn}.
    g_p=<,>_p is the tensor field at p


    Y1:=X1

    Y2:=X2 - [ <X1,Y1>_p/<X1,X1>_p]X1

    .
    .

    So that we would get Yj=Xj at every step. And I think this means that each term
    in Gram Schmidt after the first, would then be zero, i.e, given :


    Yk:=Xk - [....]


    We would have [....] == 0.


    Thanks, Happy Nw Year, Merry Christma-Hannu -Kwanza-Dan, or Happy Festivus-- if that is possible :)
     
  7. Jan 1, 2008 #6
    For the sphere one can try to convert the azimuth ([itex]\theta[/itex]) and zenith ([itex]\phi[/itex]) coordinate system into an ortonormal system. Since the lengths of the basis vectors are

    [tex]|\mathbf{e}_\theta| = \frac{1}{\sin\phi}\qquad|\mathbf{e}_\phi| = 1[/tex]

    one replaces these with a normalised frame field

    [tex]\mathbf{e}_u=\sin\phi\,\mathbf{e}_\theta\qquad\mathbf{e}_v=\mathbf{e}_\phi[/tex]

    This means that one is trying to define new coordinates [itex](u,v)[/itex] as

    [tex] u = \int \frac{1}{\sin\phi}d\theta\qquad v = \phi[/tex]

    To see if this works, calculate [itex]\Delta u[/itex] around a "rectangle" with corners [itex](\theta_0,\phi_0)[/itex] and [itex](\theta_0 + 1,\phi_0 + 1)[/itex]:

    [tex]\Delta u = \mathop{\int_\text{rectangle}}_\text{boundary}\frac{1}{\sin\phi}d\theta=\frac{1}{\sin\phi_0} - \frac{1}{\sin (\phi_0 +1)}\neq 0[/itex]

    This means that [itex]u[/itex] does not work as a coordinate function, it is path-dependent.


    The other way of saying the same thing is to note that

    [tex]\left[\sin\phi\,\partial_\theta,\partial_\phi\right] = \sin\phi\,\partial_\theta\partial_\phi - \partial_\phi\sin\phi\,\partial_\theta = -\cos\phi\,\partial_\theta[/tex]

    So the normalised vector fields no longer commute.


    Yet another way of understanding this is to see that given [itex]du=d\theta/\sin\phi[/itex], then

    [tex] d^2u = -\frac{\cos\phi}{\sin^2\phi}d\phi\wedge d\theta \neq 0[/tex]

    Hence [itex]du[/itex] is not closed.
     
  8. Jan 1, 2008 #7

    WWGD

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    Very Helpful. Thanks for your patience and thoroughness, Order of Things.
     
  9. Jan 1, 2008 #8
    Thanks, diff. geometry is great stuff. :smile:
     
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