Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to get the normal bivector to a surface?

  1. Apr 18, 2007 #1
    How to get the normal bivector to a surface??

    Given a surface [tex]f(x)=0[/tex] in a manifold, for examle [tex]R^n[/tex], its normal vector can be constructed as [tex]{\partial}^{\mu}f[/tex].

    But i don't know how to construct the so-called normal bivector (or binormal) to this surface. It has the form as [tex]{\epsilon}_{\mu \nu}[/tex]...
  2. jcsd
  3. Apr 18, 2007 #2

    Chris Hillman

    User Avatar
    Science Advisor

    You need to be more specific, I think. Is this in the context of Frenet-Serret theory in E^3? Or in say the context of minimal surfaces in E^4?
  4. Apr 19, 2007 #3
    Thank you, i think i've got it.

    For example, given a D-dimensional Lorentzian(or Riemann) manifold as [tex](M,g_{ab})[/tex] as
    [tex] ds^2 = g_{tt}dt^2 + g_{rr}dr^2 + f(r)d{\Omega}^2_{D-2}[/tex],
    then the hypersurface with constant t and r is a (D-2)-dimensional sphere. The point is that this sphere is of co-dimension 2. So its normal is a bivector.

    We may get its binormal as follows:
    First we choose the coordinates as [tex]\{ t,r,x^1,\cdots,x^{D-2} \}[/tex], in which [tex]\{x^1,\cdots,x^{D-2}\}[/tex] are coordinates on [tex]S^{D-2}[/tex]. Then we have the nature volume element on the (D-2)-sphere as
    [tex]\tilde{\epsilon}=\frac{1}{(D-2)!}\sqrt{h}{\epsilon}_{a_1 \cdots a_{D-2}}dx^{a_1}\wedge \cdots \wedge dx^{a_{D-2}}[/tex].
    Here h is the determinant of metric on the sphere.
    Then we can use the standard "Hodge dual" to get the binormal as
    [tex]*(\tilde{\epsilon}) = *\left( \frac{1}{(D-2)!}\sqrt{h}{\epsilon}_{a_1 \cdots a_{D-2}}dx^{a_1} \wedge \cdots \wedge dx^{a_{D-2}} \right)= \cdots = \frac{1}{2} {\omega}_{\mu \nu} dx^{\mu} \wedge dx^{\nu}[/tex]
    For [tex] x^{a_1},\cdots,x^{a_{D-1}}[/tex] is limited on the sphere, so the non-zero component of the above equation is [tex] \mu,\nu = t,r[/tex].
    Finally we can get a 2-form as
    [tex] \omega \equiv \frac{1}{2} {\omega}_{\mu \nu}dx^{\mu} \wedge dx^{nu} \equiv {\omega}_{t r} dt \wedge dr [/tex].
    This is just the so-called binormal to the sphere [tex]S^{D-2}[/tex]. Of course, some normalization may be made.

    Last edited: Apr 19, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook