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

Star Operator

  1. Dec 22, 2006 #1
    I am reading some books about differential forms. I dont quite understand what is the geometrical meaning of star (hodge) operator. Can anyone give me a hand please?

  2. jcsd
  3. Dec 23, 2006 #2

    Chris Hillman

    User Avatar
    Science Advisor

    In d dimensions, the Hodge star sets up a duality between p-forms at (d-p)-forms, via
    [itex] \alpha \wedge \beta = ({}^\star\alpha , \beta) \, \omega[/itex]
    where [tex]\omega[/tex] is the volume form (a basis form in the one dimensional space of d-forms), and where [tex]\alpha[/tex] is a p-form, and where [tex]\beta, \; {}^\star\alpha [/tex] are (d-p)-forms.

    So for example, when d=4, we can describe a current using a one-form J or a dual three-form *J, and given a two-form describing the EM field, F, we can take its dual to find another two-form, *F. Half of Maxwell's equations are then seen to be trivial: [tex]dF = 0[/tex] automatically since [tex]F=dA[/tex]. The other half are then [tex]d {}^\star F = 4 \pi \, {}^\star J[/tex].

    The book Gravitation by Misner, Thorne, and Wheeler contains some nice discussion. For example, from the simple two-form [tex]F = -q/r^2 \, dt \wedge dr[/tex] (which describes a Coloumb-type purely radial electrostatic field), we obtain the simple two-form [tex]{}^\star F= q \sin(\theta) d\theta \wedge d\phi[/tex], which is suitable for integration over a sphere. The duality becomes even more vivid if you use the coframe
    [itex] \sigma^{\hat{0}} = -dt, \; \sigma^{\hat{1}} = dr, \; \sigma^{\hat{2}} = r \, d\theta, \; \sigma^{\hat{3}} = r \sin(\theta) \, d\phi[/tex]
    Then we obtain
    [itex] F = q/r^2 \, \sigma^{\hat{0}} \wedge \sigma^{\hat{1}} , \; {}^\star F = q/r^2 \, \sigma^{\hat{2}} \wedge \sigma^{\hat{3}} [/itex]
    which we could have obtained even faster using
    [itex] {}^\star \left( \sigma^{\hat{0}} \wedge \sigma^{\hat{1}} \right) =\sigma^{\hat{2}} \wedge \sigma^{\hat{3}} [/itex]
    However, the "coordinate coframe" is usually more convenient when we are actually integrating. Note that I am talking about Minkowski spacetime here, not a curved spacetime. The volume form is simply the wedge product of the four covectors making up the coframe. However, this can be readily generalized to work on curved spacetimes.

    (This example is potentially misleading, since in general, F will be a "rank two" two-form, meaning that it contains both [tex]dt \wedge dr[/tex] and [tex]d\theta \wedge d\phi[/tex] terms. A rank one exterior form is often called a "simple" form. Don't confuse this "rank" with "second rank tensor"! In our example, our "rank one" or "simple" two-form F corresponds to an antisymmetric "second rank" tensor.)

    Another good book is Burke, Applied Differential Geometry.
    Last edited: Dec 23, 2006
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Star Operator
  1. Star shaped object (Replies: 6)

  2. Hodge Star Operator (Replies: 5)