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Components of a 2-form.

  1. Mar 7, 2013 #1
    Hi I'm a bit confused about what actually define the components of a form. I just saw an argument where one found that

    [tex] \underline{d \omega}^\rho (\vec e_\mu \wedge e_\nu) = - c_{\mu \nu}^\rho [/tex]

    and then the author wrote that this implied that

    [tex] \underline{d \omega}^\rho = - \frac{1}2 c_{\mu \nu}^\rho \underline{\omega}^\mu \wedge \underline{\omega}^\nu [/tex]

    so if the components of a p-form is defined as

    [tex] \underline{\alpha} = \frac{1}{p!} \alpha_{\mu_1 \ldots \mu_p} \underline{\omega}^{\mu_1} \wedge \ldots \wedge \underline{\omega}^{\mu_p} [/tex]

    where [itex]\alpha_{\mu_1 \ldots \mu_p}[/itex] are the components, it seems the argument above implies that one can find these by applying the p-form to the basis p-vectors.

    However I tried this with a two form [itex]\underline{\alpha} = 1/2 \alpha_{\mu \nu} \underline{\omega}^\mu \wedge \underline{\omega}^\nu [/itex] using the definition of the wedge product

    [tex] \underline{\alpha}(\vec{e}_\alpha \wedge \vec{e}_\beta) = \frac{1}{2} \alpha_{\mu \nu} \underline{\omega}^\mu \wedge \underline{\omega}^\nu (\vec{e}_\alpha \wedge \vec{e}_\beta) \\
    = \frac{1}{2} \alpha_{\mu \nu} 4 \underline \omega^{[\mu} \underline \omega^{\mu ]}( \vec e_{[\alpha} \vec e_{\beta]}) \\
    = 2\alpha_{\mu \nu} \delta^{[\mu}_{[\alpha} \delta^{\nu ]}_{\beta]} \\
    = 2 \alpha_{\alpha \beta} [/tex]

    Where I have used that [itex] \underline \omega ^\mu \underline \omega^\nu = 2! \underline \omega^\mu \underline \omega^\nu[/itex]. But should I not get [itex]\alpha_{\alpha \beta}[/itex] here? Is the caculation wrong or is my assumption of what defines the components wrong?
     
  2. jcsd
  3. Mar 7, 2013 #2

    fzero

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    Your notation is a bit confusing, which might be the reason for the extra factor of 2 that you get later. The [itex]e_\mu[/itex] are a basis for vector fields, so we do not take a wedge product of them. It's typical to write that expression as

    [tex] \underline{d \omega}^\rho ( e_\mu , e_\nu) = - c_{\mu \nu}^\rho, [/tex]

    where it's understood that

    [tex](\alpha \wedge \beta) (v,w) = \alpha(v) \beta(w) - \beta(w) \alpha(v)[/tex]

    pointwise. The pair of terms here is of course responsible for the factor of 2. However it's clear that in your expression

    you're introducing an extra factor of 2 from the erroneous manipulation of the vector fields.
     
  4. Mar 8, 2013 #3

    Check out page 59 in

    http://www.google.no/url?sa=t&rct=j...uSWMaT_Vs9fO_AucIQGIMOA&bvm=bv.43287494,d.Yms

    Here 'p-vectors' are introduced where wedge products of the vector basis is taken.
    The argument I referred to is at page 131. Check out the equations (6.175) and (6.176).

    The that argument then fallacious?
     
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