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A Maxwell's equation in differential forms formalism

  1. Feb 3, 2013 #1
    1. The problem statement, all variables and given/known data
    This is not actually a homework but a personal work. Here it is:
    Using the differential forms:

    [itex]F=\tfrac{1}{2!}{{F}_{\mu \nu }}d{{x}^{\mu }}\wedge d{{x}^{\nu }}[/itex] and [itex]J=\tfrac{1}{3!}{{J}^{\mu }}{{\varepsilon }_{\mu \alpha \beta \gamma }}d{{x}^{\alpha }}\wedge d{{x}^{\beta }}\wedge d{{x}^{\gamma }}[/itex]

    proof that the Maxwell equation: [itex]{{F}^{\mu \nu }}_{,\mu }={{\mu }_{0}}{{J}^{\nu }}[/itex] can be written as:

    [itex]d*F={{\mu }_{0}}J[/itex]

    2. Relevant equations

    [itex]*F\equiv \tfrac{1}{4}{{F}^{\mu \nu }}{{\varepsilon }_{\mu \nu \rho \sigma }}d{{x}^{\rho }}\wedge d{{x}^{\sigma }}[/itex]

    [itex]d*F=\tfrac{1}{4}{{F}^{\mu \nu }}_{,a}{{\varepsilon }_{\mu \nu \rho \sigma }}d{{x}^{\alpha }}\wedge d{{x}^{\rho }}\wedge d{{x}^{\sigma }}[/itex]

    3. The attempt at a solution

    I have tried several ways to do this but they either fail (not giving the desired result) or get to a point where I cannot proceed any further.

    Here are some attempts:
    A) I multiplied both sides of the given equation with [itex]{{\varepsilon }_{\nu \alpha \beta \gamma }}d{{x}^{\alpha }}\wedge d{{x}^{\beta }}\wedge d{{x}^{\gamma }}[/itex] in order to make the current 3-form to appear in the RHS and then I tried to manipulate the LHS of the equation in order to make the [itex]d*F[/itex] to appear. This is the process:

    [itex]{{F}^{\mu \nu }}_{,\mu }={{\mu }_{0}}{{J}^{\nu }}\Rightarrow {{F}^{\mu \nu }}_{,\mu }{{\varepsilon }_{\nu \alpha \beta \gamma }}d{{x}^{\alpha }}\wedge d{{x}^{\beta }}\wedge d{{x}^{\gamma }}=3!J[/itex]

    As you can see, only one of the indices of Levi-Civita Symbol (LCS) contracts with the one indices of the field tensor (in contrast with the [itex]d*F[/itex] expression where two indices contract with the two indices of the field tensor) and also in the [itex]d*F[/itex] expression the derivative contracts with a basis 1-form while in the above equation the derivative contracts with one indice of the field tensor. In order to bypass the problem, I tried to decouple those disturbing contractions using Kronecker deltas:

    [itex]{{F}^{\mu \nu }}_{,\mu }{{\varepsilon }_{\nu \alpha \beta \gamma }}d{{x}^{\alpha }}\wedge d{{x}^{\beta }}\wedge d{{x}^{\gamma }}={{\delta }_{\mu }}^{\rho }{{\delta }_{\alpha }}^{\sigma }{{F}^{\mu \nu }}_{,\rho }{{\varepsilon }_{\nu \sigma \beta \gamma }}d{{x}^{\alpha }}\wedge d{{x}^{\beta }}\wedge d{{x}^{\gamma }}[/itex]

    I don’t know how to proceed further, in order to get the desired contractions. I tried to represent the deltas with LCSs, like this:

    [itex]{{\varepsilon }_{\rho \sigma \mu \alpha }}{{\varepsilon }^{\rho \sigma \mu \nu }}=3!{{\delta }_{\alpha }}^{\nu }[/itex]

    but then the equation gets more complex. I suppose that I have to use some property of the LCSs in order to manipulate the 5 LCS that appear.

    B) I tried to connect [itex]*d*F[/itex] with [itex]*J[/itex] , a process that involves simpler manipulations, but I come up with an equation that has an opposite sign of the desired. Here is what I did:

    [itex]*d*F=\tfrac{1}{24}{{\varepsilon }^{\mu \nu \beta \gamma }}{{F}_{\mu \nu }}^{,\alpha }{{\varepsilon }_{\alpha \beta \gamma \delta }}d{{x}^{\delta }}=\tfrac{1}{6}{{F}_{\mu \nu }}^{,\mu }d{{x}^{\nu }}\Rightarrow 6*d*F={{F}_{\mu \nu }}^{,\mu }d{{x}^{\nu }}[/itex]

    [itex]*{{J}_{}}=\tfrac{1}{3!}\tfrac{1}{3!}{{J}_{\mu }}{{\varepsilon }^{\mu \alpha \beta \gamma }}{{\varepsilon }_{\alpha \beta \gamma \delta }}d{{x}^{\delta }}=-\tfrac{1}{3!}{{J}_{\nu }}d{{x}^{\nu }}\Rightarrow -6*{J}={{J}_{\nu }}d{{x}^{\nu }}[/itex]

    (I used the identity [itex]{{\varepsilon }_{\mu \nu \alpha \beta }}{{\varepsilon }^{\mu \nu \rho \sigma }}=2\left( {{\delta }_{\alpha }}^{\rho }{{\delta }_{\beta }}^{\sigma }-{{\delta }_{\beta }}^{\rho }{{\delta }_{\alpha }}^{\sigma } \right) [/itex] in the first equation and [itex]{{\varepsilon }_{\rho \sigma \mu \alpha }}{{\varepsilon }^{\rho \sigma \mu \nu }}=3!{{\delta }_{\alpha }}^{\nu }[/itex] in the second)

    So, according to the given equation, the above 1-forms are related like this:

    [itex]{{F}_{\mu \nu }}^{,\mu }={{\mu }_{0}}{{J}_{\nu }}\Rightarrow *d*F=-{{\mu }_{0}}*{J}\Rightarrow d*F=-{{\mu }_{0}}{J}[/itex]

    That minus sign shouldn’t be there!!!

    Please check this last process for mistakes; it is the one that gets me closer to the result. If it correct, then apparently what I want to proof is the result with minus sign. But in the literature this equation has always a positive sign. I checked the definitions of the forms and the Hodge duality that are used in literature (which are the same as I those presented above), but the result they get has a positive sign (for reference check here http://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field , in the “Differential forms approach” section)
     
  2. jcsd
  3. Feb 4, 2013 #2
    it is better to go with one form current.
    j=jadxa
    with the definition of F,we have
    *F=1/4(εabcd Fcd dxaΔdxb)
    we can employ riemann normal coordinates in case of curved spacetime,
    g=η,zero christoffel symbols this all means dε=0
    d*F=1/4 d(εabcd Fcd dxaΔdxb)=1/4 ε[ab cde]Fcd dxeΔdxaΔdxb
    this is three form with components
    (d*F)eab=3/2 ε[ab cde]Fcd
    then component of *d*F are
    (*d*F)p=1/6 εpeab(3/2 ε[ab cde]Fcd)=1/4 εabpe εabcdeFcd=-δ[c|p ge|d]eFcd
    =-δcpgedeFcd=-∂d Fpd=∂d Fdp
    which from maxwell eqn
    aFab=-jb gives what is required.your minus sign is ok.they are just using different faraday tensor ie. the equation used is ∇aFba=jb .that's all.see wiki page on electromagnetic tensor.
     
    Last edited: Feb 4, 2013
  4. Feb 4, 2013 #3
    Thank's!
     
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