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Constant of motion, Maxwell's equations

  1. Dec 31, 2012 #1

    WannabeNewton

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    1. The problem statement, all variables and given/known data
    Let [itex]F_{ab}[/itex] be the Faraday Tensor and [itex]\xi ^{a}[/itex] a killing vector field. Suppose that the lie derivative [itex]\mathcal{L} _{\xi }F_{ab} = \xi ^{c}\triangledown _{c}F_{ab} + F_{cb}\triangledown _{a}\xi ^{c} + F_{ac}\triangledown _{b}\xi ^{c} = 0[/itex]. Show that [itex]F_{ab}\xi ^{b} = \triangledown _{a}\varphi [/itex] for some smooth scalar field [itex]\varphi [/itex] and that [itex]I = mu^{a}\xi _{a} + q\varphi [/itex] is a constant of motion for a charged particle of mass m, charge q, and 4 - velocity u.

    2. Relevant equations
    Maxwell's equations: [itex]\triangledown ^{a}F_{ab} = 0[/itex], [itex]\triangledown _{[a}F_{bc] } = 0[/itex]. Equations of motion for charged particle / Lorentz force: [itex]mu^{a}\triangledown _{a}u^{b} = qF^{b}_{c}u^{c}[/itex].


    3. The attempt at a solution
    In order to show that [itex]F_{ab}\xi ^{b} = \triangledown _{a}\varphi [/itex] for some smooth scalar field [itex]\varphi [/itex], we want to show that the one - form field [itex]F_{ab}\xi ^{b}[/itex] is closed i.e. [itex]\triangledown _{a}(F_{bc}\xi ^{c}) - \triangledown _{b}(F_{ac}\xi ^{c}) = 0[/itex]. We find that [itex]\triangledown _{a}(F_{bc}\xi ^{c}) - \triangledown _{b}(F_{ac}\xi ^{c}) = (\xi ^{c}\triangledown _{a}F_{bc} - \xi ^{c}\triangledown _{b}F_{ac} - F_{cb}\triangledown _{a}\xi ^{c} - F_{ac}\triangledown _{b}\xi ^{c})[/itex]. Using the second of Maxwell's equations, [itex]\triangledown _{[a}F_{bc] } = 0[/itex], we find that [itex]\triangledown _{a}(F_{bc}\xi ^{c}) - \triangledown _{b}(F_{ac}\xi ^{c}) = (-\xi ^{c}\triangledown _{c}F_{ab} - F_{cb}\triangledown _{a}\xi ^{c} - F_{ac}\triangledown _{b}\xi ^{c}) = 0[/itex] because this is the expression of the lie derivative which we are told vanishes. Thus we can conclude that [itex]F_{ab}\xi ^{b} = \triangledown _{a}\varphi [/itex] for some smooth scalar field [itex]\varphi [/itex]. In order to show that [itex]I = mu^{a}\xi _{a} + q\varphi [/itex] is a constant of motion along the worldline of this charged particle subject to the Lorentz force, we must show [itex]\triangledown _{u}I = u^{b}\triangledown _{b}I = 0[/itex]. We find that [itex]u^{b}\triangledown _{b}I = u^{b}\triangledown _{b}(mu^{a}\xi _{a} + q\varphi ) = mu^{b}u^{a}\triangledown _{b}\xi _{a} + mu^{b}\xi _{a}\triangledown _{b}u^{a} + qu^{b}\triangledown _{b}\varphi [/itex]. Since [itex]u^{b}u^{a}[/itex] is symmetric in the indices and [itex]\triangledown _{b}\xi _{a}[/itex] is anti - symmetric in the indices, on account of [itex]\xi [/itex] being a killing field, their contraction must vanish so we get [itex]u^{b}\triangledown _{b}I = mu^{b}\xi _{a}\triangledown _{b}u^{a} + qu^{b}\triangledown _{b}\varphi = mu^{b}\xi _{a}\triangledown _{b}u^{a} + qu^{b}F_{bc}\xi ^{c} = mu^{b}\xi _{c}\triangledown _{b}u^{c} - qu^{b}F_{cb}\xi ^{c} = \xi _{c}(mu^{b}\triangledown _{b}u^{c} - qu^{b}F_{b}^{c}) = 0[/itex]. Can anyone check if this is the right way to do it because I'm not too sure; it seems fine however I never had to use anywhere the first of Maxwell's equations [itex]\triangledown ^{a}F_{ab} = 0[/itex] even though we are given it. Maybe it is superfluous information for the purposes of this problem? Thanks again!
     
    Last edited by a moderator: Dec 31, 2012
  2. jcsd
  3. Dec 31, 2012 #2
    It looks ok! Maybe the unused equation was given for sake of completness, letting you decide if it will be used or not.
     
  4. Dec 31, 2012 #3

    WannabeNewton

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    Ok thank you mate!
     
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