Constant of motion, Maxwell's equations

[WannabeNewton]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

In order to show that F_{ab}\xi ^{b} = \triangledown _{a}\varphi for some smooth scalar field \varphi, we want to show that the one - form field F_{ab}\xi ^{b} is closed i.e. \triangledown _{a}(F_{bc}\xi ^{c}) - \triangledown _{b}(F_{ac}\xi ^{c}) = 0. We find that \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}). Using the second of Maxwell's equations, \triangledown _{[a}F_{bc] } = 0, we find that \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 because this is the expression of the lie derivative which we are told vanishes. Thus we can conclude that F_{ab}\xi ^{b} = \triangledown _{a}\varphi for some smooth scalar field \varphi. In order to show that I = mu^{a}\xi _{a} + q\varphi is a constant of motion along the worldline of this charged particle subject to the Lorentz force, we must show \triangledown _{u}I = u^{b}\triangledown _{b}I = 0. We find that 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. Since u^{b}u^{a} is symmetric in the indices and \triangledown _{b}\xi _{a} is anti - symmetric in the indices, on account of \xi being a killing field, their contraction must vanish so we get 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. 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 \triangledown ^{a}F_{ab} = 0 even though we are given it. Maybe it is superfluous information for the purposes of this problem? Thanks again!

 

[cosmic dust]

It looks ok! Maybe the unused equation was given for sake of completness, letting you decide if it will be used or not.

 

[WannabeNewton]

Ok thank you mate!