# Constant of motion, Maxwell's equations

1. Dec 31, 2012

### WannabeNewton

1. The problem statement, all variables and given/known data
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.

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

3. 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!

Last edited by a moderator: Dec 31, 2012
2. Dec 31, 2012

### 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.

3. Dec 31, 2012

### WannabeNewton

Ok thank you mate!

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