Static, spherically symmetric Maxwell tensor

[jdougherty]

Homework Statement

Show that a static, spherically symmetric Maxwell tensor has a vanishing magnetic field.

Homework Equations

Consider a static, spherically-symmetric metric $g_{ab}$. There are four Killing vector fields: a timelike $\xi^{a}$ satisfying
$$\xi_{[a}\nabla_{b}\xi_{c]} = 0$$
and three vectors $(\ell_{i})^{a}$ orthogonal to $\xi^{a}$ that generate rotations and commute with $\xi^{a}$:
$$[\xi, \ell_{i}]^{a} = 0, \qquad [\ell_{i}, \ell_{j}]^{a} = \epsilon^{ijk}(\ell_{k})^{a}$$
Given a two-form $F_{ab}$ that satisfies the source-free Maxwell's equations
$$\nabla_{[a}F_{bc]} = 0, \qquad \nabla^{a}F_{ab} = 0,$$
Let $\eta^{a} = (-\xi^{b}\xi_{b})^{-1/2}\xi^{a}$, and define the electric and magnetic fields as
$$E^{a} = {F^a}_{b}\eta^{b}, \qquad B^{a} = \tfrac{1}{2}\epsilon^{abcd}\eta_{b}F_{cd}$$
where $\epsilon_{abcd}$ is a volume form, so
$$F_{ab} = E_{[a}\eta_{b]} + \epsilon_{abcd}\eta^{c}B^{d}.$$
In order for $F_{ab}$ to share the symmetries of the spacetime, its Lie derivative along any of the Killing fields must vanish. Since it's closed, that means
$$0 = \nabla_{[a}(F_{b]c}\lambda^{c}).$$
for $\lambda^{a}$ any of the Killing fields.

This ought to be enough information to show that $B^{a} = 0$, but I can't quite get there.

The Attempt at a Solution

Let $\kappa = \xi^{a}\xi_{a}$. Expanding the Lie derivatives a bit, and using the fact that $\xi^{a}$ is hypersurface-orthogonal, you can show
$$\xi^{c}\nabla_{[a}F_{b]c} = -\kappa^{-1}\xi^{c}F_{c[a}\nabla_{b]}\kappa + \kappa^{-1}F_{c[a}\xi_{b]}\nabla^{c}\kappa \\ (\ell_{i})^{c}\nabla_{[a}F_{b]c} = \tfrac{1}{2}F_{cb}\nabla_{a}(\ell_{i})^{c} - \tfrac{1}{2}F_{ca}\nabla_{b}(\ell_{i})^{c}$$
The RHS of that second one is begging to be contracted against some $(\ell_{j})^{a}(\ell_{k})^{b}$ to make use of the commutation relations, but I can't find an appropriate selection of indices. I also tried taking the Lie derivative of $F_{ab}$ expressed in terms of $E^{a}$ and $B^{a}$, and the problem then comes down to calculating the commutator of the Killing vector and each of $E^{a}$ and $B^{a}$. However, it wasn't particularly illuminating.

Thinking about it loosely, the spherical symmetry should be what forces $B^{a} = 0$, so you'd think that throwing $B^{a}$ and the second Lie derivative expression together and shaking sufficiently vigorously would do it. The problem I have there is that there's no obvious (to me) way to mix those two elements together.

 

[WannabeNewton]

I think you are making it much too complicated for yourself. Try to show/argue that in the coordinates adapted to $\xi^{a}$ and the spherical symmetry, the only non-zero independent component of an $F_{ab}$ which shares the aforementioned symmetries is $F_{tr}$ if we assume a vanishing magnetic monopole. Then since $\xi^{\mu} = \delta^{\mu}_{t}$, $B^{a}$ must vanish identically in this coordinate representation (use the expression you wrote down for $B^{a}$ in terms of $F_{ab}$ and the 4-velocity of the observers measuring the EM field which in this case are the static observers) but if it vanishes identically in one coordinate system then it must vanish identically in all. Note that the key assumption here is the fact that $F_{ab}$ is static in a static space-time (so that there exists a family of observers which can remain "at rest" with respect to the static electromagnetic field and we know from regular EM that observers who are at rest with respect to a static EM field will see a pure E field).

 

[jdougherty]

I am making it more complicated than it needs to be, but that's intentional. In coordinates the problem is relatively easy, but I'm interested in whether it can be done without introducing the coordinate vectors on top of the Killing vectors already provided.

I have another question about the magnetic monopole part. Doesn't a magnetic monopole mean that $dF \not= 0$, since it has to equal the magnetic current? Which means that magnetic monopoles are already ruled out. Wald also makes a point of this (Prob. 6.5), which confused me, since I could show that
$$F_{ab} = 2A(r)\, (e_{0})_{[a}(e_{1})_{b]} + 2B(r)\, (e_{2})_{[a}(e_{3})_{b]}$$
only if $B(r) = 0$, but he also points out that $B(r) \not= 0$ means magnetic monopoles.