# Maxwell's Equations in Curved Space-time

1. Jun 26, 2013

### Infrared

Can one show that strict charge conservation $\nabla_{a}J^{a} = 0$ follows directly from $\nabla_{a}F^{ab} = 4\pi J^{b}$ alone?

Also, how does $d^{\star}F = 4\pi ^{\star}J$ follow directly from that same equation where $\star$ is the Hodge dual operator?

2. Jun 26, 2013

### WannabeNewton

Yes and it is extremely easy to show (almost trivial in fact). Starting with $\nabla^{a}F_{ab} = 4\pi j_{b}$, we have $\nabla^{b}\nabla^{a}F_{ab} = \nabla^{a}\nabla^{b}F_{ba} = -\nabla^{a}\nabla^{b}F_{ab}= 4\pi \nabla^{b}j_{b}$ i.e. $\nabla^{b}\nabla^{a}F_{ab} -\nabla^{a}\nabla^{b}F_{ab}= 8\pi \nabla^{b}j_{b}$. Now $\nabla_{b}\nabla_{a}F^{ab} -\nabla_{a}\nabla_{b}F^{ab}= -R_{bae}{}{}^{a}F^{eb} - R_{bae}{}{}^{b}F^{ae} = -R_{be}F^{eb} + R_{ae}F^{ae} = 0$ hence $\nabla^{a}j_{a} = 0$.

It is also very easy to show that $d(^{\star }F) = 4\pi(^{\star }j)$. We have $(^{\star}F)_{ab} = \frac{1}{2}\epsilon_{abcd}F^{cd}$ so $\epsilon^{abef}\nabla_{e}(^{\star}F)_{ab} = \frac{1}{2}\epsilon^{abef}\epsilon_{abcd}\nabla_{e}F^{cd} = -2\nabla_{e}F^{ef} = -8\pi j^{f}$. Hence $\epsilon_{fjki}\epsilon^{feab}\nabla_{e}(^{\star}F)_{ab} = -6\nabla_{[j}(^{\star}F)_{ki]}= -8\pi\epsilon_{fjki} j^{f} = -8\pi(^{\star}j)_{jki}$ therefore $3\nabla_{[a}(^{\star}F)_{bc]} = d(^{\star}F)_{abc} = 4\pi(^{\star}j)_{abc}$ i.e. $d(^{\star}F) = 4\pi(^{\star}j)$.

Note the implications of this. Because $\nabla^{a}j_{a} = 0$ in any space-time, we can apply Stokes' theorem to a space-time region $\Omega \subseteq M$ bounded by two space-like hypersurfaces $\Sigma, \Sigma'$ from a single foliation and find that $\int _{\Omega}\nabla^{a}j_{a} = 0 = \int _{\Sigma}j_{a}n^{a} -\int _{\Sigma'}j_{a}n^{a}$ i.e. the total charge $Q = -\int _{\Sigma}j_{a}n^{a}$ is conserved (here $n^{a}$ is the outward unit normal field to the space-like foliation that $\Sigma,\Sigma'$ belong to; the negative sign is to compensate for the negative sign that comes out of the inner product in the integral).

3. Jun 26, 2013

### Infrared

Thanks WannabeNewton! Your answer helps me a lot. I also have a related question. I read that a killing vector field always solves Maxwell's equations in curved (vacuum) space-time. Can you explain why this is the case?

Last edited: Jun 26, 2013
4. Jun 26, 2013

### WannabeNewton

Well let's say we have a killing vector field $\xi^{a}$ and use it as a 4-potential so that $F_{ab} = 2\nabla_{[a}\xi_{b]} = 2\nabla_{a}\xi_{b}$. Then $\nabla^{a}F_{ab} =2\nabla^{a}\nabla_{a}\xi_{b}$.
Since $R_{abcd}\xi^{d} = \nabla_{a}\nabla_{b}\xi_{c} - \nabla_{b}\nabla_{a}\xi_{c} = -\nabla_{a}\nabla_{c}\xi_{b} + \nabla_{b}\nabla_{c}\xi_{a}\\ = R_{cabd}\xi^{d} - \nabla_{c}\nabla_{a}\xi_{b} + R_{bcad}\xi^{d} + \nabla_{c}\nabla_{b}\xi_{a} = 2\nabla_{c}\nabla_{b}\xi_{a} + R_{cabd}\xi^{d} + R_{bcad}\xi^{d}$
we have $2\nabla_{c}\nabla_{b}\xi_{a}=(R_{abcd} - R_{cabd} - R_{bcad})\xi^{d}$. Using the first Bianchi identity, we find that $R_{abcd} + R_{bcad} + R_{cabd} = 0\Rightarrow R_{abcd} =- R_{cabd} -R_{bcad}$ i.e. $\nabla_{c}\nabla_{b}\xi_{a} = R_{abcd}\xi^{d}$.

In particular, $\nabla^{a}\nabla_{a}\xi_{b} = - R_{ab}{}{}^{a}{}{}_{d}\xi^{d} = -R_{bd}\xi^{d}$. If we are in vacuum space-time then $R_{ab} = 0$ so $\nabla^{a}F_{ab} = \nabla^{a}\nabla_{a}\xi_{b} = 0$.

Now $\nabla_{[a}F_{bc]} = 2\nabla_{[a}\nabla_{b}\xi_{c]} = 2R_{[abc]d}\xi^{d} = 0$ by virtue of the first Bianchi identity. Hence $F_{ab} = 2\nabla_{a}\xi_{b}$ solves Maxwell's equations in vacuum.

Such solutions aren't always of physical interest but here is a paper that gives an example of a killing vector field solution to Maxwell's equations that is of physical interest: http://prd.aps.org/abstract/PRD/v10/i6/p1680_1

5. Jul 2, 2013

### Infrared

"I will spare you the details of the calculations involved in showing these two relations hold; you can, for now, take my word that I have indeed shown them to be true." Could you please show the details of that calculation? I am still new to these kind of tensor calculus manipulations and seeing more examples would be nice, especially another related to EM in curved space-time.

6. Jul 2, 2013

### WannabeNewton

Sure! Let's first show that Gauss's law for electricity holds on the space-like Cauchy surface. Keep in mind that $n^{a}n_{a} = -1$, which implies that $n^{a}\nabla_{b}n_{a} = 0$, and that $n_{[a}\nabla_{b}n_{c]} = 0$ since the unit normal field is hypersurface orthogonal to the space-like foliation. I will denote the derivative operator associated with the spatial metric $h_{ab}$ by $\tilde{\nabla}_{a}$.

We have $\tilde{\nabla}_{a}E^{a} = h_{a}{}{}^{b}h^{a}{}{}_{c}\nabla_{b}(F^{c}{}{}_{d}n^{d})\\ = (\delta^{bc} + n^{b}n^{c})(n^{d}\nabla_{b}F_{cd} + F_{cd}\nabla_{b}n^{d})\\ = n^{d}\nabla^{c}F_{cd} + F_{cd}\nabla^{c}n^{d} + n^{b}n^{c}n^{d}\nabla_{b}F_{cd} + n^{b}n^{c}F_{cd}\nabla_{b}n^{d}$.

Now $n^{c}n^{d}\nabla_{b}F_{cd} = n^{d}n^{c}\nabla_{b}F_{dc} = -n^{c}n^{d}\nabla_{b}F_{cd}\Rightarrow n^{c}n^{d}\nabla_{b}F_{cd} = 0$

and $n_{[a}\nabla_{b}n_{c]} = 0\Rightarrow n^{b}n^{c}F_{cd}\nabla_{b}n^{d} - n^{b}n^{d}F_{cd}\nabla_{b}n^{c}= 2n^{b}n^{c}F_{cd}\nabla_{b}n^{d}\\ = F_{cd}\nabla^{d}n^{c} - F_{cd}\nabla^{c}n^{d} = - 2F_{cd}\nabla^{c}n^{d}$

thus $\tilde{\nabla}_{a}E^{a}= n^{d}\nabla^{c}F_{cd} = -4\pi j_{d}n^{d} = 4\pi\rho$ by virtue of the inhomogeneous Maxwell equations.

Showing Gauss's law for magnetism holds on the spacelike Cauchy surface is very similar.

We have $\tilde{\nabla}_{a}B^{a} =-\frac{1}{2}\epsilon^{cdef} h_{a}{}{}^{b}h^{a}{}{}_{c}\nabla_{b}(F_{de}n_{f})\\ = -\frac{1}{2}\epsilon^{cdef} (n_{f}\nabla_{c}F_{de} + F_{de}\nabla_{c}n_{f} + n^{b}n_{c}n_{f}\nabla_{b}F_{de} + n^{b}n_{c}F_{de}\nabla_{b}n_{f})$.

Now $\epsilon^{cdef}n_{c}n_{f} = 0$ because the volume form is totally antisymmetric and just as before we have $n_{[a}\nabla_{b}n_{c]} = 0 \Rightarrow \epsilon^{cdef} n^{b}n_{c}F_{de}\nabla_{b}n_{f} - \epsilon^{cdef} n^{b}n_{f}F_{de}\nabla_{b}n_{c}= 2\epsilon^{cdef} n^{b}n_{c}F_{de}\nabla_{b}n_{f}\\ = \epsilon^{cdef} F_{de}\nabla_{f}n_{c} - \epsilon^{cdef}F_{de}\nabla_{c}n_{f} = -2 \epsilon^{cdef}F_{de}\nabla_{c}n_{f}$

so we are left with $\tilde{\nabla}_{a}B^{a} = -\frac{1}{2}\epsilon^{cdef} n_{f}\nabla_{c}F_{de}$. But $\epsilon^{cdef}\nabla_{c}F_{de} = -\epsilon^{cdef}\nabla_{d}F_{ce} = \epsilon^{cdef}\nabla_{e}F_{cd}$ hence $3\epsilon^{cdef}\nabla_{c}F_{de} = 3\epsilon^{cdef}\nabla_{[c}F_{de]} = 0$ by virtue of the homogeneous Maxwell equations thus we have the desired result $\tilde{\nabla}_{a}B^{a} = 0$.