If we are studying in a spherically symmetrical spacetime we will have the following metric,

\begin{equation} \text{d}s^2 = -f(r) \ \text{d}t^2 + f^{-1}(r) \ \text{d}r^2 + r^2 (\text{d} \theta^2 + \sin^{2}{\theta} \ \text{d}\phi^2). \end{equation}

A generic (non-time varying) magnetic field in this spacetime can be described as:

\begin{equation} B^r = F(r, \theta, \phi), B^\theta = G(r, \theta, \phi), B^\phi = H(r, \theta, \phi), \\ E^r = E^\theta = E^\phi = 0, \end{equation}

and knowing that $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$, how can we define this $A_\mu$?

What I'm having difficulty with is because, out of general relativity, in curved coordinates it would be easy:

\begin{equation} \vec{E} = -\vec{\nabla} \Phi - \partial_t \vec{A}, ~\vec{B} = \vec{\nabla} \times \vec{A} \end{equation}

$\Phi$ would be $0$, and $\vec{A} = \vec{A}(q_1, q_2, q_3)$. Then the $\vec{\nabla} \times \vec{A}$ would be defined by:

\begin{equation} \vec{\nabla} \times \vec{A} = \frac{1}{h_2 h_3}\left[ \partial_2 (h_3 A_3) - \partial_3 (h_2 A_2) \right]\hat{e_1} + ... \end{equation}

where: ${h_i}^2 = g_{ii}$.

But, in general relativity, I don't know if it's right to use this relation because considering the whole metric (with the temporal part) $h_0$ would be imaginary.

So, maybe, the real issue is understanding right how to properly define the differential operators in general relativity.


1 Answer 1


It's more useful to work with an explicitly covariant formulation of electrodynamics. Working with the mostly-minus signature $(+---)$, we define the 4-potential $A^\mu = (\Phi/c, \vec A)$ and its covector partner $A_\mu = g_{\mu\nu} A^\nu$. The Faraday tensor is given by $F_{\mu\nu} = \partial_\mu A_\nu -\partial_\nu A_\mu$, and the electric and magnetic fields are defined from the Faraday tensor as $$E^i/c = F^{i0} = g^{i\mu}\partial_\mu A^0-g^{0\mu}\partial_\mu A^i$$ $$B^i = -\frac{1}{2}\epsilon^{ijk} F_{jk}= -\frac{1}{2}\epsilon^{ijk}\bigg[\partial_j\big(g_{k\mu}A^\mu\big)- \partial_k\big(g_{j\mu} A^\mu\big)\bigg] $$

It's not hard to show that in an orthogonal coordinate chart $(ct,x^1,x^2,x^3)$ in which the metric takes the form $g=\mathrm{diag}(h_0,h_1,h_2,h_3)$, the electric and magnetic fields are given by $$E^i = h_i \frac{\partial \Phi}{\partial x^i} - h_0 \frac{\partial A^i}{\partial t}$$ $$B^i = -\epsilon^{ijk} \bigg[\frac{\partial h_k}{\partial x^j} A^k + h_k \frac{\partial A^k}{\partial x^j}\bigg]$$ where we've abandoned the Einstein summation convention. In cartesian coordinates, we have $h_0=1$ and $h_1=h_2=h_3=-1$ and so the electric and magnetic fields reduce to their familiar expressions. In spherical coordinates, we have $h_0=1,h_1=-1,h_2=-(x^1)^2$, and $h_3=-(x^1)^2\sin^2(x^2)$ where $(x^1,x^2,x^3)\equiv (r,\theta,\phi)$, and the electric and magnetic fields can be extracted with a bit of algebra.

All of this raising and lowering of indices emerges from our desire to make contact with elementary electromagnetism, which is not explicitly relativistically covariant$^\dagger$ and which takes place in cartesian coordinates. One could argue that our lives would be made easier by adopting a more elegant formulation from the outset.

In terms of differential forms, the fundamental physical entity is the 2-form Faraday tensor $F$. The force $f$ on a moving particle is given by $$f_\mu = qF_{\mu\nu} u^\nu$$ where $q$ and $u$ are the particle's electric charge and 4-velocity, respectively. The Maxwell equations are given by $$ \mathrm dF = 0$$ $$ \mathrm d(\star F) = J$$ where $\star$ is the hodge operator and $J$ is the current 3-form. The homogeneous equation $\mathrm dF=0$ implies that locally there exists some 1-form $A$ such that $F = \mathrm dA \iff F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$; in terms of this $A$, the Maxwell equations reduce to $$ \mathrm d\star \mathrm d A = \mathrm J$$

As is the case in the elementary formulation, this has the advantage of taking the homogeneous Maxwell equations into account automatically ($\mathrm d^2 = 0 \implies \mathrm d( \mathrm dA) = 0$). On the other hand, it possesses a gauge redundancy - for any 0-form $\chi$, $A+ \mathrm d\chi$ is another valid solution to the equations of motion - and if the underlying spacetime is not $\mathbb R^4$, then we may encounter singularities in $A$ which require us to solve the Maxwell equations in patches and then relate the corresponding $A$'s by gauge transformations.

$^\dagger$Electromagnetism is relativistically covariant, but the elementary formulation does not make this immediately obvious.


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