# How to find the potential 4-vector in a curved spacetime context?

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

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

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

$$$$B^r = F(r, \theta, \phi), B^\theta = G(r, \theta, \phi), B^\phi = H(r, \theta, \phi), \\ E^r = E^\theta = E^\phi = 0,$$$$

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:

$$$$\vec{E} = -\vec{\nabla} \Phi - \partial_t \vec{A}, ~\vec{B} = \vec{\nabla} \times \vec{A}$$$$

$$\Phi$$ 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:

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

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.

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.