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I have the 2-form $*F=q\sin \theta d\theta \wedge d\phi$, how can I determine the eletric and magnetic fields from that?

I have tried wrtting F in the vector potential form for them finding the magnetic and eletric fields, but I have not gota anywhere

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  • $\begingroup$ I'm assuming you are using the same conventions as in wikipedia, so that $*F$ is their $G$. Since you have purely spatial terms, there is no electric field, and going in polar coordinates, you can recognise an electric monopole (Coulomb field). $\endgroup$
    – LPZ
    Commented Mar 2, 2023 at 19:07
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    $\begingroup$ We need more detail on what you don't understand. Do you know what the $*$ is for? Do you know how to remove it to just have $F$? Do you know how this would relate to the tensor form $F_{\mu\nu}$? $\endgroup$
    – Mike
    Commented Mar 2, 2023 at 21:18

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I'm assuming we're in Minkowski space, and the representation of the Hodge dual of the electromagnetic field strength tensor, $*F$, is in spherical coordinates.

The components of the electromagnetic field strength tensor consist of the components of the electric and magnetic fields. First apply the Hodge dual to both sides of the provided equation.

$$*(*F) = *(q\sin\theta d\theta\wedge d\phi)$$

As we are working on a 3+1 Lorentzian manifold, the Hodge dual of the Hodge dual of the 2-form $F$ is

$$*(*F) = -F$$

and the Hodge dual on the basis two-form in spherical coordinates $(t,r,\theta,\phi)$ is

$$ *(d\theta\wedge d\phi) = \frac{1}{\sqrt{-g}}dt\wedge dr = \frac{1}{r^2\sin\theta} dt\wedge dr $$

So

$$F = \frac{-q}{r^2}dt\wedge dr$$

In this particular coordinate basis, the electromagnetic field strength tensor has components

$$F_{\mu\nu} = \begin{pmatrix} 0 & -E_r & -E_\theta & -E_\phi \\ E_r & 0 & B_\phi & -B_\theta \\ E_\theta & -B_\phi & 0 & B_r \\ E_\phi & B_\theta & -B_r, & 0 \\ \end{pmatrix}$$

where $E_i$ and $B_i$ are the components of the electric and magnetic fields respectively. In our case, the only (two) nonzero components are

$$F_{tr} = -F_{rt} = \frac{-q}{r^2}$$

Therefore there is only the radial electric field from a point charge at the origin, and no magnetic field.

$$\vec{E} = \frac{q}{r^2}\vec{e}_r \\ \vec{B} = \vec{0}$$

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Hint: the differential 1-forms $\{dt,dr,r\,d\theta,r\sin\theta\,d\phi\}$ are 'orthonormal' with respect to the (dual of) Minkowski inner product $\eta=-dt^2+dr^2+r^2\,d\theta^2+r^2\sin^2\theta\,d\phi^2$. So, try writing your $*F$ in terms of these 4 basic 1-forms. Then, the Hodge dual is (up to sign) just a matter of replacing the two 1-forms which appear by those which do not. Finally, recall that $F=dt\wedge E+B$, so once you figure out $F$, you get $E$ and $B$ by inspection.

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