# EM tensor dual in spherical coordinates?

I'm trying to find the dual of the EM tensor in spherical coordinates. Defining the metric of the sign (-+++) $$ds^2 = - dt^2 + dr^2 + r^2 (d\theta^2 + \sin(\theta)^2d\phi^2)$$ I identified the EM tensor as $$F^{\mu\nu} = \left( \begin{array}{cccc} 0 & \text{Er} & \frac{\text{E\theta }}{r} & \frac{\text{E\phi } \csc (\theta )}{r} \\ -\text{Er} & 0 & \frac{\text{B\phi }}{r} & -\frac{\text{B\theta } \csc (\theta )}{r} \\ -\frac{\text{E\theta }}{r} & -\frac{\text{B\phi }}{r} & 0 & \frac{\text{Br} \csc (\theta )}{r^2} \\ -\frac{\text{E\phi } \csc (\theta )}{r} & \frac{\text{B\theta } \csc (\theta )}{r} & -\frac{\text{Br} \csc (\theta )}{r^2} & 0 \\ \end{array} \right)$$ where $$\csc(\theta) = 1/\sin(\theta)$$. Now finding the dual of the tensor as $$\star F^{\mu\nu} = \epsilon^{abcd} F_{cd} = \epsilon^{abcd} g_{ij} g_{kl}F^{jl}$$ ,where $$\epsilon$$ is the levi-civita symbol. I get $$\star F^{\mu\nu}= \left( \begin{array}{cccc} 0 & r^2 \sin (\theta ) \text{Br}(t,r,\theta ,\phi ) & r \sin (\theta ) \text{B\theta }(t,r,\theta ,\phi ) & r \text{B\phi }(t,r,\theta ,\phi ) \\ -r^2 \sin (\theta ) \text{Br}(t,r,\theta ,\phi ) & 0 & -r \sin (\theta ) \text{E\phi }(t,r,\theta ,\phi ) & r \text{E\theta }(t,r,\theta ,\phi ) \\ -r \sin (\theta ) \text{B\theta }(t,r,\theta ,\phi ) & r \sin (\theta ) \text{E\phi }(t,r,\theta ,\phi ) & 0 & -\text{Er}(t,r,\theta ,\phi ) \\ -r \text{B\phi }(t,r,\theta ,\phi ) & -r \text{E\theta }(t,r,\theta ,\phi ) & \text{Er}(t,r,\theta ,\phi ) & 0 \\ \end{array} \right)$$ But I was told that this isn't ok, where was my mistake? Is this definition of dual right even in non Minkowski metric?

• When I tried to replicate the question in cylindrical coordinates, my $\star F^{\mu\nu}$ differed by a factor of $1/\sqrt{-g}$ from the "right one". Apr 10, 2022 at 12:08
• Your metric has an error: it should be $\sin^2 \theta$ not $\sin \theta$ Apr 10, 2022 at 12:33
• @Eletie I think the word "typo" is better than "error". When people writes equation with mathjax, they become blind so do I. Apr 10, 2022 at 12:44
• yes a typo, I edited it, thank you. Apr 10, 2022 at 13:21
• $\epsilon^{abcd}=\varepsilon^{abcd}/\sqrt{-g}$ and $\epsilon_{abcd}=\sqrt{-g}\varepsilon_{abcd}$, where $\varepsilon$ is the symbol.
– Kosm
Apr 10, 2022 at 15:47