Gravitation By Charles Misner, Kip Thorene, John Wheeler page 109.
$\displaystyle \varphi=\arctan\frac{y}{x},\cos\theta=\frac{z}{r} \Rightarrow d\varphi=\frac{xdy-ydx}{x^2+y^2}, -d(\cos\theta)=\frac{-dz}{r}+\frac{z}{r^3}(xdx+ydy+zdz)$ and $*F=-ed(\cos\theta)\wedge d\varphi$ was given.(here $d$ represent the form.)
It claimed that $*F=(e/r^3)(xdy\wedge dz+ydz\wedge dx+zdx\wedge dy)$.
However, I've been trying for hours, even with Mathematica, but I couldn't get the right number.
Could you show me how to do this calculation please?
You have $$-d(\cos\theta)=\frac{z(x\,dx+y\,dy)-(x^2+y^2)dz}{r^3}$$ $$(x\,dx+y\,dy)\wedge(x\,dy-y\,dx)=(x^2+y^2)dx\wedge dy$$ $$z(x\,dx+y\,dy)\wedge\frac{x\,dy-y\,dx}{x^2+y^2}=z\,dx\wedge dy$$ $$-(x^2+y^2)dz\wedge\frac{x\,dy-y\,dx}{x^2+y^2}=x\,dy\wedge dz+y\,dz\wedge dx$$ Now add the last two equations and divide by $r^3$.