A p-form has the definition:
$$ \mathbf{\omega} =\omega_{|i_1…i_p|} \mathbf{d}x^{i_1} \wedge…\wedge \mathbf{d}x^{i_p} \tag{1}$$
Suppose a 2-form in Cartesian coordinates takes the form
$$\mathbf{^*F}=E_{x} \mathbf{d}y \wedge \mathbf{d}z \tag{2}$$
Then its corresponding components in spherical coordinates are
$$\mathbf{^*F}=E_{r} (r\mathbf{d} \theta) \wedge (rsin\theta\mathbf{d} \phi) \tag{3}$$
But I do not see why the $r$ and $rsin \theta$ factors are needed here, shouldn't (2) simply be
$$\mathbf{^*F}=E_{r} \tag{4}$$
because all these "additional angular variable factors" for transforming from rectangular coordinates to spherical coordinates have already been taken into account by the definition of $E_r$ alone?
If $(3)$ is the correct definition then shouldn't $(1)$ be
$$\mathbf{\omega} =\omega_{|i_1…i_p|} (\sqrt{|g_{i_1i_1}|}\mathbf{d}x^{i_1}) \wedge…\wedge (\sqrt{|g_{i_pi_p}|}\mathbf{d}x^{i_p}) \tag{5}$$
(where $g_{kn}$ is the metric tensor)
instead?
In other words is a 2-form whose components are described using spherical coordinates defined as, e.g.(with all other components=$0$)
$$\omega= \omega_{23} (x^1\mathbf{d} x^2) \wedge (x^1sinx^2\mathbf{d} x^3)$$
where $x^1=r, x^2=\theta, x^3=\phi $?
Best Answer
This is not coming from a change of coordinates from the $xyz$ coordinates to spherical. They're writing the orthonormal basis for the $1$-forms in spherical coordinates as $$dr, r\,d\theta, \quad\text{and}\quad r\sin\theta\,d\phi.$$ (For example, the Euclidean metric in spherical coordinates is given by the $2$-tensor $$dr\otimes dr + r^2\,d\theta\otimes d\theta + (r\sin\theta)^2\,d\phi\otimes d\phi.$$ We then read off the orthonormal basis from this.)
Remember that to compute the Hodge star, it is easiest to compute with an orthonormal coframe. So if $\mathbf F = E_x\,dx + E_y\,dy + E_z\,dz$, then $\star\mathbf F = E_x\,dy\wedge dz+ \dots$. Now in spherical coordinates, they're writing $\mathbf F = F_r\,dr + F_\theta(r\,d\theta) + F_\phi(r\sin\theta\,d\phi)$, and then they're applying $\star$ to this. In particular, $\star dr = (r\,d\theta)\wedge (r\sin\theta\,d\phi)$.