# How to properly construct the electromagnetic tensor in curved space-time? (Part II)

In this question, I am testing what was previously discussed. I can't seem to get my results to match D'Inverno's electromagnetic tensor for a charged point (page 239 of his book - Introducing Einstein's Relativity).

Here are D'Inverno's steps:

• The line element in spherical coordinates is ($\eta$ and $\lambda$ are functions of $r$ only) $$\mathrm{d}s^2 = \mathrm{e}^\eta \mathrm{d}t^2 - \mathrm{e}^\lambda \mathrm{d}r^2 - r^2 (\mathrm{d}\theta^2 + \sin^2\!\theta\ \mathrm{d}\phi^2).$$

• He defines this covariant electromagnetic field tensor: $$F_{\mu\nu} = E(r) \begin{pmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$$

• He then proceeds to find the electric field, and consequently the electromagnetic field tensor by using the source-free Maxwell equations: \begin{align} \nabla_\nu F^{\mu\nu} & = 0 \\ \partial_{[\lambda} F_{\mu\nu]} & = 0. \end{align}

• Solving the differential equation that appears from the equations above, he finds the electric field: $$E(r) = \mathrm{e}^{(\eta+\lambda)/2} \varepsilon/r^2.$$

• He then notes that this field is that of a point charge at infinity ($\eta$ and $\lambda$ go to zero at infinity) where $\varepsilon$ is the electric charge. I managed to reproduce these steps.

Now, here are my steps, using the four-potential procedure (the line element is the same):

• I define my contravariant four-potential (there is just the first element which is the electric potential of a point charge, just as D'Inverno found): $$A^\mu = (\varepsilon/r, 0, 0, 0).$$

• Then I lower the index of this four-potential to find the covariant one: $$A_\mu = (\mathrm{e}^\eta \varepsilon/r, 0, 0, 0).$$

• Finally I apply this equation to build the covariant electromagnetic tensor: $$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu.$$

• The result is $$F_{\mu\nu} = \frac{\mathrm{e}^\eta \varepsilon}{r^2}\! \left(r\frac{\partial\eta}{\partial r}-1\right) \begin{pmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$$

• Where:

$$\frac{\mathrm{e}^\eta \varepsilon}{r^2}\! \left(r\frac{\partial\eta}{\partial r}-1\right) = E(r)$$

And this is different from D'Inverno's electric field. I don't know what I am doing wrong. The calculations are not difficult for this simple case.

The question is, due to these calculations:

Do my contravariant four-potential needs to contain my metric funcions in some way? I was assuming it is just the four-potential for a electric charge in the flat space:

$$A^\mu = (\varepsilon/r, 0, 0, 0).$$

If everything is right, the wrong assumption must be here.