# How to properly construct the electromagnetic tensor in curved space-time?

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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 all 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. asked Jul 23, 2014

The key is to understanding why $F_{\mu\nu}=e^{(\eta+\lambda)/2}\ \varepsilon/r^2$ must hold. This is a consequence of Gauss' law as you are looking for a solution with charge $\varepsilon$: $\int_{S^2_\infty} *F=4\pi \varepsilon$. Your choice for $A^\mu$ doesn't do that.

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There is no reason to assume that the contravariant four-potential is the same as in flat space and it is not if you look to the final answer for the electric field.

As mentioned in the question, to compute the electric field one just has to solve Maxwell equations (and the answer is rather obvious if one uses the well-known formula for the divergence of some tensor $V_i$: $\nabla^i V_i = \frac{1}{\sqrt{det(g)}} \partial_i (\sqrt{det(g)} g^{ij} V_j)$).

answered Jul 24, 2014 by (5,120 points)
edited Jul 24, 2014

As I suspected. Most books don't leave this explicitly... they just go on defining an electromagnetic tensor with the field components and that's all. I am just starting to learn GR. Thank you for your reply.

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