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The Stress-Energy Tensor for electromagnetism is given by: $$ T_{\mu \nu} = F_{\mu}\,^{\alpha}F_{\nu\alpha}-\frac{1}{4}g_{\mu\nu}F_{\alpha\beta}F^{\alpha\beta} $$

How can I find $F_{\mu\nu}$ in terms of $T_{\mu\nu}$?

Rewriting the above equation using: $$ T_{\mu\nu}=- F_{\mu \alpha} g^{\alpha\beta} F_{\beta\nu} + \frac{1}{4} g_{\mu \nu}g^{\alpha\beta}F_{\beta\delta}g^{\delta\gamma} F_{\gamma\alpha}$$ from which we can write the following $4\times4$ matrix equation for the three matrices $T,\,F,\,g$, where $T$ is symmetric, $F$ is anti-symmetric and $g$ is symmetric and invertible: $$ T = -F g^{-1} F+\frac{1}{4}\left(\mathrm{Tr}\, \left[g^{-1}Fg^{-1}F\right]\right)\,g$$

The only way I can think of is writing down 10 equations (as there are free components in $T^{\mu\nu}$) and then trying to find the 6 unknowns (as there are free components of $F^{\mu\nu}$).

Is there a better way to do this?

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    $\begingroup$ It is not linear algebra since it involves a quadratic equation of matrices. $\endgroup$ – Valter Moretti Jun 12 '14 at 17:16
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    $\begingroup$ Is there a case where you know $T$ and not $F$? $\endgroup$ – Kyle Kanos Jun 12 '14 at 17:38
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    $\begingroup$ There is, if you change $F$ to $-F$, $T$ remains fixed...Is it the only case? $\endgroup$ – Valter Moretti Jun 12 '14 at 17:42
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    $\begingroup$ @Psycho_pr: there is no gauge freedom in choosing $F$. The whole point of gauge symmetry is that you can change gauge in $A$ all you want, and it wont affect the value of $F$ $\endgroup$ – Jerry Schirmer Jun 12 '14 at 20:29
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    $\begingroup$ @PPR One way to find $F$ from the metric appears in Wald Phys. Rev. D 10, 6 1974 Black hole in a uniform magnetic field. A Killing vector is enough to find an $F$ that solves Maxwell's equations without the metric, but the specific result requires the metric. $\endgroup$ – auxsvr Jun 13 '14 at 6:56
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See Edit below, the original answer is not completely correct.

There is no gauge freedom in $F$. $F$ is gauge invariant.

In fact, $F$ is completely measurable. It's components are the Electric and Magnetic fields, so you just go out with a set of test charges and measure $E$ and $B$ and you've got $F$.

One hint that $T$ and $F$ do not contain the same amount of information is that they have different numbers of independent components. $F$ has 6 independent components as an antisymmetric tensor, while $T$ has 10 as a symmetric one. This isn't a proof of anything, but a hint that they are capturing different things.

If you are working locally (ie, at a point), the simple way to see this explicitly is to use Lorentz transformations. The stress energy tensor has $10$ independent components since it is a symmetric tensor, we can use the $6$ Lorentz transformations to diagonalize $T$. Then we have 4 equations

\begin{eqnarray} T_{00} &=& \frac{1}{2}\left(E^2 + B^2\right) \\ T_{ii} &=& (E_i^2 - \frac{1}{2}E^2) + (B_i^2 - \frac{1}{2}B^2) \end{eqnarray} There is no sum over $i$ implied in the second equation, it's just a quick way of writing the 3 spatial equations.

You can see that there is no way to solve these. For one thing, there are more components in $E$ and $B$ than there are in $T$ in this frame. For another, since the fields appear squared, there is no way to determine the sign of any of the components of $E$ or $B$.

Additionally you can't tell the difference between $E$ and $B$ (ie, given $T_{00}$, who is to say whether you had $E^2=0$ or $B^2=0$ or neither)? This last point is a consequence of the electromagnetic duality: in the absence of matter, the physics of E/M is invariant under $E\rightarrow B$, $B\rightarrow -E$.

EDIT:

The above is not quite correct in detail (though I think the conclusion is correct). For whatever reason I neglected the fact that there are always 10 components of $T_{\mu\nu}$, so there are always 10 equations, even in the frame in which $T$ is diagonal. In particular, there are also conditions like \begin{eqnarray} 0 &=& E_x E_y + B_x B_y \\ 0 &=& E_x B_y - E_y B_x \end{eqnarray} So my counting argument, "There are more variables than equations," was incorrect. This fits with the idea that $T$ has more components than $E$--if anything based on counting you would think that computing $T$ from $E$ was the harder thing to do. (In fact this is generically true--the stress energy tensors you get from field theory are not the most general stress energy tensors you can write down. There are plenty of stress energy tensors you can write down that won't come from a lagrangian).

The real reason this won't work, as far as I can tell, is the electromagnetic duality as well as the fact that everything is squared. There just isn't a way to distinguish $E$ from $B$ if you write out all the components. In other words, the duality means that the equations have a degeneracy, so there are fewer equations than it naively appears, so you can't solve for all the components.

On the other hand, if you know $T$ everywhere, not just locally, that is a totally different story. That's because (1) if you know $T$ everywhere you can differentiate it, and (2) $\partial_\mu T^{\mu\nu}=0$ is just maxwells equations $\partial_\mu T^{\mu\nu}=\partial_\mu F^{\mu\nu}$, possibly up to an overall factor. So then, up to the usual caveats about needing to know the boundary conditions, if you know $T$ everywhere you can solve maxwell's equations to obtain $F$.

Moral: don't believe everything you read on the internet.

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    $\begingroup$ I cannot understand your statement stating that, since $T$ is (real) symmetric, it can be written into a diagonal form in a Minkowskian reference frame. As a matter of fact, as $T$ is real and symmetric there is $R \in O(4)$ such that $RTR^{-1} = T_0$ where $T_0$ is in diagonal form. However there is no guarantee for having $R\in O(1,3)$! In other words the said basis is non Minkowskian one and thus identifying the components of $F$, in that basis, with $E$ and $B$ makes no sense, in general. $\endgroup$ – Valter Moretti Jun 13 '14 at 9:26
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The easiest way I can think of in Minkowski space, short of doing the algebra in terms of matrices, is to use $$\begin{split} f^a &= \rho E^a + \epsilon^{abc} J_b B_c = \partial_b T^{ab} - \epsilon_0 \mu_0 \partial_t S^a\\ \frac{\partial T^{00}}{\partial t} &= - \vec{J}\cdot \vec{E}- \vec{\nabla} \cdot \vec{S} \end{split},$$ with $S^a \equiv \frac{1}{\mu_0} \epsilon^{abc} E_b B_c = T^{0a},$ $a,b \in \{1,2,3\}$, and hope that the fields will be easy to discern.

Perhaps the more symmetric form $$\frac{1}{2} (F_\mu{}^\rho F_{\nu\rho} + \star F_\mu{}^\rho \star\! F_{\nu\rho}),$$ with the star denoting the Hodge dual, will prove easier to handle in curved spacetime, if you manage to break $T_{\mu\nu}$ into a sum of matrices of similar structure.

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  • $\begingroup$ Your symmetric form is nice, it makes the duality manifest. That is basically the point, given $T$ there is no way to know if you are solving for $F$ or $\star F$. $\endgroup$ – Andrew Jun 13 '14 at 0:40
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    $\begingroup$ @Andrew That is beside the point, since locally a Lorentz transformation can convert from one field into another. The point is that you can find equations that give $F_{\mu\nu}$ from $T_{\mu\nu}$, but this map won't be 1-1 and there's no need for it to be so. $\endgroup$ – auxsvr Jun 13 '14 at 6:39

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