Given a metric, for example

$$ ds^2 = -A(r)dt^2 + B(r)dr^2 + C(r)d\theta^2 + D(r) d\phi^2, $$ and assuming that the fields go as $$ \textbf{E} = E(r)\hat{r} \quad \text{and} \quad \textbf{B}=0, $$ how does one compute the EM field tensor from this? I keep trying to express the "standard" field tensor, $$ F^{\mu \nu }={\begin{bmatrix}0&-E_{x}&-E_{y}&-E_{z}\\E_{x}&0&-B_{z}&B_{y}\\E_{y}&B_{z}&0&-B_{x}\\E_{z}&-B_{y}&B_{x}&0\end{bmatrix}} $$ in spherical coordinates but find myself just ignoring the metric. I've also tried going back to the definition, $$ F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu $$ but I am unsure how to express the four-derivatives and four-potentials. It seems like I'm missing something fairly fundamental. Any help would be appreciated.

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    $\begingroup$ Why do you expect that you would be able to recover the EM field tensor from the metric alone? $\endgroup$ – knzhou May 5 '20 at 22:55
  • $\begingroup$ My apologies. I've updated the question. $\endgroup$ – Nick M May 5 '20 at 23:29
  • $\begingroup$ This post is actually $3$ questions. Since this an EM problem, the space is Minkowski, i.e., the line element in Cartesian is $ds^2 = -dt^2 + dx^2 + dy^2 + dz^2.$ To convert the spatial part of the Minkowsk line element $dx^2 + dy^2 + dz^2$ to spherical coordinates use the spherical coordinate transformation. Regarding the tensor, $B=0$ so you're only left with the projections of the $E(r)$ field on $(x,y,z).$ Either transform the tensor, or transform $E(r)$ to Cartesian coordinates. You can calculate $F_{\mu \nu}$ by either using the metric or the vector potentials in spherical coordinates. $\endgroup$ – Cinaed Simson May 6 '20 at 19:48
  • $\begingroup$ You should remove the "general relativity" tag and add the "special relativity" tag. $\endgroup$ – Cinaed Simson May 6 '20 at 19:54

Note that $F = dA$ is a $(0,2)$ anti-symmetric tensor field on spacetime (i.e a differential $2$-form). If you want to see how the components of this tensor field change as you perform a change of coordinates, you have to apply the tensor-field component transformation law. For example, if $(x^0(\cdot), x^1(\cdot), x^2(\cdot), x^3(\cdot))$ and $(y^0(\cdot), y^1(\cdot), y^2(\cdot), y^3(\cdot))$ are two sets of coordinates, then \begin{align} F_{(y), \mu \nu}&:= F \left(\dfrac{\partial}{\partial y^{\mu}}, \dfrac{\partial}{\partial y^{\nu}} \right) \\ &= F \left(\dfrac{\partial x^{\alpha}}{\partial y^{\mu}}\dfrac{\partial}{\partial x^{\alpha}}, \dfrac{\partial x^{\beta}}{\partial y^{\nu}}\dfrac{\partial}{\partial x^{\beta}} \right) \\ &= \dfrac{\partial x^{\alpha}}{\partial y^{\mu}}\dfrac{\partial x^{\beta}}{\partial y^{\nu}} F \left(\dfrac{\partial}{\partial x^{\alpha}}, \dfrac{\partial}{\partial x^{\beta}} \right) \\ &= \dfrac{\partial x^{\alpha}}{\partial y^{\mu}}\dfrac{\partial x^{\beta}}{\partial y^{\nu}} F_{(x), \alpha \beta} \end{align}

Here, I use the notation $F_{(y), \mu \nu}$ to mean the $\mu \nu$ component of the tensor field $F$ expressed in the $y$ coordinate system.

So, what the above equation tells you is that if you know all the components of the field strength tensor field in one coordinate system, and you wish to change coordinates, then to find the components of that tensor in the other coordinate system, you apply the above rule (summation convention used throughout).

For this, there is absolutely no need for the metric tensor field. The metric tensor only comes in if you want to "raise/lower indices". So, do you want the tensor with two lower indices? One upper, one lower index? Or two upper indices? That's the only place the metric tensor field comes into play.

Btw, in your example, you have the term $C(r)\, d \theta^2$, but for spherical coordinates, it is $r^2 \sin^2 \theta\, d \theta^2$, so it should be $C(r, \theta)\, d \theta^2$ (of course, there are slightly different conventions for which angle is which, but my point is that the metric coeffcients in spherical coordinates do not depend only on $r$).


It looks like you are convinced that the metric should determine the EM field strength tensor. This is simply not true. The electromagnetic field is not a geometric entity attached to the metric structure of spacetime itself (at least not in usual GR, you can do that in higher dimensional theories if you do right compactifications, etc. See, for example, KK compactifications).

The electromagnetic degrees of freedom are additional degrees of freedom on top of the metric structure of spacetime. You need to solve the Euler-Lagrange equations obtained from the Lagrangian $\mathcal{L}_{EM}=-\frac{1}{4}\sqrt{-g}F^{\mu\nu}F_{\mu\nu}+A_\mu J^\mu$ for the vector potential $A_\mu$ in order to obtain eventually obtain the field strength tensor. So, the EM field strength tensor would depend on the configuration of the sources $J^\mu$. The metric structure of the spacetime would absolutely influence your solutions because it changes the structure of the derivatives and the differential volume element, etc. But you cannot just get the EM field strength tensor out of the spacetime metric.

  • $\begingroup$ Thank you this makes sense. But for the Minkowski metric, one can express the EM field strength tensor in general as what I called the "standard field tensor." Is there a way to express the field tensor in general given another metric? $\endgroup$ – Nick M May 7 '20 at 19:21
  • $\begingroup$ @NickM The matrix representation that you mention is just naming of different components of the field strength tensor. Since the field strength tensor is still anti-symmetric in general relativity, you can do the same naming if you wish. It still has just $6$ independent components in $3+1$ dimensions. $\endgroup$ – Dvij D.C. May 7 '20 at 20:09
  • $\begingroup$ @NickM What I mean is that the matrix representation that you have mentioned is what defines what we can the electric field and magnetic field, i.e, that electric field is the components of the form $F^{0i}=-F^{i0}$ of the field strength tensor and that magnetic field is the components of the form $F^{ij}=-F^{ji}$ (for $i\neq j)$ of the field strength tensor. There is no non-trivial physical information there. $\endgroup$ – Dvij D.C. May 7 '20 at 20:12

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