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How does $\mathbf{F}=\mathbf{E}+i\mathbf{B}$ generalize to curved space? (where $\mathbf{F}$ is the bivector of electromagnetism).

Here is what I am struggling with:

On the one hand, I can expand $\mathbf{F}$ as follows:

$$ \mathbf{F}=E_x\gamma_0\wedge \gamma_1+E_y\gamma_0\wedge \gamma_2+E_z\gamma_0\wedge \gamma_3+B_x\gamma_2\wedge\gamma_3+B_y\gamma_1\wedge \gamma_3+B_z\gamma_1\wedge \gamma_2 \tag{1} $$

Or I can do it as follows:

$$ \mathbf{F}=(E_x+IB_x)\gamma_0\wedge \gamma_1+(E_y+IB_y)\gamma_0\wedge \gamma_2+(E_z+IB_z)\gamma_0\wedge \gamma_3 \tag{2} $$

Since $I=\gamma_0\wedge \gamma_1\wedge \gamma_2 \wedge \gamma_3$, (1) is equivalent to (2).

But if I convert (1) to curvilinear coordinates, I do not end up to the same end result as if I had started with (2). So what is the correct way to go to general curvilinear coordinates: $(\gamma_0,\gamma_1,\gamma_2,\gamma_3)\to(\mathbf{e}_0,\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3)$ ?

Replacing from (1) I get

$$ \mathbf{F}=E_x\mathbf{e}_0 \wedge \mathbf{e}_1+E_y\mathbf{e}_0 \wedge \mathbf{e}_2+E_z\mathbf{e}_1\wedge \mathbf{e}_3+B_x\mathbf{e}_2\wedge\mathbf{e}_3+B_y\mathbf{e}_1\wedge \mathbf{e}_3+B_z\mathbf{e}_1\wedge \mathbf{e}_2 \tag{3} $$

whereas if I start with (2) I instead get:

$$ \mathbf{F}=(E_x+IB_x)\mathbf{e}_0\wedge \mathbf{e}_1+(E_y+IB_y)\mathbf{e}_0\wedge \mathbf{e}_2+(E_z+IB_z)\mathbf{e}_0\wedge \mathbf{e}_3 \tag{4} $$

(3) is not the same as (4) because (4) expands as (just the x part):

$$ \mathbf{F}_x=E_x\mathbf{e}_0\wedge \mathbf{e}_1 + B_x \gamma_0\wedge \gamma_1\wedge \gamma_2 \wedge \gamma_3 \wedge \mathbf{e}_0\wedge \mathbf{e}_1 $$

In the case of (4) the $Es$ remain "orthogonal" to the $Bs$ in curved space, whereas for (3), the $Es$ are free to distort generally with respect to the $Bs$.

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Where does (2) come from? It seems to be wrong on many levels. For example, $(E_x + IB_x)$ is the addition between the base field with a 4-form; this is then "multiplied" by a base 2-form $\gamma_0\wedge\gamma_1$ and your last equality seems to suggest a mixed notion of multiplication. Observe that the exterior product of more than 4 1-forms in a space of dimension 4 is always 0.

I think one should treat the "complexification" of the Faraday tensor as just a mathematical convenience to express the complex structure on $\bigwedge^2 TM$ when $M$ is a Lorentzian manifold of dimension 4. For starters, we observe that, as vector spaces, $\bigwedge^2 TM\cong\mathbb C^3$. However, we can endow the space of 2-forms with a complex structure by observing that this is given by Hodge duality. Indeed, if we operate the usual decomposition of $F$ in terms of polar and axial components $(\mathbf E,\mathbf B)$, we have (modulo a sign) that $\star F = (\mathbf B,-\mathbf E)$. Therefore, if we evocatively write $$F = \mathbf E + i\mathbf B,$$ Hodge duality is indeed multiplication/division (depending on sign convention) by $i$. By embedding $\bigwedge^3 TM\cong\mathbb R^4$ into $\mathbb C^4$ we can express Maxwell's equations in the unified form $$\text diF +i\text dF+J=0.$$ More generally, if we distinguish between electric and magnetic currents $J_e$ and $J_m$, we can write the more general and symmetric equation

$$\text diF +i\text dF+J_e + iJ_m = 0.$$

This is just a convenient way of encoding a system of many equations, and in particular the pair $$\text d F + J_m = 0,\qquad \text d\star F + J_e = 0$$ into a "single" complex equation.

So, back to the original question, the expression, when interpreted in this way, is already generalised to arbitrary Lorentzian manifolds.

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  • $\begingroup$ It is difficult for me to understand... It sound like you are suggesting that $F=E+iB$ lives in the tangent space of a manifold, and not "within the manifold itself"? Thus, $F=E+iB$ is valid irrespective of curvature because the vectors of the tangent space are always orthogonal to each other (and thus do not require curvilinear coordinates to express them)? $\endgroup$ – Alexandre H. Tremblay Jul 5 at 0:33
  • $\begingroup$ Vector fields are sections of the tangent bundle of a manifold, they don't just live within it. More generally, any tensor field is the section of a vector bundle over a manifold. $\endgroup$ – Phoenix87 Jul 5 at 7:19
  • $\begingroup$ So then are $\mathbf{E}$ and $\mathbf{B}$ the derivatives of some curve on the manifold (or a wedge product of the derivative of two curves)? $\endgroup$ – Alexandre H. Tremblay Jul 5 at 10:56
  • $\begingroup$ No E and B are vectors in $\mathbb R^3$ and the space-time manifold is four dimensional, so they can't be tangent to curves on the manifold. Indeed the EM field is described by a 2-form $\endgroup$ – Phoenix87 Jul 5 at 15:44
  • $\begingroup$ Would you say that your description is equivalent to the one described here: en.wikipedia.org/wiki/Classification_of_electromagnetic_fields in which it is stated that "The electromagnetic field at a point p (i.e. an event) of a Lorentzian spacetime is represented by a real bivector F = Fab defined over the tangent space at p." $\endgroup$ – Alexandre H. Tremblay Jul 5 at 19:25

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