All Electromagnetic Field Tensors
by Stéphane Haussler
In this section, I systematically derive all representations of the Faraday
tensor \(F\) and its Hodge dual with metric signature \((+,-,-,-)\). I
cannot imagine there is no sign error in the calculations below. The Hodge dual
\(⋆\:F\) is also noted \(G\). The starting point is the electromagnetic
field 2-form derivated in the article Maxwell’s Equations via Differential Forms:
\[\begin{split}F^{♭♭} = \left[ \begin{aligned}
- & \E^x \: dt ∧ dx \\
- & \E^y \: dt ∧ dy \\
- & \E^z \: dt ∧ dz \\
& B^x \: dy ∧ dz \\
& B^y \: dz ∧ dx \\
& B^z \: dx ∧ dy \\
\end{aligned} \right]\end{split}\]
All expressions of the Faraday Tensor \(F\) are the same object viewed from
a different perspective. All these objects are strictly equivalent and their
naming (2–form, tensor) merely a matter of convention.
\(F^{♭♭}\)
The Faraday 2–form is our starting point and the expression already given
above is repeated for completeness:
\[\begin{split}F^{♭♭} = \left[ \begin{aligned}
- & \E^x \: dt ∧ dx \\
- & \E^y \: dt ∧ dy \\
- & \E^z \: dt ∧ dz \\
& B^x \: dy ∧ dz \\
& B^y \: dz ∧ dx \\
& B^z \: dx ∧ dy \\
\end{aligned} \right] \\\end{split}\]
From the Faraday 2–form and noting that forms are tensors, we can derive the
row/column representation of the doubly covariant Faraday tensor.
\[\begin{split}F^{♭♭} = \frac{1}{2} \begin{bmatrix}
& + \E^x \: dx ∧ dt & + \E^y \: dy ∧ dt & + \E^z \: dz ∧ dt \\
- \E^x \: dt ∧ dx & & - B^z \: dy ∧ dx & + B^y \: dz ∧ dx \\
- \E^y \: dt ∧ dy & + B^z \: dx ∧ dy & & - B^x \: dz ∧ dy \\
- \E^z \: dt ∧ dz & - B^y \: dx ∧ dz & + B^x \: dy ∧ dz & \\
\end{bmatrix}\end{split}\]
Calculations
Start from the Faraday 2–form
\[\begin{split}F^{♭♭} = \left[ \begin{aligned}
- & \E^x \: dt ∧ dx \\
- & \E^y \: dt ∧ dy \\
- & \E^z \: dt ∧ dz \\
& B^x \: dy ∧ dz \\
& B^y \: dz ∧ dx \\
& B^z \: dx ∧ dy \\
\end{aligned} \right] \\\end{split}\]
Duplicate each 2–form
We just use \(A = \frac{1}{2} (A+A)\).
\[\begin{split}F^{♭♭} = \frac{1}{2} \begin{bmatrix}
- \E^x \: dt ∧ dx & - \E^x \: dt ∧ dx \\
- \E^y \: dt ∧ dy & - \E^y \: dt ∧ dy \\
- \E^z \: dt ∧ dz & - \E^z \: dt ∧ dz \\
+ B^x \: dy ∧ dz & + B^x \: dy ∧ dz \\
+ B^y \: dz ∧ dx & + B^y \: dz ∧ dx \\
+ B^z \: dx ∧ dy & + B^z \: dx ∧ dy \\
\end{bmatrix}\end{split}\]
Flip the exterior products
The trivial operation above now permits to utilize the antisymmetry of the
exterior product \(dx^μ ∧ dx^ν = -dx^ν ∧ dx^μ\), flipping the signes as
needed.
\[\begin{split}F^{♭♭} = \frac{1}{2} \begin{bmatrix}
- \E^x \: dt ∧ dx & + \E^x \: dx ∧ dt \\
- \E^y \: dt ∧ dy & + \E^y \: dy ∧ dt \\
- \E^z \: dt ∧ dz & + \E^z \: dz ∧ dt \\
+ B^x \: dy ∧ dz & - B^x \: dz ∧ dy \\
+ B^y \: dz ∧ dx & - B^y \: dx ∧ dz \\
+ B^z \: dx ∧ dy & - B^z \: dy ∧ dx \\
\end{bmatrix}\end{split}\]
The purpose of this operation is to switch the representation of the Faraday
2–form as a single row of basis 2–forms, to a row/column representation.
Reorder into rows/column representation
From there, we conclude utilizing the free matrix representation of the
Cartan-Hodge formalism, reordering the elements into rows and columns.
\[\begin{split}F^{♭♭} = \frac{1}{2} \begin{bmatrix}
& + \E^x \: dx ∧ dt & + \E^y \: dy ∧ dt & + \E^z \: dz ∧ dt \\
- \E^x \: dt ∧ dx & & - B^z \: dy ∧ dx & + B^y \: dz ∧ dx \\
- \E^y \: dt ∧ dy & + B^z \: dx ∧ dy & & - B^x \: dz ∧ dy \\
- \E^z \: dt ∧ dz & - B^y \: dx ∧ dz & + B^x \: dy ∧ dz & \\
\end{bmatrix}\end{split}\]
With implicit bivector basis, we have the standard representation with abstract
index notation
\[\begin{split}F_{μν} = \begin{bmatrix}
& + \E^x & + \E^y & + \E^z \\
- \E^x & & - B^z & + B^y \\
- \E^y & + B^z & & - B^x \\
- \E^z & - B^y & + B^x & \\
\end{bmatrix}\end{split}\]
Where the field 2-form is related to the Faraday tensor with:
\[F^{♭♭} = \frac{1}{2} \: F_{μν} \: dx^μ ∧ dx^ν\]
For sanity, I refer to Wikipedia for a quick double check of the link between
the Faraday 2–form and the Faraday tensor.
\(F^{♭♯}\)
The starting point is the twice flat Faraday 2–form \(F^{♭♭}\). Applying
the musical ♭♯ operator \(F^{♭♯}=\left(F^{♭♭}\right)^{♭♯}\) results in:
\[\begin{split}F^{♭♯} = \left[ \begin{aligned}
& \E^x \: dt ∧ ∂_x \\
& \E^y \: dt ∧ ∂_y \\
& \E^z \: dt ∧ ∂_z \\
- & B^x \: dy ∧ ∂_z \\
- & B^y \: dz ∧ ∂_x \\
- & B^z \: dx ∧ ∂_y \\
\end{aligned} \right]\end{split}\]
Calculations
Start from the Faraday 2–form
\[\begin{split}F^{♭♭} = \left[ \begin{aligned}
- & \E^x \: dt ∧ dx \\
- & \E^y \: dt ∧ dy \\
- & \E^z \: dt ∧ dz \\
& B^x \: dy ∧ dz \\
& B^y \: dz ∧ dx \\
& B^z \: dx ∧ dy \\
\end{aligned} \right] \\\end{split}\]
Distribute the flat ♭ and sharp ♯ operators
\[\begin{split}F^{♭♯} = \left(F^{♭♭}\right)^{♭♯}
= \left[ \begin{aligned}
- & \E^x \: dt ∧ dx \\
- & \E^y \: dt ∧ dy \\
- & \E^z \: dt ∧ dz \\
& B^x \: dy ∧ dz \\
& B^y \: dz ∧ dx \\
& B^z \: dx ∧ dy \\
\end{aligned} \right]^{♭♯}
= \left[ \begin{aligned}
- & \E^x \: dt^♭ ∧ dx^♯ \\
- & \E^y \: dt^♭ ∧ dy^♯ \\
- & \E^z \: dt^♭ ∧ dz^♯ \\
& B^x \: dy^♭ ∧ dz^♯ \\
& B^y \: dz^♭ ∧ dx^♯ \\
& B^z \: dx^♭ ∧ dy^♯ \\
\end{aligned} \right]\end{split}\]
Expand the sharpened basis covectors
The \(dx^μ\) terms are already flattened, and applying the flattening
operator twice does not modify these terms: \((dx^μ)^♭=dx^μ\). The
sharpened terms are expanded with the Minkowski metric: \((dx^ν)^♯ =
η_{νμ} ∂_μ\).
\[\begin{split}F^{♭♯} = \left[ \begin{aligned}
- & \E^x \: dt ∧ η^{xμ} ∂_μ \\
- & \E^y \: dt ∧ η^{yμ} ∂_μ \\
- & \E^z \: dt ∧ η^{zμ} ∂_μ \\
& B^x \: dy ∧ η^{zμ} ∂_μ \\
& B^y \: dz ∧ η^{xμ} ∂_μ \\
& B^z \: dx ∧ η^{yμ} ∂_μ \\
\end{aligned} \right]\end{split}\]
Identify the non-zero terms
\[\begin{split}F^{♭♯} = \left[ \begin{aligned}
- & \E^x \: dt ∧ η^{xx} ∂_x \\
- & \E^y \: dt ∧ η^{yy} ∂_y \\
- & \E^z \: dt ∧ η^{zz} ∂_z \\
& B^x \: dy ∧ η^{zz} ∂_z \\
& B^y \: dz ∧ η^{xx} ∂_x \\
& B^z \: dx ∧ η^{yy} ∂_y \\
\end{aligned} \right]\end{split}\]
Apply numerical values
\[\begin{split}F^{♭♯} = \left[ \begin{aligned}
- & \E^x \: dt ∧ (-1) ∂_x \\
- & \E^y \: dt ∧ (-1) ∂_y \\
- & \E^z \: dt ∧ (-1) ∂_z \\
& B^x \: dy ∧ (-1) ∂_z \\
& B^y \: dz ∧ (-1) ∂_x \\
& B^z \: dx ∧ (-1) ∂_y \\
\end{aligned} \right] = \left[ \begin{aligned}
& \E^x \: dt ∧ ∂_x \\
& \E^y \: dt ∧ ∂_y \\
& \E^z \: dt ∧ ∂_z \\
- & B^x \: dy ∧ ∂_z \\
- & B^y \: dz ∧ ∂_x \\
- & B^z \: dx ∧ ∂_y \\
\end{aligned} \right]\end{split}\]
We derive the row/column representation of the \(F^{♭♯}\) Faraday tensor:
\[\begin{split}F^{♭♯} = \frac{1}{2} \begin{bmatrix}
& + \E^x \: dt ∧ ∂_x & + \E^y \: dt ∧ ∂_y & + \E^z \: dt ∧ ∂_z \\
+ \E^x \: dx ∧ ∂_t & & - B^z \: dx ∧ ∂_y & + B^y \: dx ∧ ∂_z \\
+ \E^y \: dy ∧ ∂_t & + B^z \: dy ∧ ∂_x & & - B^x \: dy ∧ ∂_z \\
+ \E^z \: dz ∧ ∂_t & - B^y \: dz ∧ ∂_x & + B^x \: dz ∧ ∂_y & \\
\end{bmatrix}\end{split}\]
Calculations
In this version of the calculation, we expand to matrix form using the
symmetries of the mixed exterior product in Minkowski:
Expand using symmetries
\[\begin{split}F^{♭♯} = \left[ \begin{aligned}
& \E^x \: dt ∧ ∂_x \\
& \E^y \: dt ∧ ∂_y \\
& \E^z \: dt ∧ ∂_z \\
- & B^x \: dy ∧ ∂_z \\
- & B^y \: dz ∧ ∂_x \\
- & B^z \: dx ∧ ∂_y \\
\end{aligned} \right] = \left[ \begin{aligned}
& \E^x \: \frac{1}{2} \left( dt ∧ ∂_x + dx ∧ ∂_t \right) \\
& \E^y \: \frac{1}{2} \left( dt ∧ ∂_y + dy ∧ ∂_t \right) \\
& \E^z \: \frac{1}{2} \left( dt ∧ ∂_z + dz ∧ ∂_t \right) \\
- & B^x \: \frac{1}{2} \left( dy ∧ ∂_z - dz ∧ ∂_y \right) \\
- & B^y \: \frac{1}{2} \left( dz ∧ ∂_x - dx ∧ ∂_z \right) \\
- & B^z \: \frac{1}{2} \left( dx ∧ ∂_y - dy ∧ ∂_x \right) \\
\end{aligned} \right]\end{split}\]
Reorder
\[\begin{split}F^{♭♯} = \frac{1}{2} \left[ \begin{aligned}
+ \E^x \: dt ∧ ∂_x + \E^x \: dx ∧ ∂_t \\
+ \E^y \: dt ∧ ∂_y + \E^y \: dy ∧ ∂_t \\
+ \E^z \: dt ∧ ∂_z + \E^z \: dz ∧ ∂_t \\
- B^x \: dy ∧ ∂_z + B^x \: dz ∧ ∂_y \\
- B^y \: dz ∧ ∂_x + B^y \: dx ∧ ∂_z \\
- B^z \: dx ∧ ∂_y + B^z \: dy ∧ ∂_x \\
\end{aligned} \right]\end{split}\]
Reorder in row/column convention
\[\begin{split}F^{♭♯} = \frac{1}{2} \left[ \begin{aligned}
& + \E^x \: dt ∧ ∂_x & + \E^y \: dt ∧ ∂_y & + \E^z \: dt ∧ ∂_z \\
+ \E^x \: dx ∧ ∂_t & & - B^z \: dx ∧ ∂_y & + B^y \: dx ∧ ∂_z \\
+ \E^y \: dy ∧ ∂_t & + B^z \: dy ∧ ∂_x & & - B^x \: dy ∧ ∂_z \\
+ \E^z \: dz ∧ ∂_t & - B^y \: dz ∧ ∂_x & + B^x \: dz ∧ ∂_y & \\
\end{aligned} \right]\end{split}\]
With implicit bivector basis, we have :
\[\begin{split}F_μ{}^ν = \begin{bmatrix}
& + \E^x & + \E^y & + \E^z \\
+ \E^x & & - B^z & + B^y \\
+ \E^y & + B^z & & - B^x \\
+ \E^z & - B^y & + B^x & \\
\end{bmatrix}\end{split}\]
Where the mixed electromagnetic field is related to the covariant-contravariant
Faraday tensor through:
\[F^{♭♯} = \frac{1}{2} \: F_μ{}^ν \: dx^μ ∧ ∂_ν\]
\(F^{♯♯}\)
The starting point is the twice flattened Faraday tensor \(F^{♭♭}\) to
which we apply the ♭♯ operator \(F^{♭♯}=\left(F^{♭♭}\right)^{♭♯}\) and
obtain:
\[\begin{split}F^{♯♯} = \left[ \begin{aligned}
\E^x \; ∂_t ∧ ∂_x \\
\E^y \; ∂_t ∧ ∂_y \\
\E^z \; ∂_t ∧ ∂_z \\
B^x \; ∂_y ∧ ∂_z \\
B^y \; ∂_z ∧ ∂_x \\
B^z \; ∂_x ∧ ∂_y \\
\end{aligned} \right]\end{split}\]
Calculations
Start from the Faraday 2-form
\[\begin{split}F^{♭♭} = \left[ \begin{aligned}
- & \E^x \; dt ∧ dx \\
- & \E^y \; dt ∧ dy \\
- & \E^z \; dt ∧ dz \\
& B^x \; dy ∧ dz \\
& B^y \; dz ∧ dx \\
& B^z \; dx ∧ dy \\
\end{aligned} \right]\end{split}\]
Apply the musical sharp-sharp ♯♯ operator
\[\begin{split}F^{♯♯} = \left(F^{♭♭} \right)^{♯♯}
= \left[ \begin{aligned}
- & \E^x \; dt ∧ dx \\
- & \E^y \; dt ∧ dy \\
- & \E^z \; dt ∧ dz \\
& B^x \; dy ∧ dz \\
& B^y \; dz ∧ dx \\
& B^z \; dx ∧ dy \\
\end{aligned} \right]^{♯♯}\end{split}\]
Distribute the musical operators
\[\begin{split}F^{♯♯} = \left[ \begin{aligned}
- & \E^x \; (dt ∧ dx)^{♯♯} \\
- & \E^y \; (dt ∧ dy)^{♯♯} \\
- & \E^z \; (dt ∧ dz)^{♯♯} \\
& B^x \; (dy ∧ dz)^{♯♯} \\
& B^y \; (dz ∧ dx)^{♯♯} \\
& B^z \; (dx ∧ dy)^{♯♯} \\
\end{aligned} \right]\end{split}\]
Distribute the musical operators
\[\begin{split}F^{♯♯} = \left[ \begin{aligned}
- & \E^x \; dt^♯ ∧ dx^♯ \\
- & \E^y \; dt^♯ ∧ dy^♯ \\
- & \E^z \; dt^♯ ∧ dz^♯ \\
& B^x \; dy^♯ ∧ dz^♯ \\
& B^y \; dz^♯ ∧ dx^♯ \\
& B^z \; dx^♯ ∧ dy^♯ \\
\end{aligned} \right]\end{split}\]
Apply
\[\begin{split}F^{♯♯} = \left[ \begin{aligned}
- & \E^x \; η^{tμ} ∂_μ ∧ η^{xμ} ∂_μ \\
- & \E^y \; η^{tμ} ∂_μ ∧ η^{yμ} ∂_μ \\
- & \E^z \; η^{tμ} ∂_μ ∧ η^{zμ} ∂_μ \\
& B^x \; η^{yμ} ∂_μ ∧ η^{zμ} ∂_μ \\
& B^y \; η^{zμ} ∂_μ ∧ η^{xμ} ∂_μ \\
& B^z \; η^{xμ} ∂_μ ∧ η^{yμ} ∂_μ \\
\end{aligned} \right]\end{split}\]
Identify non-zero terms
\[\begin{split}F^{♯♯} = \left[ \begin{aligned}
- & \E^x \; η^{tt} ∂_t ∧ η^{xx} ∂_x \\
- & \E^y \; η^{tt} ∂_t ∧ η^{yy} ∂_y \\
- & \E^z \; η^{tt} ∂_t ∧ η^{zz} ∂_z \\
& B^x \; η^{yy} ∂_y ∧ η^{zz} ∂_z \\
& B^y \; η^{zz} ∂_z ∧ η^{xx} ∂_x \\
& B^z \; η^{xx} ∂_x ∧ η^{yy} ∂_y \\
\end{aligned} \right]\end{split}\]
Apply numerical values
\[\begin{split}F^{♯♯} = \left[ \begin{aligned}
- & \E^x \; (+1) ∂_t ∧ (-1) ∂_x \\
- & \E^y \; (+1) ∂_t ∧ (-1) ∂_y \\
- & \E^z \; (+1) ∂_t ∧ (-1) ∂_z \\
& B^x \; (-1) ∂_y ∧ (-1) ∂_z \\
& B^y \; (-1) ∂_z ∧ (-1) ∂_x \\
& B^z \; (-1) ∂_x ∧ (-1) ∂_y \\
\end{aligned} \right]\end{split}\]
Conclude
\[\begin{split}F^{♯♯} = \left[ \begin{aligned}
\E^x \; ∂_t ∧ ∂_x \\
\E^y \; ∂_t ∧ ∂_y \\
\E^z \; ∂_t ∧ ∂_z \\
B^x \; ∂_y ∧ ∂_z \\
B^y \; ∂_z ∧ ∂_x \\
B^z \; ∂_x ∧ ∂_y \\
\end{aligned} \right]\end{split}\]
We derive the row/column representation of the \(F^{♭♯}\) Faraday tensor:
\[\begin{split}F^{♯♯} = \frac{1}{2} \begin{bmatrix}
& - \E^x \; ∂_x ∧ ∂_t & - \E^y \; ∂_y ∧ ∂_t & - \E^z \; ∂_z ∧ ∂_t \\
+ \E^x \; ∂_t ∧ ∂_x & & - B^z \; ∂_y ∧ ∂_x & + B^y \; ∂_z ∧ ∂_x \\
+ \E^y \; ∂_t ∧ ∂_y & + B^z \; ∂_x ∧ ∂_y & & - B^x \; ∂_z ∧ ∂_y \\
+ \E^z \; ∂_t ∧ ∂_z & - B^y \; ∂_x ∧ ∂_z & + B^x \; ∂_y ∧ ∂_z & \\
\end{bmatrix}\end{split}\]
Calculations
Start from
\[\begin{split}F^{♯♯} = \left[ \begin{aligned}
\E^x \; ∂_t ∧ ∂_x \\
\E^y \; ∂_t ∧ ∂_y \\
\E^z \; ∂_t ∧ ∂_z \\
B^x \; ∂_y ∧ ∂_z \\
B^y \; ∂_z ∧ ∂_x \\
B^z \; ∂_x ∧ ∂_y \\
\end{aligned} \right]\end{split}\]
Apply the symmetries of the exterior product
\[\begin{split}F^{♯♯} = \left[ \begin{aligned}
\E^x \; \frac{1}{2} & (∂_t ∧ ∂_x - ∂_x ∧ ∂_t) \\
\E^y \; \frac{1}{2} & (∂_t ∧ ∂_y - ∂_y ∧ ∂_t) \\
\E^z \; \frac{1}{2} & (∂_t ∧ ∂_z - ∂_z ∧ ∂_t) \\
B^x \; \frac{1}{2} & (∂_y ∧ ∂_z - ∂_z ∧ ∂_y) \\
B^y \; \frac{1}{2} & (∂_z ∧ ∂_x - ∂_x ∧ ∂_z) \\
B^z \; \frac{1}{2} & (∂_x ∧ ∂_y - ∂_y ∧ ∂_x) \\
\end{aligned} \right]\end{split}\]
Reorder
\[\begin{split}F^{♯♯} = \frac{1}{2} \left[ \begin{aligned}
\E^x \; ∂_t ∧ ∂_x & - \E^x \; ∂_x ∧ ∂_t \\
\E^y \; ∂_t ∧ ∂_y & - \E^y \; ∂_y ∧ ∂_t \\
\E^z \; ∂_t ∧ ∂_z & - \E^z \; ∂_z ∧ ∂_t \\
B^x \; ∂_y ∧ ∂_z & - B^x \; ∂_z ∧ ∂_y \\
B^y \; ∂_z ∧ ∂_x & - B^y \; ∂_x ∧ ∂_z \\
B^z \; ∂_x ∧ ∂_y & - B^z \; ∂_y ∧ ∂_x \\
\end{aligned} \right]\end{split}\]
Reorder and conclude
\[\begin{split}F^{♯♯} = \frac{1}{2} \begin{bmatrix}
& - \E^x \; ∂_x ∧ ∂_t & - \E^y \; ∂_y ∧ ∂_t & - \E^z \; ∂_z ∧ ∂_t \\
\E^x \; ∂_t ∧ ∂_x & & - B^z \; ∂_y ∧ ∂_x & + B^y \; ∂_z ∧ ∂_x \\
\E^y \; ∂_t ∧ ∂_y & + B^z \; ∂_x ∧ ∂_y & & - B^x \; ∂_z ∧ ∂_y \\
\E^z \; ∂_t ∧ ∂_z & - B^y \; ∂_x ∧ ∂_z & + B^x \; ∂_y ∧ ∂_z & \\
\end{bmatrix}\end{split}\]
With implicit bivector basis, we have the standard representation with
abstract index notation, which also permits
to verify the calculations here:
\[\begin{split}F_{μν} = \begin{bmatrix}
& - \E^x & - \E^y & - \E^z \\
+ \E^x & & - B^z & + B^y \\
+ \E^y & + B^z & & - B^x \\
+ \E^z & - B^y & + B^x & \\
\end{bmatrix}\end{split}\]
Where the electromagnetic field is related to the doubly contravariant Faraday
tensor through:
\[F^{♯♯} = \frac{1}{2} \: F^{μν} \: ∂_μ ∧ ∂_ν\]
\(G^{♭♭}\)
The Hodge dual \(G^{♭♭}\) of the Faraday 2-form \(F^{♭♭}\) is:
\[G^{♭♭} = ⋆ F^{♭♭}\]
Expanded, we obtain:
\[\begin{split}G^{♭♭} = \left[ \begin{alignedat}{1}
B^x \; & dt ∧ dx \\
B^y \; & dt ∧ dy \\
B^z \; & dt ∧ dz \\
\E^x \; & dy ∧ dz \\
\E^y \; & dz ∧ dx \\
\E^z \; & dx ∧ dy \\
\end{alignedat} \right]\end{split}\]
Calculations
Start from the Faraday 2-form
\[\begin{split}F^{♭♭} = \left[ \begin{aligned}
- & \E^x \; dt ∧ dx \\
- & \E^y \; dt ∧ dy \\
- & \E^z \; dt ∧ dz \\
& B^x \; dy ∧ dz \\
& B^y \; dz ∧ dx \\
& B^z \; dx ∧ dy \\
\end{aligned} \right]\end{split}\]
Take the Hodge dual
\[\begin{split}G^{♭♭} = ⋆ F^{♭♭} = ⋆ \left[ \begin{aligned}
- & \E^x \; dt ∧ dx \\
- & \E^y \; dt ∧ dy \\
- & \E^z \; dt ∧ dz \\
& B^x \; dy ∧ dz \\
& B^y \; dz ∧ dx \\
& B^z \; dx ∧ dy \\
\end{aligned} \right]\end{split}\]
Distribute the Hodge dual operator
\[\begin{split}G^{♭♭} = \left[ \begin{aligned}
- & \E^x \; ⋆ dt ∧ dx \\
- & \E^y \; ⋆ dt ∧ dy \\
- & \E^z \; ⋆ dt ∧ dz \\
& B^x \; ⋆ dy ∧ dz \\
& B^y \; ⋆ dz ∧ dx \\
& B^z \; ⋆ dx ∧ dy \\
\end{aligned} \right]\end{split}\]
Apply the Hodge dual operator
You can find the Hodge dual of each bivector basis in Minkowski space
here.
\[\begin{split}G^{♭♭} = \left[ \begin{alignedat}{2}
- & \E^x \; (-1) & dy ∧ dz \\
- & \E^y \; (-1) & dz ∧ dx \\
- & \E^z \; (-1) & dx ∧ dy \\
& B^x \; (+1) & dt ∧ dx \\
& B^y \; (+1) & dt ∧ dy \\
& B^z \; (+1) & dt ∧ dz \\
\end{alignedat} \right]\end{split}\]
Simplify
\[\begin{split}G^{♭♭} = \left[ \begin{alignedat}{1}
\E^x \; & dy ∧ dz \\
\E^y \; & dz ∧ dx \\
\E^z \; & dx ∧ dy \\
B^x \; & dt ∧ dx \\
B^y \; & dt ∧ dy \\
B^z \; & dt ∧ dz \\
\end{alignedat} \right]\end{split}\]
Reorder
\[\begin{split}G^{♭♭} = \left[ \begin{alignedat}{1}
B^x \; & dt ∧ dx \\
B^y \; & dt ∧ dy \\
B^z \; & dt ∧ dz \\
\E^x \; & dy ∧ dz \\
\E^y \; & dz ∧ dx \\
\E^z \; & dx ∧ dy \\
\end{alignedat} \right]\end{split}\]
From the dual Faraday 2-form and noting that 2-forms are tensors, we can derive
the row/column representation of the doubly covariant dual Faraday tensor.
\[\begin{split}G^{♭♭} = \begin{bmatrix}
& - B^x \; dx ∧ dt & - B^y \; dy ∧ dt & - B^z \; dz ∧ dt \\
+ B^x \; dt ∧ dx & & - \E^z \; dy ∧ dx & + \E^y \; dz ∧ dx \\
+ B^y \; dt ∧ dy & + \E^z \; dx ∧ dy & & - \E^x \; dz ∧ dy \\
+ B^z \; dt ∧ dz & - \E^y \; dx ∧ dz & + \E^x \; dy ∧ dz & \\
\end{bmatrix}\end{split}\]
Calculations
Begin with the Hodge dual in column form
\[\begin{split}G^{♭♭} = \begin{bmatrix}
B^x \; dt ∧ dx \\
B^y \; dt ∧ dy \\
B^z \; dt ∧ dz \\
\E^x \; dy ∧ dz \\
\E^y \; dz ∧ dx \\
\E^z \; dx ∧ dy \\
\end{bmatrix}\end{split}\]
Duplicate each 2-form
I am; indeed; really writing that \(A = \frac{1}{2} (A+A)\).
\[\begin{split}G^{♭♭} = \frac{1}{2} \left[ \begin{alignedat}{2}
B^x \; dt ∧ dx & \, + & B^x \; dt ∧ dx \\
B^y \; dt ∧ dy & \, + & B^y \; dt ∧ dy \\
B^z \; dt ∧ dz & \, + & B^z \; dt ∧ dz \\
\E^x \; dy ∧ dz & \, + & \E^x \; dy ∧ dz \\
\E^y \; dz ∧ dx & \, + & \E^y \; dz ∧ dx \\
\E^z \; dx ∧ dy & \, + & \E^z \; dx ∧ dy \\
\end{alignedat} \right]\end{split}\]
Flip the exterior product
The purpose of the above operation was to utilize the antisymmetry of the
exterior product and flip the signs \(dx^μ ∧ dx^ν = -dx^ν ∧ dx^μ\) as
needed.
\[\begin{split}G^{♭♭} = \frac{1}{2} \left[ \begin{alignedat}{2}
B^x \; dt ∧ dx & \, - & B^x \; dx ∧ dt \\
B^y \; dt ∧ dy & \, - & B^y \; dy ∧ dt \\
B^z \; dt ∧ dz & \, - & B^z \; dz ∧ dt \\
\E^x \; dy ∧ dz & \, - & \E^x \; dz ∧ dy \\
\E^y \; dz ∧ dx & \, - & \E^y \; dx ∧ dz \\
\E^z \; dx ∧ dy & \, - & \E^z \; dy ∧ dx \\
\end{alignedat} \right]\end{split}\]
We can now switch the representation of the dual Faraday 2-Form from a
single row of basis 2-Forms, to a row/column representation.
Reorder into rows/column representation
From there, we conclude utilizing the free matrix representation of the
Cartan-Hodge formalism, reordering the elements into rows and columns.
\[\begin{split}G^{♭♭} = \begin{bmatrix}
& - B^x \; dx ∧ dt & - B^y \; dy ∧ dt & - B^z \; dz ∧ dt \\
B^x \; dt ∧ dx & & - \E^z \; dy ∧ dx & + \E^y \; dz ∧ dx \\
B^y \; dt ∧ dy & + \E^z \; dx ∧ dy & & - \E^x \; dz ∧ dy \\
B^z \; dt ∧ dz & - \E^y \; dx ∧ dz & + \E^x \; dy ∧ dz & \\
\end{bmatrix}\end{split}\]