splitting apart space and time in the differential form representation:
Comparing $E$ and $\mathcal{E}$:
$\frac{1}{c} \alpha \mathcal{E}_i = \frac{1}{c} \alpha E_i + 2 \frac{1}{c} \frac{1}{\alpha} \beta^k \beta_{[i} E_{k]} + \frac{1}{\alpha} \epsilon_{ikl} \beta^k B^l$
$\frac{1}{c} \mathcal{E}_i = \frac{1}{c} E_i + \frac{1}{\alpha^2} (2 \frac{1}{c} \beta^k \beta_{[i} E_{k]} + \epsilon_{ikl} \beta^k B^l)$
$\frac{1}{c} \mathcal{E}_i = \frac{1}{c} (\delta^k_i + 2 \frac{1}{\alpha^2} \beta^j \beta_{[i} \delta^k_{j]}) E_k + \frac{1}{\alpha^2} \epsilon_{ikl} \beta^k B^l$

Comparing $B$ to $\mathcal{B}$:
$\epsilon_{ijk} \mathcal{B}^k = 2 \frac{1}{c} \frac{1}{\alpha} \beta_{[i} E_{j]} + \epsilon_{ijk} B^k$
$\mathcal{B}_i = B_i + \frac{1}{c} \frac{1}{\alpha} \epsilon_{ijk} \beta^j E^k$

Combining ${F^a}_b, \mathcal{E}_i$, and $\mathcal{B}_i$, we find:
${F^a}_b = \pmatrix{ \frac{1}{c} \frac{1}{\alpha} \beta^k E_k & \frac{1}{c} \frac{1}{\alpha} E_j \\ \frac{1}{c} (\alpha - \frac{1}{\alpha} \beta^2) E^i + {\epsilon^i}_{kl} \beta^k B^l & \frac{1}{c} {\epsilon^i}_{jk} B^k - \frac{1}{\alpha} E^i \beta_j }$
${F^a}_b = \pmatrix{ \frac{1}{c} \frac{1}{\alpha} E_t & \frac{1}{c} \frac{1}{\alpha} E_j \\ \frac{1}{c} \alpha \mathcal{E}^i - \frac{1}{c} \frac{1}{\alpha} \beta^i \beta_k E^k & {\epsilon^i}_{jk} \mathcal{B}^k - \frac{1}{c} \frac{1}{\alpha} \beta^i E_j }$
(Notice this $\mathcal{E}^i$ is only raised with $\gamma^{ij} \mathcal{E}_j$, and not the whole $g^{ia} \mathcal{E}_a$)
Because this one is what I'll be looking at later with the stress-energy tensor.

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Differential forms in terms of the electromagnetic 4-potential

$F \wedge \star F = \alpha^2 (\frac{1}{c} \mathcal{E} \wedge dx^0 + \star (\mathcal{B} \wedge dx^0)) \wedge \star (\frac{1}{c} \mathcal{E} \wedge dx^0 + \star (\mathcal{B} \wedge dx^0))$
$= \alpha^2 ( \frac{1}{c} \mathcal{E} \wedge dx^0 + \star (\mathcal{B} \wedge dx^0) ) \wedge ( \star (\frac{1}{c} \mathcal{E} \wedge dx^0) - (\mathcal{B} \wedge dx^0) )$
$= \alpha^2 ( \frac{1}{c} \mathcal{E} \wedge dx^0 \wedge \star (\frac{1}{c} \mathcal{E} \wedge dx^0) + \star (\mathcal{B} \wedge dx^0) \wedge \star (\frac{1}{c} \mathcal{E} \wedge dx^0) - \frac{1}{c} \mathcal{E} \wedge dx^0 \wedge (\mathcal{B} \wedge dx^0) - \star (\mathcal{B} \wedge dx^0) \wedge (\mathcal{B} \wedge dx^0) )$
$= \alpha^2 ( \frac{1}{c} \mathcal{E} \wedge dx^0 \wedge \star (\frac{1}{c} \mathcal{E} \wedge dx^0) - \star (\mathcal{B} \wedge dx^0) \wedge (\mathcal{B} \wedge dx^0) )$
$= \alpha^2 ( \frac{1}{c} \mathcal{E}_i dx^i \wedge dx^0 \wedge \frac{1}{2} \frac{1}{c} \mathcal{E}_m {\epsilon^{0m}}_{jk} dx^j \wedge dx^k - \frac{1}{2} \mathcal{B}_m {\epsilon^{0m}}_{jk} dx^j \wedge dx^k\wedge (\mathcal{B}_i dx^i \wedge dx^0) )$
$= -\frac{1}{2} \alpha^2 (\frac{1}{c^2} \mathcal{E}_i \mathcal{E}_m - \mathcal{B}_i \mathcal{B}_m) {\epsilon^{0m}}_{jk} dx^0 \wedge dx^i \wedge dx^j \wedge dx^k$
$= -2 \alpha^2 (\frac{1}{c^2} \mathcal{E}_i \mathcal{E}_m - \mathcal{B}_i \mathcal{B}_m) g^{ta} g^{mb} \epsilon_{abjk} \epsilon^{0ijk} dV$
$= 2 \alpha^2 (\frac{1}{c^2} \mathcal{E}_i \mathcal{E}_m - \mathcal{B}_i \mathcal{B}_m) g^{ta} g^{mb} \delta^{ti}_{ab} dV$
$= 2 \alpha^2 (\frac{1}{c^2} \mathcal{E}_i \mathcal{E}_m - \mathcal{B}_i \mathcal{B}_m) g^{ta} g^{mb} (\delta^t_a \delta^i_b - \delta^t_b \delta^i_a) dV$
$= 2 \alpha^2 (\frac{1}{c^2} \mathcal{E}_i \mathcal{E}_j - \mathcal{B}_i \mathcal{B}_j) (g^{tt} g^{ij} - g^{ti} g^{tj}) dV$
$= 2 \alpha^2 (\frac{1}{c^2} \mathcal{E}_i \mathcal{E}_j - \mathcal{B}_i \mathcal{B}_j) (-\frac{1}{\alpha^2} (\gamma^{ij} - \frac{1}{\alpha^2} \beta^i \beta^j) - \frac{1}{\alpha^4} \beta^i \beta^j) dV$
$= -2 \gamma^{ij} (\frac{1}{c^2} \mathcal{E}_i \mathcal{E}_j - \mathcal{B}_i \mathcal{B}_j) dV$
If we choose $\mathcal{E}_t = \mathcal{B}_t = 0$:
$= -2 g^{ab} (\frac{1}{c^2} \mathcal{E}_a \mathcal{E}_b - \mathcal{B}_a \mathcal{B}_b) dV$
$= -2 (\frac{1}{c^2} \mathcal{E}^2 - \mathcal{B}^2) dV$
...TODO show this one too
$= g^{ua} F_{ab} g^{bv} F_{uv} dV$
$ = F_{ab} F^{ab} dV$
$ = -tr(F^2) dV$

$F \wedge F = \alpha^2 (\frac{1}{c} \mathcal{E} \wedge dx^0 + \star (\mathcal{B} \wedge dx^0)) \wedge (\frac{1}{c} \mathcal{E} \wedge dx^0 + \star (\mathcal{B} \wedge dx^0))$
$= \alpha^2 ( \frac{1}{c} \mathcal{E} \wedge dx^0 \wedge \frac{1}{c} \mathcal{E} \wedge dx^0 + \star (\mathcal{B} \wedge dx^0) \wedge \frac{1}{c} \mathcal{E} \wedge dx^0 + \frac{1}{c} \mathcal{E} \wedge dx^0 \wedge \star (\mathcal{B} \wedge dx^0) + \star (\mathcal{B} \wedge dx^0) \wedge \star (\mathcal{B} \wedge dx^0) )$
$= \alpha^2 ( \star (\mathcal{B} \wedge dx^0) \wedge \frac{1}{c} \mathcal{E} \wedge dx^0 + \frac{1}{c} \mathcal{E} \wedge dx^0 \wedge \star (\mathcal{B} \wedge dx^0) )$
$= \alpha^2 ( \frac{1}{2} \mathcal{B}_i {\epsilon^{0i}}_{jk} dx^j \wedge dx^k \wedge \frac{1}{c} \mathcal{E} \wedge dx^0 + \frac{1}{c} \mathcal{E} \wedge dx^0 \wedge \frac{1}{2} \mathcal{B}_i {\epsilon^{0i}}_{jk} dx^j \wedge dx^k )$
$= -\frac{1}{c} \alpha^2 \mathcal{E}_i \mathcal{B}_l {\epsilon^{0l}}_{jk} dx^0 \wedge dx^i \wedge dx^j \wedge dx^k$
$= -4 \frac{1}{c} \alpha^2 \mathcal{E}_i \mathcal{B}_l {\epsilon^{0l}}_{jk} \epsilon^{0ijk} dV$
$= -4 \frac{1}{c} \alpha^2 \mathcal{E}_i \mathcal{B}_l g^{ta} g^{lb} \epsilon_{abjk} \epsilon^{0ijk} dV$
$= -4 \frac{1}{c} \alpha^2 \mathcal{E}_i \mathcal{B}_l g^{ta} g^{lb} \delta^{ti}_{ab} dV$
$= -4 \frac{1}{c} \alpha^2 \mathcal{E}_i \mathcal{B}_l g^{ta} g^{lb} (\delta^t_a \delta^i_b - \delta^t_b \delta^i_a) dV$
$= -4 \frac{1}{c} \alpha^2 \mathcal{E}_i \mathcal{B}_j (g^{tt} g^{ij} - g^{ti} g^{tj}) dV$
$= -4 \frac{1}{c} \alpha^2 \mathcal{E}_i \mathcal{B}_j (-\frac{1}{\alpha^2} (\gamma^{ij} - \frac{1}{\alpha^2} \beta^i \beta^j) - \frac{1}{\alpha^4} \beta^i \beta^j) dV$
$= 4 \frac{1}{c} \gamma^{ij} \mathcal{E}_i \mathcal{B}_j dV$
So $\star (F \wedge F) = 4 \frac{1}{c} \gamma^{ij} \mathcal{E}_i \mathcal{B}_j$
This looks suspiciously like $F \cdot \star F$
...