$T_{uv} = \frac{1}{\mu_0} (F_{ua} {F_v}^a - \frac{1}{4} g_{uv} F_{ab} F^{ab})$
$T_{uv}$ is in units of $(\frac{kg}{C \cdot s})^2 \cdot \frac{C^2}{kg \cdot m} = \frac{kg}{m \cdot s^2}$
as a trace-free operation:
${T^u}_v = -\frac{1}{\mu_0} ({F^u}_a {F^a}_v - \frac{1}{4} \delta^u_v {F^a}_b {F^b}_a)$
Let ${(F^2)^u}_v = {F^u}_a {F^a}_v$
${T^u}_v = -\frac{1}{\mu_0} ({(F^2)^u}_v - \frac{1}{4} \delta^u_v {(F^2)^a}_a)$
Let $(A_{uv})^{TF} = A_{uv} - \frac{1}{n} g_{uv} {A^a}_a$ be the trace-free element of $A_{uv}$, where $n = $ the dimension, such that ${(A^{TF})^a}_a = 0$
${T^u}_v = -\frac{1}{\mu_0} ({(F^2)^u}_v)^{TF}$

in special relativity:
${F^a}_b = \eta^{ac} F_{cb} = \downarrow a(i) \overset{\rightarrow b(j)}{\pmatrix{ 0 & \frac{1}{c} E_j \\ \frac{1}{c} E_i & {\bar\epsilon_{ij}}^k B_k }}$
${(F^2)^a}_b = {F^a}_c {F^c}_b = \downarrow a(i) \overset{\rightarrow b(j)}{\pmatrix{ \frac{1}{c^2} E^2 & \frac{1}{c} \bar\epsilon_{kjl} E_k B_l \\ \frac{1}{c} \bar\epsilon_{ikl} E_k B_l & \frac{1}{c^2} E_i E_j + \bar\epsilon_{ikl} \bar\epsilon_{kjm} B_l B_m }} = \downarrow a(i) \overset{\rightarrow b(j)}{\pmatrix{ \frac{1}{c^2} E^2 & -\frac{1}{c} \bar\epsilon_{jkl} E_k B_l \\ \frac{1}{c} \bar\epsilon_{ikl} E_k B_l & \frac{1}{c^2} E_i E_j + B_i B_j - \delta^i_j B^2 }}$

$tr F^2 = {(F^2)^a}_a = {F^a}_b {F^b}_a = 2 (\frac{1}{c^2} E^2 - B^2)$
${((F^2)^{TF})^a}_b = {(F^2)^a}_b - \frac{1}{4} \delta^a_b tr(F^2) = \downarrow a(i) \overset{\rightarrow b(j)}{\pmatrix{ \frac{1}{2} (\frac{1}{c^2} E^2 + B^2) & -\frac{1}{c} \bar\epsilon_{jkl} E_k B_l \\ \frac{1}{c} \bar\epsilon_{ikl} E_k B_l & \frac{1}{c^2} E_i E_j + B_i B_j -\delta^i_j \frac{1}{2} (\frac{1}{c^2} E^2 + B^2) }}$
${T^a}_b = -\frac{1}{\mu_0} {((F^2)^{TF})^a}_b = \pmatrix{ -\frac{1}{2} (\epsilon_0 E^2 + \frac{1}{\mu_0} B^2) & \frac{1}{\mu_0} \frac{1}{c} \bar\epsilon_{jkl} E_k B_l \\ -\frac{1}{\mu_0} \frac{1}{c} \bar\epsilon_{ikl} E_k B_l & \delta^i_j \frac{1}{2} (\epsilon_0 E^2 + \frac{1}{\mu_0} B^2) - \epsilon_0 E_i E_j - \frac{1}{\mu_0} B_i B_j }$
$T^{ab} = {T^a}_c \eta^{cb} = \pmatrix{ \frac{1}{2} (\epsilon_0 E^2 + \frac{1}{\mu_0} B^2) & \frac{1}{\mu_0} \frac{1}{c} \bar\epsilon_{jkl} E_k B_l \\ \frac{1}{\mu_0} \frac{1}{c} \bar\epsilon_{ikl} E_k B_l & \eta^{ij} \frac{1}{2} (\epsilon_0 E^2 + \frac{1}{\mu_0} B^2) - \epsilon_0 E_i E_j - \frac{1}{\mu_0} B_i B_j }$

in general relativity:
${F^a}_b = \pmatrix{ \frac{1}{c} \frac{1}{\alpha} E^k \beta_k & \frac{1}{c} \frac{1}{\alpha} E_j \\ \frac{1}{c} (\alpha - \frac{1}{\alpha} \beta^2) E^i + {\epsilon^i}_{kl} \beta^k B^l & -\frac{1}{c} \frac{1}{\alpha} E^i \beta_j + {\epsilon^i}_{jk} B^k }$
${(F^2)^a}_b = {F^a}_u {F^u}_b = \pmatrix{ \frac{1}{c} \frac{1}{\alpha} E^k \beta_k & \frac{1}{c} \frac{1}{\alpha} E_o \\ \frac{1}{c} (\alpha - \frac{1}{\alpha} \beta^2) E^i + {\epsilon^i}_{kl} \beta^k B^l & -\frac{1}{c} \frac{1}{\alpha} E^i \beta_o + {\epsilon^i}_{ok} B^k } \cdot \pmatrix{ \frac{1}{c} \frac{1}{\alpha} E^m \beta_m & \frac{1}{c} \frac{1}{\alpha} E_j \\ \frac{1}{c} (\alpha - \frac{1}{\alpha} \beta^2) E^o + {\epsilon^o}_{mn} \beta^m B^n & -\frac{1}{c} \frac{1}{\alpha} E^o \beta_j + {\epsilon^o}_{jm} B^m }$
${(F^2)^a}_b = {F^a}_u {F^u}_b = \pmatrix{ \frac{1}{c^2} \frac{1}{\alpha} E^k \beta_k \frac{1}{\alpha} E^m \beta_m + \frac{1}{c^2} \frac{1}{\alpha} E_o ((\alpha - \frac{1}{\alpha} \beta^2) E^o + {\epsilon^o}_{mn} \beta^m B^n) & \frac{1}{c^2} \frac{1}{\alpha} E^k \beta_k \frac{1}{\alpha} E_j + \frac{1}{c^2} \frac{1}{\alpha} E_o (-\frac{1}{\alpha} E^o \beta_j + {\epsilon^o}_{jm} B^m) \\ ((\alpha - \frac{1}{\alpha} \beta^2) \frac{1}{c} E^i + {\epsilon^i}_{kl} \beta^k B^l) \frac{1}{\alpha} \frac{1}{c} E^m \beta_m + (-\frac{1}{\alpha} \frac{1}{c} E^i \beta_o + {\epsilon^i}_{ok} B^k) ((\alpha - \frac{1}{\alpha} \beta^2) \frac{1}{c} E^o + {\epsilon^o}_{mn} \beta^m B^n) & ((\alpha - \frac{1}{\alpha} \beta^2) \frac{1}{c} E^i + {\epsilon^i}_{kl} \beta^k B^l) \frac{1}{\alpha} \frac{1}{c} E_j + (-\frac{1}{\alpha} \frac{1}{c} E^i \beta_o + {\epsilon^i}_{ok} B^k) (-\frac{1}{\alpha} \frac{1}{c} E^o \beta_j + {\epsilon^o}_{jm} B^m) }$

${(F^2)^0}_0 = \frac{1}{c^2} \frac{1}{\alpha^2} E^k \beta_k E^m \beta_m + \frac{1}{c^2} E_o E^o - \frac{1}{c^2} \frac{1}{\alpha^2} \beta^2 E_o E^o + \frac{1}{c} \frac{1}{\alpha} E_o {\epsilon^o}_{mn} \beta^m B^n $
$= \frac{1}{c^2} E^2 + \frac{1}{c} \frac{1}{\alpha} \epsilon_{klm} E^k \beta^l B^m + \frac{1}{c^2} \frac{1}{\alpha^2} ((E^k \beta_k)^2 - \beta^2 E^2) $
${(F^2)^0}_j = \frac{1}{c^2} \frac{1}{\alpha^2} E^k \beta_k E_j - \frac{1}{c^2} \frac{1}{\alpha^2} E_o E^o \beta_j + \frac{1}{c} \frac{1}{\alpha} E_o {\epsilon^o}_{jm} B^m $
$= - \frac{1}{c} \frac{1}{\alpha} \epsilon_{jkl} E^k B^l + 2 \frac{1}{c^2} \frac{1}{\alpha^2} E_{[j} \beta_{k]} E^k $
${(F^2)^i}_0 = \frac{1}{c^2} \alpha E^i \frac{1}{\alpha} E^m \beta_m - \frac{1}{c^2} \frac{1}{\alpha} \beta^2 E^i \frac{1}{\alpha} E^m \beta_m + \frac{1}{c} {\epsilon^i}_{kl} \beta^k B^l \frac{1}{\alpha} E^m \beta_m - \frac{1}{c^2} \frac{1}{\alpha} E^i \beta_o \alpha E^o + \frac{1}{c^2} \frac{1}{\alpha} E^i \beta_o \frac{1}{\alpha} \beta^2 E^o + \frac{1}{c} {\epsilon^i}_{ok} B^k \alpha E^o - \frac{1}{c} {\epsilon^i}_{ok} B^k \frac{1}{\alpha} \beta^2 E^o - \frac{1}{c} \frac{1}{\alpha} E^i \beta_o {\epsilon^o}_{mn} \beta^m B^n + {\epsilon^i}_{ok} B^k {\epsilon^o}_{mn} \beta^m B^n $
$= 2 B^{[i} \beta^{k]} B_k + \frac{1}{c} \frac{1}{\alpha} {\epsilon^i}_{kl} \beta^k B^l E^m \beta_m + \frac{1}{c} (\alpha - \frac{1}{\alpha} \beta^2) {\epsilon^i}_{kl} E^k B^l $
${(F^2)^i}_j = \frac{1}{c^2} \alpha E^i \frac{1}{\alpha} E_j - \frac{1}{c^2} \frac{1}{\alpha} \beta^2 E^i \frac{1}{\alpha} E_j + \frac{1}{c} {\epsilon^i}_{kl} \beta^k B^l \frac{1}{\alpha} E_j + \frac{1}{c^2} \frac{1}{\alpha} E^i \beta_o \frac{1}{\alpha} E^o \beta_j - \frac{1}{c} {\epsilon^i}_{ok} B^k \frac{1}{\alpha} E^o \beta_j - \frac{1}{c} \frac{1}{\alpha} E^i \beta_o {\epsilon^o}_{jm} B^m + {\epsilon^i}_{ok} B^k {\epsilon^o}_{jm} B^m $
$= \frac{1}{c^2} E^i E_j + B^i B_j - \delta^i_j B^2 - \frac{1}{c} \frac{1}{\alpha} \epsilon_{jlm} \beta^m E^i B^l + \frac{1}{c} \frac{1}{\alpha} {\epsilon^i}_{kl} \beta^k E_j B^l - \frac{1}{c} \frac{1}{\alpha} {\epsilon^i}_{kl} \beta_j E^k B^l - \frac{1}{c^2} \frac{1}{\alpha^2} \beta^2 E^i E_j + \frac{1}{c^2} \frac{1}{\alpha^2} E^i \beta_j \beta_k E^k $

${(F^2)^a}_b = {F^a}_u {F^u}_b = \pmatrix{ {(F^2)^0}_0 & {(F^2)^0}_j \\ {(F^2)^i}_0 & {(F^2)_i}_j }$

$tr (F^2) = {(F^2)^a}_a = {(F^2)^0}_0 + {(F^2)^i}_i$
$= \frac{1}{c^2} E^2 + \frac{1}{c} \frac{1}{\alpha} \epsilon_{klm} E^k \beta^l B^m + \frac{1}{c^2} \frac{1}{\alpha^2} ((E^k \beta_k)^2 - \beta^2 E^2) + \frac{1}{c^2} E^i E_i + B^i B_i - \delta^i_i B^2 - \frac{1}{c} \frac{1}{\alpha} \epsilon_{ilm} \beta^m E^i B^l + \frac{1}{c} \frac{1}{\alpha} {\epsilon^i}_{kl} \beta^k E_i B^l - \frac{1}{c} \frac{1}{\alpha} {\epsilon^i}_{kl} \beta_i E^k B^l - \frac{1}{c^2} \frac{1}{\alpha^2} \beta^2 E^i E_i + \frac{1}{c^2} \frac{1}{\alpha^2} E^i \beta_i \beta_k E^k $
$= 2 \frac{1}{c^2} E^2 - 2 B^2 + 2 \frac{1}{c} \frac{1}{\alpha^2} (E^i \beta_i)^2 - 2 \frac{1}{c^2} \frac{1}{\alpha^2} \beta^2 E^2 - 4 \frac{1}{c} \frac{1}{\alpha} \epsilon_{ijk} E^i B^j \beta^k $

${T^a}_b = \frac{1}{4} \delta^a_b tr (F^2) - {(F^2)^a}_b$
${T^0}_0 = \frac{1}{4} tr (F^2) - {(F^2)^0}_0$
$= \frac{1}{4} ( 2 \frac{1}{c^2} E^2 - 2 B^2 + 2 \frac{1}{c^2} \frac{1}{\alpha^2} (E^i \beta_i)^2 - 2 \frac{1}{c^2} \frac{1}{\alpha^2} \beta^2 E^2 - 4 \frac{1}{c} \frac{1}{\alpha} \epsilon_{ijk} E^i B^j \beta^k ) - ( \frac{1}{c^2} E^2 + \frac{1}{c} \frac{1}{\alpha} \epsilon_{klm} E^k \beta^l B^m + \frac{1}{c^2} \frac{1}{\alpha^2} ((E^k \beta_k)^2 - \beta^2 E^2) )$
$= - \frac{1}{2} \frac{1}{c^2} E^2 - \frac{1}{2} B^2 - \frac{1}{\alpha^2} \frac{1}{2} \frac{1}{c^2} (E^i \beta_i)^2 + \frac{1}{\alpha^2} \frac{1}{2} \frac{1}{c^2} \beta^2 E^2 $

${T^0}_j = - ( - \frac{1}{\alpha} \frac{1}{c} \epsilon_{jkl} E^k B^l + 2 \frac{1}{\alpha^2} \frac{1}{c^2} E_{[j} \beta_{k]} E^k )$
$= \frac{1}{c} \frac{1}{\alpha} \epsilon_{jkl} E^k B^l - 2 \frac{1}{c^2} \frac{1}{\alpha^2} E_{[j} \beta_{k]} E^k $

${T^i}_0 = -( 2 B^{[i} \beta^{k]} B_k + \frac{1}{c} \frac{1}{\alpha} {\epsilon^i}_{kl} \beta^k B^l E^m \beta_m + \frac{1}{c} (\alpha - \frac{1}{\alpha} \beta^2) {\epsilon^i}_{kl} E^k B^l )$
$ = - 2 B^{[i} \beta^{k]} B_k - \frac{1}{c} \frac{1}{\alpha} {\epsilon^i}_{kl} \beta^k B^l E^m \beta_m - \frac{1}{c} (\alpha - \frac{1}{\alpha} \beta^2) {\epsilon^i}_{kl} E^k B^l $

${T^i}_j = \frac{1}{4} \delta^i_j tr (F^2) - {(F^2)^i}_j$
$= \frac{1}{4} \delta^i_j ( 2 \frac{1}{c^2} E^2 - 2 B^2 + 2 \frac{1}{c^2} \frac{1}{\alpha^2} (E^k \beta_k)^2 - 2 \frac{1}{c^2} \frac{1}{\alpha^2} \beta^2 E^2 - 4 \frac{1}{c} \frac{1}{\alpha} \epsilon_{klm} E^k B^l \beta^m ) - ( \frac{1}{c^2} \alpha E^i \frac{1}{\alpha} E_j - \frac{1}{c^2} \frac{1}{\alpha} \beta^2 E^i \frac{1}{\alpha} E_j + \frac{1}{c} {\epsilon^i}_{kl} \beta^k B^l \frac{1}{\alpha} E_j + \frac{1}{c^2} \frac{1}{\alpha} E^i \beta_o \frac{1}{\alpha} E^o \beta_j - \frac{1}{c} {\epsilon^i}_{ok} B^k \frac{1}{\alpha} E^o \beta_j - \frac{1}{c} \frac{1}{\alpha} E^i \beta_o {\epsilon^o}_{jm} B^m + {\epsilon^i}_{ok} B^k {\epsilon^o}_{jm} B^m )$
$= \delta^i_j ( \frac{1}{2} \frac{1}{c^2} E^2 (1 - \frac{1}{\alpha^2} \beta^2) + \frac{1}{2} B^2 + \frac{1}{2} \frac{1}{c^2} \frac{1}{\alpha^2} (E^k \beta_k)^2 - \frac{1}{\alpha} \frac{1}{c} \epsilon_{klm} E^k B^l \beta^m ) - \frac{1}{c^2} E^i E_j (1 - \frac{1}{\alpha^2} \beta^2) - B^i B_j + \frac{1}{c} \frac{1}{\alpha} \epsilon_{jlm} \beta^m E^i B^l - \frac{1}{c} \frac{1}{\alpha} {\epsilon^i}_{kl} \beta^k E_j B^l + \frac{1}{c} \frac{1}{\alpha} {\epsilon^i}_{kl} \beta_j E^k B^l - \frac{1}{c^2} \frac{1}{\alpha^2} E^i \beta_j \beta_k E^k $

again, in general relativity, starting with the covariant form of $F_{ab}$...
$F_{ab} = \pmatrix{ 0 & -\frac{1}{c} \alpha \mathcal{E}_i \\ \frac{1}{c} \alpha \mathcal{E}_i & \epsilon_{ijk} \mathcal{B}^k }$
$(F^2)_{ab} = F_{au} g^{uv} F_{bv} = \pmatrix{ 0 & -\frac{1}{c} \alpha \mathcal{E}_k \\ \frac{1}{c} \alpha \mathcal{E}_i & \epsilon_{ikl} \mathcal{B}^l } \pmatrix{ -1/\alpha^2 & \beta^m / \alpha^2 \\ \beta^k / \alpha^2 & \gamma^{km} - \beta^k \beta^m / \alpha^2 } \pmatrix{ 0 & -\frac{1}{c} \alpha \mathcal{E}_j \\ \frac{1}{c} \alpha \mathcal{E}_m & \epsilon_{mjn} \mathcal{B}^n }$
$= \pmatrix{ 0 & -\frac{1}{c} \alpha \mathcal{E}_k \\ \frac{1}{c} \alpha \mathcal{E}_i & \epsilon_{ikl} \mathcal{B}^l } \pmatrix{ \frac{1}{c} \frac{1}{\alpha} \beta^m \mathcal{E}_m & \frac{1}{c} \frac{1}{\alpha} \mathcal{E}_j + \frac{1}{\alpha^2} \epsilon_{jmn} \mathcal{B}^m \beta^n \\ \frac{1}{c} \alpha \mathcal{E}^k - \frac{1}{c} \frac{1}{\alpha} \beta^k \beta^m \mathcal{E}_m & - \frac{1}{c} \frac{1}{\alpha} \beta^k \mathcal{E}_j - {\epsilon_j}^{kn} \mathcal{B}_n - \frac{1}{\alpha^2} \beta^k \epsilon_{jmn} \mathcal{B}^m \beta^n }$
$= \pmatrix{ -\frac{1}{c^2} \alpha^2 \mathcal{E}^2 + \frac{1}{c^2} (\mathcal{E}_k \beta^k)^2 & \frac{1}{c^2} \mathcal{E}_j \mathcal{E}_k \beta^k + \frac{1}{c} \alpha \epsilon_{jkl} \mathcal{E}^k \mathcal{B}^l + \frac{1}{c} \frac{1}{\alpha} \mathcal{E}_k \beta^k \epsilon_{jlm} \mathcal{B}^l \beta^m \\ \frac{1}{c^2} \mathcal{E}_i \beta^m \mathcal{E}_m + \frac{1}{c} \alpha \epsilon_{ikl} \mathcal{E}^k \mathcal{B}^l + \frac{1}{c} \frac{1}{\alpha} \epsilon_{ikl} \mathcal{B}^k \beta^l \mathcal{E}_m \beta^m & \frac{1}{c^2} \mathcal{E}_i \mathcal{E}_j + \mathcal{B}_i \mathcal{B}_j - \gamma_{ij} \mathcal{B}^2 + \frac{1}{c} \frac{1}{\alpha} \mathcal{E}_i \epsilon_{jkl} \mathcal{B}^k \beta^l + \frac{1}{c} \frac{1}{\alpha} \mathcal{E}_j \epsilon_{ikl} \mathcal{B}^k \beta^l + \frac{1}{\alpha^2} \epsilon_{ikl} \mathcal{B}^k \beta^l \epsilon_{jmn} \mathcal{B}^m \beta^n }$

$\frac{1}{4} tr F^2 = \frac{1}{4} g^{ab} (F^2)_{ab}$
$ = + \frac{1}{2} \frac{1}{c^2} \mathcal{E}^2 - \frac{1}{2} \mathcal{B}^2 + \frac{1}{2} \frac{1}{\alpha^2} \beta^2 \mathcal{B}^2 - \frac{1}{2} \frac{1}{\alpha^2} (\beta^i \mathcal{B}_i)^2 + \frac{1}{c} \frac{1}{\alpha} \epsilon_{ijk} \mathcal{E}^i \mathcal{B}^j \beta^k $

$T_{ab} = \frac{1}{4} g_{ab} tr (F^2) - (F^2)_{ab}$
$= ( \frac{1}{2} \frac{1}{c^2} \mathcal{E}^2 - \frac{1}{2} \mathcal{B}^2 + \frac{1}{2} \frac{1}{\alpha^2} \beta^2 \mathcal{B}^2 - \frac{1}{2} \frac{1}{\alpha^2} (\beta^i \mathcal{B}_i)^2 + \frac{1}{c} \frac{1}{\alpha} \epsilon_{ijk} \mathcal{E}^i \mathcal{B}^j \beta^k ) \pmatrix{ -\alpha^2 + \beta_k \beta^k & \beta_j \\ \beta_i & \gamma_{ij} } - \pmatrix{ -\frac{1}{c^2} \alpha^2 \mathcal{E}^2 + \frac{1}{c^2} (\mathcal{E}_k \beta^k)^2 & \frac{1}{c^2} \mathcal{E}_j \mathcal{E}_k \beta^k + \frac{1}{c} \alpha \epsilon_{jkl} \mathcal{E}^k \mathcal{B}^l + \frac{1}{c} \frac{1}{\alpha} \mathcal{E}_k \beta^k \epsilon_{jlm} \mathcal{B}^l \beta^m \\ \frac{1}{c^2} \mathcal{E}_i \beta^m \mathcal{E}_m + \frac{1}{c} \alpha \epsilon_{ikl} \mathcal{E}^k \mathcal{B}^l + \frac{1}{c} \frac{1}{\alpha} \epsilon_{ikl} \mathcal{B}^k \beta^l \mathcal{E}_m \beta^m & \frac{1}{c^2} \mathcal{E}_i \mathcal{E}_j + \mathcal{B}_i \mathcal{B}_j - \gamma_{ij} \mathcal{B}^2 + \frac{1}{c} \frac{1}{\alpha} \mathcal{E}_i \epsilon_{jkl} \mathcal{B}^k \beta^l + \frac{1}{c} \frac{1}{\alpha} \mathcal{E}_j \epsilon_{ikl} \mathcal{B}^k \beta^l + \frac{1}{\alpha^2} \epsilon_{ikl} \mathcal{B}^k \beta^l \epsilon_{jmn} \mathcal{B}^m \beta^n }$
I think I lost a term because the $E^2$s and $B^2$s don't match...
$= \pmatrix{ \frac{1}{2} \frac{1}{c^2} \alpha^2 \mathcal{E}^2 + \frac{1}{2} \alpha^2 \mathcal{B}^2 - \frac{1}{c^2} (\mathcal{E}_k \beta^k)^2 + \frac{1}{2} \frac{1}{c^2} \beta^2 \mathcal{E}^2 - \beta^2 \mathcal{B}^2 + \frac{1}{2} (\beta^i \mathcal{B}_i)^2 - \alpha \epsilon_{ijk} \frac{1}{c} \mathcal{E}^i \mathcal{B}^j \beta^k + \frac{1}{2} \frac{1}{\alpha^2} \beta^4 \mathcal{B}^2 - \frac{1}{2} \frac{1}{\alpha^2} \beta^2 (\beta^i \mathcal{B}_i)^2 + \frac{1}{c} \frac{1}{\alpha} \beta^2 \epsilon_{ijk} \mathcal{E}^i \mathcal{B}^j \beta^k & \frac{1}{2} \frac{1}{c^2} \mathcal{E}^2 \beta_j - \frac{1}{2} \mathcal{B}^2 \beta_j + \frac{1}{2} \frac{1}{\alpha^2} \beta^2 \mathcal{B}^2 \beta_j - \frac{1}{2} \frac{1}{\alpha^2} (\beta^k \mathcal{B}_k)^2 \beta_j + \frac{1}{\alpha} \frac{1}{c} \epsilon_{lmn} \mathcal{E}^l \mathcal{B}^m \beta^n \beta_j \frac{1}{c^2} - \mathcal{E}_j \mathcal{E}_k \beta^k - \frac{1}{c} \alpha \epsilon_{jkl} \mathcal{E}^k \mathcal{B}^l - \frac{1}{c} \frac{1}{\alpha} \mathcal{E}_k \beta^k \epsilon_{jlm} \mathcal{B}^l \beta^m \\ \frac{1}{2} \frac{1}{c^2} \mathcal{E}^2 \beta_i - \frac{1}{2} \mathcal{B}^2 \beta_i + \frac{1}{2} \frac{1}{\alpha^2} \beta^2 \mathcal{B}^2 \beta_i - \frac{1}{2} \frac{1}{\alpha^2} (\beta^k \mathcal{B}_k)^2 \beta_i + \frac{1}{c} \frac{1}{\alpha} \epsilon_{lmn} \mathcal{E}^l \mathcal{B}^m \beta^n \beta_i - \frac{1}{c^2} \mathcal{E}_i \beta^m \mathcal{E}_m - \frac{1}{c} \alpha \epsilon_{ikl} \mathcal{E}^k \mathcal{B}^l - \frac{1}{c} \frac{1}{\alpha} \epsilon_{ikl} \mathcal{B}^k \beta^l \mathcal{E}_m \beta^m & - \frac{1}{c^2} \mathcal{E}_i \mathcal{E}_j - \mathcal{B}_i \mathcal{B}_j + \frac{1}{2} \frac{1}{c^2} \mathcal{E}^2 \gamma_{ij} + \frac{1}{2} \mathcal{B}^2 \gamma_{ij} + \frac{1}{2} \frac{1}{\alpha^2} \beta^2 \mathcal{B}^2 \gamma_{ij} - \frac{1}{2} \frac{1}{\alpha^2} (\beta^i \mathcal{B}_i)^2 \gamma_{ij} + \frac{1}{c} \frac{1}{\alpha} \epsilon_{ijk} \mathcal{E}^i \mathcal{B}^j \beta^k \gamma_{ij} - \frac{1}{c} \frac{1}{\alpha} \mathcal{E}_i \epsilon_{jkl} \mathcal{B}^k \beta^l - \frac{1}{c} \frac{1}{\alpha} \mathcal{E}_j \epsilon_{ikl} \mathcal{B}^k \beta^l - \frac{1}{\alpha^2} \epsilon_{ikl} \mathcal{B}^k \beta^l \epsilon_{jmn} \mathcal{B}^m \beta^n }$

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Einstein Field Equations of electromagnetic stress-energy in vacuum:
${G^u}_v = 8 \pi {T^u}_v \frac{G}{c^4}$
For $G$ the gravitational constant, with units $\frac{m^3}{kg \cdot s^2}$
${G^u}_v$ has units $\frac{1}{m^2}$
${G^u}_v = -2 ({(F^2)^u}_v)^{TF}$
Notice, $G_{uv}$ is the trace-reversal of $R_{uv}$, so $G_{uv} = R_{uv} - \frac{2}{n} g_{uv} {R^a}_a$
This means that (contract to find) ${G^a}_a = -{R^a}_a$
Therefore if $T_{uv}$ is trace-free, so is $G_{uv}$, and so is $R_{uv}$ such that $G_{uv} = R_{uv}$
${R^u}_v = -2 ({(F^2)^u}_v)^{TF}$

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Einstein Field Equations of electromagnetic stress-energy in vacuum, in normal coordinates:
$R_{uv} = {R^a}_{uav}$
${R^a}_{bcd} = 2 ({\Gamma^a}_{b[d,c]} + {\Gamma^a}_{e[c} {\Gamma^e}_{d]b})$ in a holonomic, torsion-free coordinate system.
For normal coordinates, let $g_{ab,c} = 0$, so ${\Gamma^a}_{bc} = 0$
So ${R^a}_{bcd} = 2 g^{au} \Gamma_{ub[d,c]}$ in a holonomic, torsion-free coordinate system.
${R^a}_{bcd} = g^{au} (g_{ub,[dc]} + g_{u[d|,b|c]} - g_{b[d|,u|c]})$
${R^a}_{bcd} = g^{au} (g_{u[d,c]b} - g_{b[d,c]u})$
${R^a}_{bcd} = 2 g^{au} g_{[u|[d,c]|b]}$
${R^a}_{bcd} = \frac{1}{2} g^{au} (g_{ud,cb} - g_{uc,db} - g_{bd,cu} + g_{bc,du})$
$R_{uv} = {R^a}_{uav} = \frac{1}{2} g^{ab} (g_{bv,au} - g_{ab,uv} - g_{uv,ab} + g_{ua,bv})$
and the EFE is:
$\frac{1}{2} g^{ab} (g_{bv,au} - g_{ab,uv} - g_{uv,ab} + g_{au,bv}) = -2 ((F^2)_{uv})^{TF}$
$g^{ab} (g_{bv,au} - g_{ab,uv} - g_{uv,ab} + g_{au,bv}) = -4 ((F^2)_{uv})^{TF}$

Trace again, to get the trace-free constraint on the metric:
$g^{ab} g^{uv} (g_{bv,au} - g_{ab,uv} - g_{uv,ab} + g_{ua,bv}) = 0$
$g^{ab} g^{uv} g_{au,bv} = g^{ab} g^{uv} g_{ab,uv}$

If $R_{uv} = 8 \pi T_{uv} = -2 (F^2)^{TF}_{uv}$ then $R_{uv}$ must be trace-free as well...
Then ${R^a}_a = 0$
But if ${R^a}_b = \lambda \delta^a_b$ then ${R^a}_b$ is purely trace... and for it to be purely trace and trace-free, it must equal zero.
For ${R^a}_b$ to merely be trace-free would mean the sum of the trace elements must be zero, which means any asymmetry in the elements will be preserved.
In fact if Ricci is trace-free then ${R^a}_b$ for $a \ne b$ can be any arbitrary value. Trace-free is only concerned with the diagonal elements.

$\frac{1}{2} g^{ab} g_{pq,rs} ( \delta^p_b \delta^q_v \delta^r_a \delta^s_u - \delta^p_a \delta^q_b \delta^r_u \delta^s_v - \delta^p_u \delta^q_v \delta^r_a \delta^s_b + \delta^p_u \delta^q_a \delta^r_b \delta^s_v ) = -2 (F^2)_{ps} (\delta^p_u \delta^s_v - \frac{1}{4} g_{uv} g^{ps})$
...using $(F^2)_{ab} = F_{au} g^{uv} F_{vb} = (-F_{ua}) g^{uv} (-F_{bv}) = F_{bv} g^{vu} F_{ua} = F_{bu} g^{uv} F_{va} = (F^2)_{ba}$...
$\frac{1}{2} g_{pq,rs} ( g^{qr} ( \delta^p_v \delta^s_u + \delta^p_u \delta^s_v ) - g^{pq} \delta^r_u \delta^s_v - g^{rs} \delta^p_u \delta^q_v ) = -2 F_{pq} F_{rs} ( \frac{1}{2} g^{qr} ( \delta^p_u \delta^s_v + \delta^p_v \delta^s_u ) - \frac{1}{4} g^{qr} g_{uv} g^{ps}) $
$g_{pq,rs} \cdot \frac{1}{2} g^{qr} ( \delta^p_v \delta^s_u + \delta^p_u \delta^s_v ) - \frac{1}{2} g_{pq,rs} ( g^{pq} \delta^r_u \delta^s_v + g^{rs} \delta^p_u \delta^q_v ) = (-2 F_{pr} F_{qs}) \cdot \frac{1}{2} g^{qr} ( \delta^p_u \delta^s_v + \delta^p_v \delta^s_u ) - (-2 F_{pr} F_{qs}) \frac{1}{4} g^{qr} g_{uv} g^{ps} $

Equating the first term on the lhs with the first term on the rhs: $g_{pq,rs} = -2 F_{pr} F_{qs}$
(Notice the $g^{qr}$ indices on the rhs could be swapped, but this would equate the antisymmetric indexes of $F_{pq}$ and $F_{rs}$ with the symmetric indices of $g_{pq}$ and $g_{rs}$, which wouldn't work.)
Then, starting with the terms we have equated:
$g_{pq,rs} g^{qr} = -2 F_{pr} g^{rq} F_{qs} = -2 (F^2)_{ps}$
$\frac{1}{2} g_{pq,rs} g^{qr} (\delta^p_u \delta^s_v + \delta^p_v \delta^s_u) = -2 (F^2)_{ps} \cdot \frac{1}{2} (\delta^p_u \delta^s_v + \delta^p_v \delta^s_u) = -((F^2)_{uv} + (F^2)_{vu}) = -2 (F^2)_{(uv)} = -2 (F^2)_{uv}$
Next come the terms we haven't equated, first those on the left:
$g_{pq,rs} g^{pq} = -2 g^{pq} F_{pr} F_{qs} = 2 F_{rp} g^{pq} F_{qs} = 2 (F^2)_{rs}$
And $g_{pq,rs} g^{rs} = -2 g^{rs} F_{pr} F_{qs} = 2 F_{pr} g^{rs} F_{sq} = 2 (F^2)_{pq}$
And $-\frac{1}{2} g_{pq,rs} ( g^{pq} \delta^r_u \delta^s_v + g^{rs} \delta^p_u \delta^q_v ) = -\frac{1}{2} ( 2 (F^2)_{rs} \delta^r_u \delta^s_v + 2 (F^2)_{pq} \delta^p_u \delta^q_v ) = - 2 (F^2)_{uv}$
So the total on the lhs would be $-4 (F^2)_{uv}$
On the right: $- (-2 F_{pr} F_{qs}) \frac{1}{4} g^{qr} g_{uv} g^{ps} = \frac{1}{2} {(F^2)^a}_a g_{uv}$

So the equation becomes:
$-4 (F^2)_{uv} = -2 (F^2)_{uv} + \frac{1}{2} {(F^2)^a}_a g_{uv}$
$-4 (F^2)_{uv} = {(F^2)^a}_a g_{uv}$
${(F^2)^u}_v = -\frac{1}{4} {(F^2)^a}_a \delta^u_v$
...and we find the mixed form of $F^2$ is completely in the trace.
...which would mean that $(F^2)^{TF}$ is zero...
...which would mean the Ricci curvature is zero (since $R_{ab} = -2 (F^2)^{TF}_{ab}$)....
...which would only work if the $F_{uv}$ was zero... this doesn't help.

Try again but with $g_{pq,rs} = -2 a F_{pr} F_{qs}$
$\frac{1}{2} g_{pq,rs} g^{qr} (\delta^p_u \delta^s_v + \delta^p_v \delta^s_u) = -2 a (F^2)_{ps} \cdot \frac{1}{2} (\delta^p_u \delta^s_v + \delta^p_v \delta^s_u) = -2 a (F^2)_{uv}$
$-\frac{1}{2} g_{pq,rs} (g^{pq} \delta^r_u \delta^s_v + g^{rs} \delta^p_u \delta^q_v) = -\frac{1}{2} (-2 F_{pr} F_{qs}) (g^{pq} \delta^r_u \delta^s_v + g^{rs} \delta^p_u \delta^q_v) = -2 a (F^2)_{uv}$
$- (-2 F_{pr} F_{qs}) \frac{1}{4} g^{qr} g_{uv} g^{ps} = \frac{1}{2} {(F^2)^a}_a g_{uv}$ So we get:
$4 (1 - 2 a) {(F^2)^u}_v = {(F^2)^a}_a \delta^u_v$
Well what happens when $a = \frac{1}{2}$ ?
Then we get a constraint on the trace of $F^2$ must equal zero, but no information on what exactly it is.
I think, in fact, we just get $F_{uv} = 0$ again.

Then comes ... under some choice of gauge, $R_{uv} = -\frac{1}{2} \Box g_{uv}$, especially weak field and Donder-Fock gauge.
That leaves you with $\Box g_{uv} = 4 (F^2)^{TF}_{uv}$
Then solve accordingly.

While I'm rambling, I keep hearing that, for Kahler-Einstein metrics, that $R_{uv} = \lambda g_{uv}$.
If this is the case then maybe this would put second-derivatives of the metric at the same footing as the metric themselves.
If this is the case then maybe $d^2 g_{uv} = g_{uv}$ or something.
So what's the antiderivative of $F_{uv}$? $A_u$. And what's an antiderivative of $A_u$? $exp(\int A_u dx^u)$
So then those $g_{ab} = h_{ab} exp(\int A_u dx^u)$ metrics would be on the same footing as something that might look like (the Kahler metric) $g_{ab} = \hat{g}_{ab} + i F_{ab}$?
I don't know, I haven't thought this through at all. More on it later.

TODO look up more on Kahler metrics, and Kahler-Einstein manifolds.