TODO worksheet #1: special relativity. Then show $A = A_u dx^u, F = dA, dF = d^2 A = 0, \delta F = J$
worksheet #2: same thing except in GR with ADM metric defining E and B as one-forms.
worksheet #3: same thing except in GR with ADM metric defining E and B as vectors.
electromagnetic four-potential:
$A = A_u dx^u = A_0 dx^0 + A_i dx^i$
For $A_u = \left[ \begin{matrix} \frac{1}{c} \phi \\ A_i \end{matrix} \right]$ in units of $\frac{kg \cdot m}{C \cdot s}$.
$A_i$ is the magnetic potential vector in units of $\frac{kg \cdot m}{C \cdot s}$.
$\phi$ is the electric potential in units of $\frac{kg \cdot m^2}{C \cdot s^2}$.
Faraday tensor:
$F = dA$
$F = dx^a \wedge \partial_a (A_b dx^b) = \partial_a A_b dx^a \wedge dx^b$
$F = \partial_{[a} A_{b]} dx^a \wedge dx^b$ by antisymmetry of wedge product and by relabeling opposite sum indexes.
$F = 2 \partial_{[a} A_{b]} dx^a \otimes dx^b$ with the 2 included due to the fact that $a,b$ span all indexes and the summed term is antisymmetric, as is the wedge product, so each unique $a,b$ pair will be summed twice.
$F = F_{ab} dx^a \otimes dx^b$
So $F_{ab} = 2 \partial_{[a} A_{b]}$
Also $F = \frac{1}{2} F_{ab} dx^a \wedge dx^b$
$F_{ab}$ has units of $\frac{kg}{C \cdot s}$
Faraday tensor in differential forms, separating space and time:
$F = dA$
$= dx^a \wedge \partial_a (A_b dx^b)$
$= dx^a \wedge \partial_a (A_0 dx^0 + A_j dx^j)$
$= dx^0 \wedge \partial_0 (A_0 dx^0 + A_j dx^j) + dx^i \wedge \partial_i (A_0 dx^0 + A_j dx^j)$
$= \partial_0 A_0 dx^0 \wedge dx^0 + \partial_0 A_j dx^0 \wedge dx^j + \partial_i A_0 dx^i \wedge dx^0 + \partial_i A_j dx^i \wedge dx^j$
$= (\partial_i A_0 - \partial_0 A_i) dx^i \wedge dx^0 + \frac{1}{2} (\partial_i A_j - \partial_j A_i) dx^i \wedge dx^j$
The $\frac{1}{2}$ is next to the $dx^i \wedge dx^j$ and not next to the $dx^i \wedge dx^0$ due to the fact that summing over all the $dx^i \wedge dx^j$ terms counts each twice (except the cancelled diagonal terms), while summing over $dx^i \wedge dx^0$ counts each only once.
$= (\partial_i A_0 - \partial_0 A_i) (dx^i \otimes dx^0 - dx^0 \otimes dx^i) + \frac{1}{2} (\partial_i A_j - \partial_j A_i) (dx^i \otimes dx^j - dx^j \otimes dx^i)$
$= (\partial_i A_0 - \partial_0 A_i) dx^i \otimes dx^0 - (\partial_i A_0 - \partial_0 A_i) dx^0 \otimes dx^i + (\partial_i A_j - \partial_j A_i) dx^i \otimes dx^j$
$F_{i0} = -F_{0i} = \partial_i A_0 - \partial_0 A_i = \partial_i (\frac{1}{c} \phi) - (\frac{1}{c} \partial_t) A_i = \frac{1}{c} (\phi_{,i} - A_{i,t})$
$F_{jk} = \partial_i A_j - \partial_j A_i$
In ADM components we have:
$F_{ab} = \left[\begin{matrix}
0 & F_{0j} \\
F_{i0} & F_{ij}
\end{matrix}\right] = \left[\begin{matrix}
0 & 2 \partial_{[0} A_{j]} \\
2 \partial_{[i} A_{0]} & 2 \partial_{[i} A_{j]}
\end{matrix}\right] = \left[\begin{matrix}
0 & A_{j,0} - A_{0,j} \\
A_{0,i} - A_{i,0} & A_{j,i} - A_{i,j}
\end{matrix}\right]= \left[\begin{matrix}
0 & \frac{1}{c} (A_{j,t} - \phi_{,j}) \\
-\frac{1}{c} (A_{i,t} - \phi_{,i}) & A_{j,i} - A_{i,j}
\end{matrix}\right]$
Covariant form of electric and magnetic fields:
$F = F_{i0} dx^i \wedge dx^0 + \frac{1}{2} F_{ij} dx^i \wedge dx^j$
$F = F_{i0} dx^i \otimes dx^0 + F_{jk} dx^j \otimes dx^k$
Let $F = \alpha (\frac{1}{c} \mathcal{E} \wedge dx^0 + \star (\mathcal{B} \wedge dx^0))$
Units of $E^i$ and $\mathcal{E}_i$ are in $\frac{N}{C} = \frac{kg \cdot m}{C \cdot s^2}$.
Units of $B^i$ and $\mathcal{B}_i$ are in $\frac{N \cdot s}{C \cdot m} = \frac{kg}{C \cdot s}$
Let $\star (\mathcal{B} \wedge dx^0) = \frac{1}{2} \frac{1}{\alpha} \mathcal{B}^i \epsilon_{ijk} dx^j \wedge dx^k - \frac{1}{\alpha} \mathcal{B}^j \beta^k \epsilon_{ijk} dx^i \wedge dx^0$
Let $F_{i0} dx^i \otimes dx^0 = \alpha \frac{1}{c} \mathcal{E}_i + \frac{1}{\alpha} \epsilon_{ijk} \mathcal{B}^j \beta^k$
Let $F_{jk} dx^j \otimes dx^k = \mathcal{B}^i \epsilon_{ijk} dx^j \otimes dx^k$
So this is looking really ugly. It looks like the magnetic field needs to be treated like a vector in order for the extra terms to not appear. But that leaves us with the electric
field of a one-form and a magnetic field of a vector ... which is backwards what the spatial concept of electric and magnetic fields are.
Let $\mathcal{B} = \mathcal{B}_i dx^i$ is the covariant magnetic field.
Let $F_{ij} = \epsilon_{ijk} \mathcal{B}^k$ be the covariant form of the dual of the magnetic field.
So $A_{j,i} - A_{i,j} = \epsilon_{ijk} \mathcal{B}^k$
$\epsilon^{ijm} (A_{j,i} - A_{i,j}) = \epsilon^{ijm} \epsilon_{ijk} \mathcal{B}^k$
$= \delta^{mij}_{kij} \mathcal{B}^k$
$= \delta^{mi}_{ki} \mathcal{B}^k$
$= (\delta^m_k \delta^i_i - \delta^m_i \delta^i_k) \mathcal{B}^k$
$= (3 - 1) \delta^m_k \mathcal{B}^k$
$= 2 \mathcal{B}^m$
So $2 \mathcal{B}^i = \epsilon^{ijk} (A_{k,j} - A_{j,k}) = 2 \epsilon^{ijk} A_{k,j}$
So $\mathcal{B}^i = \epsilon^{ijk} A_{k,j} = \frac{1}{\gamma} \bar\epsilon^{ijk} A_{k,j}$
Let $\mathcal{E} = \mathcal{E}_i dx^i$ is the covariant electric field.
$F_{i0} = \frac{1}{c} \alpha \mathcal{E}_i + \frac{1}{\alpha} \mathcal{B}^j \beta^k \epsilon_{ijk}$
So $\frac{1}{c} \alpha \mathcal{E}_i + \frac{1}{\alpha} \epsilon_{ijk} \mathcal{B}^j \beta^k = (\partial_i A_0 - \partial_0 A_i) = \frac{1}{c} (\phi_{,i} - A_{i,t})$
So $\mathcal{E}_i = \frac{1}{\alpha} (-A_{i,t} + \phi_{,i}) + c \frac{1}{\alpha^2} \epsilon_{ijk} \mathcal{B}^j \beta^k$
$= \frac{1}{\alpha} (-A_{i,t} + \phi_{,i}) + c \frac{1}{\alpha^2} \epsilon_{ijk} \epsilon^{jmn} A_{n,m} \beta^k$
$= \frac{1}{\alpha} (-A_{i,t} + \phi_{,i}) - c \frac{1}{\alpha^2} (\delta_i^m \delta_k^n - \delta_k^m \delta_i^n) A_{n,m} \beta^k$
$= \frac{1}{\alpha} (-A_{i,t} + \phi_{,i}) - c \frac{1}{\alpha^2} (A_{k,i} - A_{i,k}) \beta^k$
$\mathcal{E}_t$ and $\mathcal{B}_t$ can be chosen arbitrarily, since any values that are used will be cancelled. I am going to set them to 0.
$F_{uv} = \left[\begin{matrix}
0 & -\frac{1}{c} \alpha \mathcal{E}_j - \frac{1}{\alpha} \epsilon_{jmn} \mathcal{B}^m \beta^n \\
\frac{1}{c} \alpha \mathcal{E}_i + \frac{1}{\alpha} \epsilon_{imn} \mathcal{B}^m \beta^n & \epsilon_{ijk} \mathcal{B}^k
\end{matrix}\right]$
In 3+1:
$F_{uv} = \left[\begin{matrix}
0 &
-\frac{1}{c} \alpha \mathcal{E}_1 - \frac{1}{\alpha \sqrt\gamma} (\mathcal{B}^2 \beta^3 - \mathcal{B}^3 \beta^2) &
-\frac{1}{c} \alpha \mathcal{E}_2 - \frac{1}{\alpha \sqrt\gamma} (\mathcal{B}^3 \beta^1 - \mathcal{B}^1 \beta^3) &
-\frac{1}{c} \alpha \mathcal{E}_3 - \frac{1}{\alpha \sqrt\gamma} (\mathcal{B}^1 \beta^2 - \mathcal{B}^2 \beta^1)
\\
\frac{1}{c} \alpha \mathcal{E}_1 + \frac{1}{\alpha \sqrt\gamma} (\mathcal{B}^2 \beta^3 - \mathcal{B}^3 \beta^2) & 0 & \sqrt{\gamma} B^3 & -\sqrt{\gamma} B^2 \\
\frac{1}{c} \alpha \mathcal{E}_2 + \frac{1}{\alpha \sqrt\gamma} (\mathcal{B}^3 \beta^1 - \mathcal{B}^1 \beta^3) & -\sqrt{\gamma} B^3 & 0 & \sqrt{\gamma} B^1 \\
\frac{1}{c} \alpha \mathcal{E}_3 + \frac{1}{\alpha \sqrt\gamma} (\mathcal{B}^1 \beta^2 - \mathcal{B}^2 \beta^1) & \sqrt{\gamma} B^2 & -\sqrt{\gamma} B^1 & 0
\end{matrix}\right]$
$E^i$ and $B^i$ are associated with the contravariant definition $F^{ab}$. This means $E^t = B^t = 0$.
$\mathcal{E}_i$ and $\mathcal{B}_i$ are associated with the covariant definition $F_{ab}$. This means $\mathcal{E}_t = \mathcal{B}_t = 0$.
In special relativistic manifold, where $g_{ab} = \eta_{ab}$ and ${\Gamma^a}_{bc} = 0$, we will find that $E_\mu = \mathcal{E}_\mu$ and $B_\mu = \mathcal{B}_\mu$, but not when dealing with manifolds of arbitrary metrics.
In special relativity, where $\alpha = 1, \beta^i = 0, \gamma_{ij} = \delta_{ij}$, this becomes:
$F_{\alpha\beta} = \alpha \downarrow \overset{\rightarrow \beta}{ \left[\begin{matrix}
0 & -\frac{1}{c} \mathcal{E}_x & -\frac{1}{c} \mathcal{E}_y & -\frac{1}{c} \mathcal{E}_z \\
\frac{1}{c} \mathcal{E}_x & 0 & \mathcal{B}_z & -\mathcal{B}_y \\
\frac{1}{c} \mathcal{E}_y & -\mathcal{B}_z & 0 & \mathcal{B}_x \\
\frac{1}{c} \mathcal{E}_z & \mathcal{B}_y & -\mathcal{B}_x & 0
\end{matrix}\right] }$
Faraday tensor is immune to choice of covariant derivative:
$F_{ab} = 2 \partial_{[a} A_{b]} = 2 \nabla_{[a} A_{b]}$ (notice the connections cancel)
TODO finishme
Dual of Faraday tensor in $\mathcal{E}$ and $\mathcal{B}$:
$\star F = \star \alpha (\frac{1}{c} \mathcal{E} \wedge dx^0 + \star (\mathcal{B} \wedge dx^0))$
$= \frac{1}{c} \alpha \star (\mathcal{E} \wedge dx^0) + \alpha \star^2 (\mathcal{B} \wedge dx^0)$
$= -\alpha (\mathcal{B} \wedge dx^0) + \frac{1}{c} \alpha \star (\mathcal{E} \wedge dx^0)$
In individual components:
$(\star F)_{uv} = \left[\begin{matrix}
0 & \alpha \mathcal{B}_j + \frac{1}{c} \frac{1}{\alpha} \epsilon_{jmn} \mathcal{E}^m \beta^n \\
-\alpha \mathcal{B}_i - \frac{1}{c} \frac{1}{\alpha} \epsilon_{imn} \mathcal{E}^m \beta^n & \frac{1}{c} \epsilon_{ijk} \mathcal{E}^k
\end{matrix}\right]$
$(\star F)_{uv} = \left[\begin{matrix}
0 &
\alpha \mathcal{B}_1 + \frac{1}{c} \frac{1}{\alpha \sqrt\gamma} (\mathcal{E}^2 \beta^3 - \mathcal{E}^3 \beta^2) &
\alpha \mathcal{B}_2 + \frac{1}{c} \frac{1}{\alpha \sqrt\gamma} (\mathcal{E}^3 \beta^1 - \mathcal{E}^1 \beta^3) &
\alpha \mathcal{B}_3 + \frac{1}{c} \frac{1}{\alpha \sqrt\gamma} (\mathcal{E}^1 \beta^2 - \mathcal{E}^2 \beta^1)
\\
-\alpha \mathcal{B}_1 - \frac{1}{c} \frac{1}{\alpha \sqrt\gamma} (\mathcal{E}^2 \beta^3 - \mathcal{E}^3 \beta^2) & 0 & \frac{1}{c} \sqrt{\gamma} E^3 & -\frac{1}{c} \sqrt{\gamma} E^2 \\
-\alpha \mathcal{B}_2 - \frac{1}{c} \frac{1}{\alpha \sqrt\gamma} (\mathcal{E}^3 \beta^1 - \mathcal{E}^1 \beta^3) & -\frac{1}{c} \sqrt{\gamma} E^3 & 0 & \frac{1}{c} \sqrt{\gamma} E^1 \\
-\alpha \mathcal{B}_3 - \frac{1}{c} \frac{1}{\alpha \sqrt\gamma} (\mathcal{E}^1 \beta^2 - \mathcal{E}^2 \beta^1) & \frac{1}{c} \sqrt{\gamma} E^2 & -\frac{1}{c} \sqrt{\gamma} E^1 & 0
\end{matrix}\right]$
Gauss-Faraday law / second exterior derivative of four-potential / exterior derivative of Faraday tensor:
$d F = d^2 A = 0$
becomes:
$0 = (F_{23,1} + F_{31,2} + F_{12,3} - F_{32,1} - F_{13,2} - F_{21,3}) dx^1 \wedge dx^2 \wedge dx^3$
$0 = (F_{23,0} + F_{30,2} + F_{02,3} - F_{32,0} - F_{03,2} - F_{20,3}) dx^0 \wedge dx^2 \wedge dx^3$
$0 = (F_{13,0} + F_{30,1} + F_{01,3} - F_{31,0} - F_{03,1} - F_{10,3}) dx^0 \wedge dx^1 \wedge dx^3$
$0 = (F_{12,0} + F_{20,1} + F_{01,2} - F_{21,0} - F_{02,1} - F_{10,2}) dx^0 \wedge dx^1 \wedge dx^2$
becomes:
$(\sqrt\gamma \mathcal{B}^1)_{,1} + (\sqrt\gamma \mathcal{B}^2)_{,2} + (\sqrt\gamma \mathcal{B}^3)_{,3} = 0$
$(\sqrt\gamma \mathcal{B}^1)_{,0} + (\frac{1}{c} \alpha \mathcal{E}_3 + \frac{1}{\alpha \sqrt\gamma} (\mathcal{B}^1 \beta^2 - \mathcal{B}^2 \beta^1))_{,2} - (\frac{1}{c} \alpha \mathcal{E}_2 + \frac{1}{\alpha \sqrt\gamma} (\mathcal{B}^3 \beta^1 - \mathcal{B}^1 \beta^3))_{,3} = 0$
$(\sqrt\gamma \mathcal{B}^2)_{,0} + (\frac{1}{c} \alpha \mathcal{E}_1 + \frac{1}{\alpha \sqrt\gamma} (\mathcal{B}^2 \beta^3 - \mathcal{B}^3 \beta^2))_{,3} - (\frac{1}{c} \alpha \mathcal{E}_3 + \frac{1}{\alpha \sqrt\gamma} (\mathcal{B}^1 \beta^2 - \mathcal{B}^2 \beta^1))_{,1} = 0$
$(\sqrt\gamma \mathcal{B}^3)_{,0} + (\frac{1}{c} \alpha \mathcal{E}_2 + \frac{1}{\alpha \sqrt\gamma} (\mathcal{B}^3 \beta^1 - \mathcal{B}^1 \beta^3))_{,1} - (\frac{1}{c} \alpha \mathcal{E}_1 + \frac{1}{\alpha \sqrt\gamma} (\mathcal{B}^2 \beta^3 - \mathcal{B}^3 \beta^2))_{,2} = 0$
The Gauss law, in index notation:
$(\sqrt\gamma \mathcal{B}^i)_{,i} = 0$
$(\sqrt\gamma)_{,i} \mathcal{B}^i + \sqrt\gamma {\mathcal{B}^i}_{,i} = 0$
$\frac{\sqrt\gamma}{\sqrt\gamma} (\sqrt\gamma)_{,i} \mathcal{B}^i + \sqrt\gamma {\mathcal{B}^i}_{,i} = 0$
$\sqrt\gamma \frac{(\sqrt\gamma)_{,i}}{\sqrt\gamma} \mathcal{B}^i + \sqrt\gamma {\mathcal{B}^i}_{,i} = 0$
$\sqrt\gamma {(\Gamma^\perp)^j}_{ij} \mathcal{B}^i + \sqrt\gamma {\mathcal{B}^i}_{,i} = 0$
${(\Gamma^\perp)^j}_{ij} \mathcal{B}^i + {\mathcal{B}^i}_{,i} = 0$
$\nabla^\perp_i \mathcal{B}^i = 0$
TODO FIXME from here on I forgot the $dx^0$ terms in the $\star (B \wedge dx^0)$ terms of the exterior derivative definition of the Maxwell tensor, and I need to re-add it...
The Faraday law, in index notation:
$(\sqrt\gamma \mathcal{B}^i)_{,t} + \sqrt\gamma \epsilon^{ijk} (\alpha \mathcal{E}_k)_{,j} = 0$
$(\sqrt\gamma)_{,t} \mathcal{B}^i + \sqrt\gamma {\mathcal{B}^i}_{,t} + \sqrt\gamma \epsilon^{ijk} \alpha_{,j} \mathcal{E}_k + \sqrt\gamma \epsilon^{ijk} \alpha \mathcal{E}_{k,j} = 0$
$\frac{\alpha (\sqrt\gamma)_{,t}}{\alpha \sqrt\gamma} \mathcal{B}^i + {\mathcal{B}^i}_{,t} + \epsilon^{ijk} \alpha_{,j} \mathcal{E}_k + \alpha \epsilon^{ijk} \mathcal{E}_{k,j} = 0$
$\frac{(\alpha \sqrt\gamma)_{,t}}{\alpha \sqrt\gamma} \mathcal{B}^i
- \frac{\alpha_{,t} \sqrt\gamma}{\alpha \sqrt\gamma} \mathcal{B}^i
+ {\mathcal{B}^i}_{,t}
+ \epsilon^{ijk} \alpha_{,j} \mathcal{E}_k
+ \alpha \epsilon^{ijk} \mathcal{E}_{k,j} = 0$
$\frac{g_{,t}}{g} \mathcal{B}^i
- \frac{\alpha_{,t}}{\alpha} \mathcal{B}^i
+ {\mathcal{B}^i}_{,t}
+ \epsilon^{ijk} \alpha_{,j} \mathcal{E}_k
+ \alpha \epsilon^{ijk} \mathcal{E}_{k,j} = 0$
${\Gamma^\mu}_{t\mu} \mathcal{B}^i
- \frac{\alpha_{,t}}{\alpha} \mathcal{B}^i
+ {\mathcal{B}^i}_{,t}
+ \epsilon^{ijk} \alpha_{,j} \mathcal{E}_k
+ \alpha \epsilon^{ijk} \mathcal{E}_{k,j} = 0$
$\nabla_t \mathcal{B}^i
- \frac{\alpha_{,t}}{\alpha} \mathcal{B}^i
+ \epsilon^{ijk} \alpha_{,j} \mathcal{E}_k
+ \alpha \epsilon^{ijk} \mathcal{E}_{k,j} = 0$
TODO finish me. Make this look better.
in special relativity this becomes:
$B_{1,1} + B_{2,2} + B_{3,3} = 0$
$B_{1,t} + E_{3,2} - E_{2,3} = 0$
$B_{2,t} + E_{1,3} - E_{3,1} = 0$
$B_{3,t} + E_{2,1} - E_{1,2} = 0$
Gauss-Ampere law / co-differential of exterior derivative of four-potential / co-differential of Faraday tensor:
$\delta F = J$
$\star d \star F = J$
$\star d \star d A = J$
$d \star F = -\star J$
TODO check signs
becomes:
$J_0 \star dx^0 = \frac{1}{2} ((\star F)_{23,1} + (\star F)_{31,2} + (\star F)_{12,3} - (\star F)_{32,1} - (\star F)_{13,2} - (\star F)_{21,3}) (dx^1 \wedge dx^2 \wedge dx^3)$
$J_1 \star dx^1 = \frac{1}{2} ((\star F)_{23,0} + (\star F)_{30,2} + (\star F)_{02,3} - (\star F)_{32,0} - (\star F)_{03,2} - (\star F)_{20,3}) (dx^0 \wedge dx^2 \wedge dx^3)$
$J_2 \star dx^2 = \frac{1}{2} ((\star F)_{13,0} + (\star F)_{30,1} + (\star F)_{01,3} - (\star F)_{31,0} - (\star F)_{03,1} - (\star F)_{10,3}) (dx^0 \wedge dx^1 \wedge dx^3)$
$J_3 \star dx^3 = \frac{1}{2} ((\star F)_{12,0} + (\star F)_{20,1} + (\star F)_{01,2} - (\star F)_{21,0} - (\star F)_{02,1} - (\star F)_{10,2}) (dx^0 \wedge dx^1 \wedge dx^2)$
becomes:
$J_0 \star dx^0 = ((\sqrt\gamma \mathcal{E}^1)_{,1} + (\sqrt\gamma \mathcal{E}^2)_{,2} + (\sqrt\gamma \mathcal{E}^3)_{,3}) (dx^1 \wedge dx^2 \wedge dx^3)$
$J_1 \star dx^1 = ((\sqrt\gamma \mathcal{E}^1)_{,0} - \frac{1}{c} ((\alpha \mathcal{B}_3)_{,2} - (\alpha \mathcal{B}_2)_{,3})) (dx^0 \wedge dx^2 \wedge dx^3)$
$J_2 \star dx^2 = ((\sqrt\gamma \mathcal{E}^2)_{,0} - \frac{1}{c} ((\alpha \mathcal{B}_1)_{,3} - (\alpha \mathcal{B}_3)_{,1})) (dx^0 \wedge dx^3 \wedge dx^1)$
$J_3 \star dx^3 = ((\sqrt\gamma \mathcal{E}^3)_{,0} - \frac{1}{c} ((\alpha \mathcal{B}_2)_{,1} - (\alpha \mathcal{B}_1)_{,2})) (dx^0 \wedge dx^1 \wedge dx^2)$
TODO fixme...
becomes:
$c \rho_{free} \star dx^0 = c (E_{1,1} + E_{2,2} + E_{3,3}) \star dx^0$
$J_1 \star dx^1 = ( -c E_{1,0} + B_{3,2} - B_{2,3}) \star dx^1$
$J_2 \star dx^2 = ( -c E_{2,0} + B_{1,3} - B_{3,1}) \star dx^2$
$J_3 \star dx^3 = ( -c E_{3,0} + B_{2,1} - B_{1,2}) \star dx^3$
becomes:
$E_{1,1} + E_{2,2} + E_{3,3} = \rho_{free}$
$E_{1,t} - B_{3,2} + B_{2,3} = -J_1$
$E_{2,t} - B_{1,3} + B_{3,1} = -J_2$
$E_{3,t} - B_{2,1} + B_{1,2} = -J_3$
TODO explain this more:
in a vacuum, in Minkowski metric:
$\nabla \cdot F = J$
TODO somewhere in here...
I've mentioned Maxwell's laws in SR, but not in GR:
${F^{ab}}_{;b} - {R^a}_b A^b = J^a$
Which brings me to another question, what is the Ricci curvature in ADM metric decomposition?
$J^a = \pmatrix{c \rho \\ J^i}$ is in $\frac{kg}{m^2 \cdot s}$
$\rho$ is in $\frac{C}{m^3}$
$J^i$ is in $\frac{C}{m^3} \frac{m}{s} = \frac{C}{m^2 \cdot s}$
${F^{ab}}_{,b}$ is in $\frac{kg}{C \cdot m \cdot s}$
$\frac{1}{\mu_0} {F^{ab}}_{,b}$ is in $\frac{C}{m^2 \cdot s}$
${F^{ab}}_{,b} = \mu_0 J^a$ (eqn 3.36)
This becomes:
$\frac{1}{c} E_{j,j} = \mu_0 c \rho$
$E_{j,j} = \frac{1}{\epsilon_0} \rho$
$-\frac{1}{c^2} E_{i,t} + {\bar\epsilon_i}^{jk} B_{k,j} = \mu_0 J^i$
$-\epsilon_0 E_{i,t} + \frac{1}{\mu_0} {\bar\epsilon_i}^{jk} B_{k,j} = J^i$
TODO also put $\delta F = \star d \star F = J$ here maybe
$\star F_{ab} = \downarrow a \overset{\rightarrow b}{\pmatrix{
0 & -B_j \\
B_i & -\frac{1}{c} {\bar\epsilon_{ij}}^k E_k
}}$
$\star F^{ab} = \downarrow a \overset{\rightarrow b}{\pmatrix{
0 & B_j \\
-B_i & -\frac{1}{c} {\bar\epsilon_{ij}}^k E_k
}}$
${(\star F)^{ab}}_{,b} = 0$
${B^j}_{,j} = 0$
$B_{i,t} + {\bar\epsilon_i}^{jk} E_{k,j} = 0$
TODO also put $dF = d^2 A = 0$ here maybe