The following worksheet is in special relativity with metric $\eta_{\mu\nu} = diag(-1, 1, 1, 1)$
This means $A_0 = -A^0$ and $A_i = A^i$
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 $F = \frac{1}{2} F_{ab} dx^a \wedge dx^b$
In index notation:
$F = dA$
$= d (A_\mu dx^\mu)$
$= \partial_\nu A_\mu \delta^{\nu\mu}_{\alpha\beta} dx^\alpha \otimes dx^\beta$
$= \partial_\nu A_\mu \cdot 2! \delta^\nu_{[\alpha} \delta^\mu_{\beta]} dx^\alpha \otimes dx^\beta$
$F_{\alpha\beta} dx^\alpha \otimes dx^\beta = 2 \partial_{[\alpha} A_{\beta]} dx^\alpha \otimes dx^\beta$
So $F_{\alpha\beta} = 2 \partial_{[\alpha} A_{\beta]}$
$F_{ab}$ 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 1+3 components we have:
$F_{ab} = \left[\begin{matrix}
0 & F_{0j} \\
F_{i0} & F_{ij}
\end{matrix}\right] = \left[\begin{matrix}
0 & 2 \partial_{[0} A_{j]} \\
2 \partial_{[i} A_{0]} & 2 \partial_{[i} A_{j]}
\end{matrix}\right] = \left[\begin{matrix}
0 & A_{j,0} - A_{0,j} \\
A_{0,i} - A_{i,0} & A_{j,i} - A_{i,j}
\end{matrix}\right]= \left[\begin{matrix}
0 & \frac{1}{c} (A_{j,t} - \phi_{,j}) \\
-\frac{1}{c} (A_{i,t} - \phi_{,i}) & A_{j,i} - A_{i,j}
\end{matrix}\right]$
Don't forget that the wedge product of spatial basis one-forms covers all of them twice, so I include the $\frac{1}{2}$
$F = F_{i0} dx^i \otimes dx^0 + F_{jk} dx^j \otimes dx^k = F_{i0} dx^i \wedge dx^0 + \frac{1}{2} F_{ij} dx^i \wedge dx^j$
Electric and magnetic fields:
Let $F = \frac{1}{c} E \wedge dx^0 + \star (B \wedge dx^0)$
Units of $E_i$ are in $\frac{N}{C} = \frac{kg \cdot m}{C \cdot s^2}$.
Units of $B_i$ are in $\frac{N \cdot s}{C \cdot m} = \frac{kg}{C \cdot s}$
Let $\star (B \wedge dx^0) = \frac{1}{2} B^i \epsilon_{ijk} dx^j \wedge dx^k$
Let $F_{i0} dx^i \otimes dx^0 = \frac{1}{c} E_i$
Let $F_{jk} dx^j \otimes dx^k = B^i \epsilon_{ijk} dx^j \otimes dx^k$
Let $B = B_i dx^i$ is the covariant magnetic field.
Let $F_{ij} = \epsilon_{ijk} B^k$ be the covariant form of the dual of the magnetic field.
So $A_{j,i} - A_{i,j} = \epsilon_{ijk} B^k$
$\epsilon^{ijm} (A_{j,i} - A_{i,j}) = \epsilon^{ijm} \epsilon_{ijk} B^k$
$= \delta^{mij}_{kij} B^k$
$= \delta^{mi}_{ki} B^k$
$= (\delta^m_k \delta^i_i - \delta^m_i \delta^i_k) B^k$
$= (3 - 1) \delta^m_k B^k$
$= 2 B^m$
So $2 B^i = \epsilon^{ijk} (A_{k,j} - A_{j,k}) = 2 \epsilon^{ijk} A_{k,j}$
So $B^i = \epsilon^{ijk} A_{k,j}$
Let $E = E_i dx^i$ is the covariant electric field.
$F_{i0} = \frac{1}{c} E_i$
So $\frac{1}{c} E_i = (\partial_i A_0 - \partial_0 A_i) = \frac{1}{c} (\phi_{,i} - A_{i,t})$
So $E_i = \phi_{,i} - A_{i,t}$
$E_t$ and $B_t$ can be chosen arbitrarily, since any values that are used will be cancelled. I am going to set them to 0.
$F_{uv} = \left[\begin{matrix}
0 & -\frac{1}{c} E_j \\
\frac{1}{c} E_i & \epsilon_{ijk} B^k
\end{matrix}\right]$
In 3+1:
$F_{uv} = \left[\begin{matrix}
0 & -\frac{1}{c} E_1 & -\frac{1}{c} E_2 & -\frac{1}{c} E_3 \\
\frac{1}{c} E_1 & 0 & B^3 & -B^2 \\
\frac{1}{c} E_2 & -B^3 & 0 & B^1 \\
\frac{1}{c} E_3 & B^2 & -B^1 & 0
\end{matrix}\right]$
Faraday tensor raised by our metric $\eta^{uv}$:
$F^{uv} = \left[\begin{matrix}
0 & \frac{1}{c} E_j \\
-\frac{1}{c} E_i & \epsilon_{ijk} B^k
\end{matrix}\right]$
In 3+1:
$F_{uv} = \left[\begin{matrix}
0 & \frac{1}{c} E_1 & \frac{1}{c} E_2 & \frac{1}{c} E_3 \\
-\frac{1}{c} E_1 & 0 & B^3 & -B^2 \\
-\frac{1}{c} E_2 & -B^3 & 0 & B^1 \\
-\frac{1}{c} E_3 & B^2 & -B^1 & 0
\end{matrix}\right]$
Dual of Faraday tensor in $E$ and $B$:
$\star F = \star (\frac{1}{c} E \wedge dx^0 + \star (B \wedge dx^0))$
$= \star (\frac{1}{c} E \wedge dx^0) + \star^2 (B \wedge dx^0)$
$= -(B \wedge dx^0) + \star (\frac{1}{c} E \wedge dx^0)$
In individual components:
$(\star F)_{uv} = \left[\begin{matrix}
0 & B_j \\
-B_i & \frac{1}{c} \epsilon_{ijk} E^k
\end{matrix}\right]$
In 1+3 components:
$(\star F)_{uv} = \left[\begin{matrix}
0 & B_1 & B_2 & B_3 \\
-B_1 & 0 & \frac{1}{c} E^3 & -\frac{1}{c} E^2 \\
-B_2 & -\frac{1}{c} E^3 & 0 & \frac{1}{c} E^1 \\
-B_3 & \frac{1}{c} E^2 & -\frac{1}{c} E^1 & 0
\end{matrix}\right]$
Raised by our metric $\eta^{\mu\nu}$
$\star F^{uv} = \downarrow a \overset{\rightarrow b}{ \left[ \begin{matrix}
0 & -B_j \\
B_i & \frac{1}{c} {\epsilon_{ij}}^k E_k
\end{matrix} \right] }$
In 1+3 components:
$(\star F)^{uv} = \left[\begin{matrix}
0 & -B_1 & -B_2 & -B_3 \\
B_1 & 0 & \frac{1}{c} E^3 & -\frac{1}{c} E^2 \\
B_2 & -\frac{1}{c} E^3 & 0 & \frac{1}{c} E^1 \\
B_3 & \frac{1}{c} E^2 & -\frac{1}{c} E^1 & 0
\end{matrix}\right]$
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:
${B^1}_{,1} + {B^2}_{,2} + {B^3}_{,3} = 0$
${B^1}_{,0} + \frac{1}{c} (E_{3,2} - E_{2,3}) = 0$
${B^2}_{,0} + \frac{1}{c} (E_{1,3} - E_{3,1}) = 0$
${B^3}_{,0} + \frac{1}{c} (E_{2,1} - E_{1,2}) = 0$
becomes:
$B_{1,1} + B_{2,2} + B_{3,3} = 0$
$B_{1,t} + E_{3,2} - E_{2,3} = 0$
$B_{2,t} + E_{1,3} - E_{3,1} = 0$
$B_{3,t} + E_{2,1} - E_{1,2} = 0$
The Gauss law, in index notation:
$B_{i,i} = 0$
The Faraday law, in index notation:
$B_{i,t} + \epsilon_{ijk} E_{k,j} = 0$
$F_{\mu\nu} dx^\mu \otimes dx^\nu = (dA)_{\mu\nu} = \partial_\alpha A_\beta \delta^{\alpha\beta}_{\mu\nu} dx^\mu \otimes dx^\nu$
$dF = (dF)_{\rho\mu\nu} dx^\rho \otimes dx^\mu \otimes dx^\nu$
$= \partial_\gamma (\partial_\alpha A_\beta \delta^{\alpha\beta}_{\delta\epsilon}) \delta^{\gamma\delta\epsilon}_{\rho\mu\nu} dx^\rho \otimes dx^\mu \otimes dx^\nu$
$= 2 A_{\beta,\alpha\gamma} \delta^{\gamma\alpha\beta}_{\rho\mu\nu} dx^\rho \otimes dx^\mu \otimes dx^\nu$
$= 0$ by symmetry of second derivative of A and antisymmetry of the generalized Kronecker delta.
$\star dF = (\star dF)_\sigma dx^\sigma$
$= \frac{1}{6} (dF)_{\rho\mu\nu} {\epsilon^{\rho\mu\nu}}_\sigma dx^\sigma$
$= \frac{1}{6} 2 A_{\beta,\alpha\gamma} \delta^{\gamma\alpha\beta}_{\rho\mu\nu} {\epsilon^{\rho\mu\nu}}_\sigma dx^\sigma$
$= 2 A_{\beta,\alpha\gamma} {\epsilon^{\gamma\alpha\beta}}_\sigma dx^\sigma$
$= 0$ by symmetry of second derivative of A and antisymmetry of the Levi-Civita connection.
Similarly,
${(\star F)^{\alpha\beta}}_{;\beta} \partial_\alpha$
$= (\frac{1}{2} F_{\mu\nu} \epsilon^{\mu\nu\alpha\beta})_{;\beta} \partial_\alpha$
$= (A_{[\nu;\mu]} \epsilon^{\mu\nu\alpha\beta})_{;\beta} \partial_\alpha$
$= (A_{\nu;\mu} \epsilon^{\mu\nu\alpha\beta})_{;\beta} \partial_\alpha$ by the rule that $A_{[\mu,\nu]} = A_{[\mu;\nu]}$
$= A_{\nu;\mu;\beta} \epsilon^{\mu\nu\alpha\beta} \partial_\alpha$
$= A_{\mu;\nu;\alpha} \epsilon^{\mu\nu\alpha\beta} g_{\beta\gamma} dx^\gamma$
$= A_{\mu;[\nu;\alpha]} \epsilon^{\mu\nu\alpha\beta} g_{\beta\gamma} dx^\gamma$ by antisymmetry of the Levi-Civita tensor
$= \frac{1}{2} {{R_\mu}^\rho}_{\nu\alpha} A_\rho \epsilon^{\mu\nu\alpha\beta} g_{\beta\gamma} dx^\gamma$
$= -\frac{1}{2} A_\rho {R^\rho}_{\mu\nu\alpha} \epsilon^{\mu\nu\alpha\beta} g_{\beta\gamma} dx^\gamma$
$= 0$ by the first Bianchi identity (assuming no torsion).
Therefore...
${(\star F)_{\alpha\beta}}^{;\beta} = 0$
...is similar in its statement that $d F = d^2 A = 0$, yet this uses covariant derivatives instead of partial derivatives.
Gauss-Ampere law / co-differential of exterior derivative of four-potential / co-differential of Faraday tensor:
$J^a = \left[ \begin{matrix}
c \rho \\ J^i
\end{matrix} \right]$ is the four-current, in units $\frac{C}{m^2 \cdot s}$
$J^i$ is the current density, in units $\frac{C}{m^2 \cdot s}$
$\rho$ is the charge density, in units $\frac{C}{m^3}$
$J_a = \left[ \begin{matrix}
-c \rho \\ J_i
\end{matrix} \right]$
$\mu_0 = 1.25663706212 \cdot 10^{-6} \frac{kg \cdot m}{C^2}$ is the vacuum permeability.
$\frac{1}{\mu_0} {F^{ab}}_{,b}$ is in units $\frac{C}{m^2 \cdot s}$
${F^{ab}}_{,b} = \mu_0 J^a$ is in units $\frac{kg}{C \cdot m \cdot s}$
$\frac{1}{2} \delta F = \mu_0 J$
$\frac{1}{2} \star d \star F = \mu_0 J$
$\frac{1}{2} \star d \star d A = \mu_0 J$
$\frac{1}{2} d \star F = -\star \mu_0 J$
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:
$J_0 (dx^1 \wedge dx^2 \wedge dx^3) = ((E^1)_{,1} + (E^2)_{,2} + (E^3)_{,3}) (dx^1 \wedge dx^2 \wedge dx^3)$
$J_1 (dx^0 \wedge dx^2 \wedge dx^3) = ((E^1)_{,0} - \frac{1}{c} ((B_3)_{,2} - (B_2)_{,3})) (dx^0 \wedge dx^2 \wedge dx^3)$
$J_2 (dx^0 \wedge dx^3 \wedge dx^1) = ((E^2)_{,0} - \frac{1}{c} ((B_1)_{,3} - (B_3)_{,1})) (dx^0 \wedge dx^3 \wedge dx^1)$
$J_3 (dx^0 \wedge dx^1 \wedge dx^2) = ((E^3)_{,0} - \frac{1}{c} ((B_2)_{,1} - (B_1)_{,2})) (dx^0 \wedge dx^1 \wedge dx^2)$
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:
$E_{1,1} + E_{2,2} + E_{3,3} = \rho_{free}$
$E_{1,t} - B_{3,2} + B_{2,3} = -J_1$
$E_{2,t} - B_{1,3} + B_{3,1} = -J_2$
$E_{3,t} - B_{2,1} + B_{1,2} = -J_3$
In index notation:
$\frac{1}{c} E_{j,j} = \mu_0 c \rho$
$E_{j,j} = \frac{1}{\epsilon_0} \rho$
$-\frac{1}{c^2} E_{i,t} + {\epsilon_i}^{jk} B_{k,j} = \mu_0 J^i$
$\epsilon_0 E_{i,t} - \frac{1}{\mu_0} {\epsilon_i}^{jk} B_{k,j} = -J^i$
TODO like above, explain how $\frac{1}{2} \delta F = J$ is the same as ${F^{ab}}_{,b} = J^a$
Next let's look at $d \star d \star A$
$\star A = A_\mu \star dx^\mu = A_\mu {\epsilon^\mu}_{\alpha\beta\gamma} dx^\alpha \otimes dx^\beta \otimes dx^\gamma$
$d \star A = \partial_\nu (A_\mu {\epsilon^\mu}_{\alpha\beta\gamma}) \delta^{\nu\alpha\beta\gamma}_{\delta\epsilon\rho\sigma} dx^\delta \otimes dx^\epsilon \otimes dx^\rho \otimes dx^\sigma$
$\star d \star A = \frac{1}{4!} \partial_\nu (A_\mu {\epsilon^\mu}_{\alpha\beta\gamma}) \delta^{\nu\alpha\beta\gamma}_{\delta\epsilon\rho\sigma} \epsilon^{\delta\epsilon\rho\sigma}$
$= \partial_\nu (A_\mu {\epsilon^\mu}_{\alpha\beta\gamma}) \epsilon^{\nu\alpha\beta\gamma}$
$= \partial_\nu (A_\mu g^{\mu\epsilon} \epsilon_{\epsilon\alpha\beta\gamma}) \epsilon^{\nu\alpha\beta\gamma}$
$= \partial_\nu (A_\mu g^{\mu\epsilon} \sqrt{|g|} \bar\epsilon_{\epsilon\alpha\beta\gamma}) \epsilon^{\nu\alpha\beta\gamma}$
$= (
A_{\mu,\nu} g^{\mu\epsilon} \sqrt{|g|}
+ A_\mu {g^{\mu\epsilon}}_{,\nu} \sqrt{|g|}
+ A_\mu g^{\mu\epsilon} (\sqrt{|g|})_{,\nu}
) \bar\epsilon_{\epsilon\alpha\beta\gamma} \epsilon^{\nu\alpha\beta\gamma}$
$= (
{A^\epsilon}_{,\nu} \sqrt{|g|}
+ A^\epsilon {\Gamma^\mu}_{\nu\mu} \sqrt{|g|}
) \bar\epsilon_{\epsilon\alpha\beta\gamma} \epsilon^{\nu\alpha\beta\gamma}$
$= (
{A^\epsilon}_{,\nu}
+ A^\epsilon {\Gamma^\mu}_{\nu\mu}
) \epsilon_{\epsilon\alpha\beta\gamma} \epsilon^{\nu\alpha\beta\gamma}$
$= -3! (
{A^\epsilon}_{,\nu}
+ A^\epsilon {\Gamma^\mu}_{\nu\mu}
) \delta^\nu_\epsilon$
$= -3! ({A^\nu}_{,\nu} + A^\nu {\Gamma^\mu}_{\nu\mu})$
$= -3! {A^\nu}_{;\nu}$
...for constant metric...
$= -3! {A^\mu}_{,\mu}$
The Lorentz gauge condition says to set this to ${A^\mu}_{,\mu} = 0$
Last is $d \star d \star A = -3! {A^\mu}_{,\mu\alpha} dx^\alpha$
...so as long as ${A^\mu}_{,\mu} = const$ we find $d \star d \star A = 0$
Combine $d \star d \star A = 0$ with $\frac{1}{2} \star d \star d A = \mu_0 J$ to find
$\frac{1}{2} (d \star d \star + \star d \star d) A = \mu_0 J$
$\frac{1}{2} \Delta A = \mu_0 J$
...for $\Delta A$ the Hodge-Laplacian operator
So the Gauss/Faraday side of Maxwell's laws are summed up in $d^2 A = 0$
And the Gauss/Ampere side of Maxwell's laws are summed up in $\frac{1}{2} \Delta A = \mu_0 J$