index notation convention:
$\alpha - \omega$ = time+space.
$ijklmn$ = space.
the rest of $a - z$ = time+space+electromagnetism.
unit coordinate convention: $dx^0 = c dt$
This is using natural units, were
$1 = c =$ $\frac{m}{s} =$ the speed of light.
$1 = G =$ $\frac{m^3}{kg \cdot s^2} =$ gravitational constant
$1 = k_e =$ $\cdot \frac{kg \cdot m^3}{C^2 \cdot s^2} =$ {{Coulomb_constant_in_kg_m3_per_C2_s2}} $\frac{kg}{C} \cdot \frac{m^3}{C \cdot s^2} =$ Coulomb's constant.
$1 = \sqrt{\frac{k_e}{G}} =$ {{C_in_kg = Math.sqrt(Coulomb_constant_in_kg_m3_per_C2_s2 / gravitational_constant_in_m3_per_kg_s2)}} $\cdot \frac{kg}{C}$,
so $1 C =$ {{C_in_kg}} $kg$,
and $1 kg =$ {{kg_in_C = 1 / C_in_kg}} $C$.
$A_u =$ electromagnetic four-potential, in units of $\frac{kg \cdot m}{C \cdot s}$
$A_5 = 1$ but has units $\frac{kg \cdot m}{C \cdot s}$, so $A_5 = c \cdot \sqrt{\frac{k_e}{G}} =$ {{speed_of_light_in_m_per_s * C_in_kg}} $ \cdot \frac{kg \cdot m}{C \cdot s}$.
$\phi > 0 =$ scalar field, in units of $\frac{C \cdot s}{kg \cdot m}$
$\tilde{g}_{ab} =$ metric of spacetime plus electromagnetism in 4+1 is in units $\left[ 1 \right]$.
$\tilde{g}_{ab}
= {}_{\downarrow a(\alpha)} \overset{\rightarrow b(\beta)}{
\left[
\begin{matrix}
\tilde{g}_{\alpha\beta} & \tilde{g}_{\alpha 5} \\
\tilde{g}_{5\beta} & \tilde{g}_{55}
\end{matrix}
\right]}$
$= \left[ \begin{matrix}
g_{\alpha\beta} + \phi^2 A_\alpha A_\beta &
\phi^2 A_\alpha A_5 \\
\phi^2 A_\beta A_5 &
\phi^2 (A_5)^2
\end{matrix} \right]$
${\tilde{\Gamma}^a}_{bc}$ is in units of $\frac{1}{m}$
${\tilde{\Gamma}^a}_{bc} = \tilde{g}^{ad} \cdot \tilde{\Gamma}_{dbc}$
$=\overset{\downarrow c [ \downarrow a \rightarrow b]}{\left[ \begin{matrix}
\left[ \begin{matrix}
g^{\alpha\delta} (
\Gamma_{\delta\beta\gamma} + \phi^2 (
A_\delta A_{(\beta,\gamma)}
+ A_{(\beta} F_{\gamma)\delta}
)
) - g^{\alpha\mu} A_\mu \phi^2 A_{(\beta,\gamma)} &
g^{\alpha\delta} \frac{1}{2} \phi^2 F_{\gamma\delta} A_5 \\
-\frac{1}{A_5} g^{\delta\mu} A_\mu (
\Gamma_{\delta\beta\gamma} + \phi^2 (
A_\delta A_{(\beta,\gamma)}
+ A_{(\beta} F_{\gamma)\delta}
)
) + \frac{1}{A_5} (g^{\mu\nu} A_\mu A_\nu + \frac{1}{\phi^2}) \phi^2 A_{(\beta,\gamma)} &
- g^{\delta\mu} A_\mu \frac{1}{2} \phi^2 F_{\gamma\delta}
\end{matrix} \right] \\
\left[ \begin{matrix}
g^{\alpha\delta} \frac{1}{2} \phi^2 F_{\beta\delta} A_5 &
0 \\
- g^{\delta\mu} A_\mu \frac{1}{2} \phi^2 F_{\beta\delta} &
0
\end{matrix} \right]
\end{matrix} \right] }$
$=\overset{\downarrow c [ \downarrow a \rightarrow b]}{\left[ \begin{matrix}
\left[ \begin{matrix}
{\Gamma^\alpha}_{\beta\gamma}
+ \phi^2 A_{(\beta} {F_{\gamma)}}^\alpha &
\frac{1}{2} \phi^2 A_5 {F_\gamma}^\alpha \\
\frac{1}{A_5} (
- A_\mu (
{\Gamma^\mu}_{\beta\gamma}
+ \phi^2 A_{(\beta} {F_{\gamma)}}^\mu
)
+ A_{(\beta,\gamma)}
) &
-\frac{1}{2} \phi^2 A_\mu {F_\gamma}^\mu
\end{matrix} \right] \\
\left[ \begin{matrix}
\frac{1}{2} \phi^2 A_5 {F_\beta}^\alpha &
0 \\
-\frac{1}{2} \phi^2 A_\mu {F_\beta}^\mu &
0
\end{matrix} \right]
\end{matrix} \right] }$
Let $F_{\alpha\beta} = 2 A_{[\beta,\alpha]}$ the Faraday tensor, is in units of $\frac{kg}{C \cdot s}$
This means $F_{a5} = F_{5a} = 0$, and this means $F^{ab}$ will have the same value when raised by $g^{ab}$ or $\tilde{g}^{ab}$.
position and velocity:
$x^a$ is in units of $m$
$u^a = \partial_\tau x^a$ is in units of $1$
$\partial_\tau u^a$ is in units of $\frac{1}{m}$ (notice that, to get a time value, you must use $c \partial_\tau$, which is in units of $\frac{1}{s}$).
using the low-velocity approximation $u^0 \approx 1$, $u^i \approx \frac{v^i}{c}$
and using the Newtonian connection approximation ${\Gamma^i}_{00} = -\frac{G \cdot M}{c^2 \cdot r^2}$ (so ${\Gamma^i}_{tt} = -\frac{G \cdot M}{r^2}$) and ${\Gamma^\mu}_{\alpha\beta} = 0$ otherwise.
spatial term:
$c \partial_\tau v^i =
- c^2 {\Gamma^i}_{00}
+ c \frac{q}{M} {F^i}_0
+ \frac{q}{M} {F^i}_j v^j
+ \frac{G}{k_e} (A_0 + \frac{1}{c} A_j v^j) ({F^i}_0 + \frac{1}{c} {F^i}_k v^k )
$
Let ${F^i}_0 \approx \frac{1}{c} E^i$ and ${F^i}_j \approx {\epsilon^i}_{jk} B^k$
Let $A_0 = \frac{1}{c} \Phi$ be the electromagnetic potential.
$c \partial_\tau v^i =
\frac{GM}{r^2}
+ \frac{q}{M} (E^i + {\epsilon^i}_{jk} v^j B^k)
+ \frac{1}{c^2} \frac{G}{k_e} (\Phi + A_j v^j) (E^i + {\epsilon^i}_{kl} v^k B^l )
$
$c \partial_\tau v^i = \frac{GM}{r^2} + (\frac{q}{M} + \frac{1}{c^2} \frac{G}{k_e} (\Phi + A_j v^j))(E^i + {\epsilon^i}_{kl} v^k B^l )$
...and we see that, in the Lorentz force law component, the charge should be offset by the electromagnetic potential field, which itself grows with distance. Wouldn't that mean infinite charge? How do the units pan out? What if you plug in a potential field at a distance of the size of the known universe?
$\tilde{T}_{ab} = c^2 \rho_M u_a u_b + P \tilde{g}_{ab}$
so $\tilde{T}_{5\mu} = c^2 \rho_M u_5 u_\mu + P \tilde{g}_{5\mu}
= c^2 \rho_M \cdot \frac{q}{M} u_\mu + P \tilde{g}_{5\mu}
= c^2 \rho_q u_\mu
= J_\mu$ (with some conversions I left out... TODO add them back in).
so $G_{5\mu} = 8 \pi T_{5\mu}$ should reproduce the Gauss-Amepere law ${F_{\mu\nu}}^{;\nu} = \mu_0 J_\mu$.
TODO's:
- straighten out the $M$'s ... which is our mass, and which is the other particle's mass (creating the field this particle is interacting with).
- for some known 4-potential (current through a wire), calculate the magnitude of $A \cdot F$ and $A^2$, and see if it should be measurable.
- for $\partial_\tau u^5$, for low-velocity Newtonian gravitation approximation, see if the charge can change.