Do you remember what the sun looks like?

Is my math correct? {R_{\mu \nu }-{\tfrac {1}{2}}R\,g_{\mu \nu }+\Lambda g_{\mu \nu }={\frac {8\pi G}{c^{4}}}T_{\mu \nu }\,.} {T_{\mu \nu }^{\mathrm {(vac)} }=-{\frac {\Lambda c^{4}}{8\pi G}}g_{\mu \nu }\,.} {\rho _{\mathrm {vac} }={\frac {\Lambda c^{2}}{8\pi G}}} If the energy-momentum tensor Tμν is that of an electromagnetic field in free space, i.e. if the electromagnetic stress–energy tensor {T^{\alpha \beta }=\,-{\frac {1}{\mu _{0}}}\left({F^{\alpha }}^{\psi }{F_{\psi }}^{\beta }+{\tfrac {1}{4}}g^{\alpha \beta }F_{\psi \tau }F^{\psi \tau }\right)} -is used, then the Einstein field equations are called the Einstein–Maxwell equations (with cosmological constant Λ, taken to be zero in conventional relativity theory): Additionally, the covariant Maxwell equations are also applicable in free space: Fαβ;β=0 F[αβ;γ]=1/3(Fαβ;γ+Fβγ;α+Fγα;β)=1/3(Fαβ,γ+Fβγ,α+Fγ, β)=0 {{\begin{aligned}{F^{\alpha \beta }}_{;\beta }&=0\\F_{[\alpha \beta ;\gamma ]}&={\tfrac {1}{3}}\left(F_{\alpha \beta ;\gamma }+F_{\beta \gamma ;\alpha }+F_{\gamma \alpha ;\beta }\right)={\tfrac {1}{3}}\left(F_{\alpha \beta ,\gamma }+F_{\beta \gamma ,\alpha }+F_{\gamma \alpha ,\beta }\right)=0.\end{aligned}}} where the semicolon represents a covariant derivative, and the brackets denote anti-symmetrization. The first equation asserts that the 4-divergence of the two-form F is zero, and the second that its exterior derivative is zero. From the latter, it follows by the Poincaré lemma that in a coordinate chart it is possible to introduce an electromagnetic field potential Aα such that {{\begin{aligned}{F^{\alpha \beta }}_{;\beta }&=0\\F_{[\alpha \beta ;\gamma ]}&={\tfrac {1}{3}}\left(F_{\alpha \beta ;\gamma }+F_{\beta \gamma ;\alpha }+F_{\gamma \alpha ;\beta }\right)={\tfrac {1}{3}}\left(F_{\alpha \beta ,\gamma }+F_{\beta \gamma ,\alpha }+F_{\gamma \alpha ,\beta }\right)=0.\end{aligned}}} F α\β=A\α;\β−A\β;\α=A\α,\β−A\β,\α {F_{\alpha \beta }=A_{\alpha ;\beta }-A_{\beta ;\alpha }=A_{\alpha ,\beta }-A_{\beta ,\alpha }} in which the comma denotes a partial derivative. This is often taken as equivalent to the covariant Maxwell equation from which it is derived. However, there are global solutions of the equation which may lack a globally defined potential. First, the determinant of the metric in 4 dimensions can be written: det(g)=[1][24]]\ε^{\α\β\γ\δ\ε^{\κ\λ\μ\ν}g{α\κ}g{\β\λ}g{\γ\μ}g{\δ\ν}\,} {\displaystyle \det(g)={\tfrac {1}{24}}\varepsilon ^{\alpha \beta \gamma \delta }\varepsilon ^{\kappa \lambda \mu \nu }g_{\alpha \kappa }g_{\beta \lambda }g_{\gamma \mu }g_{\delta \nu }\,} using the Levi-Civita symbol; and the inverse of the metric in 4 dimensions can be written as: g^{α\κ}={{\{{\{1}{6}}\ε^{\α\β\γ\δ\}\ε^{\κ\λ\μ\ν}g{\β\λ}g{\γ\μ}g{\δ\ν}}{\det(g)}}\,.} {\displaystyle g^{\alpha \kappa }={\frac {{\tfrac {1}{6}}\varepsilon ^{\alpha \beta \gamma \delta }\varepsilon ^{\kappa \lambda \mu \nu }g_{\beta \lambda }g_{\gamma \mu }g_{\delta \nu }}{\det(g)}}\,.} Substituting this definition of the inverse of the metric into the equations then multiplying both sides by det(g) until there are none left in the denominator results in polynomial equations in the metric tensor and its first and second derivatives. The action from which the equations are derived can also be written in polynomial form by suitable redefinitions of the fields. In differential geometry, the Einstein tensor (named after Albert Einstein; also known as the trace-reversed Ricci tensor) is used to express the curvature of a pseudo-Riemannian manifold. In general relativity, it occurs in the Einstein field equations for gravitation that describe spacetime curvature in a manner consistent with energy and momentum conservation.

What is the airspeed velocity of an unladen swallow?

Just looked at your characters... really nice. Must have took some work getting all that at 600. You have some nice gear