Back Einstein Field Equations Recall the contracted Bianchi identity:
G μ ν ; μ = ( R μ ν − 1 2 R g μ ν ) ; μ = 0 , G^{\mu \nu}{}_{; \mu} = \left(R^{\mu \nu} - \frac{1}{2} R g^{\mu \nu}\right)_{; \mu} = 0, G μν ; μ = ( R μν − 2 1 R g μν ) ; μ = 0 , and the conservation of energy-momentum:
T μ ν ; μ = 0. T^{\mu \nu}{}_{; \mu} = 0. T μν ; μ = 0. Because of metric compatibility (the covariant derivative of metric tensor is zero), we can add a term:
( R μ ν − 1 2 R g μ ν + Λ g μ ν ) ; μ = 0 , \left(R^{\mu \nu} - \frac{1}{2} R g^{\mu \nu} + \Lambda g^{\mu \nu}\right)_{; \mu} = 0, ( R μν − 2 1 R g μν + Λ g μν ) ; μ = 0 , where Λ \Lambda Λ
R μ ν − 1 2 R g μ ν + Λ g μ ν = κ T μ ν , R^{\mu \nu} - \frac{1}{2} R g^{\mu \nu} + \Lambda g^{\mu \nu} = \kappa T^{\mu \nu}, R μν − 2 1 R g μν + Λ g μν = κ T μν , where κ \kappa κ
taking the low velocity limit taking the weak gravity limit assuming time-independent metric and match it with Newton-Cartan theory.
In low velocity limit, the proper time approaches the coordinate time and the spatial velocities approach zero:
τ → t , U → [ 1 0 0 0 ] , \begin{align*} \tau &\to t, \\ \boldsymbol{U} &\to \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \end{align*} τ U → t , → 1 0 0 0 , this means that we can use the Newton-Cartan theory.
In weak gravity limit, the metric approaches the Minkowski metric plus a very small error:
g μ ν → η μ ν + h μ ν , η μ ν = diag ( − 1 , 1 , 1 , 1 ) , ∣ h μ ν ∣ ≪ 1. \begin{align*} g_{\mu \nu} &\to \eta_{\mu \nu} + h_{\mu \nu}, \\ \eta_{\mu \nu} &= \textrm{diag}(-1, 1, 1, 1), \\ |h_{\mu \nu}| &\ll 1. \end{align*} g μν η μν ∣ h μν ∣ → η μν + h μν , = diag ( − 1 , 1 , 1 , 1 ) , ≪ 1. This means that when doing summations, h μ ν h_{\mu \nu} h μν
A μ α g α ν = A μ α ( η α ν + h α ν ) = A μ α η α ν , A^{\mu \alpha} g_{\alpha \nu} = A^{\mu \alpha} (\eta_{\alpha \nu} + h_{\alpha \nu}) = A^{\mu \alpha} \eta_{\alpha \nu}, A μα g αν = A μα ( η αν + h αν ) = A μα η αν , but when taking derivatives, η μ ν \eta_{\mu \nu} η μν
g μ ν , σ = η μ ν , σ + h μ ν , σ = h μ ν , σ . g_{\mu \nu, \sigma} = \eta_{\mu \nu, \sigma} + h_{\mu \nu, \sigma} = h_{\mu \nu, \sigma}. g μν , σ = η μν , σ + h μν , σ = h μν , σ . Time-independent metric means the following:
g μ ν , t = 0 , g_{\mu \nu, t} = 0, g μν , t = 0 , implying:
Γ σ μ ν , t = 0. \Gamma^{\sigma}{}_{\mu \nu, t} = 0. Γ σ μν , t = 0. A final assumption is that the cosmological constant is small enough to be ignored and that all gravity is caused by mass. Meaning the energy-momentum tensor is that of a dust and the only nonzero component is:
T t t = ρ , T^{tt} = \rho, T tt = ρ , and the component is the same with its both indices lowered:
T t t = T μ ν g μ t g ν t = T t t η t t η t t = T t t ( − 1 ) ( − 1 ) = T t t . T_{tt} = T^{\mu \nu} g_{\mu t} g_{\nu t} = T^{tt} \eta_{tt} \eta_{tt} = T^{tt} (-1) (-1) = T^{tt}. T tt = T μν g μ t g ν t = T tt η tt η tt = T tt ( − 1 ) ( − 1 ) = T tt . The Einstein field equations look like this:
R μ ν − 1 2 R g μ ν = κ T μ ν , R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = \kappa T_{\mu \nu}, R μν − 2 1 R g μν = κ T μν , from the spatial components, we can calculate the spatial components of the Ricci tensor:
R i j − 1 2 R g i j = κ T i j , R i j − 1 2 R η i j = 0 , R i j − 1 2 R δ i j = 0 , R i j = 1 2 R δ i j , \begin{align*} R_{ij} - \frac{1}{2} R g_{ij} &= \kappa T_{ij}, \\ R_{ij} - \frac{1}{2} R \eta_{ij} &= 0, \\ R_{ij} - \frac{1}{2} R \delta_{ij} &= 0, \\ R_{ij} &= \frac{1}{2} R \delta_{ij}, \end{align*} R ij − 2 1 R g ij R ij − 2 1 R η ij R ij − 2 1 R δ ij R ij = κ T ij , = 0 , = 0 , = 2 1 R δ ij , and this may be substituted into the equation for the Ricci scalar to calculate the time-time component of the Ricci tensor:
R = R μ ν g μ ν = R μ ν η μ ν = − R t t + R 11 + R 22 + R 33 = − R t t + 3 2 R , R t t = 1 2 R , \begin{align*} R &= R_{\mu \nu} g^{\mu \nu} \\ &= R_{\mu \nu} \eta^{\mu \nu} \\ &= -R_{tt} + R_{11} + R_{22} + R_{33} \\ &= -R_{tt} + \frac{3}{2} R, \\ R_{tt} &= \frac{1}{2} R, \end{align*} R R tt = R μν g μν = R μν η μν = − R tt + R 11 + R 22 + R 33 = − R tt + 2 3 R , = 2 1 R , which is consistent with the spatial components of the Ricci tensor:
R μ ν = 1 2 R δ μ ν . R_{\mu \nu} = \frac{1}{2} R \delta_{\mu \nu}. R μν = 2 1 R δ μν . Substituting this into the Einstein field equations, we get (off-diagonal terms give us 0 = 0 0 = 0 0 = 0
R μ ν − 1 2 R g μ ν = κ T μ ν , 1 2 R δ μ ν − 1 2 R η μ ν = κ T μ ν , 1 2 R δ μ μ − 1 2 R η μ μ = κ T μ μ , 1 2 R δ t t − 1 2 R η t t = κ T t t , 1 2 R + 1 2 R = κ ρ , R = κ ρ , R i i − 1 2 R g i i = κ T i i , R i i − 1 2 R η i i = 0 , R i i − 1 2 R δ i i = 0 , 1 2 R − 1 2 R = 0 , 0 = 0 , \begin{align*} R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} &= \kappa T_{\mu \nu}, \\ \frac{1}{2} R \delta_{\mu \nu} - \frac{1}{2} R \eta_{\mu \nu} &= \kappa T_{\mu \nu}, \\ \frac{1}{2} R \delta_{\mu \mu} - \frac{1}{2} R \eta_{\mu \mu} &= \kappa T_{\mu \mu}, \\[3ex] \frac{1}{2} R \delta_{tt} - \frac{1}{2} R \eta_{tt} &= \kappa T_{tt}, \\ \frac{1}{2} R + \frac{1}{2} R &= \kappa \rho, \\ R &= \kappa \rho, \\[3ex] R_{ii} - \frac{1}{2} R g_{ii} &= \kappa T_{ii}, \\ R_{ii} - \frac{1}{2} R \eta_{ii} &= 0, \\ R_{ii} - \frac{1}{2} R \delta_{ii} &= 0, \\ \frac{1}{2} R - \frac{1}{2} R &= 0, \\ 0 &= 0, \end{align*} R μν − 2 1 R g μν 2 1 R δ μν − 2 1 R η μν 2 1 R δ μμ − 2 1 R η μμ 2 1 R δ tt − 2 1 R η tt 2 1 R + 2 1 R R R ii − 2 1 R g ii R ii − 2 1 R η ii R ii − 2 1 R δ ii 2 1 R − 2 1 R 0 = κ T μν , = κ T μν , = κ T μμ , = κ T tt , = κ ρ , = κ ρ , = κ T ii , = 0 , = 0 , = 0 , = 0 , so the only relevant term is the time-time component of the equations which results in:
R = κ ρ . R = \kappa \rho. R = κ ρ . Now, remember from the Newton-Cartan theory:
R t t = 4 π ρ , R_{tt} = 4 \pi \rho, R tt = 4 π ρ , where we found that
R t t = 1 2 R = 1 2 κ ρ . R_{tt} = \frac{1}{2} R = \frac{1}{2} \kappa \rho. R tt = 2 1 R = 2 1 κ ρ . We can set these two equations equal and solve for κ \kappa κ
1 2 κ ρ = 4 π , κ = 8 π . \begin{align*} \frac{1}{2} \kappa \rho = 4 \pi, \\ \kappa = 8 \pi. \end{align*} 2 1 κ ρ = 4 π , κ = 8 π . We have solved for the Einstein gravitational constant and can now substite it into the Einstein field equation:
R μ ν − 1 2 R g μ ν + Λ g μ ν = 8 π T μ ν . R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} + \Lambda g_{\mu \nu} = 8 \pi T_{\mu \nu}. R μν − 2 1 R g μν + Λ g μν = 8 π T μν . We can also obtain a trace reversed form by multiplying both sides by g μ ν g^{\mu \nu} g μν
R μ ν g μ ν − 1 2 R g μ ν g μ ν + Λ g μ ν g μ ν = κ T μ ν g μ ν , R − 1 2 R δ μ μ + Λ δ μ μ = κ T , R − 2 R + 4 Λ = κ T , − R + 4 Λ = κ T , R = 4 Λ − κ T , \begin{align*} R_{\mu \nu} g^{\mu \nu} - \frac{1}{2} R g_{\mu \nu} g^{\mu \nu} + \Lambda g_{\mu \nu} g^{\mu \nu} &= \kappa T_{\mu \nu} g^{\mu \nu}, \\ R - \frac{1}{2} R \delta^{\mu}_{\mu} + \Lambda \delta^{\mu}_{\mu} &= \kappa T, \\ R - 2 R + 4 \Lambda &= \kappa T, \\ -R + 4 \Lambda &= \kappa T, \\ R &= 4 \Lambda -\kappa T, \end{align*} R μν g μν − 2 1 R g μν g μν + Λ g μν g μν R − 2 1 R δ μ μ + Λ δ μ μ R − 2 R + 4Λ − R + 4Λ R = κ T μν g μν , = κ T , = κ T , = κ T , = 4Λ − κ T , substituting for the Ricci scalar:
R μ ν − 1 2 ( 4 Λ − κ T ) g μ ν + Λ g μ ν = κ T μ ν , R μ ν + 1 2 κ T g μ ν − 2 Λ g μ ν + Λ g μ ν = κ T μ ν , R μ ν + 1 2 κ T g μ ν − Λ g μ ν = κ T μ ν , R μ ν = κ ( T μ ν − 1 2 T g μ ν ) + Λ g μ ν , = 8 π ( T μ ν − 1 2 T g μ ν ) + Λ g μ ν , \begin{align*} R_{\mu \nu} - \frac{1}{2} (4 \Lambda - \kappa T) g_{\mu \nu} + \Lambda g_{\mu \nu} &= \kappa T_{\mu \nu}, \\ R_{\mu \nu} + \frac{1}{2} \kappa T g_{\mu \nu} - 2 \Lambda g_{\mu \nu} + \Lambda g_{\mu \nu} &= \kappa T_{\mu \nu}, \\ R_{\mu \nu} + \frac{1}{2} \kappa T g_{\mu \nu} - \Lambda g_{\mu \nu} &= \kappa T_{\mu \nu}, \\ R_{\mu \nu} &= \kappa \left(T_{\mu \nu} - \frac{1}{2} T g_{\mu \nu}\right) + \Lambda g_{\mu \nu}, \\ &= 8 \pi \left(T_{\mu \nu} - \frac{1}{2} T g_{\mu \nu}\right) + \Lambda g_{\mu \nu}, \end{align*} R μν − 2 1 ( 4Λ − κ T ) g μν + Λ g μν R μν + 2 1 κ T g μν − 2Λ g μν + Λ g μν R μν + 2 1 κ T g μν − Λ g μν R μν = κ T μν , = κ T μν , = κ T μν , = κ ( T μν − 2 1 T g μν ) + Λ g μν , = 8 π ( T μν − 2 1 T g μν ) + Λ g μν , where
T = T μ μ = T μ ν g μ ν T = T^{\mu}{}_{\mu} = T_{\mu \nu} g^{\mu \nu} T = T μ μ = T μν g μν is the trace of the energy-momentum tensor.
So we have the two forms of the equations:
R μ ν − 1 2 R g μ ν + Λ g μ ν = 8 π T μ ν , 8 π ( T μ ν − 1 2 T g μ ν ) + Λ g μ ν = R μ ν . \begin{align*} R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} + \Lambda g_{\mu \nu} &= 8 \pi T_{\mu \nu}, \\ 8 \pi \left(T_{\mu \nu} - \frac{1}{2} T g_{\mu \nu}\right) + \Lambda g_{\mu \nu} &= R_{\mu \nu}. \end{align*} R μν − 2 1 R g μν + Λ g μν 8 π ( T μν − 2 1 T g μν ) + Λ g μν = 8 π T μν , = R μν . The Einstein gravitational constant
κ ≈ 2.08 ⋅ 1 0 − 43 N − 1 , \kappa \approx 2.08 \cdot 10^{-43}\ N^{-1}, κ ≈ 2.08 ⋅ 1 0 − 43 N − 1 , meaning it takes a lot of mass and energy to curve the spacetime.