The Einstein Field Equations
The Einstein Field Equations
The Einstein field equations (EFE) are the dynamical core of general relativity, relating spacetime curvature to the energy-momentum content of matter. They are 10 coupled, nonlinear PDEs for the metric tensor. The cosmological constant \(\Lambda\) models vacuum energy and drives accelerated expansion.
Einstein Field Equations
\(G_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4}T_{\mu\nu}\), where \(G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R\) is the Einstein tensor and \(T_{\mu\nu}\) is the stress-energy tensor. Trace-reversed form: \(R_{\mu\nu} = \frac{8\pi G}{c^4}\!\left(T_{\mu\nu} - \tfrac{1}{2}g_{\mu\nu}T\right) + \Lambda g_{\mu\nu}\).
Hilbert Action & Variational Derivation
The EFE follow from varying the Einstein-Hilbert action: \(S = \frac{c^4}{16\pi G}\int (R - 2\Lambda)\sqrt{-g}\,d^4x + S_{\text{matter}}\). Varying with respect to \(g^{\mu\nu}\) yields the EFE. Conservation \(\nabla_\mu T^{\mu\nu}=0\) follows automatically from the contracted Bianchi identity.
Example 1: Perfect Fluid Stress-Energy
Write \(T_{\mu\nu}\) for a perfect fluid with density \(\rho\), pressure \(p\), and 4-velocity \(U^\mu\).
Solution: \(T_{\mu\nu} = \left(\rho + \frac{p}{c^2}\right)U_\mu U_\nu + p g_{\mu\nu}\). For dust (\(p=0\)): \(T_{\mu\nu} = \rho U_\mu U_\nu\). In the Newtonian limit: \(T^{00} = \rho c^2\).
Example 2: Newtonian Limit
Show the EFE reduce to Poisson's equation \(\nabla^2\Phi = 4\pi G\rho\) in the weak-field, slow-motion limit.
Solution: Write \(g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}\), \(|h|\ll 1\). The \((00)\) EFE linearizes to \(\nabla^2 h_{00} = -\frac{16\pi G\rho}{c^2}\). Setting \(h_{00} = -2\Phi/c^2\) recovers \(\nabla^2\Phi = 4\pi G\rho\).
Practice
- Derive the trace of the EFE: \(R = -\frac{8\pi G}{c^4}T + 4\Lambda\).
- Show \(\nabla_\mu T^{\mu\nu}=0\) from the Bianchi identity and EFE.
- Write the EFE in vacuum (\(T_{\mu\nu}=0,\,\Lambda=0\)) and interpret the result.
- How does \(\Lambda\) act as a negative-pressure fluid with equation of state \(w=-1\)?
Show Answer Key
1. Trace: $g^{\mu\nu}(R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R+\Lambda g_{\mu\nu}) = g^{\mu\nu}\frac{8\pi G}{c^4}T_{\mu\nu}$. LHS: $R-2R+4\Lambda = -R+4\Lambda$. RHS: $\frac{8\pi G}{c^4}T$. So $R = -\frac{8\pi G}{c^4}T+4\Lambda$.
2. Contracted Bianchi identity: $\nabla_\mu G^{\mu\nu} = 0$. EFE: $G^{\mu\nu}+\Lambda g^{\mu\nu} = \frac{8\pi G}{c^4}T^{\mu\nu}$. Since $\nabla_\mu g^{\mu\nu}=0$ (metric compatibility): $\nabla_\mu T^{\mu\nu} = \frac{c^4}{8\pi G}\nabla_\mu G^{\mu\nu} = 0$. Energy-momentum conservation follows automatically from the geometry. ✓
3. Vacuum EFE ($T_{\mu\nu}=0$, $\Lambda=0$): $R_{\mu\nu}=0$ (Ricci-flat). This doesn't mean spacetime is flat — the Weyl tensor (trace-free part of Riemann) can be nonzero, encoding tidal forces (gravitational waves, Schwarzschild geometry). The full Riemann tensor vanishes only in truly flat spacetime.
4. Cosmological constant as fluid: $T^{\Lambda}_{\mu\nu} = -\frac{\Lambda c^4}{8\pi G}g_{\mu\nu}$. Perfect fluid form: $T_{\mu\nu}=(\rho+P/c^2)U_\mu U_\nu + Pg_{\mu\nu}$. Comparing: $\rho_\Lambda = \Lambda c^2/(8\pi G)$, $P_\Lambda = -\rho_\Lambda c^2$, so $w = P/(\rho c^2) = -1$. Negative pressure means the Friedmann acceleration equation $\ddot{a}/a = -\frac{4\pi G}{3}(\rho+3P/c^2)$ gives $\ddot{a}>0$: accelerated expansion.