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  "body": "\n## Definition\n\nLet $M$ be a four-dimensional manifold, let $\\Lambda \\in \\R$ be a constant, the cosmological constant, and let $\\kappa \u003e 0$ be a constant.\nLet $T$ be a given symmetric $(0,2)$-[[tensor field|WX302T93TW]] on $M$, the stress–energy tensor.\nA [[Lorentzian metric|BY4QYYS788]] $g$ on $M$ satisfies the *Einstein field equations* if\n$$\\mathrm{Ric}_g - \\tfrac{1}{2} R_g \\, g + \\Lambda g = \\kappa T.$$\nHere $\\mathrm{Ric}_g$ is the [[Ricci curvature|4GTNM0WHDH]] and $R_g$ is the [[scalar curvature|ZHX2VYQDZ3]] of $g$.\nIn physical units $\\kappa = 8 \\pi G / c^4$, with $G$ the gravitational constant and $c$ the speed of light.\nThe *vacuum Einstein equations* are the case $T = 0$ and $\\Lambda = 0$.\n\n## Remarks\n\nTaking the trace with respect to $g$ in dimension four shows that the vacuum equations are equivalent to $\\mathrm{Ric}_g = 0$.\nIn local coordinates the equations read $R_{\\mu\\nu} - \\tfrac{1}{2} R g_{\\mu\\nu} + \\Lambda g_{\\mu\\nu} = \\kappa T_{\\mu\\nu}$, ten equations for the ten components of $g$.\n\nExample: Minkowski space, $\\R^4$ with $g = -dt^2 + dx^2 + dy^2 + dz^2$, is flat, so $\\mathrm{Ric}_g = 0$ and it solves the vacuum equations.\n\nNon-example: de Sitter space has $\\mathrm{Ric}_g = \\Lambda_0 g$ with a constant $\\Lambda_0 \u003e 0$, so it does not solve the vacuum equations.\nIt solves the equations with $T = 0$ and cosmological constant $\\Lambda = \\Lambda_0$.\n\nThe left side has zero divergence by the contracted Bianchi identity, which forces the conservation law $\\nabla^\\mu T_{\\mu\\nu} = 0$.\nYvonne Choquet-Bruhat proved that the vacuum equations have a well-posed initial value problem for suitable initial data.\nThe Riemannian analogue of $\\mathrm{Ric}_g = \\lambda g$ defines an [[Einstein manifold|41S2EZ0KR3]].\n",
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