Newtonian limit

Weak Field Approximation

We know that the Minkowski metric, $\eta_{\mu\nu}$, describes the flat spacetime. For a slightly curved spactime, the metric can be given by the Minkowski metric $$g_{\mu\nu}=\eta_{\mu\nu} + h_{\mu\nu},$$ where $|h_{\mu\nu}|\ll 1$. The Riemann tensor can then be calculated, to first order in $h_{\mu\nu}$, $$R_{\alpha\beta\mu\nu}=\frac{1}{2}(h_{\alpha\nu,\beta\mu}+h_{\beta\mu,\alpha\nu}-h_{\alpha\mu,\beta\nu}-h_{\beta\nu,\alpha\mu}).$$ Define $$\begin{align} h^\mu_\beta &:= \eta^{\mu\alpha}h_{\alpha\beta},\\ h^{\mu\nu} &:= \eta^{\nu\beta}h^\mu_\beta,\\ h &:= h^\alpha_\alpha\\ \bar{h}^{\alpha\beta} &:= h^{\alpha\beta} - \frac{1}{2}\eta^{\alpha\beta}h. \end{align}$$ With these definition and the Riemann tensor we have got, the Einstein tensor is $$G_{\alpha\beta}=-\frac{1}{2}\left[\bar{h}_{ \alpha\beta,\mu}^{,\mu} + \eta_{\alpha\beta}\bar{h}_{\mu\nu}^{,\mu\nu} - \bar{h}_{\alpha\mu,\beta}^{,\mu} - \bar{h}_{\beta\mu,\alpha}^{,\mu} + O(h^2_{\alpha\beta}) \right]$$

Supppse we make a small change in coordinates of the form $$x^{\alpha'} = x^\alpha + \xi^\alpha(x^\beta),$$ where $|\xi^\alpha_{,\beta}|\ll 1$. The transformation matrix of this transformation is given by $$\begin{align} \Lambda^{\alpha'}_\beta &= \frac{\partial x^\alpha{'}}{\partial x^\beta} = \delta^\alpha_\beta + \xi^\alpha_{,\beta},\\ \Lambda^\alpha_{\beta'} &= \delta^\alpha_\beta - \xi^\alpha_{,\beta} + O(|\xi^\alpha_{,\beta}|^2). \end{align}$$ Under this transformation, the metric becomes $$g_{\alpha'\beta'}=\eta_{\alpha\beta} + h_{\alpha\beta} - \xi_{\alpha,\beta} - \xi_{\beta,\alpha},$$ where $$\xi_{\alpha}:=\eta_{\alpha\beta}\xi^\beta.$$ This is equivalent to transforming $$h_{\alpha\beta}^{\text{(new)}} = h_{\alpha\beta}-\xi_{\alpha,\beta}-\xi_{\beta,\alpha}.$$ Since $|\xi^\alpha_{,\beta}|\ll 1$, the metric is still in the form of Minkowski metric plus a small perturbation term. This is known as the gauge transformation. Under such a transformation, $$\bar{h}^{\text{(new)}}_{\mu\nu} = \bar{h}^{\text{(old)}}_{\mu\nu} - \xi_{\mu,\nu} - \xi_{\nu,\mu} + \eta_{\mu\nu}\xi^\alpha_{,\alpha}.$$ Hence, $$\bar{h}^{\text{(new)}\mu\nu}_{,\nu} = \bar{h}^{\mu\nu}_{,\nu}-\xi^{\mu,\nu}_{,\nu}.$$ If we want a gauge transformation such that $$\bar{h}^{\text{(new)}\mu\nu}_{,\nu} = 0,$$ $\xi^\mu$ can be determined by $$\Box\xi^\mu = \xi^{\mu,\nu}_{,\nu} = \bar{h}^{\mu\nu}_{,\nu},$$ where $\Box$ is the four-dimensional Laplacian, or D’Alembertian or wave operator, $$\Box f := f^{,\mu}_{,\mu} = \eta^{\mu\nu}f_{,\mu\nu}=\left(-\frac{\partial^2}{\partial t^2}+\nabla^2\right)f.$$ After applying such a gauge transformation, the second, third and fourth terms of the Einstein tensor are zero. Ignoring higher order terms, we have $$G^{\alpha\beta} = -\frac{1}{2}\Box\bar{h}^{\alpha\beta}.$$ The weak-field Einstein equations are $$\bbox[5px, border: 2px solid #fff]{ \Box \bar{h}^{\mu\nu} = -2K T^{\mu\nu}. }$$

Weak Field Metric

Newtonian gravity is gravity that is too weak to produce velocities near the spped of light $|\vec{v}|\ll 1$. Non-relativistic also means $T^{00}\approx \rho$. So, $$\Box \bar{h}^{00} = -2K\rho.$$ Also, since $\partial/\partial t$ is of the order $v \partial/\partial x$, $$\Box = \nabla^2 + O(v^2\nabla^2).$$ Ignoring higher order terms, $$\nabla^2\bar{h}^{00}=-2K\rho.$$ Comparing to the Newtonian equation, $$\nabla^2\phi=4\pi\rho,$$ Let $$\bar{h}^{00}=2k\phi,$$ for some real constant $k$. That is $$8k\pi=-2K \implies K = -4k\pi.\tag{1}$$ Then, $$\begin{align} h &= h^\alpha_\alpha = -\bar{h}^\alpha_\alpha = \bar{h}^{00}\\ h^{00} &= -k\phi \\ h^{xx} &= h^{yy} = h^{zz} = -k\phi. \end{align}$$ Hence, the metric becomes $$ds^2 = -(1-k\phi)dt^2 + (1+k\phi)(dx^2 + dy^2 + dz^2)$$

The geodesic is $$p^\alpha p^\gamma_{,\alpha} + \Gamma^\gamma_{\alpha\beta}p^\alpha p^\beta = 0.$$ The three-component geodesic equations are $$p^\alpha p^i_{,\alpha} + \Gamma^i_{\alpha\beta}p^\alpha p^\beta = 0$$ For non-relativistic particles, $p^0 \ll p^i$ and $(p^0)^2\approx m^2$, so $$\begin{align} p^0p^i_{,0} + \Gamma^i_{00}(p^0)^2&=0\\ \frac{dp^i}{d\tau} &= -m\Gamma^i_{00}. \end{align}$$ The Christoffel symbols are $$\begin{align} \Gamma^i_{00} &= \frac{1}{2}g^{i\alpha}(g_{\alpha 0,0} + g_{\alpha 0,0} - g_{00,\alpha})\\ &= \frac{1}{2}g^{ii}(-g_{00,i}). \end{align}$$ Substitute back, we have got $$\frac{dp^i}{d\tau} = \frac{m}{2}g^{ii}g_{00,i} = \frac{mk}{2}(1+k\phi)^{-1}\phi_{,i}.$$ Ignoring higher order terms, $$\frac{dp^i}{d\tau}\approx m\frac{k}{2}\phi_{,i}.$$ Comparing to the Newtonian equation of motion, $$\frac{dp^i}{d\tau} = -m\phi_{,i},$$ we conclude $$k=-2,$$ and $$\bbox[5px, border: 2px solid #fff]{ ds^2 = -(1+2\phi)dt^2 + (1-2\phi)(dx^2 + dy^2 + dz^2) }$$

Einstein Field Equations

Substitute $k=-2$ into (1), we got $$K = (-2)(-4\pi)=8\pi.$$ Hence, $$\bbox[5px, border: 2px solid #fff]{ R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R =8\pi T_{\mu\nu} }$$

Reference: A first course in general relativity by Bernard F. Schutz