%Given to GPE to check, 11.9.00 \documentstyle[12pt]{article} \setlength{\textwidth}{6.5truein} \setlength{\oddsidemargin}{0truein} \setlength{\topmargin}{-0.7truein} \setlength{\textheight}{9.5truein} \setlength{\parindent}{0.0truein} \setlength{\parskip}{0.15truein} \def\lesssim{\mathrel{\hbox{\rlap{\hbox{\lower5pt\hbox{$\sim$}}}\hbox{$<$}}}} \def\gtrsim{\mathrel{\hbox{\rlap{\hbox{\lower5pt\hbox{$\sim$}}}\hbox{$>$}}}} \def\ni{\noindent} \def\bfn{\mbox{\boldmath$\nabla$}} \begin{document} \addtocounter{section}{10} \addtocounter{page}{67} \section*{Lecture Eleven: Theory of Relativity} \section{The Gravitational Field Equations} We are now in a position to talk the problem of deriving a set of field equations for gravity. Let's recall a few results from previous lectures: \begin{itemize} \item We showed that in a weak gravitational field $$g_{00} = (1+2\phi/c^2)$$ (Henceforth in these lecture notes I will (usually) set $c=1$.) \item We have constructed a tensor $$T^{\mu\nu},\ T_{\mu\nu}$$ with a $T_{00}$ component which, for `dust', measures the mass density $$T_{00} = \rho$$ \item The gravitational field equation of Newtonian gravity is $$\nabla^2 \phi = 4\pi G\rho$$ {\it i.e.}, for a weak gravitational field $$\nabla^2g_{00} = 8\pi G T_{00}. \eqno(1) $$ \item The last equation suggests that the field equations are tensorial equations of rank 2, of the form $$G_{\mu\nu} = 8\pi G T_{\mu\nu} \eqno(2)$$ where the tensor $G_{\mu \nu}$ must be consistent with the Newtonian limit (1). \end{itemize} But what is the form of $G_{\mu\nu}$? We have a few clues: \begin{itemize} \item $G_{\mu\nu}$ must involve second derivatives of the metric tensor so that in the weak field limit $G_{00} \rightarrow \nabla^2g_{00}$. \item The vanishing of a gravitational field is equivalent to the vanishing of the curvature tensor, which as we know involves the second derivatives of $g_{\mu \nu}$. This suggests that $G_{\mu \nu}$ is related to rank two tensors constructed from the curvature tensor. \item $T_{\mu\nu}$ is symmetric and so $G_{\mu\nu}$ must be symmetric. \item $T_{\mu\nu}$ is conserved, $T_{\mu\nu ; \nu} = 0$, hence $G_{\mu\nu}$ must also be conserved $G_{\mu\nu ; \nu} = 0$. \end{itemize} As a guess, let's write down a rank 2 tensor constructed from the curvature tensor: \begin{center} \begin{tabular}{lcl } \hspace{3.9cm}Ricci & & \hspace{0.7cm}Curvature \\ \hspace{3.9cm}tensor & & \hspace{0.7cm}Scalar\\ \hspace{3.9cm}$\downarrow$ & & \hspace{0.7cm}$\downarrow$\\ \multicolumn{3}{c}{$G_{\mu\nu}\ \ =\ \ \alpha \ R_{\mu\nu}\ +\ \beta \ g_{\mu\nu} \ R$}\\ \hspace{3.5cm}$\nwarrow$ & &\hspace{-1cm}$\nearrow$\\ \multicolumn{3}{c}{constants}\\ %& constants & \\ \end{tabular} \end{center} This guess automatically satisfies the first three conditions. The tensor $G_{\mu \nu}$ involves just the metric tensor and its first two derivatives. The tensor $G_{\mu \nu}$ is symmetric and it vanishes if all components of the curvature tensor vanish. Now let's make use of the condition that $G_{\mu \nu}$ satisfies a conservation equation. Raise an index \begin{eqnarray*} G^\kappa_\nu & = & g^{\kappa\mu} G_{\mu\nu} = \alpha g^{\kappa\mu} R_{\mu\nu} + \beta g^{\kappa\mu} g_{\mu\nu} R\\ & = & \alpha R^\kappa_\nu + \beta g^{\kappa_\nu} R, \end{eqnarray*} and take the covariant derivative, $$ G^\kappa_{\nu ;\kappa} = \alpha R^\kappa_{\nu ; \kappa} + \beta \delta^\kappa_\nu R_{; \kappa}. $$ Now recall the Bianchi identity $$ \left ( R^\mu_\eta - {1\over 2} \delta^\mu_\eta R \right )_{;\mu} = 0, $$ this gives $$ G^\kappa_{\nu \;\kappa} = \left ( {1 \over 2} \alpha + \beta \right) R_{;\nu}, $$ and so if $G_{\mu \nu}$ satisfies a conservation equation $$ \begin{array}{c} \Rightarrow \hspace{.5cm} G^\kappa_{\nu \; \kappa} = 0\hspace{.5cm} \parbox{2cm}{so either} \end{array} \left \{ \begin{array}{l} \beta = - {1 \over 2} \alpha, \\ \\ R_{;\nu} = 0 . \end{array} \right. $$ We can easily rule out the idea that $R_{;\nu}$ is always equal to zero starting from the field equations (2). \begin{eqnarray*} G^\kappa_\kappa & = & 8\pi G T^\kappa_\kappa \\ & = & \alpha R^\kappa_\kappa + \beta^{\delta\kappa}_\kappa R \\ & = & \alpha R + \beta R \\ & = & \alpha^\prime R, \qquad \qquad \qquad {\rm where} \;\; \alpha^\prime = \alpha + \beta. \end{eqnarray*} So, taking the covariant derivative of this last relation $$\alpha^\prime R_{;\nu} = 8\pi G T^\kappa_{\kappa ; \nu} = 0, \eqno(3)$$ But (3) requires that $${\partial T^\kappa_\kappa \over \partial x^\nu} = 0 \hspace{.5cm}\parbox[t]{8cm}{always, which cannot be satisfied in general since it requires ${\partial\rho_0/ \partial x = 0}$. }$$ We must therefore select the solution $$ \beta = - {1 \over 2 } \alpha$$ and so the field equations must look like $$\alpha\left (R_{\mu\nu} - { 1 \over 2} g_{\mu\nu}R\right ) = 8\pi G T_{\mu\nu}. \eqno(4) $$ To fix the constant $\alpha$, we can look at the weak field limit of equations (4). For non-relativistic matter, \[ \begin{array}{rcl} T_{ij} & \ll & T_{00} \end{array} \left ( \begin{array}{rcl} T_{00} & \sim & \rho_0 c^2 \\ T_{ij} & \sim & \rho_0 v^2 \end{array} \right ) \] and for weak gravitational fields $$g_{\mu\nu} \approx \eta_{\mu\nu},$$ {\it i.e.} the metric tensor is nearly that of Minkowski space-time. \end{document} Now the field equations look like \begin{center} \begin{tabular}{llllcr} & & \multicolumn{4}{r}{$\alpha \left (\underbrace{R_{\mu\nu} - 1/2 g_{\mu\nu}R}\right ) = 8\pi G T_{\mu\nu} $}\\ & & & & $\downarrow$ & \\ So, if & $T_{ij}$ & $\approx$ & $0$ & $\downarrow$ &\\ & & & & $\downarrow$ & \\ & $R_{ij}$ & $\approx$ & $1/2 g_{ij}R$$\longleftarrow$$\longleftarrow$$\longleftarrow\longleftarrow \longleftarrow \longleftarrow \longleftarrow$&\hspace{-.5cm}$\longleftarrow$$\longleftarrow\leftarrow\leftarrow$ \hspace{2cm} \\ & & & & & \\ & & $\approx$ & $1/2 \eta_{ij} R$ %$\Longrightarrow$$\Longrightarrow$ $\Longrightarrow$$\Longrightarrow$ $\Longrightarrow$ $\Longrightarrow$ $\Longrightarrow$ & \hspace{-1cm}$\Longrightarrow$ $\Longrightarrow$ $\Longrightarrow$ $\Longrightarrow$$\Longrightarrow$ & \hspace{-1cm}$\Longrightarrow$ $\Longrightarrow$$\Longrightarrow$$\Longrightarrow$\\ & & & & &$\Downarrow$\\ \multicolumn{5}{l}{The definition of a curvature scalar is}& $\Downarrow$\\ & & & & & $\Downarrow$\\ &$R$ & = & $g^{\mu\kappa} R_{\mu\kappa}$ & & $\Downarrow$\\ & & & & & $\Downarrow$\\ & & = & $R_{oo} - R_{ii}$ & & $\Downarrow$ \\ & & & & & $\Downarrow$\\ & & = & $R_{00} + 3/2 R$ %$\Longleftarrow$ $\Longleftarrow$ $\Longleftarrow$ $\Longleftarrow$$\Longleftarrow$ $\Longleftarrow$ $\Longleftarrow$% $\Longleftarrow$ & $\Longleftarrow$ $\Longleftarrow$ $\Longleftarrow$ $\Longleftarrow$ $\Longleftarrow$ & $\Longleftarrow$ $\Longleftarrow$\\ & & & & & \\ So, & & & & \\ & & & & & \\ & $R$ & = & $-2R_{00}$ & \\ and since & & & & \\ & $G_{\mu\nu}$ & = &$\alpha (R_{\mu\nu} - 1/2 g_{\mu\nu} R)$ & & \\ & & & & & \\ & & = & $\alpha (R_{00} + 1/2 \cdot 2 R_{00})$ & & \\ & & & & & \\ & & = & $2\alpha R_{00}$ \end{tabular} \end{center} The curvature tensor is, by definition, $$R_{\lambda\mu\nu\kappa} = {1\over 2} \left [ {\partial^2 g\lambda\nu \over \partial x^\kappa \partial x^\mu} - {\partial^2 g_{\mu\nu} \over \partial x^\kappa \partial x^\lambda} - {\partial^2 g_{\lambda\kappa} \over \partial x^\nu\partial x^\mu} + {\partial^2 g_{\mu\kappa} \over \partial x^\nu \partial x^\kappa }\right ] + g{\eta\sigma} \left [\Gamma^\eta_{\nu\lambda} \Gamma^\sigma_{\kappa\lambda} \Gamma^\sigma_{\mu\nu} \right ]$$ in a static weak gravitational field, all time derivatives vanish, products of $\Gamma$'s vanish and $$G_{00} = (1+2\phi)$$ so, \[ \begin{array}{llcl} & R_{\mu\nu} & = & g^{\lambda\kappa} R_{\lambda\mu\kappa\nu}\\ & & &\\ & R_{00} & = & g^{\lambda\kappa} R_{\lambda 0 \kappa 0} = R_{0000} - R_{ioio}\\ & & &\\ {\rm but} & & \\ %& & & \\ & R_{0000} & \approx & 0, \ \ R_{iojo} = 1/2 {\partial^2 g_{00} \over \partial x^{i\partial x j}} \end{array} \] So, for a static weak gravitational field $$G_{00} = 2\alpha R_{00} = - \alpha \nabla^2 g_{00}$$ and so $$-\alpha \nabla^2 g_{00} = 8\pi G T_{00}$$ {\bf But}, in a weak gravitational field $$\nabla^2 g_{00} = 8\pi G T_{00}$$ hence $$\alpha = - 1$$ and we derive the field equations $$(R_{\mu\nu} - 1/2 g_{\mu\nu} R ) = - 8\pi G T_{\mu\nu}$$ These are Einstein's field equations of General Relativity. \begin{center} \begin{tabular}{llrll} \multicolumn{5}{c} {$\underbrace{G^\kappa_{\nu j\kappa}} = \alpha {\cal R}^\kappa_{\nu j\kappa} + \beta \delta^\kappa_\nu Rj\kappa$}\\ & &$\downarrow$ & & \\ & & \hspace{1.7cm}$\downarrow$ & \multicolumn{2}{l}{but, recall the Bianchi identity,}\\ & & $\downarrow$& & \\ & & $\downarrow$ & $\Rightarrow$ & $(R^\mu_\eta - 1/2 \delta^\mu_\eta R )j\mu = 0 $\\ & & $\downarrow$ & & \\ & & $\downarrow$ & so, & \\ & & $\downarrow$ & & \\ \multicolumn{5}{r} {$\overbrace{G^\kappa_{\nu j\kappa}} = (1/2 \alpha + \beta) R j\nu$} \end{tabular} \end{center} \end{document}