% File: curved.tex % Section: ? % Title: Curved space-time % Last modified: 12.08.2003 % \documentclass[a4paper,12pt]{article} \usepackage{amsmath,amssymb} \textwidth 16cm \textheight 25cm \oddsidemargin 0cm \topmargin -1.5cm \pagestyle{empty} \begin{document} \begin{center} \large\textbf{CURVED SPACE-TIME} \end{center} \vspace{.1cm} General coordinate transformation $x^\mu \rightarrow \tilde{x}^\mu(x)$ induces tensors to transform as follows (we distinguish between lower and upper indices)\,: \begin{equation} \label{tensors} d\tilde{x}^\mu = \frac{\partial\tilde{x}^\mu}{\partial x^\nu}dx^\nu\,, \ \ \ \ \tilde{g}_{\mu\nu}(\tilde{x}) = \frac{\partial x^\alpha}{\partial\tilde{x}^\mu} \frac{\partial x^\beta}{\partial\tilde{x}^\nu}g_{\alpha\beta}\,, \ \ \ \ \tilde{A}_{\nu}^{\mu}(\tilde{x}) = \frac{\partial\tilde{x}^\mu}{\partial x^\alpha} \frac{\partial x^\beta}{\partial\tilde{x}^\nu}A_{\beta}^{\alpha}(x)\,. \end{equation} A (squared) space-time interval is assumed to be scalar (i.e., invariant), \begin{equation} \label{} ds^2 = g_{\mu\nu}dx^\mu dx^\nu = \tilde{g}_{\mu\nu}d\tilde{x}^\mu d\tilde{x}^\nu\,. \end{equation} To construct an invariant volume element we use the determinant of metrics: \begin{gather} g_{\mu\lambda}g^{\lambda\nu} = \delta_{\mu}^{\nu}\,, \ \ g \doteq \left|\det(g_{\mu\nu})\right| = \gamma\det(g_{\mu\nu})\,, \ \ g^{-1} = \gamma\det(g^{\mu\nu})\,, \ \ \gamma = \text{sgn}\det(g_{\mu\nu})\,, \\ d^n\tilde{x} = \Bigl|\det\Bigl(\frac{\partial\tilde{x}}{\partial x}\Bigr)\Bigr|d^nx\,, \ \ \ \ \ \ \ \tilde{g}={\det}^2\Bigl(\frac{\partial x}{\partial\tilde{x}}\Bigr)g\,, \ \ \ \ \Rightarrow \ \ \ \ \sqrt{\tilde{g}}\:d^n\!\tilde{x} = \sqrt{g}\:d^n\!x\,. \end{gather} Let us denote by $\epsilon$ and $E$ the antisymmetric tensors of the flat and curved spaces, respectively, and let $\epsilon_{1\ldots n}=1\,, \ \ \epsilon^{\mu\ldots\nu}=\gamma\epsilon_{\mu\ldots\nu}$. Then \begin{gather} \epsilon_{\mu_1\ldots\mu_n}A_{\mu_1\nu_1}\ldots A_{\mu_n\nu_n} = (\det A)\,\epsilon_{\nu_1\ldots\nu_n}\,, \ \ \ \epsilon_{\mu_1\ldots\mu_k\nu_{k+1}\ldots\nu_n} \epsilon_{\mu_1\ldots\mu_k\lambda_{k+1}\ldots\lambda_n} = k!\det(\delta_{\nu\lambda})\,, \\ E_{\mu_1\ldots\mu_n} = \sqrt{g}\:\epsilon_{\mu_1\ldots\mu_n} = \gamma\sqrt{g}\:\epsilon^{\mu_1\ldots\mu_n} = \gamma g E^{\mu_1\ldots\mu_n}\,. \end{gather} Covariant derivative in curved space: \begin{equation} \label{covar} \nabla_\alpha A_{\nu}^{\mu} = \partial_\alpha A_{\nu}^{\mu} + \Gamma_{\alpha\beta}^{\mu}A_{\nu}^{\beta} - \Gamma_{\alpha\nu}^{\beta}A_{\beta}^{\mu}\,, \ \ \ \ \ \nabla(AB) = (\nabla\!A)B + A\,\nabla\!B\,. \end{equation} Assuming that covariant differentiation respects metrics, $\nabla_\alpha g_{\mu\nu} = \nabla_\alpha g^{\mu\nu} = 0$, one obtains \begin{gather} g_{\mu\nu}\Gamma_{\alpha\beta}^{\nu} = \frac{1}{2}\, (\partial_\alpha g_{\beta\mu} + \partial_\beta g_{\alpha\mu} -\partial_\mu g_{\alpha\beta})\,, \ \ \ \ \Gamma_{\alpha\beta}^{\mu} = \Gamma_{\beta\alpha}^{\mu} \,, \\ \partial_\alpha g_{\mu\nu} = g_{\mu\beta}\Gamma_{\alpha\nu}^{\beta} + g_{\nu\beta}\Gamma_{\alpha\mu}^{\beta}\,, \\ \partial_\alpha g^{\mu\nu} = -g^{\mu\beta}\Gamma_{\alpha\beta}^{\nu} - g^{\nu\beta}\Gamma_{\alpha\beta}^{\mu}\,. \end{gather} Curvature: \begin{gather} R_{\beta\mu\nu}^{\alpha} = \partial_\mu\Gamma_{\beta\nu}^{\alpha} - \partial_\nu\Gamma_{\beta\mu}^{\alpha} + \Gamma_{\lambda\mu}^{\alpha}\Gamma_{\beta\nu}^{\lambda} - \Gamma_{\lambda\nu}^{\alpha}\Gamma_{\beta\mu}^{\lambda}\,, \\ R_{\mu\nu} = R_{\nu\mu} = R_{\mu\alpha\nu}^{\alpha}\,, \ \ \ \ \ \ R = g^{\mu\nu}R_{\mu\nu}\,, \\ (\nabla_\alpha\nabla_\beta - \nabla_\beta\nabla_\alpha)A_{\mu}^{\nu} = R_{\lambda\alpha\beta}^{\nu}A_{\mu}^{\lambda} - R_{\mu\alpha\beta}^{\lambda}A_{\lambda}^{\nu}\,. \end{gather} Parametrizing $\tilde{x}^\mu = x^\mu + \varepsilon^\mu(x)$ and using $\delta(\det A) = \det\!A\ \text{tr}(A^{-1}\delta A)$\,, we find \begin{gather} \delta(\sqrt g) = \frac{\sqrt g}{2}g^{\mu\nu}\delta g_{\mu\nu} = -\frac{\sqrt g}{2}g_{\mu\nu}\,\delta g^{\mu\nu}\,, \\ \partial_\mu \sqrt g = -\frac{\sqrt g}{2}g_{\lambda\nu}\,\partial_\mu g^{\lambda\nu} = \sqrt g\ \Gamma_{\mu\nu}^{\nu}\ , \ \ \ \ \ \ \ \ \partial_\mu(\sqrt g A^\mu) = \sqrt g\ \nabla_\mu A^\mu\,, \\ \bar\delta g_{\mu\nu} = -\varepsilon^\lambda\partial_\lambda g_{\mu\nu} - g_{\mu\lambda}\partial_\nu\varepsilon^\lambda - g_{\nu\lambda}\partial_\mu\varepsilon^\lambda = -\nabla_\mu\varepsilon_\nu - \nabla_\nu\varepsilon_\mu\ , \\ \bar\delta g^{\mu\nu} = -\varepsilon^\lambda\partial_\lambda g^{\mu\nu} + g^{\mu\lambda}\partial_\lambda\varepsilon^\nu + g^{\nu\lambda}\partial_\lambda\varepsilon^\mu = \nabla^\mu\varepsilon^\nu + \nabla^\nu\varepsilon^\mu\,, \\ \bar\delta(\sqrt g) = -\frac{\sqrt g}{2}g_{\mu\nu}\bar\delta g^{\mu\nu} = \frac{\sqrt g}{2}g_{\mu\nu}\,\varepsilon^\lambda\partial_\lambda g^{\mu\nu} - \sqrt g\,\partial_\lambda\varepsilon^\lambda = - \partial_\lambda(\sqrt g\,\varepsilon^\lambda)\,. \end{gather} \newpage \begin{center} \large\textbf{Tetrads} \end{center} \vspace{.1cm} A tetrad, or vielbein, is a local orthonormal set of basis vectors $e_{a}^{\mu}(x)$ which serve to switch from curved space (Greek indices) to locally flat one (Latin indices) and back, \begin{equation} \label{switch} A_{\mu}^{\nu} = e_{\mu}^{a}\,e_{b}^{\nu}\,A_{a}^{b}\,, \ \ \ \ A_{a}^{b} = e_{a}^{\mu}\,e_{\nu}^{b}\,A_{\mu}^{\nu}\,, \ \ \ \ \partial_\mu = e_{\mu}^{a}\,\partial_a\,, \ \ \ \ \partial_a = e_{a}^{\mu}\,\partial_\mu\,, \end{equation} due to orthonormality conditions \begin{equation} \label{ortho} e_{a}^{\mu}\,e_{\nu}^{a} = \delta_{\nu}^{\mu}\,, \ \ \ e_{\mu}^{a}\,e_{b}^{\mu} = \delta_{b}^{a}\,, \ \ \ g_{\mu\nu}\,e_{a}^{\mu}\,e_{b}^{\nu} = \eta_{ab}\,, \ \ \ g_{\mu\nu} = e_{\mu}^{a}\,e_{\nu}^{b}\,\eta_{ab}\,, \end{equation} with $\eta_{ab}$ being a flat (quasi)Euclidean metric tensor. In a sense, one views a tetrad as the ``square root" of the metrics $g_{\mu\nu}$\,. We may treat $e_{a}^{\mu}$ as independent variables and $e_{\mu}^{a}$ as their inverse, whence \begin{equation} \label{} \delta e_{\mu}^{a} = -e_{\nu}^{a}\,\delta(e_{b}^{\nu})\,e_{\mu}^{b}\,, \ \ \ \ \delta(\sqrt g) = -\sqrt g\,e_{\mu}^{a}\,\delta e_{a}^{\mu} = \sqrt g\,e_{a}^{\mu}\,\delta e_{\mu}^{a}\,. \end{equation} Transformation properties of tensor with indices of any kind consist of usual coordinate transformations for Greek and local Lorentz ones for Latin indices: \begin{equation} \label{transf} \tilde{T}_{b'\nu'}^{a'\mu'}(\tilde{x}) = \Lambda_{a}^{a'}(x)\,\Lambda_{b'}^{b}(x)\, \frac{\partial\tilde{x}^{\mu'}}{\partial x^\mu}\, \frac{\partial x^\nu}{\partial\tilde{x}^{\nu'}}\, T_{b\nu}^{a\mu}(x)\,, \ \ \ \ \ \ \Lambda_{a}^{c}\,\Lambda_{c}^{b} = \Lambda_{c}^{b}\,\Lambda_{a}^{c} = \delta_a^b\,. \end{equation} We see that tensors with only flat (Latin) indices are coordinate scalars. Covariant differentiation of flat indices is governed by the `spin connection' $\omega_\mu{}^a{}_b$\,: \begin{equation} \label{omega} \nabla_\mu A_{b}^{a} = \partial_\mu A_{b}^{a} +\omega_\mu{}^a{}_c\,A_{b}^{c} - \omega_\mu{}^c{}_b\,A_{c}^{a}\,. \end{equation} Christoffel symbols, governing covariant differentiation of curved indices, are related with $w$'s through the `tetrad postulate': \begin{equation} \label{} \nabla_\mu e_{a}^{\nu} = 0 \ \ \ \ \Rightarrow \ \ \ \ \Gamma_{\mu\lambda}^{\nu} = e_{a}^{\nu}\,\partial_\mu e_{\lambda}^{a} + e_{a}^{\nu}\,e_{\lambda}^{b}\,\omega_\mu{}^a{}_b\,, \ \ \ \ \omega_\mu{}^a{}_b = e_{\nu}^{a}\,e_{b}^{\lambda}\,\Gamma_{\mu\lambda}^\nu - e_{b}^{\lambda}\,\partial_\mu e_{\lambda}^{a}\,. \end{equation} These relations hold for any connection. In case of \ $\nabla g = 0$ \,we additionally have \begin{equation} \label{} \nabla_\mu\,\eta_{ab} = 0 \ \ \ \ \Rightarrow \ \ \ \ \omega_\mu{}^a{}_b = -\omega_{\mu b}{}^a \end{equation} as well as the following explicit expression for $w$'s in terms of tetrads: \begin{equation} \label{} \omega_{\mu ab} = \frac{1}{2}\,e_{a}^{\nu}\,(\partial_\mu e_{b\nu} - \partial_\nu e_{b\mu} + e_{b}^{\lambda}\,e_{c\mu}\, \partial_\lambda e_{\nu}^{c}) \ - \ (\ a\leftrightarrow\,b\ )\,, \end{equation} or, equivalently, \begin{equation} \label{} \omega_\mu{}^a{}_b\,e_{\nu}^{b} - \omega_\nu{}^a{}_b\,e_{\mu}^{b} + \partial_\mu e_{\nu}^a - \partial_\nu e_{\mu}^a = 0\,. \end{equation} One also may rewrite the metrical definition of the energy-momentum tensor in terms of tetrads: \begin{equation} \label{} T_{\mu\nu} \ \doteq\ \frac{2}{\sqrt{g}}\,\frac{\delta S}{\delta g^{\mu\nu}}\ = \ \frac{1}{2\sqrt{g}}\,\left(\frac{\delta S}{\delta e_a^{\nu}}e_{a\mu} + \frac{\delta S}{\delta e_a^{\mu}}e_{a\nu}\right). \end{equation} \end{document}