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\ihead[]{General Relativity WS 2005/06}
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\title{Script\thanks{written by Tobias Seifen and Jan Hartmann, drawings by Florian Johann}\\General Relativity and Cosmology\\WS 2005/06\\ University Bonn}
\author{Prof. H.-R. Petry}

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\tableofcontents

\chapter{Basics (from electrodynamics)}

Special units: $c = 1$
\section{Action functional of electrodynamics} \index{Action functional}
\begin{eqnarray}
	 S(A, \{x_{i}\}) & = & -\sum_{i=1}^{N}m_{i}\int\eta \Bigl(\dot x_{i}(\tau_{i}),\dot x_{i}(\tau_{i})\Bigr)^{\frac{1}{2}}\,\dd\tau_{i} \nonumber\\
 	& & - \sum_{i=1}^{N}q_{i}\int A_{\mu} \bigl (x_{i}(\tau_{i})\bigr)\dot x^{\mu}(\tau_{i})\,\dd\tau_{i} \nonumber\\
 	& & - \frac{1}{16\pi}\int \dd^{4}\!x \, F_{\mu\nu}(x)F^{\mu\nu}(x)
\end{eqnarray}
$A$ is the vector potential of the electromagnetic field and $x_{i}$ are the particle trajectories
\[F_{\mu\nu}(x) = \del{\mu}A_{\nu}(x)-\del{\nu}A_{\mu}(x) \qquad (F=\dd A) \]
\[F^{\mu\nu}(x)=\eta^{\mu\alpha}\,\eta^{\nu\beta}\,F_{\alpha\beta} \]
\[ \eta\Bigl(\dot x_i(\tau_i),\dot x_i(\tau_i)\Bigr)=\eta_{\mu\nu}\,\dot x^{\mu}(\tau_i)\,\dot x^{\nu}(\tau_i) \]
$\eta^{\mu\nu}$ is the matrix inverse to $\eta_{\mu\nu}$ in cartesian coordinates $x^{0}=t$, $x^{i}$ euclidean space-time coordinates
$\qquad\eta_{\mu\nu}=\varepsilon_{\mu}\delta_{\mu\nu} \qquad \varepsilon_{0}=1, \ \varepsilon_{i}=-1 \quad (i=1,2,3) $
\section{Symmetries}
\begin{enumerate}
	\item Reparametrization: $\qquad \tau_{i} \rightarrow \tau_{i}(\tau')$
	\item Gauge invariance: $\qquad A_{\mu} \rightarrow A_{\mu}'=A_{\mu}+\del{\mu}\lambda $
	\item Poincar\'e invariance: \index{Poincar\'e invariance} \\
		Let $p:\mathbb{R}^{4}\rightarrow\mathbb{R}^{4}$ be a Poincar\'e transformation
		\[p(x)^{\mu}=\Lambda_{\nu}^{\mu}\,x^{\nu}+a^{\mu}\]
		\[A_{\mu}(x) \rightarrow A_{\mu}'(x)=A_{\alpha}\bigl(p\,(x)\bigr)\Lambda_{\mu}^{\alpha}\]
		\[x_{i}(\tau_{i})p^{-1}\bigl(x_{i}(\tau_{i})\bigr) \qquad\qquad \Rightarrow S\textnormal{ is invariant}\] 
\end{enumerate}

\chapter{Introduction to General Relativity}
\section{How to incorporate gravitation (on the classical level)?}
Einstein: Gravitation is just described be replacing $\eta_{\mu\nu} \rightarrow g_{\mu\nu}(x)$\\
In addition: $\qquad \dd^{4}\!x \rightarrow \sqrt{\,\mid\det\,g_{\mu\nu}\mid}\,\dd^{4}\!x$ \\ Action functional after Einstein's replacements: $\qquad \eta_{\mu\nu}\rightarrow g_{\mu\nu}(x)$
\index{Action functional}
\begin{eqnarray}
 	S(A, \{x_{i}\}) & = & -\sum_{i=1}^{N}\int\left[ m_{i}\bigl(g_{\mu\nu}\,\dot x_{i}^{\mu}(\tau_{i})\,\dot x_{i}^{\nu}(\tau_{i})\bigr)^{\frac{1}{2}}+q_{i}\,A_{\mu}\bigl(x_{i}(\tau_{i})\bigr)\dot x^{\mu}(\tau_{i})\right]\dd\tau_{i}\nonumber\\
	& & - \frac{1}{16\pi}\int \dd^{4}\!x \sqrt{\,\mid\det\,g\mid}\, F_{\mu\nu}F^{\mu\nu}
\end{eqnarray}
\[F^{\mu\nu}=g^{\mu\alpha}g^{\nu\beta}F_{\alpha\beta}\]
Signature of $g_{\mu\nu}(x)$:\\
Linear Algebra: There is always a basis of $\mathbb{R}^{4}\ \bigl(e_{i}(x)\bigr)$ s.t.
\[g_{\mu\nu}(x)e_{i}^{\mu}(x)e_{k}^{\nu}(x)=\varepsilon_{i}\,\delta_{ik}\qquad \textnormal{with }\varepsilon=\pm1 \textnormal{ and }e_{i}^{\mu}(x)\textnormal{ being components of }e_{i}(x)\]
Signature is just the collection $(\varepsilon_{0}, \varepsilon_{1}, \varepsilon_{2}, \varepsilon_{3})$\\
Requirement: $\varepsilon_{0}=1,\, \varepsilon_{i}=-1\quad (i=1,2,3)$\\
$\dot x^\mu$ components of a tangent vector (time-like)\\
$g_{\mu\nu}(x)\,\dot x^{\mu}\,\dot x^{\nu}$ length of the tangent at point $x$\\ $x$-dependence $\Rightarrow$ length and angels change with space and time \bigskip \\
Volume integration of a function $\rho(x)$\\
$\int\!\rho(x)$ is limit of a Riemann sum  $\ \sum_{i}\rho(x_{i})\,\Delta V(x_{i})$\\
How is $\Delta V(x_{i})$ determined:\\
Consider in $\mathbb{R}^{4}$ a bilinear nondegenerate metric $f$\\
$\Rightarrow f(a,b)$ is a bilinear function (in $a,b \in \mathbb{R}^{n}$). $f(a,a)$ is the length of a square. The volume spanned by $n$ linear independent vectors $e_{\mu}\ (\mu=1,\ldots,n)$ is given as follows
\[\textnormal{If }V=\left\{x\textnormal{; with }x=\sum_{\mu=1}^{n}\lambda^{\mu}e_{\mu}\textnormal{ and }0\leq\lambda_{\mu}\leq\Delta x_{\mu}\right\}\]
Then $\Delta V$ (the Volume of $V$) is given by $\Delta V^{2}=\,\mid\!\! \textnormal{det}\,f(e_{\mu},e_{\nu})\!\mid^{2}\Delta x^{1}\Delta x^{2}\ldots\Delta x^{n}$\\
In $\mathbb{R}^{2}$, choose just the standard euclidean metric. Let $e_{1},e_{2}$ be the standard basis $\qquad \textnormal{det}\,f\left(e_{\mu},e_{\nu}\right)=1$
\section{Particle motion}\index{Particle motion}
Given by Euler-Lagrange-equations (for one particle only):
\[\frac{\dd}{\dd\tau}\left(\frac{mg_{\mu\alpha}\dot x^{\alpha}}{N}+qA_{\mu}\right) = \frac{m}{2}\left(\del{\mu}g_{\alpha\beta}\right)\frac{\dot x^{\alpha}\dot x^{\beta}}{N}+q\left(\del{\mu}A_\alpha\right)\dot x^{\alpha}; \quad N=(g_{\mu\nu}\dot x^{\mu}\dot x^{\nu})^\frac{1}{2}\]
Requirement $N=1$ (Possible because there is invariance under reparametrizations)
\[\Rightarrow \frac{\dd}{\dd\tau}(m\,g_{\mu\alpha}\,\dot x^{\alpha})-\frac{m}{2}\frac{\partial}{\partial x^{\mu}}\,g_{\alpha\beta}\,\dot x^{\alpha}\,\dot x^{\beta}=qF_{\mu\alpha}\,\dot x^{\alpha}\]
\[\Rightarrow m\Bigl(\ddot x^{\mu}+\underbrace{\Gamma_{\alpha\beta}^{\mu}\,\dot x^{\alpha}\,\dot x^{\beta}}_{\textnormal{gravitational force}}\Bigr)=qF^{\mu}_{\nu}\,\dot x^{\nu}\]
If $q=0$ the mass drops out: $\Rightarrow$ particle trajectory is the same for all mass value. For $q=0$ particle trajectories are curves of minimal length (called geodesics).
\begin{equation}
\fbox{$\displaystyle \ddot x^{\mu}+\Gamma_{\alpha\beta}^{\mu}\,\dot x^{\alpha}\,\dot x^{\beta}=0$} \qquad\qquad \textnormal{(geodesic equation)}
\end{equation} \index{Geodesic equation}
\begin{equation}
\label{eq:christoffel}
\Gamma_{\alpha\beta}^{\mu}=\frac{1}{2}g^{\mu\delta}\left(g_{\delta\alpha,\beta}+g_{\delta\beta,\alpha}-g_{\alpha\beta,\delta}\right)\qquad\qquad \textnormal{(Christoffel symbols)}
\end{equation} \index{Christoffel symbols}
\[\qquad \textnormal{with} \quad g_{\alpha\beta,\delta}=\del{\delta}\,g_{\alpha\beta} \]
\\
Beetle crawling on $\mathbb{R}^{2}$ following a curve $\bigl(y^{1}(\tau),y^{2}(\tau)\bigr)$\\
\begin{figure}[ht]
	\centering
	\includegraphics{beetle}
	\caption{beetle following a curve}
	\label{beetle}
\end{figure}
\[\Delta l=\mbox{const}=1 \qquad\qquad\qquad L=\int_{\tau_{1}}^{\tau_{2}}\Bigl((\dot y^{1})^{2}+(\dot y^{2})^{2}\Bigr)^{\frac{1}{2}} \dd\tau\]
$\Delta l$ will depend on temperature, temperature should vary spatially.
\[\Delta l=\frac{1}{1+\mid y\mid^{2}} \qquad\qquad\qquad L=\int\frac{1}{1+\mid y(\tau)\mid^{2}}\Bigl((\dot y^{1})^{2}+(\dot y^{2})^{2}\Bigr)^{\frac{1}{2}}\dd\tau\]
Trick to compute $L$: New coordinates: $y^{1}=\dfrac{x^{1}}{1-x^{3}}; \quad y^{2}=\dfrac{x^{2}}{1-x^{3}}\quad$ with $\|x\|=1$ Stereographic projection \index{Stereographic projection} (Fig. \ref{stereo})
\begin{equation}
\Rightarrow L=\int\frac{1}{2}\left(\left(\dot x^1\right)^2+\left(\dot x^2\right)^2+\left(\dot x^3\right)^2\right)^{\frac{1}{2}}\dd\tau
\end{equation}
\[\Rightarrow x\in S^{3}\qquad (\mbox{2-Sphere in }\mathbb{R}^{3})\]
$x(\tau)$ is curve of shortest length in $S^{2} \Rightarrow x(\tau)$ are segments of great circles on $S^{2}$\\
\begin{figure}[ht]
	\centering
	\includegraphics{proj}
	\caption{stereographic projection}
	\label{stereo}
\end{figure}

Great circles on $S^{2}$ are mapped into circles on $\mathbb{R}^{2}$ or straight lines through the origin.\\
$\Rightarrow$ Topology of space-time is changed.\\
$\Rightarrow$ Introduction of a space-time dependent metric leads to curved 4-dimensional surfaces (manifolds).

\section{Symmetries of the action} \index{Symmetries}
Reparametrizations, gauge tramsformations, \emph{General coordinate transformations:}\\
Let $f:\mathbb{R}^{4}\to\mathbb{R}^{4}$ be a diffeomorphism (invertable and infinitely many times differentiable) (``a smooth map'' in the future)\\
Replace in $S: \quad x_i^\mu(\tau_i) \to \left(f^{-1}\right)^\mu\bigl(x_i(\tau_i)\bigr) \quad A_\mu \to \bigl(f^*A\bigr)^{\mu}(x)=A_\alpha \bigl(f(x)\bigr)\ddel{\mu}f^\alpha(x)$\\
Replace in addition: $g_{\mu\nu}(x)$ by $\bigl(f^*g\bigr)_{\mu\nu}(x)=g_{\alpha\beta}\bigl(f(x)\bigr)\ddel{\mu}f^\alpha(x) \ddel{\nu}f^\beta(x)$\\
$\Rightarrow S$ is invariant.

If $g_{\mu\nu}=\eta_{\mu\nu}$ equal and $f(x)=\Lambda x+a=p(x)$ ($\Lambda$ is a Lorentz transformation) the transformation agrees with our rule for transforming with respect to Poincar\'e transformations. \\
Properties: $ \left(f\circ g\right)^* = g^* \circ f^* $

\chapter{Tensor Calculus}
Mathematical property behind Tensor Calculus: definitions, computational rules, transformation properties

\section{Definitions}

\begin{description}
\item[Linear Algebra:] Let $V$ be a real $n$-dimensional vector space $\mathbb{R}^n$. $V^\star$ denotes the space of linear functions on $V$ (again a $n$-dimensional vector space). $V^{\star\star}$ (the space of linear functions on $V^\star$) is canonically isomorphic to $V$:\\
$\alpha \in V^{\star\star}, \quad   y \in V^\star, \quad \alpha(y)=y(h) \quad \textnormal{with}\  h \in V$ \\
$\Rightarrow \alpha$ can alway be identified with $h$
\item[Def. 1:] Let $\omega : \underbrace{V\times\ldots\times V}_{k\textnormal{-times}}\times\underbrace{V^\star\times\ldots\times V^\star}_{l\textnormal{-times}} \rightarrow \mathbb R$ be a multilinear map.
\begin{eqnarray*}
\Rightarrow \omega \left(h_1,\ldots , h_k,h'_1, \ldots , h'_l\right)\ \ \textnormal{is linear in each argument.} \quad & h_1,\ldots , h_k \in V \\
	& h'_1,\ldots , h'_l \in V^\star
\end{eqnarray*}
If $e_1, \ldots, e_n$ is a basis of $V$ and $e'^1, \ldots, e'^n$ is the canonically defined dual basis of $V^*$ $\bigl( e'^\mu(e_\nu) = \delta^\mu_\nu \bigr)$ then the so-called components
\begin{equation}
\omega\,^{\mu_1 \ldots \mu_l}_{\nu_1 \ldots \nu_k} = \omega\left(e_{\nu_1}, \ldots, e_{\nu_k}, e'^{\mu_1}, \ldots, e'^{\mu_l}\right)
\end{equation}
determine $\omega$ completely.
The set of all such functions ($k, l$ fixed) forms a vector space $V^{(k,l)}$. A tensor field $\omega$ is by definition a map $\omega: V\rightarrow V^{(k,l)}$ (or an open subset $U\subset V$).  \\
$x\in V \Rightarrow \omega(x) \in V^{(k,l)} \Rightarrow \omega(x)$ is completely specified by $\omega(x)\,^{\mu_1 \ldots \mu_l}_{\nu_1 \ldots \nu_k}$ \\
For $(k,l)$ fixed we call $\omega$ a $(k,l)$-tensor field. \index{Tensor field}
\end{description}
\begin{enumerate}
\item[(a)] $(k,l)$ tensor fields can be added, and multiplied by functions $\lambda(x),\quad x \in V$
\item[(b)] They can be multiplied: Let $\omega$ be a $(k,l)$-tensor field, $\omega'(x)$ a $(k',l')$-tensorfiled. $\omega \cdot \omega'$ is defined by
\begin{eqnarray*}
\lefteqn{\left(\omega \cdot \omega'\right)(x)\left(h_1,\ldots,h_{k+k'},h'_1,\ldots,h'_{l+l'}\right) } \\ 
& & \qquad = \omega(x)\left(h_1,\ldots,h_k,h'_1,\ldots,h'_l\right) \cdot \omega(x)\left(h_{k+1},\ldots,h_{k+k'},h'_{l+1},\ldots,h'_{l+l'}\right)
\end{eqnarray*}
$\Rightarrow \left(\omega\cdot\omega'\right)(x)\,^{\mu_1\ldots\mu_{l+l'}}_{\nu_1\ldots\nu_{k+k'}} = \omega(x)\,^{\mu_1\ldots\mu_{l}}_{\nu_1\ldots\nu_{k}} \cdot \omega(x)\,^{\mu_1\ldots\mu_{l'}}_{\nu_1\ldots\nu_{k'}} \qquad$ (components multiply)
\item[(c)] Let $\sigma \in S_k$ be a permutation of $k$ objects,  $\sigma' \in S_k'$ be a permutation of $k'$ objects. $P_{\sigma\sigma'}$ is defined by the formula
\[ \bigl(P_{\sigma\sigma'} \omega(x)\bigr)\left( h_1,\ldots,h_k,h'_1,\ldots,h'_l\right) = \omega(x)\left( h_{\sigma(1)},\ldots,h_{\sigma(k)},h'_{\sigma'(1)},\ldots,h'_{\sigma'(l)}\right)\]
$\Rightarrow P_{\sigma\sigma'}\ \omega$ is again a $(k,l)$-tensor field.

$\Rightarrow P_{\sigma\sigma'}\ \omega(x)\,^{\mu_1\ldots\mu_l}_{\nu_1\ldots\nu_k} = \omega(x)\,^{\mu_{\sigma'(1)}\ldots\mu_{\sigma'(l)}}_{\nu_{\sigma(1)}\ldots\nu_{\sigma(k)}}$
\item[(d)] Contraction: \index{Contraction} if $\omega$ is a $(k,l)$-tensor field then  a $(k-1,l-1)$-tensor field is defined by
\[\dot\omega(x)\left( h_1,\ldots,h_{k-1},h'_1,\ldots,h'_{l-1} \right) = \omega(x)\left( e_\mu, h_1,\ldots,h_{k-1},e'^\mu, h'_1,\ldots,h'_{l-1} \right) \]
\[\dot\omega\,^{\mu_1\ldots\mu_{l-1}}_{\nu_1\ldots\nu_{k-1}}(x) =  \omega\,^{\mu\,\mu_1\ldots\mu_{l-1}}_{\mu\,\nu_1\ldots\nu_{k-1}}(x) \qquad\qquad\qquad\qquad\textnormal{(summation convention!)}\]
\item[(e)] Let $V \in \mathbb R^4$, $\ g(x)(h,k)$ be given by components $g_{\mu\nu}(x)$. \\
$g(x)$ defines a linear (even invertible) map $g(x): V\rightarrow V^\star$ by the formula
\[g(x)(h)(k)=g(x)(h,k) \qquad\qquad(h,k \in V)\]
$\Rightarrow g$ defines a map assigning to a $(k,l)$-tensor field a $(k+1,l-1)$-field $\tilde\omega$
\[\tilde\omega(x)\left( h_1,\ldots,h_{k+1},h'_1,\ldots,h'_{l-1}\right) = \omega(x) \left( h_2,\ldots,h_{k+1},g(x)h_1, h'_1,\ldots,h'_{l-1}\right)\]
Components: $\tilde\omega\,^{\mu_1\ldots\mu_{l-1}}_{\nu_1\ldots\nu_{k+1}} = g_{\nu_1\alpha} \ \omega\,^{\alpha\mu_1\ldots\mu_{l-1}}_{\nu_2\ldots\nu_{k+1}}$
\end{enumerate}

\section{Transformation rule} \index{Transformation rule}
Let $f: \mathbb R^4 \to \mathbb R^4$ be a diffeomorphism, let $\omega$ and $f^*\omega$ be a $(k,l)$-tensor field
\[\omega \to f^* \omega \]
\textbf{Definition:}
\begin{eqnarray}
\lefteqn{f^*\omega(x) \left(h_1,\ldots , h_k,h'_1, \ldots , h'_l\right) \qquad \qquad \qquad h_i \in \mathbb R^4, h'_i \in \mathbb R^{4^*}} \nonumber \\
& & = \omega\bigl(f(x)\bigr)\left( Df(x)h_1, \ldots, Df(x)h_k,\left(Df(x)^{-1}\right)^th'_1, \ldots ,\left(Df(x)^{-1}\right)^th'_l \right)
\end{eqnarray}
in components:
\begin{equation}
\omega\,^{\mu_1\ldots\mu_{l}}_{\nu_1\ldots\nu_{k}} \to \left(f^*\omega\right)^{\mu_1\ldots\mu_{l}}_{\nu_1\ldots\nu_{k}} =
\omega\bigl(y(x)\bigr)^{\beta_1\ldots\beta_{l}}_{\alpha_1\ldots\alpha_{k}}\cdot \frac{\partial y^{\alpha_1}}{\partial x^{\nu_1}} \cdots \frac{\partial y^{\alpha_k}}{\partial x^{\nu_k}} \cdot \frac{\partial x^{\mu_1}}{\partial y^{\beta_1}} \cdots \frac{\partial x^{\mu_l}}{\partial y^{\beta_l}}
\end{equation}
with $y(x)^\mu = f(x)^\mu$

This transformation rule commutes with \emph{all} operations defined before.

\section{The gradient of a tensor field} \index{Gradient}
\subsection{Definition}
Notation: $\omega \to \nabla \omega$  \\
$\nabla \omega$ should be a $(k+1,l)$-tensor field.

Components:
\begin{eqnarray}
\nabla \omega\,^{\mu_1\ldots\mu_{l}}_{\nu_1\ldots\nu_{k}} = \omega\,^{\mu_1\ldots\mu_{l}}_{\nu_1\ldots\nu_{k};\mu}
 =  \del{\mu} \omega\,^{\mu_1\ldots\mu_{l}}_{\nu_1\ldots\nu_{k}} \!\!\! & + \!\!\!& \Gamma^{\mu_1}_{\alpha \mu}\ \omega\,^{\alpha\mu_2\ldots\mu_{l}}_{\nu_1\ldots\nu_{k}} + \ldots + \Gamma^{\mu_l}_{\alpha \mu}\ \omega\,^{\mu_1\ldots\mu_{l-1}\alpha}_{\nu_1\ldots\nu_{k}} \\
& - \!\!\! & \Gamma^{\alpha}_{\nu_1 \mu}\ \omega\,^{\mu_1\ldots\mu_{l}}_{\alpha\nu_2\ldots\nu_{k}} - \ldots - \Gamma^{\alpha}_{\nu_k \mu}\ \omega\,^{\mu_1\ldots\mu_{l}}_{\nu_1\ldots\nu_{k-1}\alpha} \nonumber
\end{eqnarray}
with $\Gamma^\alpha_{\beta\gamma}$ denoting the Christoffel symbols (see also eq. \ref{eq:christoffel}): \index{Christoffel symbols}
\[\Gamma^\alpha_{\beta\gamma} =  \frac{1}{2} g^{\alpha \delta} \left( \del{\beta} g_{\delta \gamma} + 
\del{\gamma} g_{\delta \beta} - \del{\delta} g_{\beta\gamma}  \right)\]

\subsection{Properties}
\begin{itemize}
\item $\nabla$ commutes with all tensor operations and the product rule holds for the tensor product.
\item For a $(0,0)$-tensor field, $\nabla \varphi$ is the normal gradient ($\varphi$ is a function) \\
components: $\del{\mu} \varphi$
\item In particular: \fbox{$g_{\alpha \beta; \mu} = 0$}
\end{itemize}

\subsection{Special cases}
Suppose $\omega\,_{\nu_1\ldots\nu_k}$ are the components of a $(k,0)$-tensor field.

$\omega\,_{\nu_1\ldots\nu_k;\,\nu_0}$ antisymmetrized:
\begin{eqnarray*}
\lefteqn{\omega_{\nu_1\ldots\nu_k;\,\nu_0} - \omega_{\nu_0\nu_2\ldots\nu_k;\,\nu_1} + \omega_{\nu_0\nu_1\nu_3\ldots\nu_k;\,\nu_2} - \ldots \pm \omega_{\nu_0\ldots\nu_{k-1};\,\nu_k}} \\
& & =  \del{\nu_0} \omega_{\nu_1\ldots\nu_k} - \del{\nu_1} \omega_{\nu_0\nu_2\ldots\nu_k} + \ldots
\end{eqnarray*}
Second simplification for $(0,k)$-tensor fields:
\[{\omega\,^{\mu\nu_1\ldots\nu_{k-1}}}_{;\mu} = \frac{1}{ \sqrt{|g|}}\ \del{\mu} \sqrt{|g|}\ \omega\,^{\mu\nu_1\ldots\nu_{k-1}}
\qquad \qquad g = \det g_{\alpha\beta}\]
Example: ${F^{\mu\nu}}_{;\mu} = \dfrac{1}{ \sqrt{|g|}}\ \ddel{\mu} \sqrt{|g|}\ F^{\mu\nu}$

\section{Transformation rule and integration}
Suppose $\varphi : \mathbb R^4 \to \mathbb R^4$ is a real function. Consider
\[ A = \int_{\mathbb R^4} \varphi(y)\sqrt{|g'|}\ \dd^4 x\]
with $y^\mu = y^\mu(x)$ a diffeomorphism, $\quad g'_{\alpha\beta} = g_{\mu\nu}\bigl(y(x)\bigr)\ \dfrac{\partial y^\mu}{\partial x^\alpha} \dfrac{\partial y^\nu}{\partial x^\beta}$
\[ \Rightarrow \sqrt{|g'|} = \sqrt{|g(y)|}\ \left| \det \frac{\partial y^\mu}{\partial x^\alpha} \right| \qquad \Rightarrow A =  \int_{\mathbb R^4} \varphi(y)\sqrt{|g(y)|}\ \dd^4 y\]
\[ \Rightarrow A = \int_{\mathbb R^4} \varphi\quad \textnormal{is invariant under coordinate transformations} \]

\section{The curvature tensor}
Let $Y^\mu$ be the components of a vector field.
\begin{eqnarray}
\lefteqn{{Y^\mu}_{;\alpha\beta} - {Y^\mu}_{;\beta\alpha} = R^\mu_{\gamma\alpha\beta} Y^\gamma} \nonumber\\
\lefteqn{R^\mu_{\gamma\alpha\beta} = \del{\alpha} \Gamma^\mu_{\gamma\beta} - \del{\beta} \Gamma^\mu_{\gamma\alpha} + \Gamma^\mu_{\delta\alpha} \Gamma^\delta_{\gamma\beta} - \Gamma^\mu_{\delta\beta} \Gamma^\delta_{\gamma\alpha}}
\end{eqnarray}
R is a $(3,1)$-tensor field. \index{Curvature tensor}

$\Gamma = \Gamma(g) \Rightarrow R(g)$ depends on g and its first and second derivatives.

$R\left(f^* g\right) = f^* R(g)$

R is called the curvature tensor associated to g.

\subsection{Riemann tensor}
The $(4,0)$-tensor field with components
\begin{equation}
	R_{\mu\gamma\alpha\beta} = g_{\mu \delta}\ R^\delta_{\gamma\alpha\beta}
\end{equation}
is called the Riemann tensor. \index{Riemann tensor}

\subsubsection*{Symmetry properties:}
\[R_{\mu\gamma\alpha\beta} = - R_{\mu\gamma\beta\alpha} = - R_{\gamma\mu\alpha\beta} \]
\[R_{\mu\gamma\alpha\beta} = R_{\alpha\beta\mu\gamma} \]

\subsection{Ricci tensor} \index{Ricci tensor}
\begin{equation}
R_{\gamma\beta} = g^{\mu\alpha} R_{\mu\gamma\alpha\beta}
\end{equation}
are components of a $(2,0)$-tensor field called the Ricci tensor.
\[ R_\mathrm{ic} = g^{\gamma\beta} R_{\gamma\beta} \textnormal{ is a function!} \]
\[ R_\mathrm{ic} \left( f^* g \right) = f^* R(g) \]

\subsubsection*{Remark:}
$R_\mathrm{ic}$ is the only function (up to real numbers) with this property which is linear in the second derivative. $R_\mathrm{ic}$ is called the Ricci scalar. \index{Ricci scalar}

\section{Action functional} \index{Action functional}
Our physical action is still to be supplemented by an action part for $g_{\alpha\beta}$. Call it $S_\mathrm{gr.}$

To keep symmetries:
\begin{equation}
S_\mathrm{gr.} = \int \dd^4x \sqrt{|g|}\ F(g) \qquad\qquad\qquad \textnormal{with } F\left(f^* g\right) = f^* F(g)
\end{equation}
such that $S_\mathrm{gr.}$ is invariant under any diffeomorphisms $f$.

Simple requirement: $F(g)$ is linear in the second derivative of $g$.
\[\Rightarrow F(g) = \alpha \cdot R_\mathrm{ic}(g) + \beta \qquad\qquad\qquad\textnormal{Einstein: }\alpha = - \frac{1}{16\pi G}, \quad \beta = 0 \]

\chapter{Special coordinates}

\section{Riemannian normal coordinates} \index{Riemannian normal coordinates}
Particle  motion: solve
\begin{equation}
\ddot y^\alpha + \Gamma^\alpha_{\beta\gamma}\, \dot y^\beta \dot y^\gamma = 0
\label{eq:part_motion}
\end{equation}
\[g_{\alpha\beta}\, \dot y^\alpha \dot y^\beta = 1 \qquad\qquad \textnormal{drop this condition} \]
The solution is uniquely determined by fixing initial values.
\[ y(0) = y_0 \qquad\qquad \dot y(0) = 0 \qquad\qquad \textnormal{Fix } y_0.\]
\[y(\tau) = y(h,\tau) \]
We write $f(h) = y(h,1)$ \bigskip\\
\includegraphics[width=60mm]{riemann} \medskip

Math. property of $f$:\\
$f$ is a diffeomorphism of an open set $U$ into an open set $V$, $\qquad U,V\subset\mathbb R^4$

We have:$\qquad h = 0 \qquad y(0,1) = y_0 \qquad f(0) = y_0 \qquad 0\in U, y_0 \in V$ \\
$y(h,\alpha\tau)$ is again a solution of \eqref{eq:part_motion} $\qquad y(h,0) = y_0$
\[\frac{\dd}{\dd\tau} y(h,\alpha\tau) \Big \vert _{\tau = 0} = \alpha h \]
\[\qquad \Rightarrow y(h,\alpha\tau) = y(\alpha h,\tau) \]
\[\qquad \Rightarrow f(\alpha h) = y(h, \alpha) \qquad\qquad
\underbrace{\bigl\{\alpha h\bigr\}}_\textnormal{straight lines} \stackrel{f}{\longrightarrow} \underbrace{\bigl\{y(h,\alpha)\bigr\}}_\textnormal{geodesics}\]

\begin{figure}[ht]
\includegraphics{explosion}
\caption{Explosion}
\end{figure}

\subsubsection*{Definition of normal coordinates:}
\[ y \to f(x) \qquad\qquad x=h \qquad \textnormal{(construction as before)}\]
\[ \ddot x^\alpha + \Gamma^\alpha_{\beta\gamma}\, \dot x^\beta \dot x^\gamma = 0 \qquad (\textnormal{geodesic equation in coordinates }x) \]
solution with $x(0) = 0 : x(\tau) = x \cdot \tau$
\[ \Rightarrow \Gamma^\alpha_{\beta\gamma}(x \tau)\, x^\beta x^\gamma = 0 \quad\stackrel{\tau=0}{\Longrightarrow}\quad \Gamma^\alpha_{\beta\gamma}(0)\, x^\beta x^\gamma = 0 \quad\Rightarrow\quad \Gamma^\alpha_{\beta\gamma} = 0\]
\[g_{\beta\gamma;\alpha} = \del{\alpha}\,g_{\beta\gamma}(x) - \Gamma^\delta_{\beta\alpha}(x)\,g_{\delta\gamma}(x) - \Gamma^\delta_{\gamma\alpha}(x)\,g_{\delta\gamma}(x) = 0 \quad\stackrel{x=0}{\Longrightarrow} \left(\del{\alpha}\,g_{\beta\gamma}\right)(0) = 0 \]
Taylor expansion: $g_{\alpha\beta}(x) = g_{\alpha\beta}(0) + \left(\ddel{\gamma}\,g_{\alpha\beta}(0)\right)x^\gamma + \mathcal O(x^2)$

$g_{\alpha\beta}(0)$ can be brought to the form $\eta_{\alpha\beta}$ by a \emph{linear} transformation \\
$\Rightarrow $without loss of generality, we can assume that this is already done.
\begin{equation}
g_{\alpha\beta}(x) = \eta_{\alpha\beta} + \mathcal O(x^2)
\end{equation}
Exercise: $g_{\alpha\beta}(x) = \eta_{\alpha\beta} - \frac{1}{3} R_{\alpha\mu\beta\nu}(0)\,x^\mu x^\nu + \mathcal O(x^3)$
\begin{description}
\item[Corollary] The only scalar invariant under arbitrary coordinate transformations which can be formed out of $g_{\alpha\beta}$, its first and second derivatives, which is linear in the second derivative, is the \emph{curvature scalar} (up to scalar multiplication).
\end{description}
Use normal coordinates: \quad at $x = 0 \qquad \left( \del{\alpha}\,g_{\beta\gamma}(0)\right) = 0$ \\
second derivatives are proportional to the Riemann tensor $R_{\alpha\mu\beta\nu}(0)$ \\
$\Rightarrow$ the scalar must be proportional to these curvature components \\
$\Rightarrow$ only possible way: $\quad\lambda \eta^{\alpha\beta}\eta^{\mu\nu}R_{\alpha\mu\beta\nu}(0) = R(0) \qquad\qquad \eta^{\alpha\beta} = g^{\alpha\beta}(0)$ \\
$\Rightarrow R(y) = \lambda g^{\alpha\beta}g^{\mu\nu}R_{\alpha\mu\beta\nu}(y) \quad$ for any point $y$. $\qquad\lambda\in\mathbb R$

\section{Comoving coordinates} \index{Comoving coordinates}
Let $g$ be a metric on space-time, coordinates $y^\alpha$ are called ``comoving'' if the metric takes the form:
\[g_{00}=1, \quad g_{0i}=0 \qquad i=1,2,3\]
$\Rightarrow$ The geodesic equation reads as follows:
\begin{eqnarray*}
\frac{\dd}{\dd\tau}\,\dot y^0&=&\frac{1}{2}\frac{\partial}{\partial y^0}\,g_{ik}\,\dot y^i\dot y^k \qquad\qquad\ \, k,i=1,2,3\\
\frac{\dd}{\dd\tau}\,g_{ik}\,\dot y^k&=&\frac{1}{2}\left(\frac{\partial}{\partial y^i}\,g_{mn}\right)\dot y^m\dot y^n\qquad m,n=1,2,3
\end{eqnarray*}
\begin{equation}
\fbox{in addition: $(\dot y^0)^2+g_{ik}\,\dot y^i\dot y^k=1$}
\label{eq:cogeo}
\end{equation}
Special trivial solutions: $y^k=a^k=\const$
\[\Rightarrow \dot y^k=0 \qquad \ddot y^0=0 \Rightarrow y^0=\alpha\tau+\beta \  \stackrel{\eqref{eq:cogeo}}{\Longrightarrow}\  y^0=\tau+\beta\]

Existance of comoving coordinates:\\
Let $g_{\mu\nu}(x)$ be a metric in any set of coordinates; there is always a coordinate transformation such that the metric takes the form $g_{00}=1, \  g_{0i}=0$\\
Proof relies on the Jacobi methode in mechanics.
\subsection*{Reminder} \index{Jacobi methode}
Let $\mathcal{L}$ be the Lagrangian for the motion of a particle.
\[\mathcal{L}=\mathcal{L}(x,\dot x,t), \quad x,\dot x \in \mathbb R^3; \quad\qquad p_i=\frac{\partial \mathcal L}{\partial \dot x^i}, \quad \mathcal H=\sum_{i=1}^3 p_i\,\dot x^i-\mathcal L\]
Jacobi:
\[p_i=\frac{\partial W}{\partial x^i},\qquad W=W(x,Q),\qquad Q \in \mathbb R^3\]
satisfies Jacobi equation $\frac{\partial W}{\partial t}+\mathcal H(x,\stackrel{x}{\nabla}W)=0$\\
The particle trajectories are obtained by solving the equations:
\[P=\stackrel{Q}{\nabla}W(x,Q,t)\qquad\qquad P,Q\in\mathbb R^3\quad \const \qquad \textnormal{for }x=x(P,Q,t)\]\\
Gravitation 
\begin{eqnarray*}
S&=&-m\int\dd\tau \bigl(g_{\alpha\beta}(x)\dot x^\alpha\dot x^\beta\bigr)^{\frac{1}{2}}\\
& &\int\dd x^0\underbrace{\bigl(g_{00}+2g_{0i}\,\dot x^i+g_{ik}\,\dot x^i\dot x^k\bigr)^\frac{1}{2}}_{=\mathcal L}\\
p_i&=&-m\dfrac{g_{0i}+g_{ik}\,\dot x^k}{\bigl(g_{00}+2g_{0i}\,\dot x^i+g_{ik}\,\dot x^i\dot x^k\bigr)^\frac{1}{2}}=\frac{\partial W}{\partial x^i}\\
\mathcal H&=&m\dfrac{g_{00}+g_{0i}\,\dot x^i}{\bigl(g_{00}+2g_{0i}\,\dot x^i+g_{ik}\,\dot x^i\dot x^k\bigr)^\frac{1}{2}}=-\frac{\partial W}{\partial x^0}
\end{eqnarray*}
\begin{equation}
\Rightarrow \fbox{$g^{\mu\nu}\ddel{\mu}W\ddel{\nu}W=m^2$}\qquad\textnormal{Jacobi equation in general relativity}
\label{eq:jacobi} \index{Jacobi equation}
\end{equation}
Comoving coordinates? consider the following functions: $y^0=W;\qquad y^i=\frac{\partial}{\partial Q^i}W$\\
Claim: These functions constitute a set of comuving coordinates!\\
Proof: Put $a_\mu^\alpha=\dell{\mu}{y^\alpha} \qquad\qquad$ \fbox{$A_\mu^\nu=\tilde g_{\alpha\beta}\,a_\mu^\alpha b^{\beta\nu}$}\\
with $\tilde g_{00}=1,\quad\tilde g_{ik}$ equal to the inverse matrix of $a_{i\mu}b_k^\mu\qquad i,k=1,2,3$\\
$b^{\beta\mu}=a_\alpha^\beta g^{\alpha\mu}$ and $\tilde g_{0i}=\tilde g_{i0}=0$\\
Claim: $A_\mu^\nu=\delta_\mu^\nu$\\
Sketch of proof: This is true if $A_\mu^\nu h_{\alpha\nu}=h_{\alpha\mu}$ for any set of 4 linearly independent vectors $h\quad (\alpha=0,\ldots,3)$ Put $h_{\alpha\mu}=\del{\mu}y^\alpha=a_\mu^\alpha$\\
For the proof use \eqref{eq:jacobi}
\[\Rightarrow g_{\mu\nu}=\tilde g_{\alpha\beta}\,a_\mu^\alpha a_\nu^\beta=\tilde g_{\alpha\beta}\dell{\mu}{y^\alpha}\dell{\mu}{y^\beta}\]
$\Rightarrow g_{\alpha\beta}$ is related by a coordinate transformation to $\tilde g_{\alpha\beta}$ with $\tilde g_{00}=1,\ \tilde g_{0i}=\tilde g_{i0}=0$\\
$\Rightarrow y^\alpha$ are comoving coordinates

\chapter{Field equations of the gravitational field} \index{Gravitational field equation}\index{Field equations}
\section{Variation of the action}
\begin{eqnarray}
S&=&-\sum_{i=1}^N m_i\int\!\left(g_{\alpha\beta}\,\dot x^\alpha(\tau_i)\dot x^\beta(\tau_i)\right)\!\dd\tau_i-\sum_{i=1}^N q_i\int\dd\tau_i\,A_\mu\dot x^\mu(\tau_i)\nonumber\\
& &-\frac{1}{16\pi}\int\dd^4x\sqrt{|g|}\ F_{\alpha\beta}F^{\alpha\beta}-\frac{1}{16\pi G} \int\dd^4x\sqrt{|g|}\ R
\end{eqnarray}
$g=\det g_{\mu\nu},\qquad F_{\alpha\beta}=\del{\alpha}A_\beta-\del{\beta}A_\alpha,\qquad G$: gravitational constant

\subsection{Variation with respect to the vector potential}
Procedure: Put $A(s)=A+s\,\delta A \qquad (\delta A_\mu$ has compact support$)$\\
$\qquad F\to F(s);\qquad S \to S\bigl(A(s)\bigr)$\\
Extremum is looked for by asking for equation: $\frac{\dd}{\dd s}S\bigl(A(s)\bigr)=0\ $ for any $\delta A$\\
Hamiltonian's Principle \index{Hamiltonian's Principle}
\begin{eqnarray}
\Rightarrow \frac{\dd S}{\dd s} &=& -\sum_{i=1}^N q_i\int\!\delta A_\mu\dot x^\mu\,\dd\tau_i-\frac{1}{16\pi}\int\!\dd^4x\sqrt{|g|}\ \left(\del{\mu}\delta A_\nu\right)F^{\mu\nu}\cdot4\nonumber\\
& &\int\!\dd^4x\sqrt{|g|}\ \left(\ddel{\mu}\delta A_\nu\right)F^{\mu\nu}=\underbrace{\int\!\dd^4x\del\mu\delta A_\nu F^{\mu\nu}\sqrt{|g|}}_{= 0 \textnormal{ Gauss theorem}}-\int\!\dd^4x\,\delta A_\nu\nonumber\\
\frac{\dd S}{\dd s}&=&-\sum q_i\int\!\delta A_\mu\dot x^\mu\dd\tau_i +\frac{1}{4\pi}\int\!\dd^4x\sqrt{|g|}\underbrace{\left(\dfrac{1}{\sqrt{|g|}}\del\mu\sqrt{|g|}\, F^{\mu\nu}\right)}_{=F^{\mu\nu}{}_{;\mu}}\delta A_\nu\nonumber\\
\frac{\dd}{\dd s} S&=&\int\!\dd^4x\sqrt{|g|}\left(-J^\nu(x)+\frac{1}{4\pi}F^{\mu\nu}{}_{;\mu}\right)\delta A_\nu
\end{eqnarray}
\[\textnormal{with }J^\mu(x)=\int\!\sum_{i=1}^N q_i\dot x_i^\mu\dfrac{\delta\bigl(x-x_i(\tau_i)\bigr)}{\sqrt{|g|}}\,\dd\tau_i\] \bigskip \\
Principle of least action $\Rightarrow \frac{\dd}{\dd s}S=0$ for any $\delta A_\nu$
\begin{equation}
\Rightarrow \fbox{$F^{\mu\nu}{}_{;\nu}=4\pi J^\mu(x)$}\qquad J^\mu(x) \textnormal{ electromagnetic current}
\label{eq:df}
\end{equation}
\[\qquad\dd F=0\qquad\qquad \left(\textnormal{consequence of }F_{\mu\nu}=\del\mu A_\nu-\del\nu A_\mu\right)\] %\bigskip \\
%Next step: Replace $g_{\alpha\beta}$ by $g_{\alpha\beta}(s)=g_{\alpha\beta}+s\,\delta g_{\alpha\beta}$\\
%$S\to S\bigl(g(s)\bigr),\qquad$ Compute $\dfrac{\dd}{\dd s}S$

\subsection{Variation with respect to the metrical tensor}
\begin{eqnarray}
S &=& S_\mathrm p + S_\mathrm{em} + S_\mathrm{gr} \\
S_\mathrm p &=& \sum_{i=1}^N \left(m_i \int\!\left( g_{\mu\nu}\,\dot x^\mu \dot x^\nu\right)^\frac{1}{2}\dd\tau_i + q_i \int\!A_\mu \dot x^\mu\,\dd\tau_i \right) \nonumber\\
S_\mathrm{em} &=& -\frac{1}{16\pi} \int\dd^4x \sqrt{|g|}\ F_{\mu\nu}F^{\mu\nu} \qquad\qquad \left(F_{\mu\nu} = \del{\mu}A_\nu - \del{\nu}{\mu}\right) \nonumber \\
S_\mathrm{gr} &=& -\frac{1}{16\pi G } \int\dd^4x \sqrt{|g|}\ R(g) \nonumber
\end{eqnarray}
Put $g_{\mu\nu}(\alpha) = g_{\mu\nu} + \alpha\,\delta g_{\mu\nu} \qquad (\delta g_{\mu\nu}$ has compact support$)$ \\
Replace in $S$ $g_{\mu\nu}$ by $g_{\mu\nu}(\alpha)$ \\
$S \to S(\alpha)\qquad$ Extrema are given by $\frac{\dd}{\dd\alpha} S(\alpha) = 0$
\subsubsection{Needed:}
\begin{itemize}
\item $\delta\sqrt{|g|} = \dfrac{\dd}{\dd\alpha}\sqrt{|g(\alpha)|}\,\Big|_{\alpha = 0} \qquad\qquad$exercise $\Rightarrow \dfrac{1}{\sqrt{|g|}}\,\delta\sqrt{|g|} =  \dfrac{1}{2}\,\delta g_{\mu\nu}\,g^{\mu\nu}$
\item $\delta g^{\mu\nu} = \dfrac{\dd}{\dd\alpha}\, g^{\mu\nu}(\alpha)\Big|_{\alpha=0} =\ ?$ \medskip \\
Use $g^{\mu\nu}(\alpha)\,g_{\mu\lambda}(\alpha) = \delta^\nu_\lambda \qquad \Rightarrow \delta g^{\mu\nu}\,g_{\mu\lambda} + g^{\mu\nu}\,\delta g_{\mu\lambda} = 0$
\[\Rightarrow \delta g^{\mu\nu} = -g^{\mu\alpha}g^{\nu\beta}\,\delta g_{\alpha\beta}\]
\item $\delta R = \delta\left(g^{\mu\nu} R_{\mu\nu}\right) = \delta g^{\mu\nu}\,R_{\mu\nu} + g^{\mu\nu}\,\delta R_{\mu\nu}$ \medskip\\
exercise: $g^{\mu\nu}\,\delta R_{\mu\nu} = {J^\alpha}_{;\alpha} \qquad\qquad J^\mu = \delta\Gamma^\mu_{\alpha\beta}\,g^{\alpha\beta} - g^{\mu\alpha}\,\delta\Gamma^\beta_{\alpha\beta}$
\item $\displaystyle \delta S_\mathrm p = -\sum\limits_{i=1}^N\int m_i \frac{1}{2}\,\frac{\delta g_{\mu\nu}\, \dot x^\mu \dot x^\nu}{\left(g_{\alpha\beta}\, \dot x^\alpha \dot x^\beta \right)^\frac{1}{2}}\,\dd\tau_i$
\item $\displaystyle \delta S_\mathrm{em} = - \frac{1}{2}\int\dd^4x \sqrt{|g|}\ T^{\mu\nu}_\mathrm{em}$ \medskip\\
$T^{\mu\nu}_\mathrm{em} = \dfrac{1}{4\pi}\left(-F^{\mu\alpha}F^\nu_\alpha + \dfrac{1}{4}\, g^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\right) \qquad$ (energy-momentum tensor of em field) \index{Energy-momentum tensor}
\item $\displaystyle \delta S_\mathrm{gr} =\frac{1}{16\pi G}\int\dd^4x \sqrt{|g|}\bigl(G^{\mu\nu}\,\delta g_{\mu\nu}\bigr) - \underbrace{ \frac{1}{16\pi G}\int\dd^4x \sqrt{|g|}\ {J^\mu}_{;\nu}}_{(*)}$ \\
$G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}\, g_{\mu\nu}\,R \qquad$ (Einstein tensor) \index{Einstein tensor} \medskip\\
${J^\mu}_{;\mu} = \dfrac{1}{\sqrt{|g|}}\ \ddel{\mu} \sqrt{|g|}\ J^\mu\quad$ Transform into a surface integral ($R$ large) $\Rightarrow (*) = 0$
\end{itemize}
Write $\displaystyle \delta S_\mathrm p = \frac{1}{2} \int\dd^4x \sqrt{|g|}\ \delta g_{\mu\nu}(x) T^{\mu\nu}_\mathrm m(x)$
\[T^{\mu\nu}_\mathrm m (x) = \sum_{i=1}^N \int \frac{\delta^4\bigl(x - x_i(\tau_i)\bigr)}{\sqrt{|g|}}\, m_i\, \dot x_i^\mu \dot x_i^\nu\,\dd\tau_i  \]
\subsubsection{End result:}
\[\delta S = \frac{1}{2} \int\dd^4x \sqrt{|g|} \left(\delta g_{\mu\nu}\left(T^{\mu\nu} - \frac{1}{8\pi G}\,G^{\mu\nu}\right) \right) \qquad\qquad\textnormal{with } T^{\mu\nu} = T^{\mu\nu}_\mathrm m + T^{\mu\nu}_\mathrm{em}\]
\[\delta S = 0 \textnormal{ for any choice of } \delta g_{\mu\nu}\]
\[\qquad\Rightarrow \fbox{$G^{\mu\nu} = 8\pi G\ T^{\mu\nu}$} \]

\section{Field equations} \index{Field equations}
\begin{eqnarray}
G^{\mu\nu} = 8\pi G\ T^{\mu\nu} \qquad\qquad\qquad\ \, & T^{\mu\nu} = T^{\mu\nu}_\mathrm m + T^{\mu\nu}_\mathrm{em} & \\
{F^{\mu\nu}}_{;\mu} = 4\pi J^\nu_\mathrm e \qquad\qquad\qquad\quad\ \, & & \textnormal{(Maxwell)} \index{Maxwell equations} \\
m_i\left(\ddot x^\mu + \Gamma^\mu_{\alpha\beta}\,\dot x^\alpha \dot x^\beta \right) = q_i\,F^\mu_\nu \dot x^\nu & g_{\alpha\beta}\, \dot x^\alpha \dot x^\beta = 1 & \textnormal{(particle trajectories)} \index{Particle trajectories}
\end{eqnarray}
${G^{\mu\nu}}_{;\nu} = 0$ hold for any $g_{\mu\nu}$ by construction (Bianchi identity) \index{Bianchi identity}
\begin{enumerate}
	\item Bianchi identities
	\item Particle energy-momentum tensor
\end{enumerate}

\subsection{Bianchi identity}
Let $g$ be any spacetime metric (components $g_{\mu\nu}$). Then the Einsteintensor $G^{\mu\nu}(g)$ always
satisfies the identities:
\[G^{\mu\nu}{}_{;\nu}=0\]
\subsubsection{Derivation and interpretation}
Let B a vectorfield of compact support (components $B^{\mu}$). Consider the differential equation for curves $y(\tau)$:
\[\dot{y}^\mu(\tau)=B^\mu(\tau)\]
Let $y(\tau,x)$ be the \textit{unique} solution with the initial value $y(0)=x$. $\Rightarrow f_\tau(x)=y(\tau,x)$ is a
diffeomorphism and $f_\tau$ is family of diffeomorphisms such that $f_0=\text{id}$.\\
Consider $f^*_\tau g$ and compute:\\
\textbf{Claim:}
\begin{equation}
\frac{\dd}{\dd\tau}f^*_\tau g_{\mu\nu}(x)\Big\vert_{\tau=0} = B_{\alpha;\beta}+B_{\beta;\alpha}
\end{equation}
\textbf{Proof:}
\begin{align*}
  f^*_\tau g_{\alpha\beta}(x) =&\ g_{\alpha'\beta'}(f_\tau(x))\cdot \del{\alpha}y^{\alpha'}(\tau,x)\cdot \del{\beta}y^{\beta'}(\tau,x)\\
  \Rightarrow \frac{\dd}{\dd\tau}f^*_\tau g_{\alpha\beta}(x)\Big\vert_{\tau=0} = &\ \frac{\dd}{\dd\tau}y^\mu(\tau,x)\Big\vert_{\tau=0}\cdot \del{\mu} g_{\alpha\beta}(x)+g_{\alpha'\beta}(x)\del{\alpha}\frac{\dd}{\dd\tau}y^{\alpha'}(\tau,x)\Big\vert_{\tau=0}\\
  &\ +g_{\alpha\beta'}\del{\beta}\frac{\dd}{\dd\tau}y^{\beta'}(\tau,x)\Big\vert_{\tau=0}\\
  \Rightarrow \frac{\dd}{\dd\tau}y(\tau,x)\Big\vert_{\tau=0}=&\ B^\mu(x)\\
  \Rightarrow \frac{\dd}{\dd\tau}f^*_\tau g_{\alpha\beta}(x)\Big\vert_{\tau=0}=&\ B^\mu(x)\del{\mu}g_{\alpha\beta}(x)+g_{\alpha'\beta}(x)\del{\alpha}B^{\alpha'}(x) + g_{\alpha\beta'}(x)\del{\beta}B^{\beta'}(x)\\
  =&\ B^\mu(x)g_{\alpha\beta;\mu}(x) + g_{\alpha'\beta}(x)B^{\alpha'}(x) + g_{\alpha\beta'}(x)B^{\beta'}(x) 
\end{align*}
All terms with Cristoffel symbols cancel!\\
$g_{\alpha\beta;\mu}=0$ implies
\[\frac{\dd}{\dd\tau}f^* g_{\alpha\beta}\Big\vert_{\tau = 0}=B_{\alpha;\beta}+B_{\beta;\alpha}\]
\bigskip

The gravitational actions
\[S_\mathrm{gr}=-\frac{1}{16\pi G} \int \dd^4x\sqrt{|g|}R(g)\]
satisfies
\[S_\mathrm{gr}(f^* g)=S_\mathrm{gr}(g)\]
In particular
\[S_\mathrm{gr}(f^*_\tau g)=S_\mathrm{gr}(g)\]
\[ \Rightarrow 0 = \frac{\dd}{\dd\tau}S_\mathrm{gr}(f^*_\tau g)\Big\vert_{\tau = 0}=\frac{\dd}{\dd\tau}S_\mathrm{gr}(g+\tau \delta g)\Big\vert_{\tau = 0}\]
where $\delta g_{\mu\nu}=(\frac{\dd}{\dd\tau}f^*_\tau g)_{\mu\nu}$
\begin{align*}
  \Rightarrow 0=\frac{\dd}{\dd\tau}S_\mathrm{gr}(f^*_\tau g)\Big\vert_{\tau = 0}=&\frac{1}{16\pi G}\int \dd^4x\sqrt{|g|}G^{\mu\nu}\,\delta g_{\mu\nu}\\
  =&\frac{1}{16\pi G}\int \dd^4x\sqrt{|g|}G^{\mu\nu}(B_{\mu;\nu}+B_{\nu;\mu})\\
  =&\frac{1}{8\pi G}\int \dd^4x\sqrt{|g|}G^{\mu\nu}B_{\mu;\nu}\\
  =&\frac{1}{8\pi G}\int \dd^4x\sqrt{|g|}(\underbrace{(G^{\mu\nu}B_{\mu}}_{=:K^\nu}){}_{;\nu}- G^{\mu\nu}{}_{;\nu}\,B_{\mu})\\
  =&\frac{1}{8\pi G}\int \dd^4x\sqrt{|g|}(K^\nu{}_{;\nu}- G^{\mu\nu}{}_{;\nu}\,B_{\mu})
\end{align*}
Assume B has compact support:
\[\Rightarrow K^\nu{}_{;\nu}=\frac{1}{\sqrt{|g|}}\del{\nu}\sqrt{|g|}\,K^\nu\]
and
\[\int \dd^4x \sqrt{|g|}\,K^\nu{}_{;\nu}=\int \dd^4x \del{\nu}( \sqrt{|g|}\,K^\nu)=0\]
by Gauss theorem.
\[ \Rightarrow 0 = \int \dd^4x \sqrt{|g|}\,G^{\mu\nu}{}_{;\nu}\,B_\mu \]
$B$ arbitrary $\Rightarrow \fbox{$G^{\mu\nu}{}_{;\nu}=0$}$\bigskip \\
\textbf{Einstein's equations:} \index{Einstein equation}
\begin{equation}
\fbox{$\displaystyle G^{\mu\nu}=8\pi\cdot G\cdot T^{\mu\nu}$}
\end{equation}
$\Rightarrow G^{\mu\nu}{}_{;\nu}=0 $ implies $ T^{\mu\nu}{}_{;\nu}=0$\bigskip \\
\textbf{Remark:} This leads only to conserved quantities on $G$ if $f^*_\tau g = g$

\section{Equation of state} \index{Equation of state}
\begin{equation}
T_\mathrm m^{\mu\nu} = \sum_{i=1}^N m_i\int \dot x_i^\mu(\tau_i) \dot x_i^\nu(\tau_i) \dfrac{\delta^4\bigl(x-x_i(\tau_i)\bigr)}{\sqrt{|g|}}\,\dd\tau_i \\
g_{\mu\nu}\dot x^\mu\dot x^\nu = 1 
\label{eq:T_m}
\end{equation}
One particle only $\Rightarrow$ Einstein's equation: 
\begin{equation}
G^{\mu\nu}=8\pi G m \int\dfrac{\delta^4\bigl(x-x(\tau)\bigr)}{\sqrt{|g|}}\,\dot x^\mu(\tau)\dot x^\nu(\tau)\,\dd\tau
\end{equation}
No immediate solution: Therefore study the case of a massive body, smooth particle distribution.\\
Turning $T_\mathrm m^{\mu\nu}$ into an energy-momentum tensor for a massive body. $T_\mathrm m^{\mu\nu}$ is first studied in terms of Riemannian coordinates $y$.\\
$x \in \mathbb R^4$ is mapped into $y=0$
\begin{center}
\includegraphics[height=40mm]{map}
\end{center}

First simplification: $\sqrt{|g|}=1 \qquad\qquad g_{\mu\nu}(y)=\eta_{\mu\nu}$\\
$\Rightarrow$ Integration in \eqref{eq:T_m} can be done:
\[T_\mathrm m^{\mu\nu}=\sum_{i=1}^N m_i\dfrac{\dot y^\mu(\tau_i)\dot y\nu(\tau_i)}{\dot y^0(\tau_i)}\,\delta^3\bigl(\vec y-\vec y_i(\tau_i)\bigr) \qquad\qquad y_i^0(\tau_i)=y^0\]

\subsubsection{Remark:} $m\ \dot y^\mu(\tau_i)'=p^\mu\quad$ 4-momentum of a particle \index{4-momentum of a particle}
\[\eta_{\mu\nu}\,p_i^\mu p_i^\nu=m_i{}^2 \qquad p_i^0=\sqrt{\vec p_i{}^2+m_i{}^2} \quad\textnormal{energy of particle}\]
\[\Rightarrow T_\mathrm m^{\mu\nu}(y)=\int \dfrac{\dd^3p}{p^0}\,p^\mu p^\nu\rho(\vec p, y)\qquad\text{where }\rho(\vec p,y)=\sum_{i=1}^N\delta^3\bigl(\vec p_i-m_i\vec y_i(\tau_i)\bigr)\delta^3\bigl(\vec y-y_i(\tau_i)\bigr)\]

\subsubsection{Symmetry assumptions:}
$\rho$ can be replaced by a smooth particle distribution which is homogeneous (independant of $y$) and rotationally invariant.
\begin{eqnarray}
\Rightarrow \ \,\qquad\rho &=& \rho(|\vec p|^2)\nonumber\\
\Rightarrow T_\mathrm m^{\mu\nu}(y) &=& \int \dfrac{\dd^3p}{p^0}p^\mu p^\nu \rho(|\vec p|^2)\nonumber\\
T_\mathrm m^{00}(y) &=& \int\dd^3p\,\bigl(\vec p^2+m^2\bigr)^\frac{1}{2}\rho(|\vec p|^2)=\varepsilon \qquad \text{energy density}\label{eq:energy}\\
T_\mathrm m^{0i}(y) &=& 0\qquad\qquad\qquad\qquad\qquad\qquad\qquad\, i,j=1,2,3\nonumber\\
T_\mathrm m^{ij}(y) &=& \int\dfrac{\dd^3p}{p^0}\,p^ip^j\rho(|\vec p|^2)=\delta^{ij}P \label{eq:pressure}\\
P &=& \frac{1}{3}\int\dfrac{\dd^3p}{p^0}\,|\vec p|^2\rho(|\vec p|^2) \ \qquad\qquad\qquad \text{pressure}\nonumber\\
\eqref{eq:energy},\eqref{eq:pressure}\Rightarrow P &=& f(\varepsilon) \qquad\qquad\qquad\qquad\qquad\qquad \text{(equation of state)} \index{Equation of state}
\end{eqnarray}
All particle masses equal to $m$:
\begin{eqnarray*}
\varepsilon=\varepsilon_\mathrm m &=& \int\!\rho_\mathrm m (p,x)p^0\ \dd^3p\\
P=P_\mathrm m &=& \frac{1}{3}\int\!\rho_\mathrm m (p,x)\frac{|\vec p|^2}{p^0}\ \dd^3p \qquad\qquad\text{(in normal coordinates)}
\end{eqnarray*}
$\rho_\mathrm m$: density of particles with momentum $p$ ($\Rightarrow \int\dd^3p\rho_\mathrm m(p,x)$ is the particle density $n_\mathrm m$ itself)

\subsubsection{Example:}
Replace $\rho_\mathrm m$ by a smooth particle distribution\\
Fermions (spin $\frac{1}{2}$) with mass $m$ confined to a box (lenght $L$)\\
quantum-mechanical wavefunction: (in units $\hbar=c=1$)
\[\Psi_k(y)=\dfrac{e^{i k_\mu y^\mu}}{\sqrt{L^3}} \qquad\qquad \vec k=\frac{2\pi}{L}\left(\begin{array}{c}n_1\\n_2\\n_3\end{array}\right) \qquad n_i\in\mathbb Z \qquad\qquad k^0=(\vec k^2+m^2)^\frac{1}{2}\]
The probability $\rho$ of finding a particle with momentum $p$ at $y$ is
\begin{eqnarray*}
\rho &=& \sum_i\delta^3(\vec p-\vec k_i)|\Psi_{k_i}(y)|^2=\sum_i\dfrac{\delta^3(\vec p-\vec k_i)}{L^3}\\
&=& \mathop{\sum_{k_i}}_{|k|^2<k_F{}^2}\dfrac{\delta^3(\vec p-\vec k_i)}{L^3} \,\qquad\qquad\qquad k_F \text{ fermi momentum for groundstate}\\
&\cong& \frac{1}{(2\pi)^3}\int_{|k|<k_F}\dd^3k\ \delta^3(\vec p-\vec k_i) \qquad\qquad \text{(converting sum to integral)}
\end{eqnarray*}

\subsubsection{Result:}
\[\rho(|k^2|)=\frac{1}{(2\pi)^3}\ \Theta(k_F-|\vec p|)\,s \qquad\quad \Theta(x)=\begin{cases}1,\ x>0\\0,\ x<0\end{cases}\quad s\text{ is the number of spin states}\]
\smallskip

for spin $\displaystyle \frac{1}{2}: \quad \rho(|k|^2)=\frac{2}{(2\pi)^3}\,\Theta(k_F-|\vec p|)$
\begin{eqnarray}
\varepsilon_\mathrm m &=& \frac{2}{(2\pi)^3}\int_{|p|<k_F}\!\!\!\!\!\!\!\!\!\!\!\!\dd^3p\ (p^2+m^2)^\frac{1}{2}=\frac{1}{\pi^2}\int_0^{k_F}(p^2+m^2)^\frac{1}{2}p^2\dd p=\varepsilon(k_F)  \\
P_\mathrm m &=& \frac{1}{3}\frac{1}{\pi^2}\int_0^{k_F}\!\!\!\!\!\dfrac{p^4 \dd p}{(p^2+m^2)^\frac{1}{2}}=P(k_F)  \\
\Rightarrow n_\mathrm m &=& \frac{1}{3\pi^2}k_F^3 
\end{eqnarray}
\subsubsection{Remark:}
Suppose the body consists of particle with $m=m_i$
\[\varepsilon=\sum_i\varepsilon_{m_i} \qquad\qquad P=\sum_i P_{m_i}\]
\smallskip

So far $T_\mathrm m^{00}(y)=\varepsilon \qquad\qquad T_\mathrm m^{0i}=0 \qquad\qquad T_\mathrm m^{ij}=P\delta^{ij}$
\[\Rightarrow T_\mathrm m^{\mu\nu}=(\varepsilon+P)j^\mu j^\nu-P\eta^{\mu\nu} \qquad\qquad j^0=1,\ j^i=0\]
Conclusion: 
\[T_\mathrm m^{\mu\nu}(x)=\Bigl(\varepsilon(x)+P(x)j^\mu(x)j^\nu(x)-P(x)g^{\mu\nu}(x)\Bigr)\]
holds in arbitrary coordinates $x$. $j^\mu$ however is only restricted by $g_{\mu\nu}(x)j^\mu(x)j^\nu(x)=1$\bigskip

The form of the energy-momentum tensor of a massive body is \index{Energy-momentum tensor!- of a massive body}\index{Energy-momentum tensor}
\[T_\mathrm m^{\mu\nu}(x)=\Bigl(\varepsilon(x)+P(x)\Bigr)j^\mu(x)j^\nu(x)-g^{\mu\nu}(x)P(x)\]
$\varepsilon, P$ energy density and pressure functions which must supplemented by an equation of state $P=f(\varepsilon)$ (externally given) \smallskip\\
In addition $T_\mathrm m^{\mu\nu}{}_{;\nu}=0$

\chapter[Static spherically symmetric solutions]{Static (time-independent) spherically symmetric solutions of Einstein's equation} %bessere Überschrift
\index{Solution of Einstein's equation!- static spherically symmetric}
Coordinates $\quad x^0=t; \qquad x^1=r; \qquad x^2=\vartheta; \qquad x^3=\varphi$
\[\Rightarrow g_{00}=1; \qquad g_{11}=-1; \qquad g_{22}=-r^2; \qquad g_{33}=-r^2\sin^2\vartheta\]
\\
Generalized Ansatz for Einstein's equation:
\[g_{00}\to g_{00}=\mathrm e^{\nu(r)}, \quad g_{11}\to -\mathrm e^{\lambda(r)}; \qquad j^i=0 \qquad i=1,2,3\]
\[\Rightarrow j^0=\mathrm e^{-\frac{\nu}{2}}; \quad \varepsilon=\varepsilon(r), \quad P=P(r);\quad T^{\mu\nu}{}_{;\nu}=0; \quad P=P(\varepsilon); \quad \nu,\lambda \text{: functions}\]
\\
Christoffel symbols (nonvanishing components only):
\[ \begin{array}{llll}
\Gamma^1_{11}=\frac{\lambda}{2},\quad & \Gamma^0_{10}=\frac{\nu'}{2},\quad & \Gamma^2_{33}=-\sin\vartheta\cos\vartheta,\quad & \Gamma^1_{22}=-r\mathrm e^{-\lambda} \medskip\\
\Gamma^1_{00}=\frac{\nu}{2}\,\mathrm e^{\nu-\lambda},\quad & \Gamma^2_{12}=\Gamma^3_{13}=\frac{1}{r},\quad & \Gamma^2_{23}=\cot\vartheta,\quad & \Gamma^1_{33}=-r\sin^2\vartheta\mathrm e^{-\lambda}
\end{array} \]
\[+\text{ expressions arising from }\Gamma^\gamma_{\alpha\beta}=\Gamma^\gamma_{\beta\alpha}\]
\\
Einstein tensor (nonvanishing components only):
\[G_0^0=-\mathrm e^{-\lambda}\left(\frac{1}{r^2}-\frac{\lambda'}{r}\right)+\frac{1}{r^2};\qquad G_1^1=-\mathrm e^{-\lambda}\left(\frac{\nu'}{r}+\frac{1}{r^2}\right)+\frac{1}{r^2};\qquad\]
\[G_2^2=G_3^3=-\frac{1}{2}\mathrm e^{-\lambda}\left(\nu''+\frac{\nu'}{2}+\frac{\nu'-\lambda'}{2}-\frac{\nu'\lambda'}{2}\right)\]
\\
energy momentum tensor (nonvanishing components only):
\[T_0^0=\varepsilon;\qquad T_1^1=T_2^2=T_3^3=-P\]
\begin{equation}
0=T^{\mu\nu}{}_{;\nu}\quad\Rightarrow\quad\fbox{$P'+\frac{\nu'}{2}(\varepsilon+P)=0$} \label{eq:pprime}
\end{equation}
\[G_\beta^\alpha=-8\pi G T_\beta^\alpha \qquad\qquad \text{(3 equations)} +\ \eqref{eq:pprime} \]
\begin{equation}
\fbox{$G_0^0=8\pi G\varepsilon; \qquad G_1^1=-8\pi G P; \qquad +\eqref{eq:pprime}$} 
\end{equation}
\begin{itemize}
\item only equations to be solved together with $P=P(\varepsilon)$
\item imply $\quad G_2^2=-8\pi G T_2^2$
\end{itemize}

\section{Solving the equations}
\begin{align}
G^0_0 =\ & \fbox{$\displaystyle -\e^{-\lambda} \left(\frac{1}{r^2} - \frac{\lambda'}{r}\right) + \frac{1}{r^2} = 8\pi G \varepsilon$} \tag{a}\label{eq:einst_a}\\
G^1_1 =\ & \fbox{$\displaystyle -\e^{-\lambda} \left(\frac{\nu'}{r} + \frac{1}{r^2}\right) + \frac{1}{r^2} = -8\pi GP$} \tag{b}\label{eq:einst_b}\\
& \fbox{$\displaystyle \left(\varepsilon + P\right) \frac{\nu'}{2} + P' = 0$}  \tag{c}\label{eq:einst_c}
\end{align}

\begin{enumerate}
\item[(a)] $\cdot\,r^2: -\dfrac{\dd}{\dd r}\,\e^{-\lambda} r + 1 = 8\pi G\varepsilon r^2$
\begin{align*}
\Rightarrow \e^{-\lambda} = 1 - \frac{2 GM(r)}{r} \qquad\qquad&\text{where } M(r) = 4\pi \int_0^r {r'}^2 \dd r' \varepsilon(r') \\
& (\text{energy content of a ball with radius } r)
\end{align*}
\[\Rightarrow g_{11} = -\frac{1}{1 - \frac{2GM(r)}{r}} \]
\item[(b)] $\dfrac{\nu'}{r} = \left(8\pi GP + \dfrac{1}{r^2}\right) \e^\lambda - \dfrac{1}{r^2}$
\[\Rightarrow e^\nu = \exp \left( - \int\limits_r^\infty \dd r'\ \frac{1}{r'} - \frac{\left(8\pi G r' P + \frac{1}{r'}\right)}{1 - \frac{2GM(r')}{r'}}  \right) \]
\[\Rightarrow g_{00} = \e^\nu \qquad \text{ is expressed again by } \varepsilon(r) \quad\text{(the energey density)}\]
If the equation \eqref{eq:einst_c} is solved, we have a complte solution for $\varepsilon, \nu, \lambda$
\item[(c)] 1. Remark: There is always a trivial solution $\varepsilon = P = 0$ \\
2. The nontrivial solution is given by use of the equation of state $P = P(\varepsilon)$
\[ \Rightarrow\ \fbox{$P' = f(\varepsilon) \cdot \varepsilon'$}\  (*)\qquad\qquad f(\varepsilon) = \frac{\partial P}{\partial\varepsilon}\ \bigl(\varepsilon(r)\bigr) \]
\end{enumerate}
Use the equation \eqref{eq:einst_c} together wirth $(*)$, the equation \eqref{eq:einst_b}, and the expression for $\lambda$ to optain an expression for $\varepsilon'$
\begin{eqnarray}
\lefteqn{\varepsilon' = -\frac{r}{2f\bigl(\varepsilon(r)\bigr)}\ \Bigl( \varepsilon(r) + P\bigl(\varepsilon(r)\bigr)\Bigr) \left(\frac{P\bigl(\varepsilon(r)\bigr)}{1 - \frac{2GM(r)}{r}} + \frac{1}{r^3}\ \frac{2GM(r)}{1 - \frac{2GM(r)}{r}} \right) < 0} \label{eq:solution1}\\
\lefteqn{\frac{\dd M}{\dd r} = 4\pi r^2 \varepsilon}\label{eq:solution2}
\end{eqnarray}
\eqref{eq:solution1} and \eqref{eq:solution2} constitue a system of first order differential equations for $\varepsilon$ and $M$. The Solution is unique once you have specified initial values $\varepsilon(0)$ (central energy density) together with $M(0) = 0$.
\[\Rightarrow \text{solution depends only on }\varepsilon(0) \]
A general solution for $\varepsilon$ with realistic equations of state is only obtained numerically. Still some analytical results of general nature can be obtained: the equation for $\varepsilon'$ \eqref{eq:solution1} shows that $\varepsilon(r)$ will decrease for any initial value $\varepsilon(r)$. If $\varepsilon(r)$ becomes zero at $R$, continue according to ${T^{\mu\nu}}_{;\nu} = 0$ with the trivial solution $\varepsilon = P = 0$ \bigskip

\includegraphics[height=40mm]{ed} \bigskip

The mass of the body: $\displaystyle M(R) = 4\pi \int\limits_0^R {r'}^2 \dd r' \varepsilon(r')$ \\
$\Rightarrow M$ (mass of the body) and $R$ (radius of the body) are \emph{not} independent!
\[M = M\bigl(\varepsilon(0)\bigr),\qquad\quad R = R\bigl(\varepsilon(0)\bigr);\qquad\quad \varepsilon(0) \text{ central energy density} \]
$\Rightarrow$ point particle ($M$ fixed, $R \to 0$) is not possible (in a straight-forward way) \bigskip

\includegraphics[height=55mm]{limit1}\bigskip

First observation of such limits for white dwarves (composed of iron): Chandrasekhar limit \index{Chandrasekhar limit} \pagebreak[1]

If inverse $\beta$-decay is allowed: \medskip\\
\includegraphics[height=55mm]{limit2} \index{Oppenheimer-Volkov limit}\\
Details only by numerical computation

\section{The Schwarzschild exterior solution} \index{Schwarzschild exterior solution}
\[g_{11}(r) = -\e^\lambda = -\left(1 - \dfrac{2GM(r)}{r} \right)^{-1} \qquad M(r) = 4\pi \int\limits_0^r \dd r' {r'}^2 \varepsilon(r') \]
\subsubsection*{For $r > R$:}
\begin{eqnarray*}
M(r) &\!\!=\!\!& M + \underbrace{4\pi \int\limits_R^r \dd r' {r'}^2 \varepsilon(r')}_{= 0} = M \qquad\qquad\quad\ \ M := M(R)\ \text{mass of the star} \\
g_{11} &\!\!=\!\!& -\left( 1 - \frac{2GM}{r}\right)^{-1} = \left(1 - \frac{r_g}{r} \right)^{-1} \qquad\quad r_g = 2GM\ \text{Schwarzschild radius} \index{Schwarzschild radius} \\
e^\nu &\!\!=\!\!& \exp \left(- \int\limits_r^\infty \dd r' \left(\frac{1}{r'} - \frac{\frac{1}{r'}}{1 - \frac{r_g}{r}} \right) \right) = \exp \left(- \int\limits_r^\infty \dd r' \left(\frac{1}{r'} - \frac{1}{r' - r_g} \right) \right) \\
&\!\!=\!\!& \exp \left( \ln \frac{r - r_g}{r} \right) = 1 - \frac{r_g}{r}
\end{eqnarray*}
\subsubsection*{Schwarzschild exterior solution: ($r > R > r_g$)}
\begin{equation}
g_{00} = \left(1 - \frac{r_g}{r}\right), \qquad g_{11} = -\left( 1 - \frac{r_g}{r}\right)^{-1}, \qquad g_{22} = -r^2, \qquad g_{33} = -r^2 \sin^2\vartheta \label{eq:scharzschild}
\end{equation}
\begin{tabular}{lp{10cm}}
$r \to \infty : \frac{r_g}{r} \to 0 \quad  \Rightarrow$ & metric approaches the value of the Minkowski metric in polar coordinates
\end{tabular} \\
Note: $R > r_g$ in the full solution

\subsection{Testing the solution}
Testing the Schwarzschild exterior solution by studying the motion of point particles outside the star: \\
equations of motion: geodesic equation
\[ \frac{\dd}{\dd \tau}\ g_{\mu\nu}\ \dot x^\nu = \frac{1}{2}\, \del{\mu}\ g_{\alpha\beta}\ \dot x^\alpha \dot x^\beta \qquad\qquad\quad x^0 = t,\quad x^1 = r,\quad x^2 = \vartheta,\quad x^3 = \varphi\]
$g_{\mu\nu}$ given by \eqref{eq:scharzschild} and with $g_{\alpha\beta}\ \dot x^\alpha \dot x^\beta = 1$
\[ \begin{array}{ll}
\dfrac{\dd}{\dd t} \left( 1 - \dfrac{r_g}{r}\right) \dot x^0 = 0 \qquad\qquad& (\mu = 0) \medskip\\
\dfrac{\dd}{\dd \tau}\ r^2\,\dot\vartheta = r^2\,\cos\vartheta\,\sin\vartheta\ \dot\varphi^2 \qquad\qquad& (\mu = 2) \medskip\\
\dfrac{\dd}{\dd \tau}\ r^2\,\sin^2\vartheta\ \dot\varphi = 0 \qquad\qquad& (\mu = 3) \medskip\\
1 = \left(1 - \dfrac{r_g}{r}\right)\left(\dot x^0\right)^2 - \left(1 - \dfrac{r_g}{r}\right)^{-1} \dot r^2 - r\, \dot\vartheta^2 - r^2\,\sin^2\vartheta\ \dot\varphi^2
\end{array} \]
without loss of generality: $\vartheta = \frac{\pi}{2}$
\[\Rightarrow \dot\varphi = \dfrac{l}{r^2} \qquad\qquad \dot x^0 = \dfrac{a}{1 - \frac{r_g}{r}} \qquad\qquad a,l \text{ constant}\]
\begin{equation}
\Rightarrow \left( a^2 - 1\right) - \dot r^2\ \dfrac{l^2}{r^2} \underbrace{\left(1 - \dfrac{r_g}{r}\right)}_{(*)} + \dfrac{r_g}{r} = 0
\label{eq:schwarzschild}
\end{equation}
Except for the term $(*)$, this is conservation of energy in the Newtonian Kepler problem.\bigskip

Integration constants $l$, $a$:
\begin{eqnarray}
l &\!\!=\!\!& \frac{L}{m} \qquad\qquad l:\text{ angular momentum} \qquad m:\text{particle mass} \nonumber \\
a &\!\!=\!\!& \left(1-\frac{r_g}{r}\right) \dot x^0 \label{eq:schw_a}
\end{eqnarray}
\[\eqref{eq:schwarzschild}\Rightarrow \underbrace{\frac{m}{2} \left(a^2 - 1\right)}_{E_\mathrm{part}} - \underbrace{\frac{m}{2} \left(\dot r^2 + \frac{l^2}{r^2}\right)}_{T_\mathrm{kin}} + \underbrace{\frac{m}{2}\,\frac{l^2}{r^3}\,r_g}_\text{small correction} + \underbrace{\frac{GmM}{r}}_\text{pot. energy} = 0 \]
\[\Rightarrow a^2 = \dfrac{2 E_\mathrm{part}}{m} + 1 \stackrel{\text{non rel.}}{\approx} \left(1 + \dfrac{E_\mathrm{part}}{m}\right)^2 \qquad\qquad \text{non rel.: }\dfrac{E_\mathrm{part}}{m} \ll 1 \]
\[\Rightarrow a = 1 + \dfrac{E_\mathrm{part}}{m} \qquad \stackrel{\eqref{eq:schw_a}}{\Longrightarrow} \underbrace{\left(1 - \dfrac{r_g}{r}\right)}_{\approx 1} m\dot x^0 = m + E_\mathrm{part} \qquad\text{(non relativistically!)} \]
\[\Rightarrow p^0 = m\dot x^0 = m + E_\mathrm{part} \qquad\qquad p^0: \text{rel. energy} \]

Classical way of solving: express $r(\tau)$ = $r\bigl(\varphi(\tau)\bigr)$ and introduce the function $u = \frac{1}{r}$ instead of $r$.
\[\dot u = -\frac{1}{r^2}\,\dot r = -\frac{1}{r^2}\,r'\dot\varphi =  -\frac{1}{r^2}\,r'\frac{l}{r^2} = -\frac{l}{r^4}\,r' \qquad\quad('=\text{derivative with respect to }\varphi)\]
Insert into \eqref{eq:schwarzschild}. Result:
\begin{equation}
a^2 - 1 - l^2{u'}^2 - l^2 u^2 \left( 1 - r_g u\right) + r_g u = 0
\end{equation}
Differentiate once again with respect to $\varphi$ and divide by $u' l^2$. Result:
\begin{equation}
-u'' - u \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \underbrace{+ \frac{3}{2}\, r_gu^2}_{\text{new correction due to Einstein}} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! + \frac{r_g}{2l^2} = 0 \label{eq:schw_u}
\end{equation}
Non relativistically (i.e. without the correction):
\begin{equation}
-u'' - u + \frac{r_g}{2l^2} = 0
\end{equation}
\[\leadsto u = A \cos \varphi + \dfrac{r_g}{2l^2} \qquad \text{yields Kepler ellipses, motion is always circular} \]
\begin{tabular}{lll}
\!\!\!Outside the special case: & $u = \dfrac{r_g}{2l^2} + \Delta u$ & $\Delta u$ small \smallskip\\
\!\!\!use the approximation & $u^2 = \left( \dfrac{r_g}{2l^2} \right)^2 + 2\dfrac{r_g}{2l^2}\, \Delta u\quad$ & insert this into \eqref{eq:schw_u}:
\end{tabular}
\[ -\Delta u'' - \Delta u + \frac{3}{2}\,r_g\,2\beta\,\Delta u + \frac{3}{2}\, r_g\,\beta^2 = 0 \qquad\qquad \beta = \frac{r_g}{2l^2}\]
general solution:
\begin{eqnarray}
\Delta u &\!\!=\!\!& A \cdot \cos \Bigl( \left(1 - 2\alpha\right)^\frac{1}{2} \varphi\Bigr) - \beta\frac{\alpha}{1 - 2\alpha} \qquad\qquad \alpha = \frac{3}{2}\,r_g\,\beta \nonumber \\
u &\!\!=\!\!& A \cdot \cos \Bigl( \underbrace{\left(1 - 2\alpha\right)^\frac{1}{2}}_{\approx \left(1 - \alpha\right)} \varphi\Bigr) + \beta \underbrace{\left(1 - \frac{\alpha}{1 - \alpha}\right)}_{\approx \left(1 - \alpha\right)} \qquad \qquad \alpha \ll 1
\end{eqnarray}

\subsubsection*{Geometrical properties of the particle trajectories:}
\[u = \frac{1}{r} = a \cdot \cos \bigl( \left(1 - \alpha\right) \varphi \bigr) + \beta \left(1 - \alpha\right) \]
\begin{center}
\includegraphics[height=40mm]{perih}
\end{center}
The perihelion position will be reached again when $\left(1 - \alpha\right) \varphi = 2\pi$. \\
$\Delta \varphi$ (the perihelion precession angle) is found by \index{Perihelion precession}
\[ \left(1 - \alpha\right) \left(2\pi + \Delta \varphi\right) = 2\pi \]
\[ \Rightarrow \Delta \varphi = 2\pi \left(-1 + \frac{1}{1 - \alpha}\right) = 2\pi\alpha \qquad\qquad \alpha = \frac{3}{2}\,r_g\,\beta;\qquad \beta = \frac{r_g}{2l^2}; \qquad l = \frac{L}{m} \]
All quantities are measured in astronomy, e.g. for the planet mercury: \\
Mercury revolves $x$ times per century. $x\Delta\varphi$ is then found theoretically:
\[x\Delta\varphi = 43.02\ \unit{arcsec} \qquad\qquad\text{(precession in 100 years)} \]
From empirical observation, the value 42.46 arcsec in 100 years was found by Leverrier in 1859. \index{Leverrier}

\section{Deflection of light rays} \index{Deflection of light rays}
\begin{description}
\item[Assumption:] Geometrical optics work $\Rightarrow$ Electrodynamics : light rays are given by null geodesics, i.e. by geodesic curves with $g_{\mu\nu}\,\dot x^\mu\dot x^\nu = 0$.
\item[geodesic equation:] $\left( 1 - \dfrac{r_g}{r}\right) \dot x^0 = a \quad$ for a solution in plane with $\vartheta = \dfrac{\pi}{2}$
\begin{eqnarray}
& &r^2 \dot \varphi = l \qquad\qquad\qquad g_{\mu\nu}\,\dot x^\mu \dot x^\nu = 0 \nonumber \\
& &\left(1 - \frac{r_g}{r}\right) \left(\dot x^0\right)^2 - \frac{\dot r^2}{1 - \frac{r_g}{r}} - r^2 \dot\varphi^2 = 0 \label{eq:light_defl_1}\\
& & a^2 - \dot r^2 - \frac{l^2}{r^2} \Bigl(1 - \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \underbrace{\frac{r_g}{r}}_{\text{ general relativistic correction}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \Bigr) = 0 \qquad\qquad\qquad 0 < r_g \le R < r \nonumber
\end{eqnarray}
\end{description}
Write $r(\tau) = r\bigl(\varphi(\tau)\bigr)$, use $u = \dfrac{1}{r}$, $\dot r = r' \dot\varphi$. For $u$, eq. \eqref{eq:light_defl_1} yields a diff. equation
\begin{equation}
u'' + u \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\underbrace{- \frac{3}{2}\,r_g\, u^2}_\text{general relativistic correction} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!= 0 \label{eq:light_defl_2}
\end{equation}
\begin{enumerate}
\item neglecting the correction results in the solution $u = A \cdot \cos \varphi = \dfrac{1}{r}$\\ (straight line in polar coordinates with $A$: constant of integration)
\item approximate solution which takes the correction into account: \\
we write $u = A \cdot \cos \varphi + \Delta u \qquad\qquad\qquad \Delta u$: correction \\
Approximately $u^2 = A^2 \cdot \cos^2 \varphi$, valid if $\frac{r_g}{R_0}\ll 1$. Put $A = \frac{1}{R_0}$. Insertion into \eqref{eq:light_defl_2} yields diff. equation for $\Delta u$ with the solution $\Delta u = \frac{r_g}{2R_0^2}\left( \cos^2\varphi + 2\sin^2\varphi\right)$
\begin{equation}
\frac{1}{r} = u + \Delta u = \frac{1}{R_0}\,\cos \varphi + \frac{r_g}{2R_0^2}\left(\cos^2\varphi + 2\sin^2\varphi\right) \label{eq:light_defl_3}
\end{equation}
\begin{picture}(350,60)
\put(0,55){\line(1,0){150}}
\multiput(75,4)(0,4){13}{\line(0,1){2}}
\qbezier(0,50)(75,60)(150,50)
\put(170,7){deflection angles computed from \eqref{eq:light_defl_3}.}
\put(60,24){$R_0$}
\put(182,41){\vector(-4,1){30}}
\put(184,38){\begin{footnotesize}true curve\end{footnotesize}}
\linethickness{1pt}
\put(0,4){\line(1,0){150}}
\end{picture}

\end{enumerate}

Limiting angles $\varphi$ for which $r=\infty$ or $u=0: \varphi=\pm \left(\frac{\pi}{2}+\Delta\varphi\right)$. Deflection angle is $2\Delta\varphi$
\begin{equation}
u(0)=\frac{1}{R_0}\Bigl(\cos\left(\frac{\pi}{2}+\Delta\varphi\right)\Bigr)+\frac{r_g}{2R_0}\left(\cos^2\left(\frac{\pi}{2}+\Delta\varphi\right)+2\sin^2\left(\frac{\pi}{2}+\Delta\varphi\right)\right)
\end{equation}
quadratic equation for $\cos(\frac{\pi}{2}+\Delta\varphi)$
\[\cos\left(\frac{\pi}{2}+\Delta\varphi\right)=\frac{R_0}{r_g}-\frac{R_0}{r_g}\left(1+\left(\frac{r_g}{R_0}\right)^2\right)^\frac{1}{2}\approx-\frac{r_g}{R_0} \qquad\qquad \frac{r_g}{R_0} \ll 1 \]
\[\Rightarrow \Delta\varphi=\frac{r_g}{R_0} \qquad\qquad\qquad \text{Angle of deflection:}\quad 2\cdot\frac{r_g}{R_0}\]
Einstein computed the deflection angle in 1915: result for the sun $1.75''$.\\
Experiment in 1919: $1.98''$ and $1.6''$

\chapter{Black holes}
\section{Black holes and worm holes} \index{Black holes} \index{Worm holes}
How serious is the Schwarzschild singularity in the mertic at $r=r_g$?\\
In our calculation. $R$ (star radius) was always larger than $r_g$. Before the metrical coefficients $(1-\frac{r_g}{r})$ become $0$, the metric is already replaced by the completely regular interior Schwarzschild metric.\\
Suppose the exterior solution is not obtained our way, but by some unknown time-dependent process (collapse)\\
$\Rightarrow$ The exterior Schwarzschild solution
\[g_{00}=1-\frac{r_g}{r}; \quad g_{11}=\left(1-\frac{r_g}{r}\right)^{-1}; \quad g_{22}=-r^2; \quad g_{33}=-r^2\sin^2\vartheta\]
will then still describe the exterior part of the metric for $\varepsilon=P=0 \quad (T^{\mu\nu}=0)$, but only for $r>r_g$\\
How to describe the interior part $r<r_g$?

\section{The Kruskal metric} \index{Kruskal metric}
Consider the so-called Kruskal metric:
\[\text{Coordinates}\qquad y^0=v,\qquad y^1=u,\qquad y^2=\vartheta,\qquad y^3=\varphi\]
\[\text{metric}\qquad g'_{00}=-g'_{11}=\frac{4r_g^3}{r}\,\e^\frac{r}{r_g}; \qquad g'_{22}=-r^2; \qquad g'_{33}=-r^2\sin^2\vartheta \qquad\qquad r=r(u,v)\]
with $r(u,v)$ given by the condition $h(r)=\left(\frac{r}{r_g}-1\right)\e^\frac{r}{r_g}=u^2-v^2$\\ $\Bigl($i.e. $r=h^{-1}(u^2-v^2)\Bigr)\qquad$ domain of definition $u^2-v^2>-1$\\
At $u^2-v^2=-1 \Rightarrow r=0 \qquad\qquad u^2-v^2=0 \Rightarrow r=r_g$ \bigskip
\begin{center}
\includegraphics[height=45mm]{kruskal1}
\label{kruskal}
\end{center}
Consider the blue domain in figure \ref{kruskal}. Claim: There is a diffeomorphism $f$ such that the blue domain is mapped on the Schwarzschild domain with property that the Schwarzschild domain pulled back with the help of this map to the blue domain yields exactly the Kruskal metric in this domain.\\
What does $f$ look like?
\[y^2=x^2=\vartheta; \qquad y^3=x^3=\varphi; \qquad x^1=r=r(u,v); \qquad x^0=2r_g\mathrm{artanh}\left(\frac{v}{u}\right)\]
Light rays and particle trajectories in the Kruskal describtion.\\
For both we have: $g'_{\mu\nu}\dot y^\mu\dot y^\nu=c\qquad c=1$ for particles, $c=0$ for light rays
\begin{eqnarray*}
\Rightarrow \dot v^2-\dot u^2 &\!\!-\!\!& \frac{1}{g(r)}\left(r^2\dot\vartheta^2+r^2\sin^2\vartheta\dot\varphi^2+c\right)=0 \qquad\qquad g(r)=\frac{4r_g^3}{r}\,\e^{-\frac{r}{r_g}}\\
\dot v^2 &\!\!=\!\!& \dot u^2+\frac{1}{g(r)}\left(r^2\dot\vartheta^2+r^2\sin^2\vartheta\dot\varphi^2+c\right)\\
v(\tau)-v(0) &\!\!=\!\!& \int\limits_0^\tau\dd\tau'\Biggl(\dot u^2+\underbrace{\frac{1}{g(r)}\left(r^2\dot\vartheta^2+r^2\sin^2\vartheta\dot\varphi^2+c\right)}_{(*)}\Biggr)^\frac{1}{2}\qquad\qquad \text{neglect }(*)\\
v(\tau)-v(0) &\!\!\geq\!\!& \int\limits_0^\tau\!\! |\dot u|\,\dd\tau'=\pm\int\limits_0^\tau\!\! \dot u(\tau')\,\dd\tau'=\pm\bigl(u(\tau)-u(0)\bigr)
\end{eqnarray*}
\begin{center}
\includegraphics[height=62mm]{kruskal2a}
\includegraphics[height=62mm]{kruskal2b}
\end{center}
\begin{enumerate}
\item[I] particle can escape to infinity or hit the $r=0$ point; no signals reach the new domains of the Kruskal-world
\item[II] particle will always hit the singularity; no signals reach the new domains of the Kruskal-world; no signals can reach back the exterior Schwarzschild domain
\item[III] signals can reach the Schwarzschild domain
\end{enumerate}

\subsubsection*{Remark:}
The Kruskal metric satisfies everywhere $G^{\mu\nu}=0$ (vanishing energy-momentum tensor)\\
$\Rightarrow R_{\mu\nu}=0; \qquad R=0 \qquad$ But $R_{\alpha\beta\mu\nu}\neq0$\\
By computation: $R_{\alpha\beta\gamma\delta}R^{\alpha\beta\gamma\delta}=\dfrac{12r_g^2}{r^6}$
$\Rightarrow$Kruskal manifold is the maximal extension of the Schwarzschild exterior solution with $r>r_g$\\
There is another non maximal solution of the extension problem!

\subsubsection*{Eddington's solution:} \index{Eddington's solution}
coordinates: $y^0=t',\ \ y^1,\ y^2,\ y^3 \quad\qquad y=(y^0,y^1,y^2,y^3)\quad $ cartesian coordinates
\[g'_{\mu\nu}=\eta_{\mu\nu}+A_\mu A_\nu; \qquad A_0=\left(\frac{r_g}{r}\right)^\frac{1}{2}; \qquad A_i=\mp y^i\,\dfrac{r_g^\frac{1}{2}}{r^\frac{3}{2}},\quad (i=1,2,3)\]
\[\text{maps}\qquad \vec y=\left(\begin{array}{c} \sin\vartheta\sin\varphi\\ \sin\vartheta\cos\varphi\\ \cos\vartheta \end{array}\right)\qquad\qquad y^0=x^0\pm\ln\left(\frac{r}{r_g}-1\right)\qquad x^0:\text{ Schwarzschild time}\]
establish an isometric of the 2 solutions.
\begin{description}
\item[Remark:]
But: for both extensions $(\pm)$ the Eddington solution is not invariant under time-reversal (still favours the Kruskal solution)
\item[Question:]
Can a black hole be formed by a time dependent process (collapse)? Is there a \emph{simple} model for a collapsing cloud of particles?
\item[Model:]
Tolman-Oppenheimer solution for a spherical collapse of a dust cloud ($P=0$, vanishing pressure) \index{Tolman-Oppenheimer solution}
\end{description}

\section{Spherical collapse of a massive body} \index{Collapse}
Ansatz for the metric in comoving coordinates ($x^0 = t,\ \ x^1 = r,\ \ x^2 = \vartheta ,\ \ x^3 = \varphi$):
\[g_{00} = 1;\qquad g_{11} = -R(t)^2 f(r);\qquad g_{22} = -R(t)^2r^2; \qquad g_{33} = -R(t)^2 r^2\sin^2\vartheta \]
(other components of $g_{\mu\nu}$ vanish) \medskip\\
Only the interior part will be described $\qquad\Rightarrow r\le r_0$ \medskip\\
Boundary conditions: The collapse will start at $t = 0$ with $R(0) = 1,\ \dot R(0) = 0$. \medskip\\
The energy momentum tensor has the form:
\[T^{\mu\nu} = \left(\varepsilon + P\right) j^\mu j^\nu - P g^{\mu\nu}\qquad\qquad g_{\mu\nu} j^\mu j^\nu = 1 \]
In our case we can simply put $\quad j^0 = 1, \qquad j^i = 0 \qquad (i = 1,2,3)$ \bigskip\\
ideal dust (assumption): $\quad P = 0$
\begin{eqnarray*}
&\!\!\Rightarrow\!\!& {T^{\mu\nu}}_{;\nu} = 0 \qquad\text{collapses to 1 equation:} \\
& & - \frac{\dd}{\dd t}\,\varepsilon R^3 = 0 \qquad\text{(conservation of energy)} \\
&\!\!\Rightarrow\!\!& \varepsilon R^3 = a \qquad \Rightarrow \varepsilon = \frac{a}{R^3}, \quad a = \varepsilon(0) \\
&\!\!\Rightarrow\!\!& \varepsilon(t) = \frac{\varepsilon(0)}{R(t)^3}
\end{eqnarray*}
\begin{eqnarray}
G^1_1 &=& \frac{1}{R^2} \left(\frac{1}{r^2} - \frac{1}{r^2 f}\right) + 2\left(\frac{\dd}{\dd t} \frac{\dot R}{R} + \frac{3\dot R^2}{R^2}\right) = 0 \label{eq:collapse1}\\
G^0_0 &=& - \frac{1}{R^2f} \left(\frac{1}{r^2} - \frac{f'}{rf}\right) + \frac{1}{R^2 r^2} + \frac{3\dot R^2}{R^2} = 8\pi G\,\varepsilon \label{eq:collapse2}
\end{eqnarray}
\begin{eqnarray*}
\eqref{eq:collapse1} &\Rightarrow& \frac{1}{r^2 f} - \frac{1}{r^2} = -a \qquad\qquad (a: \text{new constant}) \\
&\Rightarrow& f = \frac{1}{1 - ar^2}
\end{eqnarray*}
insert this into \eqref{eq:collapse2}:
\begin{equation}
\Rightarrow \frac{3\dot R^2}{R^2} + \frac{3a}{R^2} =  8\pi G\,\varepsilon
\end{equation}
\[ \left.\begin{array}{l}
R(0) = 1 \\ \dot R(0) = 0
\end{array} \right\} \stackrel{t=0}{\Longrightarrow} \fbox{$a = \dfrac{8\pi G}{3}\, \varepsilon(0) $} \]
\[\Rightarrow \frac{3\dot R^2}{R^2} + \frac{8\pi G\, \varepsilon(0)}{R^2} - \frac{8\pi G\,\varepsilon(0)}{R^3} = 0 \]

Parametric solution: $\quad t = t(\Psi), \qquad R = R(\Psi)$
\[\Rightarrow t = \frac{\Psi + \sin \Psi}{2\sqrt{a}} \qquad\qquad R = \frac{1}{2} \left(1 + \cos \Psi\right) \]
To check this: compute $\quad \dfrac{\dd}{\dd t} R(\Psi) = \dfrac{\frac{\dd}{\dd\Psi} R(\Psi)}{\frac{\dd}{\dd\Psi} t(\Psi)}$ \bigskip\\
\begin{tabular}{lll}
$\Psi = 0$ & $\Rightarrow R = 1\quad$ & beginning of the collapse \medskip\\
$\Psi = \pi$ & $\Rightarrow R = 0\quad$ & $\Rightarrow t(\pi)$ (time for the collapse) fulfills $\ t(\pi) = T = \dfrac{\pi}{2\sqrt{a}}$ \\
& & collapse will be completed in a finite amount of time
\end{tabular} \bigskip

$\Rightarrow$ Range of the coordinates:
\[0 \le r \le r_0 \qquad r_0:\text{radius of the body in comoving coordinates} \]
\[\Rightarrow r_0^2 < \frac{1}{a} \qquad\text{(in order that the metric makes sense)}\]
Range of the allowed values of $t$:
\[0 \le t < T \qquad T:\text{lifetime of the collapsing body}\]

\subsubsection*{Deficit of this construction:}
Nothing is said about the history before the collapse and nothing is said about the world outside the collapsing body!

This deficit was remedied by Tolman and Oppenheimer. \index{Tolman} \index{Oppenheimer} The idea is the following:
\begin{enumerate}
\item The outside world is described by the Schwarzschild metric with a cerain mass (related to $\varepsilon(0)$).
\item This domain is glued to the interior solution. This involves a coordinate transformation from comoving to Schwarzschild coordinates. The coefficients of the metric can be matched together.
\end{enumerate}
Benefits: \\
We can follow the history of the collapsing body until it falls beyond its Schwarzschild radius

\subsection{Dust cloud solution} \index{Dust cloud}
Comoving coordinates $\quad x^0 = t, \quad x^1 = r, \quad x^2 = \vartheta, \quad x^3 = \varphi$ \medskip\\
non vanishing components of $g_{\mu\nu}$:
\[ g_{00} = 1, \qquad g_{11} = - \frac{R(t)^2}{1 - ar^2}, \qquad g_{22} = -R(t)^2 r^2, \qquad g_{33} = -R(t)^2 r^2\sin^2\vartheta \]
$R(t)$ decreases from the value 1 at $t = 1$ to 0 (total collapse) at $T = \frac{\frac{\pi}{2}}{\sqrt{a}}$.
\[ a = \frac{8\pi}{3} G \varepsilon(0),\qquad \varepsilon(t) = \frac{a}{R(t)^3} \]

\includegraphics[height=55mm]{dust1}

$r_0$ (the radius of the cloud) can be adjusted freely, but $r_0 < \frac{1}{\sqrt{a}}$. This will now be matched to the Schwarzschild solution. \bigskip

coordinates $\quad y^0 = \widebar t,\quad y^1 = \widebar r, \quad y^2 = \vartheta, \quad y^3 = \varphi$
\[ \widebar g_{00} = \left( 1 - \frac{r_g}{\widebar r}\right) = - \frac{1}{\widebar g_{11}}, \qquad \widebar g_{22} = - \widebar r^2, \qquad \widebar g_{33} = -\widebar r^2\sin^2\vartheta\]
domain of definition $\left(\widebar r > r_g\right)$: \smallskip\\
\includegraphics[height=55mm]{dust2} \medskip\\
$\Rightarrow$ a map from I to II will be constructed such that the transformed metrical coefficients match continously.
\begin{eqnarray}
\widebar r &=& R(t) \cdot r \label{eq:dustcloud1}\\
\widebar t &=& \left(\frac{1 - ar_0^2}{a}\right)^\frac{1}{2} \int\limits_{S(r,t)}^1 \frac{\dd\lambda}{1 - \frac{ar_0^2}{r}} \left(\frac{\lambda}{1-\lambda}\right)^\frac{1}{2} \label{eq:dustcloud2}\\
&&\text{with } S(r,t) = 1 - \left(\frac{1 - ar^2}{^ - ar_0^2}\right)^\frac{1}{2} \bigl(1 - R(t)\bigr) \nonumber
\end{eqnarray}
$\Rightarrow$ cloud metric transformed to coordinates $\widebar r$, $\widebar t$.
\[ \widetilde g_{11} = -A,\qquad \widetilde g_{00} = B \]
\begin{eqnarray*}
B(r,t) &=& \frac{R}{S} \left(\frac{1 - ar^2}{1 - ar_o^2}\right) \frac{\left(1 - \frac{ar_0^2}{S}\right)^2}{1 - \frac{ar_0^2}{R}} \\
A(r,t) &=& \left(1 - \frac{ar^2}{R}\right)^{-1}
\end{eqnarray*}
$r$, $t$ have to be expressed as functions of $\widebar r$, $\widebar t$ by inverting \eqref{eq:dustcloud1}, \eqref{eq:dustcloud2}. \\
At the border of the star: $r = r_0$. Then the values of $A$ and $B$ can be easily computed:
\begin{eqnarray*}
A(r_0,t) &=& \left(1 - \frac{ar_0^2}{R(t)}\right)^{-1} \\
B(r_0,t) &=& \left(1 - \frac{ar_0^2}{R(t)}\right) = +A^{-1}
\end{eqnarray*}
$\Rightarrow$ The transformed metric can now be continously matched to the Schwarzschild solution. At $\widebar r = r_0 \cdot R(t)$ we must have:
\[ \frac{r_g}{r} = \frac{ar_0^2}{R(t)} = \frac{r_g}{r_0 R(t)} \qquad\qquad\qquad a r_0^3 = r_g \quad\text{(matching condition)} \]
\[2 \cdot\!\!\!\!\!\!\!\! \underbrace{\frac{4\pi}{3} r_0^3 \varepsilon(0) \cdot G}_\text{mass of the dust cloud} = 2 \cdot\!\!\!\!\!\!\!\!\!\!\!\! \underbrace{M\cdot G}_\text{gravitating mass of the Schwarzschild solution} \]

Only parts near the surface of the dust cloud are mapped into the Schwarzschild world. When the Schwarzschild metric has infinite value, the surface of the dust cloud will disappear beyond the Schwarzschild radius. In comoving coordinates however, it will still collaps. But then the dust cloud and the exterior Schwarzschild domain are totally disconnected!

Thus we have computed (in comoving time) only a snapshot of the total collapse seen from outside the radius of the dust cloud. Admittedly a crude model of stellar collapse. \smallskip

\includegraphics[height=45mm]{dust3} \\
radial light ray: $\quad\vartheta =$ const, $\quad\varphi =$ const, $\qquad g_{\mu\nu}\ \dot y^\mu\dot y^\nu = 0$
\[ \Rightarrow \left(1 - \frac{r_g}{r}\right) \dot{\widebar t}^2 - \left(1 - \frac{r_g}{r}\right)^{-1} \dot{\widebar r}^2 = 0 \qquad\qquad \dot{\widebar t} = \frac{1}{1 - \frac{r_g}{r}} \dot{\widebar r} \]
\[\widebar t' - \widebar t(t) = \!\!\!\int\limits_{aR(t)}^{r'}\!\! \frac{\dd r}{1 - \frac{r_g}{r}} \qquad\qquad\qquad \widebar t' = \widebar t(t) + \!\!\!\int\limits_{aR(t)}^{r'}\!\! \frac{\dd r}{1 - \frac{r_g}{r}}\]
Let $t \to t_c$: $\displaystyle \underbrace{\widebar t'}_\text{time of absorption\qquad\qquad}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! = \underbrace{\widebar t(t_c)}_{\to\infty} + \underbrace{\int\limits_{aR(t_c)}^{r'}\!\! \frac{\dd r}{1 - \frac{r_g}{r}}}_{\to \infty}$ \\
$\Rightarrow$ Light rays between emitters and observers will need infinite times to be observed.

\section{Reissner-Nordström} \index{Reissner-Nordström}
In polar coordinates:
\[ G^{\mu\nu} = 8\pi G T^{\mu\nu}(F) \]
\[ \dd F = 0 \qquad\qquad \mathrm{div} F = 0 \qquad (\tau \neq 0) \]

solution: $\quad F = \dd A \qquad\quad A_0 = \dfrac{Q}{r}, \quad A_i = 0$
\[g_{00} = 1 - \frac{r_g}{r} + \frac{Q^2}{r^2} = - \frac{1}{g_{11}}, \qquad g_{22} = -r^2, \qquad g_{33} = -r^2\sin^2\vartheta \]
singularity: $g_{00}(r) = 0$
\[r^2 - 2GMr + Q^2 = 0 \qquad\qquad\qquad r = \frac{r_g}{2} + \sqrt{\frac{r_g^2}{4} - Q^2 } \]
condition for singularity: $\dfrac{G^2M^2}{R} > \dfrac{Q^2}{R}$

\section[How metrics are denoted in textbooks]{Some comments on how metrics are denoted in textbooks}
So far, our notation always stated:
\begin{enumerate}
\item the coordinates $x^\mu$
\item the metrical coefficients $g_{\mu\nu}$
\end{enumerate}
This can be collapsed into one formula by writing $\quad g = g_{\mu\nu}\ \dd x^\mu \otimes \dd x^\nu$ \medskip \\
Example: $\quad x^0 = t,\quad x^1 = r, \quad x^2 = \vartheta, \quad x^3 = \varphi$
\[ g = \left( 1 - \frac{r_g}{r}\right) \dd t \otimes \dd t - \left(1 - \frac{r_g}{r}\right)^{-1} \dd r \otimes \dd r - r^2\, \dd\vartheta \otimes \dd\vartheta - r^2 \sin^2\vartheta\ \dd\varphi \otimes \dd\varphi  \]
physical abbreviation (ex. $\dd\vartheta \otimes \dd\vartheta = \dd\vartheta^2$):
\[ g = \left(1 - \frac{r_g}{r}\right) \dd t^2 - \left(1 - \frac{r_g}{r}\right)^{-1} \dd r^2 - r^2\, \dd\vartheta^2 - r^2\sin^2\vartheta\ \dd\varphi^2 \]
\[\text{Warning:}\quad\dd r \otimes \dd t \neq \dd r\, \dd t \qquad\qquad\otimes \text{ non-commuting} \] %FIXME Symbole ("Warndreieck", Blitz)
\[r = f(u,v) \qquad  \dd r = \ddel[u]{}\, f\, \dd u + \ddel[v]{}\, f\, \dd v\]
\[r = s(u,v) \qquad  \dd r = \ddel[u]{}\, s\, \dd u + \ddel[v]{}\, s\, \dd v\]
$\Rightarrow$ simplifies coordinate transformations

\section{Kerr metric} \index{Kerr metric}
time-independent axially symmetric solution of $G^{\mu\nu}=0$\\
Starting point: Minkowski metric in generalized spherical coordinates (Boyer-Lindquist coordinates): \index{Boyer-Lindquist coordinates}
\[g=\left(\dd x^0\right)^2-\left(\dd x^1\right)^2-\left(\dd x^2\right)^2-\left(\dd x^3\right)^2 \qquad\qquad x \text{ cartesian coordinates}\]
\[x^1=\left(r^2+a^2\right)^\frac{1}{2}\sin\vartheta\cos\varphi \qquad x^2=\left(r^2+a^2\right)^\frac{1}{2}\sin\vartheta\sin\varphi \qquad x^3=r\cos\vartheta\]
$r=\const\Rightarrow$ (ellipsoids instead of spheres)\\
After coordinate transformation:
\[g=\dd x^2-\frac{\rho^2}{r^2+a^2}\dd r^2+-\left(r^2+a^2\right)\sin^2\vartheta\ \dd\varphi^2-\rho^2\dd\vartheta^2 \qquad \rho^2=r^2+a^2\sin^2\vartheta\]

Kerr solution \index{Kerr solution}
\begin{eqnarray*}
g' &=& \left(1-r_g\frac{1}{\rho^2}\right)\dd t^2-\frac{\rho^2}{\Delta}\dd r^2-\rho^2\dd\vartheta^2-\left(r^2+a^2+\frac{r_gra^2\sin^2\vartheta}{\rho^2}\right)\sin^2\vartheta\ \dd\varphi^2\\
&&+\frac{r_gra\sin^2\vartheta}{\rho^2}\left(\dd t\otimes\dd\varphi+\dd\varphi\otimes\dd t\right)\\
&&\rho^2=r^2+a^2\sin^2\vartheta \qquad \Delta=r^2-r_gr+a^2 \qquad r_g=\const
\end{eqnarray*}
Special cases:
\begin{enumerate}
\item $a=0$
\[G'=\left(1-\frac{r_g}{r}\right)\dd t^2-\left(1-\frac{r_g}{r}\right)^{-1}\dd r^2-r^2\left(\dd\vartheta^2+\sin^2\vartheta\dd\varphi^2\right)\]
Schwarzschild solution! $\quad r_g=2GM$ $\quad M$ field producing mass
\item $r_g=0$\\
$g'=\eta$ in generalized spherical coordinates. Interpretation of the full parameter set of solutions. Metric is produced by a energy-momentum distribution which is bonded by an ellipsoid.
\end{enumerate}
The nondiagonal piece $\sim (\dd\varphi\otimes\dd t+\dd t\otimes\dd\varphi)$ is caused by a rotation of the ellipsoid. 
\[\varphi\to\varphi+\alpha t\Rightarrow\dd\varphi\to\dd\varphi+\alpha\dd t\]
\[\dd\varphi^2\Rightarrow\dd\varphi^2+\alpha(\dd\varphi\otimes\dd t+\dd t\otimes\dd\varphi)+\alpha^2\dd t^2\]
can approximatly used to eliminate the nondiagonal piece in a rotating coordinate set. $a$ related to angular momentum of the field producing particle.

\subsubsection*{Warning:}
The Kerr solution as the limiting outside of matter solution has never been produced in analogy to the spherical case. Kerr solution has never succesfully matched to an interior solution. In particular the relation of $a$ with angular momentum is not completely clear. ($r_g=0, a\neq 0$ is always the Minkowski metric)

Domain of definition of $g'$
\begin{enumerate}
\item $\Delta=0$ at $r_0=\frac{r_g}{2}+\left(\left(\frac{r_g}{2}\right)^2-a^2\right)^\frac{1}{2} \qquad\qquad r>r_0$
\item $g^{00}=0$ at $r_1=\frac{r_g}{2}+\left(\left(\frac{r_g}{2}\right)^2-a^2\cos^2\vartheta\right)^\frac{1}{2}\quad <r_g$
\end{enumerate}
Domain of definition conveniently choosen is $r>r_g$. In this domain the motion of test particles can be studied. (Can be done exactly)

\subsection{Particle trajectories} \index{Jacobi methode}
Jacobi-methode works:
Jacobi-equation in general relativity:
\begin{equation}
g^{\mu\nu}\del{\mu}W\del{\nu}W=m^2
\label{eq:jacobi2} \index{Jacobi equation}
\end{equation}
$\Rightarrow W(x,Q_1,Q_2,Q_3)$ Trajectories are computed from the equation:
\begin{equation}
-P_i=\del{}{Q_i}W(x,Q) \qquad (i=1,2,3),\qquad P,Q=\const
\label{eq:minusP}
\end{equation}
Solve for $x_i \Rightarrow x_i(x^0,Q,P)$\\
When does the methode work practically? $W$ is seperable, i.e.
\[W=\sum_{\mu=0}^3 W_\mu(x^\mu) \qquad\qquad (W_\mu\text{ depends only on }x^\mu)\]
When $x^\mu$ is a cyclic coordinate, then $W_\mu(x^\mu)=Q_\mu x^\mu \quad Q_\mu=\const$ (no summation)\\
In our case, cyclic means $g^{\alpha\beta}$ are independent of $x^\mu$ for all $\alpha, \beta$
\[W=-Et+L_3\varphi+S_r(r)+S_\vartheta(\vartheta) \quad\text{ has to be inserted into \eqref{eq:jacobi2}}\]
\[\Rightarrow m^2=\frac{1}{\Delta\rho^2}\Bigl(-\left(r^2+a^2\right)E+L_3a\Bigr)^2 -\frac{1}{\rho^2\sin^2\vartheta}\Bigl(L_3-a\sin^2\vartheta E\Bigr)^2 -\frac{\Delta}{\rho^2}\left(\delll{r}{S_r}\right)^2-\frac{1}{\rho^2}\left(\delll{\vartheta}{S_\vartheta}\right)^2\]
Multiply with $\rho^2$ and bring everything on one side of the equation:
\begin{eqnarray*}
0 &=& \underbrace{\frac{1}{\Delta}\Bigl(-\left(r^2+a^2\right)E+L_3a\Bigr)^2-m^2r^2-\Delta\left(\delll{r}{S_r}\right)^2}_{G(r)=c}\\
&& \underbrace{-\frac{1}{\sin^2\vartheta}\left(L_3-a\sin^2\vartheta E\right)^2-m^2a^2\cos^2\vartheta-\left(\delll{\vartheta}{S_\vartheta}\right)^2}_{F(\vartheta)=-c}\\
0 &=& G(r)+F(\vartheta) \qquad\qquad c\geq 0 \const\\
c &=& \frac{1}{\Delta}\Bigl(-\left(r^2+a^2\right)E+L_3a\Bigr)^2-m^2r^2-\Delta\left(\delll{r}{S_r}\right)^2\\
-c &=& -\frac{1}{\sin^2\vartheta}\left(L_3-a\sin^2\vartheta\, E\right)^2-m^2a^2\cos^2\vartheta-\left(\delll{\vartheta}{S_\vartheta}\right)^2\\
& \Rightarrow & \delll{\vartheta}{S_\vartheta},\delll{r}{S_r} \text{ can be obtained trivially and }S_r, S_\vartheta \text{ can be computed}\
\end{eqnarray*}
Put $Q_1=E, \quad Q_2=L_3, \quad Q_3=c \Rightarrow W=W(x,Q_1,Q_2,Q_3)$ is completely specified for the Kerr solution.\\
$\Rightarrow$ Particle trajectories can be determined now completely from \eqref{eq:minusP}

Analogios problem in Newtonian mechanics:\\
Newtonian potential at large distances from a gravitating center is:
\[-\frac{GM}{r}+\frac{d\cos\vartheta}{r^2} \qquad\qquad d=\const\]
The Jacobi methode reads as follows (in polar coordinates)
\begin{eqnarray*}
0 &=& \delll{t}{W}+\frac{1}{2m}\left(\left(\delll{r}{W}\right)^2 +\frac{1}{r^2}\left(\delll{\vartheta}{W}\right)^2 +\frac{1}{r^2\sin^2\vartheta}\left(\delll{\varphi}{W}\right)^2\right) -\frac{GMm}{r}+\frac{md\cos\vartheta}{r^2}\\
W &=& -Et+L_3\varphi+S_r(r)+S_\vartheta(\vartheta)\\
&\Rightarrow & -E+\frac{1}{2m}\left(\delll{r}{S_r}\right)^2 +\frac{1}{2mr^2}\left(\left(\delll{\vartheta}{S_\vartheta}\right)^2 +\frac{1}{\sin^2\vartheta}L_3^2+2m^2d\cos\vartheta\right)-\frac{GMm}{r}=0\\
& \Rightarrow & \left(\delll{\vartheta}{S_\vartheta}\right)^2+\frac{1}{\sin^2\vartheta}L_3^2 +2m^2d\cos\vartheta=c\\
&& -E+\frac{1}{2m}\left(\delll{r}{S_r}\right)^2+\frac{1}{2mr^2}c-\frac{GMm}{r}=0
\end{eqnarray*}

\chapter{Gravitational waves} \index{Gravitational waves}
More exact solutions: gravitational waves

\section{Special case: electromagnetic wave}
Vector potential with Lorentz gauge: $\qquad A^1=A^3=A^0=0,\ A^2=f\left(x^0-x^1\right)$\\
%insert figure1
Wave solution for $G^{\mu\nu}=0$ with $G$ given by $(x^0,x^1,x^2,x^3\ \text{cartesian})$
\[g=\left(\dd x^0\right)^2-\left(\dd x^1\right)^2-\left(\dd x^2\right)^2-\left(\dd x^3\right)^2+f\left(x^1-x^0,x^2,x^3\right)\dd u^2\]
\[f \text{ disturbance of the Minkowski metric} \qquad\qquad \dd u=\dd x^0-\dd x^1\]
$G^{\mu\nu}=0$ reduces to the condition: $\left[\left(\del{2}\right)^2+\left(\del{3}\right)^2\right]f=0 \qquad (\Delta_2f=0)$ \medskip\\
%insert figure2
Pulse moves with time. $\Delta f=0$ means if $z=x^2+ix^3$
\[\frac{\partial^2}{\partial z\partial \overline z}\,f=0\]
\[\Rightarrow f=g(z)+\overline{g(z)} \qquad\qquad g(z) \text{ is analytic (if $g$ has no singularituies)}\]
\[g(z)=\sum_{i=0}^\infty \underbrace{c_i\left(x^0-x^1\right)}_{=u}z^i\]
Solutions with singularities: $g(z)=\dfrac{b(x^0-x^1)}{z-a(x^0-x^1)}$

Disturbance can be added!\\
But only if the direction of propagation is kept fixed. Strong difference to the electromagnetic case. (non linearity of the Einstein-equations in the components $g_{\mu\nu}$)\\
To obtaine more general solutions is an approximate way, write $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$ and assume $h_{\mu\nu}$ is small. Then linearize Einstein's equation to obtaine partial differential equation for $h_{\mu\nu}$.\\
General invariance is then equivalent to a replacement
\[h_{\mu\nu} \to h'_{\mu\nu}=h_{\mu\nu}+\dell{\mu}{B_\nu}+\dell{\nu}{B_\mu}\]
After ``fixing a gauge'' $G^{\mu\nu}=0$ becomes $\Box h_{\mu\nu}=0$

\section{Weak gravitational waves}
Starting point: $G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2} R\,g_{\mu\nu} = 8\pi G\,T_{\mu\nu}$
\begin{eqnarray} 
\Rightarrow &\!\!\!\!& -R = 8\pi G\,T_{\alpha\beta}\,g^{\alpha\beta} \nonumber\\
\Rightarrow &\!\!\!\!& R_{\mu\nu} = 8\pi G\,S_{\mu\nu} \qquad\qquad\qquad S_{\mu\nu} = T_{\mu\nu} - \frac{1}{2} g_{\mu\nu} \left( g^{\alpha\beta} T_{\alpha\beta}\right) \label{eq:waves1}
\end{eqnarray}
Ansatz: $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ \qquad ($h$ small) \\
Idea: insert this into \eqref{eq:waves1} and keep only terms linear in $h$.
\[ \Gamma^\alpha_{\beta\gamma} = \frac{1}{2} \left(\del{\beta}\,{h^\alpha\!}_\gamma + \del{\gamma}\,{h^\alpha\!}_\beta - \eta^{\alpha\delta} \del{\delta}\,h_{\beta\gamma} \right)\]
raising and lowering of indices done with $\eta_{\mu\nu}$ only. (This is done now for the whole lecture.)
\[\Rightarrow R_{\mu\nu} \simeq \del{\nu}\,\Gamma^\lambda_{\lambda\mu} - \del{\lambda}\,\Gamma^\lambda_{\mu\nu} \]
\eqref{eq:waves1} $\Rightarrow$ linearized Einstein equation: \index{Einstein equation!linearized} \index{Linearized Einstein equation}
\begin{equation}
\fbox{$\displaystyle \Box h_{\mu\nu} - \deldel{\lambda}{\mu}\,{h^\lambda\!}_\nu - \deldel{\lambda}{\nu}\,{h^\lambda\!}_\mu + \deldel{\mu}{\nu}\,{h^\lambda\!}_\lambda = 16\pi G\,S_{\mu\nu}$} \label{eq:waves2}
\end{equation}
Linearization produces the superposition principle of electrodynamics.
\subsubsection*{Observation:}
\begin{enumerate}
\item Only Poincar\'e invariance is kept.
\item There is a new symmetry which will be called ``gauge invariance''. \index{Gauge invariance}
\end{enumerate}
\subsubsection*{Symmetry:}
If $h_{\mu\nu}$ is a solution of \eqref{eq:waves2}, then $h'_{\mu\nu} = h_{\mu\nu} + \del{\mu} B_\nu + \del{\nu} B_\mu$ is also a solution. This symmetry is what is left over from general invariance under arbitrary diffeomorphisms (proof later). \\
$\Rightarrow$ Solutions are representing the same physical field if they differ only by a gauge transformation. \medskip

This gauge invariance principle allows to simplify \eqref{eq:waves2}. \medskip\\
Define $B_\mu$ such that $\displaystyle \Box B_\mu = \frac{1}{2} \del{\nu}{h^\nu\!}_\mu - \del{\mu}{h^\nu\!}_\nu$ \\
Define $h'_{\mu\nu}$ as above.
\[\eqref{eq:waves2} \Rightarrow \Box h'_{\mu\nu} = 16\pi G\,S_{\mu\nu} \qquad\quad \text{with}\quad \del{\nu}{h'^\nu\!}_\mu - \del{\mu}{h'^\nu\!}_\nu = 0 \quad\text{(Lorentz condition)}\] \index{Lorentz condition}
Special solution:
\begin{equation}
h'_{\mu\nu} = 4G \int\!\dd^3y\, \frac{S_{\mu\nu}\!\left(\vec y, x^0 - \left|\vec x - \vec y \right|\right)}{\left|\vec x - \vec y\right|} \qquad\qquad \text{(retarded solution)} \index{Retarded solution}
\end{equation}
To this a solution of the homogeneous equation is to be added to obtain the general solution.
\[\Box h'_{\mu\nu} = 0 \qquad\quad \text{with} \quad \del{\nu}{h'^\nu\!}_\mu - \del{\mu}{h'^\nu\!}_\nu = 0 \]

\subsection{Inspection of the corresponding situation in electrodynamics}
\[\Box A_{\mu\nu} = 0 \qquad\quad \text{with} \quad \del \mu A^\mu = 0 \]
wave solution: $A_\mu = \Re\,a_\mu\,\e^{-i k_\alpha x^\alpha}$ \qquad\quad with \quad $k_\alpha k^\alpha = 0$, \quad$a_\alpha k^\alpha = 0$ \medskip\\
Choose 4 linearly independent vectors $k_\alpha$, $\underline{k}_\alpha$ (with $\underline{k}_0 = k_0$ and $\underline{k}_i = -k_i$)\\
$e_{i\alpha}$ ($i = 1,2$) with $e_{i\alpha}\,k^\alpha = e_{i\alpha}\,\underline{k}^\alpha = e_{1\alpha}\,{e_2}^\alpha = 0$ and $e_{i\alpha}\,{e_i}^\alpha = -1$ \medskip\\
Put $e_{\pm\alpha} = e_{1\alpha} \pm i\,e_{2\alpha}$
\[\Rightarrow a_\mu = a\,k_\mu + b\,\underline{k}_\mu + a_+\,e_{+\mu} + a_-\,e_{-\mu} \]
First term corresponds to gauge transformation and can be omitted. \\
$b = 0$ because of Lorentz condition. \\
$\Rightarrow$ A wave solution in electrodynamics is a linear combination of the solutions
\[ A_{\pm\mu} = \Re\, a_\pm\,e_{\pm\mu}\ \e^{-i k_\alpha x^\alpha} \qquad\qquad e_\pm: \text{polarization vector} \]
If we perform a rotation around the $k$-axis with angle $\varphi$
\[\Rightarrow e_{\pm\mu} \to \lambda_\pm\,e_{\pm\mu} \qquad\qquad \lambda_\pm = \e^{\pm i\varphi} \quad \text{($\pm$: helicity)} \]
Electromagnetic waves have helicity $\pm 1$. \index{Helicity}

\subsection{Gravitational wave solution}
\[h_{\mu\nu} = \Re\, h^0_{\mu\nu}\,\e^{-ik_\alpha x^\alpha} \qquad\qquad k_\alpha k^\alpha = 0 \]
Lorentz condition: \quad $\displaystyle k_\mu {{h^0}^\mu\!}_\nu - \frac{1}{2} k_\nu {{h^0}^\mu\!}_\mu = 0$ \index{Lorentz condition}
\[\Rightarrow h^0_{\mu\nu} = \underbrace{k_\mu a_\nu + k_\nu a_\mu}_{(*)} + a_+ e_{+\mu}e_{+\nu} + a_- e_{-\mu}e_{-\nu} \]
$(*)$ corresponds to a gauge transformation for gravitational waves and can be omitted. \medskip

$\Rightarrow$ Up to physically irrelevant terms, gravitational wave solutions have the form:
\begin{equation}
h_{\mu\nu} = \sum\limits_\pm \Re\, a_\pm e_{\pm\mu} e_{\pm\nu}\, \e^{-ik_\alpha x^\alpha}
\end{equation}

Under rotation with angle $\varphi$ around the $k$-axis the polarization tensor $e_{\pm\mu} e_{\pm\nu}$ changes into
\[e_{\pm\mu} e_{\pm\nu} \left( \lambda_\pm\right)^2 \qquad\qquad \left( \lambda_\pm\right)^2 = \e^{\pm 2i\varphi} \]
The helicity therefore is $\pm 2$.

\subsection[General coordinate transformations and gauge transformations]{General coordinate transformations and gauge transformations within the weak field approximation}
1-parameter families of diffeomorphisms and vector fields \\
Let $X^\mu$ be a vector field. \medskip\\
Consider the differential equation
\[\dot X^\mu(\tau) = X^\mu\big(x(\tau)\big) \qquad\qquad x(\tau) = x(\tau,x_0) \text{ a curve} \qquad\qquad x_0 = x(0) \]
$x(\tau)$ is uniquely determined by the initial value $x_0$. \medskip\\
$\Rightarrow \varphi_\tau$ (a 1-parameter family of diffeomorphisms) is given by setting $\varphi_\tau(x_0) = x(\tau, x_0)$.
\begin{picture}(350,55)
\put(10,12){$\times$}
\put(3,5){$x_0$}
\put(210,15){$\times$}
\put(220,12){$x(\tau,x_0) = \varphi_\tau(x_0)$}
\qbezier(18,20)(110,70)(200,25)
\thicklines
\put(200,25){\vector(2,-1){10}}
\put(105,35){$x(\tau)$}
\end{picture}
\begin{eqnarray*}
\Rightarrow \left. \frac{\dd}{\dd\tau}\varphi_\tau^*\,g_{\mu\nu}(x)\right|_{\tau=0}\!\!\! &=& X_\mu(x)_{;\nu} + X_\nu(x)_{;\mu} \\
\varphi_\tau^*\,g_{\mu\nu} &=& g_{\mu\nu} + \tau \left( X_{\mu_;\nu} + X_{\nu;\mu} \right) + \mathcal O(\tau^2)
\end{eqnarray*}
weak field approximation:
\begin{eqnarray*}
\varphi_\tau^*\,g_{\mu\nu} &=& \eta_{\mu\nu} + h_{\mu\nu} + \tau \left( X_{\mu_,\nu} + X_{\nu,\mu} \right) + \underbrace{\tau X_\alpha \left( {h^\alpha\!}_{\mu,\nu} + {h^\alpha\!}_{\nu,\mu} - \del{\delta}\, h_{\mu\nu}\,\eta^{\delta\alpha} \right)}_{\text{small}} \\
&=& \eta_{\mu\nu} + \underbrace{h_{\mu\nu} + \tau\left( \del{\mu}\,X_\nu + \del{\nu}\,X_\mu \right)}_{=h'_{\mu\nu}}
\end{eqnarray*}
$\Rightarrow$ In the weak field approximation, ``small'' diffeomorphisms correspond exactly to gauge transformations. \medskip\\
$\Rightarrow$ Really only two helicity states exist for gravitational waves.

\chapter{Cosmology} \index{Cosmology}
The metric $g$ of the universe must be determined from Einsetin's equation:
\[ G^{\mu\nu} = 8\pi G\, T^{\mu\nu} \]
The description needs 3-spaces $M_3$ which are generated by symmetry groups where any point can be transformed into any other point by a symmetry transformation.

\begin{description}
\item[Example 0:] $M_3 = \mathbb R^3$
\begin{eqnarray*}
g_3 &=& \left(\dd x^1\right)^2 + \left(\dd x^2\right)^2 + \left(\dd x^3\right)^2  \\
&=& \dd r^2 + r^2\,\dd\vartheta^2 + r^2\sin^2\vartheta\ \dd\varphi^2
\end{eqnarray*}
symmetry group: $SO(3)$ + translations
\item[Example 1:] $M_3 = S^3 = \left\{ x \in \mathbb R^4: \left(x^0\right)^2 + \left(x^1\right)^2 + \left(x^2\right)^2 + \left(x^3\right)^2 = 1 \right\}$
\begin{eqnarray*}
g_3 &=& \left(\dd x^0\right)^2 + \left(\dd x^1\right)^2 + \left(\dd x^2\right)^2 + \left(\dd x^3\right)^2 \qquad \left( x^0 = \Bigl(1 - \sum\limits_i \left(\dd x^i \right)^2\Bigr)^\frac{1}{2} \right) \\
&=& \frac{\dd r^2}{1 - r^2} + r^2\,\dd\vartheta^2 + r^2\sin^2\vartheta\ \dd\varphi^2
\end{eqnarray*}
symmetry group: $SO(4)$
\item[Example 2:] $M_3 = \left\{ x \in \mathbb R^4: \left(x^0\right)^2 - \left(x^1\right)^2 - \left(x^2\right)^2 - \left(x^3\right)^2 = 1 \right\}=H_3$
\begin{eqnarray*}
 g_3 &=& \frac{\dd r^2}{1 + r^2} + r^2\,\dd\vartheta^2 + r^2\sin^2\vartheta\ \dd\varphi^2
\end{eqnarray*}
\end{description}
The full space time metric in comoving coordinates is called Robertson-Walker metric. It must have the form: \index{Robertson-Walker metric}
\[g = \dd t^2 - g_3 \cdot R(t)^2 \]
\[ g_3 = \frac{\dd r^2}{1 - kr^2} + r^2\,\dd\vartheta^2 + r^2\sin^2\vartheta\ \dd\varphi^2 \qquad\qquad\qquad k = \left\{\!\!\begin{array}{c}0 \\ \pm 1 \end{array} \right. \]
Symmetry group: $k=\begin{cases}1 \Rightarrow O(4)\\ 0\Rightarrow \text{Euclidean group}\\ -1\Rightarrow \text{Lorentzgroup} \end{cases}$ Always a 6-dimensional Lie-group.
\[g_{00}=1,\qquad g_{11}=\frac{-1}{1-kr^2},\qquad g_{22}=-r^2,\qquad g_{33}=-r^2\sin^2\theta\]
\[T^{\mu\nu}=(\varepsilon+P)j^\mu j^\nu-Pg^{\mu\nu} \qquad\qquad\qquad j^0=1,\qquad j^i=0 \qquad\text{(by symmetry)}\]
$\varepsilon,P$ should be functions of time $t$ only.

\section{Fundamental cosmological equations} \index{Fundamental equations of cosmology}\index{Cosmology!- fundamental equations}
Fundamental equations of cosmology
\begin{eqnarray}
&&\text{Einstein}\Rightarrow\fbox{$\dot R^2+k=\frac{8\pi G}{3}\varepsilon R^2$}\qquad\text{(see dust cloud problem)} \label{eq:cosm1}\\
&&T^{\mu\nu}{}_{;\mu}=0\quad\fbox{$\frac{\dd}{\dd t}\varepsilon R^3=-P\frac{\dd}{\dd t}R^3$} \quad (\dd E=-P\dd V) \label{eq:cosm2}
\end{eqnarray}
Unknowns: $\varepsilon,P,R\quad$ Equation of state reduces then to $\varepsilon,R\quad \left(P(\varepsilon)  \text{ must be given}\right)$
Equation of state can only be given, if the chemical composition of the universe is simple. This is indeed the case! Mass is mostly in hydrogen and He  + radiation \bigskip

Question: What experimental information is hidden in $R(t)$?\\
Guess: properties of light waves will be modified. How?\\
Light waves are solutions of Maxwell equations: $\dd F=0 \quad \text{div}F=0$
$\dd F$ does not contain metric and is valid in every coordinate set \index{Maxwell equations}
\[\text{div}F=\frac{1}{\sqrt{-g}}\del{\mu}\sqrt{-g}\ g^{\mu\alpha}g^{\nu\beta}F_{\alpha\beta}=0 \qquad g=\det g_{\mu\nu}\]
Rewrite $g=R^2(t)\left(\left(\frac{1}{R(t)}\dd t\right)^2-g_3\right)$\\
Introduce a new time variable $\tau$ (called conformed time), such that
\[\frac{1}{R(t)}\dd t=\dd\tau \qquad t'(\tau)=R\left(t(\tau)\right)\quad\left(\text{or }\frac{\dd\tau}{\dd t}=\frac{1}{R(t)}\right)\]
\[g=R(t)^2\tilde g,\qquad \tilde g=\dd\tau^2-g_3 \qquad \text{ (independent of $\tau$!)}\]
$\Rightarrow$ In Maxwell's equations $R$ drops out.\\
There exist wave solutions $\quad F^{\mu\nu}=e^{i\omega\tau}\!\!\!\!\!\!\!\!\!\!\!\underbrace{F^0_{\mu\nu}}_{\text{independent of }\tau}\!\!\!\!\!\!\!\!\!\!\!+e^{-i\omega\tau}F^0_{\mu\nu}$\\
$\Rightarrow$ Wave solution in comoving coordinates have the form: $F_{\mu\nu}=e^{i\omega\tau(t)}F'_{\mu\nu}+e^{-i\omega\tau(t)}F'_{\mu\nu}$\medskip\\
What is the observed frequency?\\
Let $t$ be a cosmological time. $\Delta t$ an observation intervall. $\frac{\Delta t}{t}\ll1$
\[\Rightarrow \tau(t+\Delta t)=\tau(t)+\Delta t \tau'(t) \qquad F_{\mu\nu}=\underbrace{e^{i\omega\tau(t)}}_{\text{phase}}\cdot e^{i\omega\tau'(t)\Delta t}F^0_{\mu\nu}\qquad \omega_{\text{obs}}=\omega\tau'(t)=\frac{\omega}{R(t)}\]
Let be $t_{\text{abs}}$ and $t_{\text{em}}$ time of absorption and emission of wave respectivly.
\[z=\frac{\omega_{\text{em}}}{\omega_{\text{abs}}}-1=\frac{\lambda_{\text{abs}}}{\lambda_{\text{em}}}-1=\frac{R(t_{\text{abs}})}{R(t_{\text{em}})}-1\qquad\text{Redshift function}\]\index{Redshift function}
$t_{\text{em}}=t_{\text{abs}}-d\qquad d$ distance between absorber and emitter, $t_{\text{abs}}=t_0$ present time
\begin{eqnarray}
z=z(d) &=& \frac{R(t_0)}{R(t_o-d)}-1=\frac{\dot R(t_0)}{R(t_0)}\ d\quad\text{in linear approximation} \nonumber\\
\frac{\dot R(t_0)}{R(t_0)} &=& H_0 \qquad \text{Hubble's constant}\index{Hubble's constant}\\
z &=& H_0d \qquad \text{Hubble's law}\index{Hubble's law}
\end{eqnarray}
Explanation by Doppler effect: $z=v \qquad\qquad v=H_0d \Rightarrow$ redshift is also explained\\
$\Rightarrow \frac{\dot R(t_0)}{R(t_0)}$ gives initial condition for our cosmological equation \medskip\\
$z(d)$ in quadratic approximation:
\[z(d)=H_0d+\left(1+\frac{q_0}{2}\right)H_0^2d^2+\mathcal O(d^3) \qquad\qquad q_0=\frac{-\ddot R(t_0)}{R(t_0)}\frac{1}{H_0^2}\]
$H_0,q_0$ can be extracted from experimental observation\bigskip\\
Fundamental cosmological equations
\[\eqref{eq:cosm1} \Rightarrow \dot R=+\left(-k+\frac{8\pi G}{3}\varepsilon R^2\right)^{\frac{1}{2}} \qquad\qquad \text{(+ sign is fixed by $H_0>0$)}\]
$\Rightarrow R$ will increase\\
Looking backward in time $\Rightarrow$ universe has started at apoint $\Rightarrow$ Big Bang! \index{Big Bang}\medskip\\
Equation of state: $P=0$ (model with dust particles in the form of galaxies)\\
working hypothesis (not too bad)\\
At present time:
\begin{eqnarray*}
\varepsilon(t_0) &=& \frac{3}{8\pi G}\left(\frac{k}{R_0^2}+H_0^2\right)\\
\Rightarrow\frac{k}{k_0^2} &\lessgtr& 0 \qquad\text{if}\quad \varepsilon(t_0)\lessgtr \rho_c=\frac{3H_0^2}{8\pi G}\\
P(t_0) &=& \frac{1}{8\pi G}\left(\frac{k}{R_0^2}+H_0^2(1-q_0)\right) \text{if } P=0\\
\frac{\varepsilon(t_0)}{\rho_c} &=& 2q_0\qquad k\lessgtr 0\qquad\text{if}\quad q_0\gtrless\frac{1}{2} 
\end{eqnarray*}
Eliminating now $\varepsilon$ and $\frac{k}{R^2}$ from the formula \eqref{eq:cosm1}
\begin{equation}
\frac{\dot R(t)}{R(t_0)}=H_0\left(1-2q_0+\frac{2q_0R(t_0)}{R(t)}\right)^\frac{1}{2}\label{eq:cosm1b}
\end{equation}
only the nonobservable number $R(t_0)$ enters $\Rightarrow \frac{R(t)}{R(t_0)}$ can be computed\medskip\\
Note: \eqref{eq:cosm1b} was derived under the hypothesis:
\[P(t)=0 \Rightarrow \frac{\dd}{\dd t}\ \varepsilon R^3=0 \Rightarrow \varepsilon(t)=\frac{\varepsilon(t_0)R(t_0)^3}{R(t)^3}\]

\begin{eqnarray*}
t &=& \frac{1}{H_0}\int_0^{\frac{R}{R(t_0)}}\left(1-2q_0+\frac{2q_0}{x}\right)^{-\frac{1}{2}}\dd x=t\left(\frac{R}{R_0}\right)\\
\Rightarrow t_0 &=& \frac{1}{H_0}\int_0^1\left(1-2q_0+\frac{2q_0}{x}\right)^{-\frac{1}{2}}\dd x\leq\frac{1}{H_0} \qquad\text{(Hubble time, age of universe)}\index{Hubble time}
\end{eqnarray*}
$q_0>\frac{1}{2}:\quad k=1\qquad$ Define an angle $\vartheta$ by
\[1-\cos\vartheta=\frac{2q_0-1}{q_0}\frac{R(t)}{R(t_0)}\Rightarrow H_0t=q_0(2q_0-1)^{-\frac{3}{2}}(\vartheta-\sin\vartheta)\]
$q_0=\frac{1}{2}:\quad k=0$
\[\frac{R}{R_0}=\left(\frac{3}{2}H_0t\right)^\frac{2}{3}\]
$0<q_0<\frac{1}{2}:\quad k=-1$
\[1-\cosh\varphi=\frac{2q_0-1}{q_0}\frac{R}{R_0} \Rightarrow H_0t=q_0(1-2q_0)^{-\frac{3}{2}}(\sinh\varphi-\varphi)\]

\section{Cosmic microwave background}\index{Cosmic microwave background}
Experimental astronomical findings: There is a background radiation in the universe (very homogeneous) whose frequency distribution very accurately given by Planck's law:\index{Planck's law}
\[\rho(x)\dd\omega=\frac{1}{\pi^2}\dfrac{\omega_{\text{obs}}^3\dd\omega_{\text{obs}}}{e^{-\frac{\omega_{\text{obs}}}{k_BT}}-1} \qquad\quad (\hbar=c=1)\qquad k_B \text{ Boltzmann's constant and $T=3.5$K}\]
$\Rightarrow$ Taking into account that these frequencies all are redshifted, since the time of creation of this blackbody radiation
\[\omega_{\text{obs}}=\frac{\omega}{R(t_0)},\quad \omega_{\text{em}}=\frac{\omega}{R(t_{\text{em}})} \Rightarrow \omega_{\text{obs}}=\frac{R(t_{\text{em}})}{R(t_0)}\omega_{\text{em}}\]
\[e^{-\frac{\omega_{\text{obs}}}{k_BT}}=e^{-\frac{\omega_{\text{em}}}{k_BT'}} \qquad T'=\frac{TR(t_0)}{R(t_{\text{em}})}\]
$\Rightarrow$ Temperature at time of emission might have been very hot in the past.\\
$\Rightarrow$ Explanation: Ion recombination in the past form a plasma (electrons and nuclei) with temperature $\sim 4000$K which then was redshifted.\\
If photon temperature is high $\Rightarrow$ energy contribution at very remote time must be dominated by radiation
\[\varepsilon_\text{ph}\gg\varepsilon_\text{matter}\quad\text{and moreover}\quad \varepsilon_\text{ph}\ \text{and}\ P_\text{ph}\quad\text{obey}\quad P_\text{ph}=\frac{1}{3}\ \varepsilon_\text{ph}\]
equation of state
\[P=\frac{\varepsilon}{3}\ \text{ and }\ \frac{\dd}{\dd t}\ \varepsilon R^2=-\frac{\varepsilon}{3}\frac{\dd}{\dd t}R^3\qquad \Rightarrow\ \varepsilon=\frac{\alpha}{R(t)^4} \qquad \alpha=\const\]
\[\dot R^2+k=\frac{8\pi}{3}G\varepsilon R^2 \quad\Rightarrow\fbox{$\dot R^2+k=\frac{8\pi G}{3}\frac{\alpha}{R^2}$}\]
For $0<t<t'$ solve the cosmological equation for radiation dominated universe.\\
For $t>t'$ solve the equation with a assumption for dust: $t'=10^{12}$s (time of ion recombination)\\
matter content (measure); matter is found by hydrogen and helium (helium abundance 24\%)\\
$\Rightarrow$ For time $t<t'$ the model can be tested against observed abundances of helium.\medskip\\
Looking more carefully at the experimentally observed Planck distribution, one finds:
\[\Rightarrow \frac{1}{4\pi^3}\dfrac{|\vec q|\dd^3q}{e^{\frac{|\vec q|}{k_bT}-1}} \qquad\qquad \omega=|\vec q| \qquad\qquad q=\left(\begin{array}{c}\omega\\ \vec q\end{array}\right) \qquad\qquad (\hbar=c=1)\]

$T$ is a function of angles. Can that be understood? \index{Angular dependence of the CMB}\\
So far, we assumed that emitter and observer of this radiation were comoving.\\
$\Rightarrow$ If observer is not comoving but moving with a relative velocity $\vec v$, we have to do a Lorentztransformation $\Lambda$
\[e^\frac{\omega}{k_bT}=e^\frac{<e_0,q>}{k_bT}\to e^\frac{<e_0,\Lambda q>}{k_bT}=e^\frac{<\Lambda^{-1},q>}{k_bT}=e^\frac{\omega}{k_b\widetilde T}\]
\[\Lambda^{-1}e_0=\left(\begin{array}{c}\sqrt{1+|\vec v|^2}\\ v\end{array}\right) \Rightarrow <\Lambda^{-1}e_0,q>=\sqrt{1+|\vec v|^2}-<\vec q,\vec v>=\omega\left(\sqrt{1+|\vec v|^2}-<\vec q,\vec v>\right)\]
\[\widetilde T=\dfrac{T}{(q+v^2)^\frac{1}{2}-<q_0,v>}=T\left(1+<q_0,v>\right) \qquad\qquad \vec q_0=\frac{\vec q}{|\vec q|}\]
$\widetilde T$: the observed temperature seems to be direction dependent.\\
Ploting the observed temperature in a directionally dependent way, gives the $\widetilde T$ as a function of angles.\\
Expanding $\widetilde T$ in terms of spherical harmonics\\
$\Rightarrow T$ is the monopole contribution and $T<v,q_0>$ is the dipole contribution.\bigskip\\
Change of Einstein's equation:\\
reappearance of a cosmological constant \index{Cosmological constant}
\[G^{\mu\nu}+\Lambda g^{\mu\nu}=8\pi GT^{\mu\nu} \quad\Rightarrow \text{ In the action a term }\Lambda\int\!\!\!\sqrt{g}\ \dd^4x \text{ is present!}\]


\chapter{Fundamental, connection and curvature form} \index{Fundamental form} \index{Connection form} \index{Curvature form}
First step: Find 4 orthogonal vectorfields $X^\mu_i$; coordinates $\mu$, ennumerate vectorfileds $(i=1,2,3,4)$
\begin{eqnarray}
&&g_{\mu\nu}X_i^\mu X_j^\nu=\eta_{ij} \label{eq:vecfield}\qquad\text{ such vectorfields are not unique}\\
&&x_i^\mu \to {x'}_i^\mu=S_i^k(x) x_k^\mu \qquad S(x) \text{ is a Lorentztransformation}\nonumber
\end{eqnarray}
Define 4 covectors $\theta_\mu^i$ by the equation
\begin{equation}
\theta_\mu^iX_j^\mu=\delta_j^i \label{eq:fund}
\end{equation}
$\Rightarrow \theta_\mu^i$ ($i$ fixed) are the components of a 1-form. Hence, the $\theta^i$ are 4 linearly independent 1-forms. Moreover
\begin{equation}
g_{\mu\nu}=\eta_{ik}\theta_\mu^i\theta_\nu^k \qquad \text{$\bigl($follows from \eqref{eq:vecfield} and \eqref{eq:fund}$\bigr)$} \label{eq:theta} 
\end{equation}
If $S_i^k(x)$ is a Lorentztransformation, the replacement $\theta_\mu^i\to{\theta'}_\mu^i=S_k^i(x)\theta_\mu^k$ leaves \eqref{eq:theta} invariant.\medskip\\
Define $\theta$ as $\theta=\left(\begin{array}{c}\theta^0\\ \theta^1\\ \theta^2\\ \theta^3 \end{array}\right)$ Componentwise $\theta_\mu=\left(\begin{array}{c}\theta_\mu^0\\ \theta_\mu^1\\ \theta_\mu^2\\ \theta_\mu^3 \end{array}\right) \Rightarrow \theta$ is a 1-form which takes values in $\mathbb R^4$. The 1-form $\theta$ is called the fundamental form. \index{Fundamental form}\bigskip\\
Consider for fixed $i$ again $\theta_\mu^i$ we can apply the covariant derivative $\nabla_\nu$
\begin{equation}
\theta_\mu^i\to\nabla_\nu\theta_\mu^i \Rightarrow \nabla_\nu\theta_\mu^i=-\omega_{k\nu}^i\theta_\mu^k \label{eq:deriv}
\end{equation}
$\omega_{k\nu}^i \Rightarrow \omega_{k\nu}^i$ defines a 1-form with values in a matrix space. Call that 1-form $\omega_\mu$, then the components are $\omega_{k\nu}^i$\\
$\Rightarrow$ \eqref{eq:deriv} can be written in the compact form $\nabla_\nu\theta=-\omega_\nu\theta$\\
Equivallantly
\begin{eqnarray*}
&&\nabla_\nu\theta_\mu=-\omega_\nu\theta_\mu\\
&\Rightarrow& \nabla_\nu\theta_\mu-\nabla_\mu\theta_\nu+\omega_\nu\theta_\mu-\omega_\mu\theta_\nu=0\\
&\Rightarrow& \del{\nu}\theta_\mu-\del{\mu}\theta_\nu+\omega_\nu\theta_\mu-\omega_\mu\theta_\nu=0\\
&\Rightarrow& \fbox{$\dd\theta+\omega\wedge\theta=0$} \qquad\text{First equation of structure} \index{First equation of structure}\index{Equation of structure !-first}
\end{eqnarray*}
We know: 
\[\nabla_\nu g_{\alpha\beta}=0 \Rightarrow \nabla_\nu\eta_{ik}\theta_\alpha^i\theta_\beta^k=0 \Rightarrow 0=\eta_{ik}\left(\nabla_\nu\theta_\alpha^i\right)\theta_\beta^k+\eta_{ik}\theta_\alpha^i\nabla_\nu\theta_\beta^k\]
inserting the formula for $\nabla_\mu\theta_\alpha^i$ from the defining equation for $\omega_{k\mu}^i$
\[\Rightarrow \omega_{ik\nu}=\eta_{ij}\omega_{k\nu}^j \text{ has the property } \omega_{ik\mu}+\omega_{ki\mu}=0\]
$\Rightarrow \omega_\mu$ takes values in the Lie-algebra of the Lorentz-group. $\omega$ is called the connection form. \index{Connection form}\bigskip\\
Consider now the quantity
\[\left(\nabla_\mu\nabla_\nu-\nabla_\nu\nabla_\mu\right)\theta_\alpha^i=-\Omega_{k\mu\nu}^i\theta_\alpha^k \qquad\text{(from the linear independence of the $\theta$)}\]
\begin{eqnarray}
\nabla_\mu\left(\nabla_\nu\theta_\alpha^i\right) &=& -\nabla_\mu\omega_{k\nu}^i\theta_\alpha^k=-\left(\nabla_\mu\omega_{k\nu}^i\right)\theta_\alpha^k-\omega_{k\nu}^i\nabla_\mu\theta_\alpha^k\nonumber\\
&=& \left(\nabla_\mu\omega_{k\nu}^i\right)\theta_\alpha^k+\omega_{k\nu}^i\omega_{l\mu}^k\theta_\alpha^l\nonumber\\
\Rightarrow\left(\nabla_\mu\nabla_\nu-\nabla_\nu\nabla_\mu\right)\theta_\alpha^i &=& -\left(\del\mu\omega_{k\nu}^i-\del\nu\omega_{k\mu}^i-\omega_{l\nu}^i\omega_{k\mu}^l+\omega_{l\mu}^i\omega_{k\nu}^l\right)\theta_\alpha^k\nonumber\\
\Rightarrow \Omega_{k\mu\nu}^i &=& \del\mu\omega_{k\nu}^i-\del\nu\omega_{k\mu}^i+\omega_{l\mu}^i\omega_{k\nu}^l-\omega_{l\nu}^i\omega_{k\mu}^l \label{eq:curv}
\end{eqnarray}
$\Rightarrow \Omega_k^i$ is a two-form. $\Omega$, with matrix elements $\Omega_k^i$, is a two form with values in the Lie-algebra of the Lorentz-group. $\Omega$ is called the curvature form. \index{Curvature form}\\
\eqref{eq:curv} can be written in matrix form
\begin{eqnarray*}
\Omega &=& \dd\omega+\omega\wedge\omega\qquad\text{or}\\
\Omega_{\mu\nu} &=& \del\mu\omega_\nu-\del\nu\omega_\mu+\left[\omega_\mu,\omega_\nu\right] \qquad\text{Second equation of structure} \index{Second equation of structure}\index{Equation of structure !-second}
\end{eqnarray*}\medskip
\begin{eqnarray*}
-R_{\alpha\mu\nu}^\beta\theta_\beta^i &=& \left(\nabla_\mu\nabla_\nu-\nabla_\nu\nabla_\mu\right)\theta_\alpha^i =-\Omega_{k\mu\nu}^i\theta_\alpha^k\\
&\Rightarrow& -R_{\alpha\mu\nu}^\beta\theta_\beta^i\theta_\gamma^k\eta_{ik}=-R_{\alpha\mu\nu}^\beta g_{\beta\gamma}=-R_{\beta\alpha\mu\nu} \quad\text{Riemman's curvature tensor}\\
R_{\gamma\alpha\mu\nu} &=& \Omega_{l\mu\nu}^i\theta_\alpha^l\theta_\gamma^k\eta_{ik}\\
&=& \Omega_{kl\mu\nu}\theta_\alpha^l\theta_\gamma^k\\
&& \text{(Formula relates coeficients of curvature form and curvature tensor)}\\
R=0 &\Leftrightarrow& \Omega=0
\end{eqnarray*}\medskip\\
Suppose the curvature tensor is identically 0.\\
Theorem: Then there exists coordinates $y^\mu$ such that the metric transformed to these coordinates has the form $g_{\mu\nu}=\eta_{\mu\nu}$\\
Proof:
\begin{eqnarray}
&& \dd\theta+\omega\wedge\theta=0; \qquad\qquad \dd\omega+\omega\wedge\omega=0 \text{ or}\nonumber\\
&& \del\mu\theta_\nu-\del\nu\theta_\mu+\omega_\mu\theta_\nu-\omega_\mu\theta_\nu=0 \label{eq:a}\\
&& \del\mu\omega_\nu-\del\nu\omega_\mu+\left[\omega_\mu,\omega_\nu\right]=0 \label{eq:b}
\end{eqnarray}
Suppose $\omega_\mu$ has the form 
\begin{equation}
\omega_\mu=S^{-1}\del\mu S \label{eq:c}
\end{equation}
and $S(x)$ is a Lorentztransformation
\begin{description}
\item{1.} $\omega_\mu$ then takes values in the Lie-algebra of the Lorentz-group
\item{2.} 
\begin{eqnarray*}
\del\mu\omega_\nu &=& S^{-1}\frac{\partial^2}{\partial x^\nu\partial x\mu}S-\left(S^{-1}\del\mu S\right)S^{-1}\del\nu S\\
\del\mu\omega_\nu-\del\nu\omega_\mu &=& -\omega_\mu\omega_\nu+\omega_\nu\omega_\mu\\
&\Rightarrow& S^{-1}\del\mu S \text{ satisfies \eqref{eq:b}}
\end{eqnarray*}
General arguments show: \eqref{eq:c} is the only form of a solution of \eqref{eq:b}
\begin{eqnarray*}
&\Rightarrow& \del\mu\theta_\nu -\del\nu\theta_\mu +\left(S^{-1}\del\mu S\right)\theta_\nu -\left(S^{-1}\del\nu S\right)\theta_\mu =0\\
&\stackrel{\cdot S}{\Rightarrow}& S\del\mu\theta_\nu-S\del\nu\theta_\mu +\left(\del\mu S\right)\theta_\nu-\left(\del\nu S\right)\theta_\mu=0\\
&\Rightarrow& \del\mu S\theta_\nu-\del\nu S\theta_\mu=0\\
&\stackrel{\text{Poincar\'e-Lemma}}{\Longrightarrow}& S\theta_\nu=\del\nu\widetilde\theta\qquad \text{with } \widetilde\theta=\left(\begin{array}{c}y^0\\y^1\\y^2\\y^3 \end{array}\right)\\
g_{\alpha\beta} &=& \eta_{ik}\theta_\alpha^i\theta_\beta^k=\eta_{ik}\left(S\theta_\alpha\right)^i\left(S\theta_\beta\right)^k=\eta_{ik}\del\alpha y^i\del\beta y^k
\end{eqnarray*}
It follows immediately that written in coordinates $y^0, y^1, y^2, y^3$ the metric takes Minkowski form.
\end{description}

\section{Redefinition of vierbein and connection form}
Vierbein vector field $X_i^\mu \qquad(i=0,1,2,3)$ \index{Vierbein}
\[\theta_\nu^iX_k^\nu = \delta_k^i\]
\[A^\mu = a^iX_i^\mu \qquad\qquad a^i\text{ just functions}\]
\[\nabla_\mu A^\nu = \left(\del\mu a^i\right)X_i^\mu+a^i\nabla_\mu X_i^\nu=\left(\del\mu a^i+\omega_{k\mu}^ia^k\right)\nabla_\mu a^i\theta_\nu^i\]
\[\nabla_\mu\theta_\nu^i = -\omega_{k\mu}^i\theta_\nu^k; \qquad \left(\nabla_\mu a^i\right)=\left(\del\mu a^i+\omega_{k\mu}^ia^k\right)\]
\[a = \left(\begin{array}{c}a^0\\ a^1\\ a^2\\ a^3 \end{array}\right) \qquad\qquad \nabla_\mu a=\del\mu a+\omega_\mu a\]
Behaviour under redefinition of the vierbein and the connection form:\\
$\theta_\mu \to {\theta'}_\mu=S\theta_\mu \qquad S(x)$ is a space-time dependent Lorentztransformation. Under such a change 
\begin{eqnarray*}
&\Rightarrow& a\to a'=S^{-1}(x)a\\
&& \omega \to \omega'=S^{-1}\del\mu S+S^{-1}\omega S\\
&& \left(\nabla_\mu a\right) \to \left(\nabla_\mu a\right)'=S\nabla_\mu a
\end{eqnarray*}
Application of this construction to spinor fields.\\
Spinor fields are functions $M\to\mathbb C^4\quad(M \text{ space-time})$ Dirac-matrices should be defiened with properties:
\begin{equation}
\left[\gamma_\mu,\gamma_\nu\right]_+=2g_{\mu\nu}\mathbb I_4 \label{eq:gamma}\index{Dirac-matrices}
\end{equation}
Solution for \eqref{eq:gamma}: $\gamma_\mu=\gamma_i\theta_\mu^i$. Check \eqref{eq:gamma}
\[\theta_\mu^i\theta_\nu^k\left[\gamma_i,\gamma_k\right]_+=2\theta_\mu^i\theta_\nu^k\eta_{ik}\mathbb I_4=2g_{\mu\nu}\mathbb I_4\]
Under changes of the $\theta_\mu$ by a Lorentztransformation, the construction of $\gamma_\mu$ changes by a similarity transformation: $\gamma_\mu\to\gamma'_\mu=\widetilde S\gamma_\mu\widetilde S^{-1}$\\
Define the covariant derivative of spinor fields by
\begin{eqnarray*}
&& \nabla_\mu\Psi=\del\mu\Psi+\omega_\mu^{ik}\gamma_i\gamma_k\Psi\\
&\Rightarrow& \Psi\to\widetilde S\Psi \qquad\qquad \omega\to S^{-1}\del\mu S+S^{-1}\omega_\mu S\\
&& \nabla_\mu\Psi\to S\nabla_\mu\Psi\\
&& \fbox{$g^{\mu\nu}\gamma_\mu\nabla_\nu\Psi=m\Psi$}\qquad\text{Dirac equation}\index{Dirac equation}
\end{eqnarray*}
Coupling in addition to an external electromagnetic field\\
Replace $\nabla_\mu\to\nabla'_\mu=\nabla_\mu+iqA_\mu \qquad A_\mu$ is the vectorpotential of the electromagnetic field.

\section{Parallel transport}\index{Parallel transport}

%insert image  parallel transport

Coordinates $x^\mu(\tau)$\\
The vector $y(\tau)$ is satisfiing the differential equation:
\[\dot y^\mu(\tau)+\Gamma_{\alpha\beta}^\mu\left(x(\tau)\right)\dot x^\alpha(\tau)y^\beta(\tau)=0\]
$\Rightarrow y(\tau)$ is uniquely specified by $y(0)=k$\medskip\\
Definition: $y(\tau)$ is called the parralel transport of $k$ to the point $x(\tau)$.\\
Let $y_k(\tau)$ and $y_{k'}(\tau)$ be two such parralel transports along $x(\tau)$ with $k$ and $k'$ respective. Consider the scalar product $g_{\mu\nu}\left(x(\tau)\right)y_k^\mu(\tau)y_{k'}^\nu(\tau)$
\begin{eqnarray*}
\frac{\dd}{\dd\tau}\ g_{\mu\nu}\left(x(\tau)\right)y_k^\mu(\tau)y_{k'}^\nu(\tau) &=& \dot x^\alpha(\tau)\del\alpha g_{\mu\nu}\left(x(\tau)\right)y_k^\mu(\tau)y_{k'}^\nu(\tau)\\
&& +g_{\mu\nu}\left(x(\tau)\right)\dot y_k^\mu y_{k'}^\nu +g_{\mu\nu}\left(x(\tau)\right)y_k^\mu\dot y_{k'}^\nu\\
&=& \dot x^\alpha g_{\mu\nu;\alpha}y_k^\mu(\tau)y_{k'}^\nu(\tau)=0\\
g_{\mu\nu}\left(x(\tau)\right)y_k^\mu(\tau)y_{k'}^\nu(\tau) &=& g_{\mu\nu}\left(x(\tau)\right)k^\mu {k'}^\nu
\end{eqnarray*}
$\Rightarrow$ Angles and length of a parallel transported vector along a given curve don't change
In general then result of parallel transports depends on the curve.

%insert figure closed curve

Let $x(\tau)$ be a closed curve $x(1)=x(0)$\\
We have: $g_{\mu\nu}\left(x(0)\right)k^\mu{k'}^\nu=g_{\mu\nu}\bigl(\underbrace{x(1)}_{=x(0)}\bigr)y_k^\mu(1)y_{k'}^\nu(1)$\\
Assume: $x^\mu$ are normal coordinates $\Rightarrow g_{\mu\nu}\left(x(0)\right)=\eta_{\mu\nu} \Rightarrow y_k(1)=\Lambda k$\\
In Riemannian normal coordinates:
\[\dot y_k^\mu(1)+\Gamma_{\alpha\beta}^\mu\left(x(1)\right)\dot x^\alpha(1)y_k(1)=0\]
\[y_k^\mu(\tau)=k^\nu+\int_0^\tau\!\!\!\!\Gamma_{\alpha\beta}^\mu\left(x(\tau')\right)\dot x^\alpha(\tau')y_k(\tau')\dd\tau'\]
Put $A_\beta^\mu(\tau)=\gamma_{\alpha\beta}^\mu\left(x(\tau)\right)\dot x^\alpha(\tau) \quad\Rightarrow y_k(\tau)=B(\tau)k$
\[B(\tau)=\mathbb I+\int_0^\tau\!\!\!\!A(\tau')\dd\tau'+\int_0^\tau\!\!\!\!\dd\tau'\int_0^{\tau'}\!\!\!\!\dd\tau''A(\tau')A(\tau'')+\ldots\]
Take $x(\tau)=\alpha x_0(\tau)$. Evaluate $B$ in quadratic approximation (up to $\alpha^2$)
\begin{eqnarray*}
A_\beta^\mu\left(x(\tau)\right) &=& \alpha\underbrace{\Gamma_{\alpha\beta}^\mu(0)}_{=0}\dot x_0^\alpha(\tau)+\alpha^2x_0^\gamma(\tau)\del\gamma\Gamma_{\alpha\beta}^\mu(0)\dot x_0^\alpha(\tau)\\
B(\tau) &=& \mathbb I+\int_0^\tau\!\!\!\!\alpha^2\Gamma_{\alpha\beta,\gamma}^\mu(0)x_0^\gamma(\tau')\dot x_0^\alpha(\tau')\dd\tau'
\end{eqnarray*}
For a closed curve with $x(0)=x(1)$
\begin{eqnarray*}
B(1) &=& \mathbb I +\alpha^2\underbrace{\int_0^1\Gamma_{\alpha\beta,\gamma}^\mu(0)\frac{x_0^\gamma(\tau')\dot x_0^\alpha(\tau')+x_0^\alpha(\tau)\dot x_0^\gamma(\tau)}{2}}_{=0}\\
&& +\Gamma_{\alpha\beta,\gamma}^\mu(0)\frac{x_0^\gamma(\tau')\dot x_0^\alpha(\tau')-x_0^\alpha(\tau)\dot x_0^\gamma(\tau)}{2}\ \dd\tau'\\
B(1) &=& \mathbb I +\frac{\alpha^2}{2}\int_0^1\!\!\!\!\dd\tau' \left(\Gamma_{\alpha\beta,\gamma}^\mu(0)-\Gamma_{\gamma\beta,\alpha}^\mu(0)\right) x_0^\gamma\dot x_0^\alpha\\
R_{\beta\alpha\gamma}^\mu(0) &=& \Gamma_{\alpha\beta,\gamma}^\mu(0)-\Gamma_{\gamma\beta,\alpha}^\mu(0)\\
\Rightarrow B_\beta^\mu(1) &=& \delta_\beta^\mu+\frac{\alpha^2}{2}\int_0^1\!\!\!\!R_{\alpha\beta\gamma}^\mu(0)x_0^\gamma(\tau')\dot x_0^\alpha(\tau')\dd\tau'\\
&=& \delta_\beta^\mu+\frac{1}{2}\int_x\!\!\!\!A_{\beta\alpha}^\mu\dd x^\alpha \qquad\quad \text{with } A_{\beta\alpha}^\mu=R_{\beta\alpha\gamma}^\mu x^\gamma; \qquad x(\tau)=\alpha x_0(\tau)
\end{eqnarray*}
Let $\mathcal F$ be the surface bounded by the closed curve $x(\tau)$\medskip\\
Gauss theorem $\Rightarrow$ \fbox{$B_\beta^\mu=\delta_\beta^\mu+\frac{1}{2}\int_\mathcal F R_{\beta\alpha\gamma}^\mu(0)\dd x^\gamma\wedge\dd x^\alpha$}\medskip\\
$\Rightarrow$ Parallel transport yields $B_\beta^\mu=\delta_\beta^\mu$ if the curvature is zero!\medskip\\
Actual test of this concept: Put a gyroscope on a satellite.


\chapter{Kaluza-Klein-theory} \index{Kaluza-Klein-theory}
An attempt to unite electromagnetism and gravitation in a 5-dimensional world.\\
Basic idea: Enlarge the 4-dimensional world (coordinates $x^0,x^1,x^2,x^3$) with metric $g_{\mu\nu}(x)$ by a fifth dimension (coordinate $x^4$). Let the metric of 5-dimensional world then be given by a metric $\gamma_{\mu\nu}$ with
\begin{eqnarray*}
&&\gamma_{\mu\nu}=g_{\mu\nu}+B_\mu B_\nu; \qquad \gamma_{4\mu}=\gamma_{\mu4}=B_\mu; \qquad \gamma_{44}=1\\
&\Rightarrow& \gamma=\left(\dd x^4+B_\mu\dd x^\mu\right)^2 +g_{\mu\nu}\dd x^\mu\dd x^\nu \qquad\qquad \nu,\mu=0,1,2,3
\end{eqnarray*}
$B_\mu$ and $g_{\mu\nu}$ should be independent of $x^4$.\medskip\\
$\gamma^{\mu\nu}$ has components
\[\gamma^{44}=1+g^{\mu\nu}B_\mu B_\nu; \qquad \gamma^{4\mu}=-g^{\mu\nu}B_\nu; \qquad \gamma^{\mu\nu}=g^{\mu\nu}; \qquad \det\gamma=\det g\]\bigskip\\
The geodesic equation in the 5-dimensional world \index{Geodesic equation!- 5-dimensional}
\[\frac{\dd}{\dd\tau}\ \gamma_{\mu\nu}\dot x^\nu=\frac{1}{2}\del\mu\ \gamma_{\alpha\beta}\dot x^\alpha\dot x^\beta \qquad\qquad \text{(indices running from 0 to 4)}\]
Inserting expressions for $\gamma_{\mu\nu}$:
\begin{eqnarray*}
\mu=4:\qquad\quad\ \frac{\dd}{\dd\tau}\left(\dot x^4+B_\alpha\dot x^\alpha\right)=0\ \qquad\qquad &\Rightarrow& \dot x^4+B_\alpha\dot x^\alpha=c=\const\\
\mu=0\dots3:\quad \frac{\dd}{\dd\tau}\left(g_{\alpha\beta}\dot x^\beta+B_\alpha\left(\dot x^4+B_\nu\dot x^\nu\right)\right) &=& \frac{1}{2}\left(\del\alpha g_{\nu\delta}\right)\dot x^\nu\dot x^\delta\\ && +\left(\del\alpha B_\nu\right)\dot x^\nu\left(\dot x^4+B_\delta\dot x^\delta\right)\\
\frac{\dd}{\dd\tau}\ g_{\mu\nu}\dot x^\nu =\frac{1}{2}\left(\del\alpha\ g_{\nu\delta}\right)\dot x^\nu\dot x^\delta+c f_{\alpha\nu}\dot x^\nu && f_{\alpha\nu}=\del\alpha\ B\nu-\del\nu\ B_\alpha
\end{eqnarray*}
Raising the index $\mu$ with $g^{\mu\nu}$:
\[\ddot x^\mu+\Gamma_{\alpha\beta}^\mu\dot x^\alpha\dot x^\beta=c f_\alpha^\mu\dot x^\alpha\]
Put $c=\frac{q}{m}\lambda^{-1}$ and identify $\lambda^{-1}f_\alpha^\mu$ with $F_\alpha^\mu$ (electromagnetic field tensor) \index{Electromagnetic field tensor}\\
Mass and charge appear as integration constant.\medskip\\
The Ricci scalar associated to $\gamma_{\mu\nu}$ \index{Ricci scalar!- in Kaluza-Klein-theory}
\[\Rightarrow P=R+\frac{1}{4}f_\alpha^\mu f_\mu^\alpha \qquad\qquad \sqrt{|\det \gamma|}\dd^5x=\dd^5x\sqrt{|\det g|}\]
The action in Einstein's theory was
\[-\frac{1}{16\pi G}\int\!\!\!\dd^4x\sqrt{|g|}R-\frac{1}{16\pi}\int\!\!\!\dd^4x\sqrt{|g|}F^{\mu\nu}F_{\mu\nu}\]
in order that $-P\sqrt{|\det\gamma|}$ and $\frac{-1}{16\pi G}\sqrt{|g|}R-\frac{1}{16\pi}\sqrt{|g|}F_{\mu\nu}F^{\mu\nu}$ are proportional, with
\[\lambda^{-1}f_\alpha^\mu=F_\alpha^\mu \Rightarrow \lambda=2\sqrt G\]\medskip\\
$\lambda^{-1}f_\alpha^\mu=F_\alpha^\mu$ implies $\lambda^{-1}B_\mu=A_\mu$\\
$\Rightarrow$ gauge transformations of $A_\mu$ implies $B_\mu\to B'_\mu=B_\mu+\del\mu\ \varphi$
\begin{eqnarray*}
\Rightarrow \gamma\to\gamma' &=& \left(\dd x^4+\left(B_\mu+\del\mu\ \varphi\right)\dd x^\mu\right)^2+g_{\mu\nu}\dd x^\mu\dd x^\nu\\
&=& \left(\dd\left(x^4+\varphi\right)+B_\mu\dd x^\mu\right)^2+g_{\mu\nu}\dd x^\mu\dd x^\nu
\end{eqnarray*}
Put $x'^4=x^4+\varphi$, this compensates the gauge transformation.

\section{Quantum mechanics with Kaluza-Klein} \index{Quantum mechanics with Kaluza-Klein}
The Klein-Gordon equation in 5 dimensions \index{Klein-Gordon equation}
\begin{equation}
\left(\frac{1}{\sqrt{|\det\gamma|}}\del\mu\sqrt{|\det\gamma|}\ \gamma^{\mu\nu}\del\nu-m_0^2\right)\Phi=0 \qquad\qquad \mu=0,\dots,4
\label{eq:kleingordon}
\end{equation}
$\det\gamma=\det g \qquad\quad \gamma^{44}=1+\gamma^{\mu\nu}\gamma_{4\mu}\gamma_{4\nu} \qquad\quad \gamma^{\mu4}=\gamma^{4\mu}=g^{\mu\nu}\gamma_{\nu4} \qquad\quad \gamma^{\mu\nu}=g^{\mu\nu}$
\[\Rightarrow \frac{1}{\sqrt{|g|}}\left(\del\mu+2\sqrt{G}A_\mu\del 4\right)\sqrt{|g|} \left(\del\nu+2\sqrt{G}A_\nu\del 4\right)\Phi -\left(m_0^2-\left(\del 4\right)^2\right)\Phi=0\]
Ansatz: $\Phi=\e^{icx^4}\widetilde\Phi(x) \qquad\qquad x=\left(x^0,x^1,x^2,x^3\right)$
\[\Rightarrow\frac{1}{\sqrt{|g|}}\left(\del\mu+ic2\sqrt{G}A_\mu\right)\sqrt{|g|} \left(\del\nu+ic2\sqrt{G}A_\nu\right)\widetilde\Phi-\left(m_0^2+c^2\right)\widetilde\Phi=0\]
\[c=\left(2\sqrt{G}\right)^{-1}q \qquad\quad q\text{ is the charge}\]
Charge is quantized in nature: $q=k e_0$\\
$k$ is an integer and $e_0$ is the quantisation unit of charge
\[\Rightarrow\frac{1}{\sqrt{|g|}}\left(\del\mu+ie_0kA_\mu\right)\sqrt{|g|}g^{\mu\nu} \left(\del\nu+ie_0kA_\nu\right)\widetilde\Phi -\left(m_0^2+k^2\frac{e_0^2}{\left(2\sqrt{G}\right)^2}\right)\widetilde\Phi=0\]
\subsubsection*{Remark 1:}
\[\Phi=\e^{ik\left(\frac{e_0}{2\sqrt{G}}x^4\right)}\widetilde\Phi\]
$x^4$ dependence is periodic with period $\dfrac{2\pi}{L}; \quad L^{-1}=\dfrac{e_0}{2\sqrt{G}}$\\
$\Rightarrow$ The new coordinate is in fact an angle. $\Rightarrow$ A circle (1-shpere) has been added to the 4-dimensional world.
\subsubsection*{Remark 2:}
The effictive mass is charge-dependent: $\quad m_0^2 \to m_0^2+\frac{k^2e_0^2}{4G}$\\
$\dfrac{k^2e_0^2}{4G}$ is a very large number in MeV
\begin{center} \underline{Not realized in nature!} \end{center}

\section{Generalizations}
\[\gamma_{44}=1 \qquad\quad \gamma_{4\mu}=\gamma_{\mu4}=B_\mu \qquad\quad \gamma_{\mu\nu}=g_{\mu\nu}+\gamma_{4\mu}\gamma_{4\nu}\]
\[\Rightarrow \del 4\gamma_{\mu\nu}=0 \rightarrow x^4\to x'^4=x^4+a \text{ is a 1-parameter symetric group of our metric}\]
Killing vector $X^\mu$ has components $X^4=1, \quad X^\mu=0$ other components equal 0\\
$\gamma_{\mu\nu}X^\mu X^\nu=1$. This condition can be relaxed. $\Rightarrow$ Theory will effictively variable gravitational constants.\medskip\\
Next generalization: Relax \eqref{eq:kleingordon} allow for a periodic dependence on $x^4$.



\newpage
\addcontentsline{toc}{chapter}{Index}
\printindex
\end{document}