%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % % Scientific Word Wrap/Unwrap Version 2.5 % % Scientific Word Wrap/Unwrap Version 3.0 % % % % If you are separating the files in this message by hand, you will % % need to identify the file type and place it in the appropriate % % directory. The possible types are: Document, DocAssoc, Other, % % Macro, Style, Graphic, PastedPict, and PlotPict. Extract files % % tagged as Document, DocAssoc, or Other into your TeX source file % % directory. Macro files go into your TeX macros directory. Style % % files are used by Scientific Word and do not need to be extracted. % % Graphic, PastedPict, and PlotPict files should be placed in a % % graphics directory. % % % % Graphic files need to be converted from the text format (this is % % done for e-mail compatability) to the original 8-bit binary format. % % % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % % Files included: % % % % "/document/hx2sol.tex", Document, 34313, 12/17/2001, 20:03:34, "" % % % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%% Start /document/hx2sol.tex %%%%%%%%%%%%%%%%%%%%% \documentclass{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Created=Wednesday, August 23, 2000 15:54:58} %TCIDATA{LastRevised=Monday, December 17, 2001 15:03:33} %TCIDATA{} %TCIDATA{} %TCIDATA{CSTFile=LaTeX article (bright).cst} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \begin{center} {\LARGE Hour Exam No.2\medskip } \end{center} Please attempt all of the following problems before the due date. All problems count the same even though some are more complex than others.% \vspace{0.5in} Several of the following problems require the connection coefficients for the connection compatible with the Schwarzschild metric: \[ \left[ \begin{array}{cccc} g_{00} & g_{01} & g_{02} & g_{03} \\ g_{10} & g_{11} & g_{12} & g_{13} \\ g_{20} & g_{21} & g_{22} & g_{23} \\ g_{30} & g_{31} & g_{32} & g_{33}% \end{array}% \right] =\left[ \begin{array}{cccc} -\left( 1-\frac{2m}{r}\right) & 0 & 0 & 0 \\ 0 & \frac{1}{1-\frac{2m}{r}} & 0 & 0 \\ 0 & 0 & r^{2} & 0 \\ 0 & 0 & 0 & r^{2}\sin ^{2}\theta% \end{array}% \right] \]% For the coordinates $x^{0}=t,x^{1}=r,x^{2}=\theta ,x^{3}=\varphi $ and the corresponding holonomic basis, the non-zero connection coefficients are as follows:% \begin{eqnarray*} \Gamma ^{0}{}_{01} &=&\Gamma ^{0}{}_{10}=\frac{m}{r\left( r-2m\right) } \\ \Gamma ^{1}{}_{00} &=&\frac{m}{r^{3}}\left( r-2m\right) \\ \Gamma ^{1}{}_{11} &=&-\frac{m}{r\left( r-2m\right) } \\ \Gamma ^{1}{}_{22} &=&-\left( r-2m\right) \\ \Gamma ^{1}{}_{33} &=&-\left( r-2m\right) \sin ^{2}\theta \\ \Gamma ^{2}{}_{12} &=&\Gamma ^{2}{}_{21}=\frac{1}{r} \\ \Gamma ^{2}{}_{33} &=&-\sin \theta \cos \theta \\ \Gamma ^{3}{}_{13} &=&\Gamma ^{3}{}_{31}=\frac{1}{r} \\ \Gamma ^{3}{}_{23} &=&\Gamma ^{3}{}_{32}=\cot \theta \end{eqnarray*}% \pagebreak {\LARGE Problem 1\medskip } For the connection given above, show that, for any vector fields $u,v$ the following conditions are satisfied:\medskip \begin{enumerate} \item[{\protect\LARGE a.}] For any function $f,$% \[ D_{u}D_{v}f=D_{v}D_{u}f \] \end{enumerate} Hint: Switch to index notation and see the result immediately.\bigskip {\large Answer 1a}{\LARGE \medskip } Note that $D_{u}D_{v}f-D_{v}D_{u}f$ is locally linear in $u,v$. Expand $u,v$ in a holonomic basis $\partial _{\alpha }$ and find% \[ D_{u}D_{v}f-D_{v}D_{u}f=u^{\alpha }v^{\beta }\left( D_{\partial _{\alpha }}D_{\partial _{\beta }}f-D_{\partial _{\beta }}D_{\partial _{\alpha }}f\right) \]% Now write out the second covariant derivative components in this basis% \[ D_{\partial _{\alpha }}D_{\partial _{\beta }}f=f_{,\beta \alpha }-f_{,\rho }\Gamma ^{\rho }{}_{\beta \alpha } \]% so that% \begin{eqnarray*} D_{u}D_{v}f-D_{v}D_{u}f &=&u^{\alpha }v^{\beta }\left( f_{,\beta \alpha }-f_{,\rho }\Gamma ^{\rho }{}_{\beta \alpha }-f_{,\alpha \beta }+f_{,\rho }\Gamma ^{\rho }{}_{\alpha \beta }\right) \\ &=&u^{\alpha }v^{\beta }\left( f_{,\beta \alpha }-f_{,\alpha \beta }-f_{,\rho }\Gamma ^{\rho }{}_{\beta \alpha }+f_{,\rho }\Gamma ^{\rho }{}_{\alpha \beta }\right) \\ &=& \\ &=&u^{\alpha }v^{\beta }f_{,\rho }\left( -\Gamma ^{\rho }{}_{\beta \alpha }+\Gamma ^{\rho }{}_{\alpha \beta }\right) \end{eqnarray*}% Notice that the connection coefficients given above have the symmetry% \[ \fbox{$\Gamma ^{\rho }{}_{\alpha \beta }=\Gamma ^{\rho }{}_{\beta \alpha }$} \]% so that% \[ \fbox{$D_{u}D_{v}f-D_{v}D_{u}f=0$.} \]% \vspace{0.5in} \begin{enumerate} \item[{\protect\LARGE b.}] For any vector field $w$% \[ D_{w}\left( u\cdot v\right) =\left( D_{w}u\right) \cdot v+u\cdot D_{w}v \] \end{enumerate} Hint: Switch to index notation and slog it out component by component using Maple to do the algebra.\bigskip {\large Answer 1b}{\LARGE \medskip } This condition is just metric compatibility which, in index notation is% \[ g_{\alpha \beta ;\delta }=0 \]% or% \[ g_{\alpha \beta ,\delta }-g_{\alpha \rho }\Gamma ^{\rho }{}_{\beta \delta }-g_{\rho \beta }\Gamma ^{\rho }{}_{\alpha \delta }=0 \]% Check it component by component, noticing that the expression is symmetric in $\alpha $ and $\beta $: $\alpha =0$ $g_{0\beta ,\delta }-g_{0\rho }\Gamma ^{\rho }{}_{\beta \delta }-g_{\rho \beta }\Gamma ^{\rho }{}_{0\delta }=0$ $g_{0\beta ,\delta }-g_{00}\Gamma ^{0}{}_{\beta \delta }-g_{\rho \beta }\Gamma ^{\rho }{}_{0\delta }=0$ $\beta =0$ $g_{00,\delta }-g_{00}\Gamma ^{0}{}_{0\delta }-g_{\rho 0}\Gamma ^{\rho }{}_{0\delta }=0$ $g_{00,\delta }-g_{00}\Gamma ^{0}{}_{0\delta }-g_{00}\Gamma ^{0}{}_{0\delta }=0$ $g_{00,\delta }-2g_{00}\Gamma ^{0}{}_{0\delta }=0$ The only non-zero $\Gamma ^{0}{}_{0\delta }$ is for $\delta =1$ and similarly the only non-zero $g_{00,\delta }$ is for $\delta =1$. Thus, we only need to check that case. \ Use Maple Expand and the Simplify to do the algebra: $g_{00,1}-2g_{00}\Gamma ^{0}{}_{01}=\frac{d}{dr}\left( -\left( 1-\frac{2m}{r}% \right) \right) -2\left( -\left( 1-\frac{2m}{r}\right) \right) \left( \frac{m% }{r\left( r-2m\right) }\right) =\allowbreak -2\frac{m}{r^{2}}-2\left( -1+2% \frac{m}{r}\right) \frac{m}{r\left( r-2m\right) }$ $g_{00,1}-2g_{00}\Gamma ^{0}{}_{01}=-2\frac{m}{r^{2}}-2\left( -1+2\frac{m}{r}% \right) \frac{m}{r\left( r-2m\right) }=\allowbreak 0$ $\beta =1$ $g_{01,\delta }-g_{00}\Gamma ^{0}{}_{1\delta }-g_{\rho 1}\Gamma ^{\rho }{}_{0\delta }=0$ $-g_{00}\Gamma ^{0}{}_{1\delta }-g_{11}\Gamma ^{1}{}_{0\delta }=0$ The only non-zero $\Gamma ^{0}{}_{1\delta }$ and also the only non-zero $% \Gamma ^{1}{}_{0\delta }$ is for $\delta =0$ so only that case needs to be checked. $-g_{00}\Gamma ^{0}{}_{10}-g_{11}\Gamma ^{1}{}_{00}=-\left( -\left( 1-\frac{% 2m}{r}\right) \right) \left( \frac{m}{r\left( r-2m\right) }\right) -\frac{1}{% 1-\frac{2m}{r}}\frac{m}{r^{3}}\left( r-2m\right) \allowbreak $ := $0$ $\beta =2$ $g_{02,\delta }-g_{00}\Gamma ^{0}{}_{2\delta }-g_{\rho 2}\Gamma ^{\rho }{}_{0\delta }=0$ $-g_{00}\Gamma ^{0}{}_{2\delta }-g_{22}\Gamma ^{2}{}_{0\delta }=0$ There are no non-zero coefficients with both 2 and 0 indexes, so this is indeed zero. $\beta =3$ $g_{03,\delta }-g_{00}\Gamma ^{0}{}_{3\delta }-g_{\rho 3}\Gamma ^{\rho }{}_{0\delta }=0$ $-g_{00}\Gamma ^{0}{}_{3\delta }-g_{22}\Gamma ^{3}{}_{0\delta }=0$ There are no non-zero coefficients with both 3 and 0 indexes, so these components are also zero. $\alpha =1$ $g_{1\beta ,\delta }-g_{1\rho }\Gamma ^{\rho }{}_{\beta \delta }-g_{\rho \beta }\Gamma ^{\rho }{}_{1\delta }=0$ $g_{1\beta ,\delta }-g_{11}\Gamma ^{1}{}_{\beta \delta }-g_{\rho \beta }\Gamma ^{\rho }{}_{1\delta }=0$ We already checked $\left( \alpha ,\beta \right) =\left( 0,1\right) $ so $% \left( \alpha ,\beta \right) =\left( 1,0\right) $ will also work. No need to check $\beta =0$. $\beta =1$ $g_{11,\delta }-g_{11}\Gamma ^{1}{}_{1\delta }-g_{\rho 1}\Gamma ^{\rho }{}_{1\delta }=0$ $g_{11,\delta }-g_{11}\Gamma ^{1}{}_{1\delta }-g_{11}\Gamma ^{1}{}_{1\delta }=0$ $g_{11,\delta }-2g_{11}\Gamma ^{1}{}_{1\delta }=0$ Both terms are zero except for $\delta =1$ so check that case. $g_{11,1}-2g_{11}\Gamma ^{1}{}_{11}=\frac{d}{dr}\left( \frac{1}{1-\frac{2m}{r% }}\right) -2\left( \frac{1}{1-\frac{2m}{r}}\right) \left( -\frac{m}{r\left( r-2m\right) }\right) =\allowbreak -2\frac{m}{\left( -r+2m\right) ^{2}}+\frac{% 2}{1-2\frac{m}{r}}\frac{m}{r\left( r-2m\right) }=\allowbreak 0.$ $\beta =2$ $g_{12,\delta }-g_{11}\Gamma ^{1}{}_{2\delta }-g_{\rho 2}\Gamma ^{\rho }{}_{1\delta }=0$ $-g_{11}\Gamma ^{1}{}_{2\delta }-g_{22}\Gamma ^{2}{}_{1\delta }=0$ The coefficients are zero except for $\delta =2$ so check that case. $-g_{11}\Gamma ^{1}{}_{22}-g_{22}\Gamma ^{2}{}_{12}=-\left( \frac{1}{1-\frac{% 2m}{r}}\right) \left( -\left( r-2m\right) \right) -r^{2}\left( \frac{1}{r}% \right) =\allowbreak 0.$ $\beta =3$ $g_{13,\delta }-g_{11}\Gamma ^{1}{}_{3\delta }-g_{\rho 3}\Gamma ^{\rho }{}_{1\delta }=0$ $-g_{11}\Gamma ^{1}{}_{3\delta }-g_{33}\Gamma ^{3}{}_{1\delta }=0$ The coefficients are zero except for $\delta =3$ so check that case. $-g_{11}\Gamma ^{1}{}_{33}-g_{33}\Gamma ^{3}{}_{13}=-\left( \frac{1}{1-\frac{% 2m}{r}}\right) \left( -\left( r-2m\right) \sin ^{2}\theta \right) -\left( r^{2}\sin ^{2}\theta \right) \frac{1}{r}=\allowbreak 0.$ $\alpha =2$ $g_{2\beta ,\delta }-g_{2\rho }\Gamma ^{\rho }{}_{\beta \delta }-g_{\rho \beta }\Gamma ^{\rho }{}_{2\delta }=0$ $g_{2\beta ,\delta }-g_{22}\Gamma ^{2}{}_{\beta \delta }-g_{\rho \beta }\Gamma ^{\rho }{}_{2\delta }=0$ No need to check $\beta =0$ or $\beta =1$ because of symmetry. $\beta =2$ $g_{22,\delta }-g_{22}\Gamma ^{2}{}_{2\delta }-g_{\rho 2}\Gamma ^{\rho }{}_{2\delta }=0$ $g_{22,\delta }-g_{22}\Gamma ^{2}{}_{2\delta }-g_{22}\Gamma ^{2}{}_{2\delta }=0$ $g_{22,\delta }-2g_{22}\Gamma ^{2}{}_{2\delta }=0$ Both terms are zero except for $\delta =1$ so check that case. $g_{22,1}-2g_{22}\Gamma ^{2}{}_{21}=\frac{d}{dr}\left( r^{2}\right) -2r^{2}% \frac{1}{r}=\allowbreak 0.$ $\beta =3$ $g_{23,\delta }-g_{22}\Gamma ^{2}{}_{3\delta }-g_{\rho 3}\Gamma ^{\rho }{}_{2\delta }=0$ $-g_{22}\Gamma ^{2}{}_{3\delta }-g_{33}\Gamma ^{3}{}_{2\delta }=0$ Both terms are zero except for $\delta =3$ so check that case. $-g_{22}\Gamma ^{2}{}_{33}-g_{33}\Gamma ^{3}{}_{23}=-r^{2}\left( -\sin \theta \cos \theta \right) -\left( r^{2}\sin ^{2}\theta \right) \cot \theta =\allowbreak 0.$ $\alpha =3$ $g_{3\beta ,\delta }-g_{3\rho }\Gamma ^{\rho }{}_{\beta \delta }-g_{\rho \beta }\Gamma ^{\rho }{}_{3\delta }=0$ $g_{3\beta ,\delta }-g_{33}\Gamma ^{3}{}_{\beta \delta }-g_{\rho \beta }\Gamma ^{\rho }{}_{3\delta }=0$ No need to check $\beta =0$ or $\beta =1$ or $\beta =2$ because of symmetry. $\beta =3$ $g_{33,\delta }-g_{33}\Gamma ^{3}{}_{3\delta }-g_{\rho 3}\Gamma ^{\rho }{}_{3\delta }=0$ $g_{33,\delta }-g_{33}\Gamma ^{3}{}_{3\delta }-g_{33}\Gamma ^{3}{}_{3\delta }=0$ $g_{33,\delta }-2g_{33}\Gamma ^{3}{}_{3\delta }=0$ Both terms are zero except for $\delta =1$ so check that case. $g_{33,1}-2g_{33}\Gamma ^{3}{}_{31}=\frac{d}{dr}\left( r^{2}\sin ^{2}\theta \right) -2\left( r^{2}\sin ^{2}\theta \right) \frac{1}{r}=\allowbreak 0.$ All components vanish and the expression is verified.\vspace{0.5in}\pagebreak {\LARGE Problem 2\medskip } The wave operator on a scalar field can be written in the form% \[ \nabla ^{2}\Phi =g^{\alpha \beta }\Phi _{;\alpha \beta } \] Use the connection coefficients given at the beginning of this exam to find the wave equation near a black hole of mass $m$.\bigskip {\large Answer 2}{\LARGE \medskip } $\Phi _{;\alpha \beta }=\Phi _{,\alpha \beta }-\Phi _{,\rho }\Gamma ^{\rho }{}_{\alpha \beta }$ $\nabla ^{2}\Phi =g^{\alpha \beta }\Phi _{;\alpha \beta }=g^{\alpha \beta }\left( \Phi _{,\alpha \beta }-\Phi _{,\rho }\Gamma ^{\rho }{}_{\alpha \beta }\right) $ $\nabla ^{2}\Phi =g^{\alpha \beta }\Phi _{,\alpha \beta }-\Phi _{,\rho }g^{\alpha \beta }\Gamma ^{\rho }{}_{\alpha \beta }$ Now work out what $g^{\alpha \beta }\Gamma ^{\rho }{}_{\alpha \beta }$ is, component by component. Note that the inverse metric tensor components are: $g^{00}=\left( g_{00}\right) ^{-1}=\left( -\left( 1-\frac{2m}{r}\right) \right) ^{-1}=\allowbreak \frac{1}{-1+2\frac{m}{r}}=\allowbreak \frac{r}{% -r+2m}=-\frac{r}{r-2m}$ is true $g^{11}=\left( g_{11}\right) ^{-1}=\left( \frac{1}{1-\frac{2m}{r}}\right) ^{-1}=\allowbreak -\frac{-r+2m}{r}=\frac{r-2m}{r}$ is true $g^{22}=\left( g_{22}\right) ^{-1}=\left( r^{2}\right) ^{-1}=\frac{1}{r^{2}}$ $g^{33}=\left( g_{33}\right) ^{-1}=\left( r^{2}\sin ^{2}\theta \right) ^{-1}=% \frac{1}{r^{2}\sin ^{2}\theta }$ so $g^{\alpha \beta }\Gamma ^{0}{}_{\alpha \beta }=g^{00}\Gamma ^{0}{}_{00}+g^{11}\Gamma ^{0}{}_{11}+g^{22}\Gamma ^{0}{}_{22}+g^{33}\Gamma ^{0}{}_{33}=0$ $g^{\alpha \beta }\Gamma ^{1}{}_{\alpha \beta }=g^{00}\Gamma ^{1}{}_{00}+g^{11}\Gamma ^{1}{}_{11}+g^{22}\Gamma ^{1}{}_{22}+g^{33}\Gamma ^{1}{}_{33}$ $g^{\alpha \beta }\Gamma ^{1}{}_{\alpha \beta }=\left( -\frac{r}{r-2m}% \right) \frac{m}{r^{3}}\left( r-2m\right) +\left( \frac{r-2m}{r}\right) \left( -\frac{m}{r\left( r-2m\right) }\right) +\frac{1}{r^{2}}\left( -\left( r-2m\right) \right) +\frac{1}{r^{2}\sin ^{2}\theta }\left( -\left( r-2m\right) \sin ^{2}\theta \right) =\allowbreak 2\frac{m-r}{r^{2}}$ is true $g^{\alpha \beta }\Gamma ^{2}{}_{\alpha \beta }=g^{00}\Gamma ^{2}{}_{00}+g^{11}\Gamma ^{2}{}_{11}+g^{22}\Gamma ^{2}{}_{22}+g^{33}\Gamma ^{2}{}_{33}$ \qquad $=\left( -\frac{r}{r-2m}\right) \Gamma ^{2}{}_{00}+\left( \frac{r-2m}{% r}\right) \Gamma ^{2}{}_{11}+\frac{1}{r^{2}}\Gamma ^{2}{}_{22}+\frac{1}{% r^{2}\sin ^{2}\theta }\Gamma ^{2}{}_{33}$ \qquad $=\left( -\frac{r}{r-2m}\right) \Gamma ^{2}{}_{00}+\left( \frac{r-2m}{% r}\right) \Gamma ^{2}{}_{11}+\frac{1}{r^{2}}\Gamma ^{2}{}_{22}+\frac{1}{% r^{2}\sin ^{2}\theta }\Gamma ^{2}{}_{33}=\frac{1}{r^{2}\sin ^{2}\theta }% \Gamma ^{2}{}_{33}=-\frac{\sin \theta \cos \theta }{r^{2}\sin ^{2}\theta }=-% \frac{\cot \theta }{r^{2}}$ is true $g^{\alpha \beta }\Gamma ^{3}{}_{\alpha \beta }=g^{00}\Gamma ^{3}{}_{00}+g^{11}\Gamma ^{3}{}_{11}+g^{22}\Gamma ^{3}{}_{22}+g^{33}\Gamma ^{3}{}_{33}=0$ and thus $\Phi _{,\rho }g^{\alpha \beta }\Gamma ^{\rho }{}_{\alpha \beta }=\Phi _{,0}g^{\alpha \beta }\Gamma ^{0}{}_{\alpha \beta }+\Phi _{,1}g^{\alpha \beta }\Gamma ^{1}{}_{\alpha \beta }+\Phi _{,2}g^{\alpha \beta }\Gamma ^{2}{}_{\alpha \beta }+\Phi _{,3}g^{\alpha \beta }\Gamma ^{3}{}_{\alpha \beta }$ \begin{eqnarray*} \Phi _{,\rho }g^{\alpha \beta }\Gamma ^{\rho }{}_{\alpha \beta } &=&2\frac{% m-r}{r^{2}}\Phi _{,1}-\frac{\cot \theta }{r^{2}}\Phi _{,2} \\ &=&2\frac{m-r}{r^{2}}\frac{\partial \Phi }{\partial r}-\frac{\cot \theta }{% r^{2}}\frac{\partial \Phi }{\partial \theta } \end{eqnarray*} $g^{\alpha \beta }\Phi _{,\alpha \beta }=g^{00}\Phi _{,00}+g^{11}\Phi _{,11}+g^{22}\Phi _{,22}+g^{33}\Phi _{,33}$ $g^{\alpha \beta }\Phi _{,\alpha \beta }=\left( -\frac{r}{r-2m}\right) \Phi _{,00}+\left( \frac{r-2m}{r}\right) \Phi _{,11}+\frac{1}{r^{2}}\Phi _{,22}+% \frac{1}{r^{2}\sin ^{2}\theta }\Phi _{,33}$ \qquad =$\left( -\frac{r}{r-2m}\right) \frac{\partial ^{2}\Phi }{\partial t^{2}}+\left( \frac{r-2m}{r}\right) \frac{\partial ^{2}\Phi }{\partial r^{2}}% +\frac{1}{r^{2}}\frac{\partial ^{2}\Phi }{\partial \theta ^{2}}+\frac{1}{% r^{2}\sin ^{2}\theta }\frac{\partial ^{2}\Phi }{\partial \varphi ^{2}}$ $\nabla ^{2}\Phi =g^{\alpha \beta }\Phi _{,\alpha \beta }-\Phi _{,\rho }g^{\alpha \beta }\Gamma ^{\rho }{}_{\alpha \beta }=\left( -\frac{r}{r-2m}% \right) \frac{\partial ^{2}\Phi }{\partial t^{2}}+\left( \frac{r-2m}{r}% \right) \frac{\partial ^{2}\Phi }{\partial r^{2}}+\frac{1}{r^{2}}\frac{% \partial ^{2}\Phi }{\partial \theta ^{2}}+\frac{1}{r^{2}\sin ^{2}\theta }% \frac{\partial ^{2}\Phi }{\partial \varphi ^{2}}-2\frac{m-r}{r^{2}}\frac{% \partial \Phi }{\partial r}+\frac{\cot \theta }{r^{2}}\frac{\partial \Phi }{% \partial \theta }$% \[ \fbox{$\nabla ^{2}\Phi =\left( -\frac{r}{r-2m}\right) \frac{\partial ^{2}\Phi }{\partial t^{2}}+\left( \frac{r-2m}{r}\right) \frac{\partial ^{2}\Phi }{\partial r^{2}}+\frac{1}{r^{2}}\frac{\partial ^{2}\Phi }{\partial \theta ^{2}}+\frac{1}{r^{2}\sin ^{2}\theta }\frac{\partial ^{2}\Phi }{% \partial \varphi ^{2}}-2\frac{m-r}{r^{2}}\frac{\partial \Phi }{\partial r}+% \frac{\cot \theta }{r^{2}}\frac{\partial \Phi }{\partial \theta }$} \] \vspace{0.5in}\pagebreak {\LARGE Problem 3}\vspace{0.5in} Suppose that an object is in free fall near a black hole (or any other spherically symmetric object) of mass $m$. The world-line of the object can be described by its Schwarzschild coordinate functions $t\left( \tau \right) ,r\left( \tau \right) ,\theta \left( \tau \right) ,\varphi \left( \tau \right) $ where $\tau $ represents proper time along that world-line. \begin{enumerate} \item[{\protect\LARGE a.}] Obtain the system of ordinary differential equations that determine these functions. These should be fully explicit relations among the derivatives of the functions $t\left( \tau \right) ,r\left( \tau \right) ,\theta \left( \tau \right) ,\varphi \left( \tau \right) $.\bigskip \end{enumerate} {\large Answer 3a}{\LARGE \medskip } Start with the geodesic equation% \[ D_{u}u=0 \]% and convert it to index notation in these coordinates. $u^{\alpha }=\frac{% dx^{\alpha }}{d\tau }=\dot{x}^{\alpha }$ where $\tau $ is the proper time. $D_{u}u=u^{\delta }u^{\alpha }{}_{;\delta }e_{\alpha }=\frac{dx^{\delta }}{% d\tau }\left( u^{\alpha }{}_{,\delta }+u^{\rho }\Gamma ^{\alpha }{}_{\rho \delta }\right) e_{\alpha }$ Note that $\frac{dx^{\delta }}{d\tau }u^{\alpha }{}_{,\delta }=\frac{% dx^{\delta }}{d\tau }\frac{\partial u^{\alpha }}{\partial x^{\delta }}=\frac{% du^{\alpha }}{d\tau }=\dot{u}^{\alpha }$ $D_{u}u=\left( \dot{u}^{\alpha }+u^{\delta }u^{\rho }\Gamma ^{\alpha }{}_{\rho \delta }\right) e_{\alpha }$ so the system of equations turns out to be% \[ \dot{u}^{\alpha }+u^{\delta }u^{\rho }\Gamma ^{\alpha }{}_{\rho \delta }=0 \]% or, using $u^{\alpha }=\frac{dx^{\alpha }}{d\tau },$% \[ \frac{d^{2}x^{\alpha }}{d\tau ^{2}}+\frac{dx^{\delta }}{d\tau }\frac{% dx^{\rho }}{d\tau }\Gamma ^{\alpha }{}_{\rho \delta }=0 \]% Write the equations out in detail: $\alpha =0$ $\frac{d^{2}t}{d\tau ^{2}}+\frac{dx^{\delta }}{d\tau }\frac{dx^{\rho }}{% d\tau }\Gamma ^{0}{}_{\rho \delta }=0$ There are only two non-zero connection coefficients with the first index equal to zero. $\frac{d^{2}t}{d\tau ^{2}}+2\frac{dx^{0}}{d\tau }\frac{dx^{1}}{d\tau }\frac{m% }{r\left( r-2m\right) }=0$ \begin{eqnarray*} \frac{d^{2}t}{d\tau ^{2}}+2\frac{dt}{d\tau }\frac{dr}{d\tau }\frac{m}{% r\left( r-2m\right) } &=&0 \\ \ddot{t}+2\dot{t}\dot{r}\frac{m}{r\left( r-2m\right) } &=&0 \end{eqnarray*} $\alpha =1$ $\frac{d^{2}x^{1}}{d\tau ^{2}}+\frac{dx^{\delta }}{d\tau }\frac{dx^{\rho }}{% d\tau }\Gamma ^{1}{}_{\rho \delta }=0$ Note that all of the nonzero $\Gamma ^{1}{}_{\rho \delta }$ have $\rho =\delta $. $\ddot{r}+\frac{dx^{0}}{d\tau }\frac{dx^{0}}{d\tau }\Gamma ^{1}{}_{00}+\frac{% dx^{1}}{d\tau }\frac{dx^{1}}{d\tau }\Gamma ^{1}{}_{11}+\frac{dx^{2}}{d\tau }% \frac{dx^{2}}{d\tau }\Gamma ^{1}{}_{22}+\frac{dx^{3}}{d\tau }\frac{dx^{3}}{% d\tau }\Gamma ^{1}{}_{33}=0$ $\ddot{r}+\dot{t}^{2}\frac{m}{r^{3}}\left( r-2m\right) +\dot{r}^{2}\left( -% \frac{m}{r\left( r-2m\right) }\right) +\dot{\theta}^{2}\left( -\left( r-2m\right) \right) +\dot{\varphi}^{2}\left( -\left( r-2m\right) \sin ^{2}\theta \right) =0$% \[ \ddot{r}+\dot{t}^{2}\frac{m}{r^{3}}\left( r-2m\right) -\dot{r}^{2}\frac{m}{% r\left( r-2m\right) }-\dot{\theta}^{2}\left( r-2m\right) -\dot{\varphi}% ^{2}\left( r-2m\right) \sin ^{2}\theta =0 \] $\alpha =2$ $\frac{d^{2}x^{2}}{d\tau ^{2}}+\frac{dx^{\delta }}{d\tau }\frac{dx^{\rho }}{% d\tau }\Gamma ^{2}{}_{\rho \delta }=0$ $\frac{d^{2}\theta }{d\tau ^{2}}+\frac{dx^{2}}{d\tau }\frac{dx^{1}}{d\tau }% \Gamma ^{2}{}_{12}+\frac{dx^{1}}{d\tau }\frac{dx^{2}}{d\tau }\Gamma ^{2}{}_{21}+\frac{dx^{3}}{d\tau }\frac{dx^{3}}{d\tau }\Gamma ^{2}{}_{33}=0$ $\ddot{\theta}+2\dot{\theta}\dot{r}\Gamma ^{2}{}_{12}+\dot{\varphi}% ^{2}\Gamma ^{2}{}_{33}=0$ $\ddot{\theta}+2\dot{\theta}\dot{r}\frac{1}{r}+\dot{\varphi}^{2}\left( -\sin \theta \cos \theta \right) =0$% \[ \ddot{\theta}+\frac{2}{r}\dot{\theta}\dot{r}-\dot{\varphi}^{2}\sin \theta \cos \theta =0 \] $\alpha =3$ $\frac{d^{2}x^{3}}{d\tau ^{2}}+\frac{dx^{\delta }}{d\tau }\frac{dx^{\rho }}{% d\tau }\Gamma ^{3}{}_{\rho \delta }=0$ $\ddot{\varphi}+2\frac{dx^{1}}{d\tau }\frac{dx^{3}}{d\tau }\Gamma ^{3}{}_{31}+2\frac{dx^{2}}{d\tau }\frac{dx^{3}}{d\tau }\Gamma ^{3}{}_{32}=0$ $\ddot{\varphi}+2\dot{r}\dot{\varphi}\Gamma ^{3}{}_{31}+2\dot{\theta}\dot{% \varphi}\Gamma ^{3}{}_{32}=0$ $\ddot{\varphi}+2\dot{r}\dot{\varphi}\frac{1}{r}+2\dot{\theta}\dot{\varphi}% \cot \theta =0$% \[ \ddot{\varphi}+\frac{2}{r}\dot{r}\dot{\varphi}+2\dot{\theta}\dot{\varphi}% \cot \theta =0 \]% Collecting the equations together,% \[ \fbox{% \begin{tabular}{l} $\ddot{t}+2\dot{t}\dot{r}\frac{m}{r\left( r-2m\right) }=0$ \\ $\ddot{r}+\dot{t}^{2}\frac{m}{r^{3}}\left( r-2m\right) -\dot{r}^{2}\frac{m}{% r\left( r-2m\right) }-\dot{\theta}^{2}\left( r-2m\right) -\dot{\varphi}% ^{2}\left( r-2m\right) \sin ^{2}\theta =0$ \\ $\ddot{\theta}+\frac{2}{r}\dot{\theta}\dot{r}-\dot{\varphi}^{2}\sin \theta \cos \theta =0$ \\ $\ddot{\varphi}+\frac{2}{r}\dot{r}\dot{\varphi}+2\dot{\theta}\dot{\varphi}% \cot \theta =0$% \end{tabular}% } \]% \vspace{0.5in} \begin{enumerate} \item[{\protect\LARGE b.}] Specialize the equations to the zero-velocity case and compare the predicted initial acceleration to what Newton's theory would predict.\bigskip \end{enumerate} {\large Answer 3b}{\LARGE \medskip } For zero velocity, we have $\dot{r}=\dot{\theta}=\dot{\varphi}=0$ and $\dot{t% }=1$. The equations then become% \begin{eqnarray*} \ddot{t} &=&0 \\ \ddot{r}+\frac{m}{r^{3}}\left( r-2m\right) &=&0 \\ \ddot{\theta} &=&0 \\ \ddot{\varphi} &=&0 \end{eqnarray*}% So there is a radial acceleration with% \[ \fbox{$\ddot{r}=-\frac{m}{r^{3}}\left( r-2m\right) =-\frac{m}{r^{2}}+\frac{% 2m^{2}}{r^{3}}.$} \]% The first term is exactly what Newton's theory predicts. The second term is a correction that becomes important at distances comparable to the Schwarzschild radius $2m$.\vspace{0.5in} \begin{enumerate} \item[{\protect\LARGE c.}] Find the conditions that must be satisfied by an object in an equatorial circular orbit around a black hole. Do not forget the constraint $u\cdot u=-1$ because you will need it.\bigskip \end{enumerate} {\large Answer 3c}{\LARGE \medskip } For an equatorial orbit, we have $\theta =\frac{\pi }{2}$ so that $\dot{% \theta}=0$. Since the orbit is circular, we also have $\dot{r}=0$. The equations then yield% \begin{eqnarray*} \ddot{t} &=&0 \\ \dot{t}^{2}\frac{m}{r^{3}}\left( r-2m\right) -\dot{\varphi}^{2}\left( r-2m\right) &=&0 \\ 0 &=&0 \\ \ddot{\varphi} &=&0 \end{eqnarray*}% So we see that both $\dot{t}$ (which is the red-shift factor, by the way) and $\dot{\varphi}$ are constant and related by $\dot{t}^{2}\frac{m}{r^{3}}-% \dot{\varphi}^{2}=0$ or% \[ \dot{\varphi}^{2}=\dot{t}^{2}\frac{m}{r^{3}} \]% These derivatives are also related by the constraint% \[ u\cdot u=-1 \]% which takes the form $g_{\alpha \beta }u^{\alpha }u^{\beta }=g_{00}\dot{t}^{2}+g_{33}\dot{\varphi}% ^{2}=-1$ $-\left( 1-\frac{2m}{r}\right) \dot{t}^{2}+r^{2}\dot{\varphi}^{2}=-1$ Plug the circular orbit condition into this constraint: $-\left( 1-\frac{2m}{r}\right) \dot{t}^{2}+r^{2}\dot{t}^{2}\frac{m}{r^{3}}% =-1 $ $-\left( 1-\frac{2m}{r}\right) \dot{t}^{2}+\dot{t}^{2}\frac{m}{r}% =\allowbreak \dot{t}^{2}\frac{-r+3m}{r}=-1$% \begin{eqnarray*} \dot{t}^{2} &=&\frac{r}{r-3m} \\ \dot{\varphi}^{2} &=&\frac{m}{r^{3}}\frac{r}{r-3m} \end{eqnarray*}% Notice that the solution exists only for \fbox{$r>3m$}. Thus, that is the radius of the smallest stable circular orbit. Objects closer than that spiral into the hole even though they are still outside the event horizon.% \vspace{0.5in}\pagebreak {\LARGE Problem 4}\vspace{0.5in} Show that the difference between two tangent-space connections $D$ and $% D^{\prime }$ on a given manifold% \[ K=D^{\prime }-D \]% can be regarded as a tensor field. \begin{enumerate} \item {\large Answer 4}{\LARGE \medskip } \end{enumerate} Insert the appropriate arguments. $D$ and $D^{\prime }$ each need a derivative vector $v$ and a vector to act on, so we have% \[ K_{v}u=D_{v}^{\prime }u-D_{v}u \]% which yields a vector field. To get a scalar field, we need to use a 1-form field $\alpha $ and obtain the scalar function% \begin{eqnarray*} K\left( \alpha ,u,v\right) &=&\alpha \left( D_{v}^{\prime }u-D_{v}u\right) \\ &=&\alpha \left( D_{v}^{\prime }u\right) -\alpha \left( D_{v}u\right) . \end{eqnarray*}% Now check that this function in locally linear by multiplying each argument by a scalar field $f$. $K\left( f\alpha ,u,v\right) =f\alpha \left( D_{v}^{\prime }u-D_{v}u\right) =fK\left( \alpha ,u,v\right) $ so the first argument works. Now check the second argument. $K\left( \alpha ,fu,v\right) =\alpha \left( D_{v}^{\prime }\left( fu\right) \right) -\alpha \left( D_{v}\left( fu\right) \right) $ \qquad $=\alpha \left( \left( D_{v}^{\prime }f\right) u+fD_{v}^{\prime }u\right) -\alpha \left( \left( D_{v}f\right) u+fD_{v}u\right) $ Leibniz's Rule \qquad $=\alpha \left( \left( D_{v}^{\prime }f\right) u\right) +\alpha \left( fD_{v}^{\prime }u\right) -\alpha \left( \left( D_{v}f\right) u\right) -\alpha \left( fD_{v}u\right) $ \qquad $=\left( D_{v}^{\prime }f\right) \alpha \left( u\right) +f\alpha \left( D_{v}^{\prime }u\right) -\left( D_{v}f\right) \alpha \left( u\right) -f\alpha \left( D_{v}u\right) $ Linearity of $\alpha $. \qquad $=f\alpha \left( D_{v}^{\prime }u\right) -f\alpha \left( D_{v}u\right) +\left( D_{v}^{\prime }f-D_{v}f\right) \alpha \left( u\right) $ However, our definition of a connection includes the condition% \[ D_{v}^{\prime }f=vf,\qquad D_{v}f=vf \]% so $K\left( \alpha ,fu,v\right) =f\alpha \left( D_{v}^{\prime }u\right) -f\alpha \left( D_{v}u\right) +\left( vf-vf\right) \alpha \left( u\right) =fK\left( \alpha ,u,v\right) $ and $K$ is locally linear in the second argument. Finally, check the third argument, using the assumed property% \[ D_{fv}^{\prime }u=fD_{v}^{\prime }u,\qquad D_{fv}u=fD_{v}u \] $K\left( \alpha ,u,fv\right) =D_{fv}^{\prime }u-D_{fv}u=fD_{v}^{\prime }u-fD_{v}u=fK\left( \alpha ,u,v\right) $ and $K$ is locally linear in the third argument and is therefore a tensor field. \vspace{0.5in}\pagebreak {\LARGE Problem 5}\vspace{0.5in} Suppose that $V_{P}$ is the space of Weyl spinors at the point $P$ and $\dot{% V}_{P}$ is the complex conjugate space. Let $E_{A}$ be basis vectors for $% V_{P}$ and let $E_{\dot{A}}$ be basis vectors for $\dot{V}_{P}$. The self-conjugate (i.e. real) spin-tensors \[ e_{A\dot{A}}=E_{A}\otimes E_{\dot{A}}+E_{\dot{A}}\otimes E_{A} \]% can be identified as basis vectors for the spacetime tangent space $T_{P}$. An orthonormal set of tangent space basis vectors $e_{\alpha }$ can be expanded in terms of these as% \[ e_{\alpha }=\gamma ^{A\dot{A}}{}_{\alpha }e_{A\dot{A}} \]% where the coefficients $\gamma ^{A\dot{A}}{}_{\alpha }$ are constants and there will be an inverse expansion% \[ e_{A\dot{A}}=\gamma _{A\dot{A}}{}^{\alpha }e_{\alpha }. \]% \bigskip \begin{enumerate} \item[{\protect\LARGE a.}] Define the connection coefficients for the spaces $V_{P}$ and $\dot{V}_{P}$. These complex functions are called the `spin connection coefficients'. How many such coefficients are there?\bigskip \end{enumerate} {\large Answer 5a}{\LARGE \medskip } The basis vectors for the space $V_{P}$ are $E_{A}$ so the connection coefficients for that space are just the components of the derivatives of those basis vectors in each tangent-space basis vector direction.% \[ D_{e_{a}}E_{A}=\Gamma ^{B}{}_{Aa}E_{B} \]% or% \[ \fbox{$\Gamma ^{B}{}_{Aa}=\Omega ^{B}\cdot D_{e_{a}}E_{A}.$} \]% Similarly, the connection coefficients for $\dot{V}_{P}$ are% \[ \fbox{$\Gamma ^{\dot{B}}{}_{\dot{A}a}=\Omega ^{\dot{B}}\cdot D_{e_{a}}E_{% \dot{A}}$} \]% With no conditions placed on these connections, the independent complex components are% \[ \Gamma ^{1}{}_{1a},\Gamma ^{1}{}_{2a},\Gamma ^{2}{}_{1a},\Gamma ^{2}{}_{2a} \]% and a similar list for the conjugate space. Thus, there are eight complex four-vectors for a total of 32 independent complex components or 64 real numbers.\vspace{0.5in} \begin{enumerate} \item[{\protect\LARGE b.}] Express the spacetime connection coefficients $% \Gamma ^{\alpha }{}_{\beta \delta }$ for an orthonormal basis $e_{\alpha }$ in terms of the spin connection coefficients, the coefficients $\gamma ^{A% \dot{A}}{}_{\alpha }$ and their inverses $\gamma _{A\dot{A}}{}^{\alpha }$% .\bigskip \end{enumerate} {\large Answer 5b}{\LARGE \medskip } The defining equation for the $\Gamma ^{\alpha }{}_{\beta \delta }$ is% \[ D_{e_{\delta }}e_{\beta }=\Gamma ^{\alpha }{}_{\beta \delta }e_{\alpha } \]% Now put in the expansion of $e_{\beta }$ in terms of spinor basis vectors and do the derivative. $e_{\beta }=\gamma ^{B\dot{B}}{}_{\beta }e_{B\dot{B}}=\gamma ^{B\dot{B}% }{}_{\beta }\left( E_{B}\otimes E_{\dot{B}}+E_{\dot{B}}\otimes E_{B}\right) $ $D_{e_{\delta }}e_{\beta }=\gamma ^{B\dot{B}}{}_{\beta }D_{e_{\delta }}\left( E_{B}\otimes E_{\dot{B}}+E_{\dot{B}}\otimes E_{B}\right) $ Use the fact that the $\gamma ^{B\dot{B}}{}_{\beta }$ are constants. $D_{e_{\delta }}e_{\beta }$ $=\gamma ^{B\dot{B}}{}_{\beta }\left( D_{e_{\delta }}E_{B}\otimes E_{\dot{B}% }+E_{B}\otimes D_{e_{\delta }}E_{\dot{B}}+D_{e_{\delta }}E_{\dot{B}}\otimes E_{B}+E_{\dot{B}}\otimes D_{e_{\delta }}E_{B}\right) $ $=\gamma ^{B\dot{B}}{}_{\beta }\left( \Gamma ^{A}{}_{B\delta }E_{A}\otimes E_{\dot{B}}+E_{B}\otimes \Gamma ^{\dot{A}}{}_{\dot{B}\delta }E_{\dot{A}% }+\Gamma ^{\dot{A}}{}_{\dot{B}\delta }E_{\dot{A}}\otimes E_{B}+E_{\dot{B}% }\otimes \Gamma ^{A}{}_{B\delta }E_{A}\right) $ $=\gamma ^{B\dot{B}}{}_{\beta }\Gamma ^{A}{}_{B\delta }E_{A}\otimes E_{\dot{B% }}+\gamma ^{B\dot{B}}{}_{\beta }\Gamma ^{\dot{A}}{}_{\dot{B}\delta }E_{B}\otimes E_{\dot{A}}$ $+\gamma ^{B\dot{B}}{}_{\beta }\Gamma ^{\dot{A}}{}_{\dot{B}\delta }E_{\dot{A}% }\otimes E_{B}+\gamma ^{B\dot{B}}{}_{\beta }\Gamma ^{A}{}_{B\delta }E_{\dot{B% }}\otimes E_{A}$ Now rename the dummy spinor indexes so that the basis tensors look the same. Just switch all the $A$ and $B$ indexes in half the terms. $D_{e_{\delta }}e_{\beta }$ $=\gamma ^{B\dot{B}}{}_{\beta }\Gamma ^{A}{}_{B\delta }E_{A}\otimes E_{\dot{B% }}+\gamma ^{A\dot{A}}{}_{\beta }\Gamma ^{\dot{B}}{}_{\dot{A}\delta }E_{A}\otimes E_{\dot{B}}$ $+\gamma ^{B\dot{B}}{}_{\beta }\Gamma ^{\dot{A}}{}_{\dot{B}\delta }E_{\dot{A}% }\otimes E_{B}+\gamma ^{A\dot{A}}{}_{\beta }\Gamma ^{B}{}_{A\delta }E_{\dot{A% }}\otimes E_{B}$ $=\left( \gamma ^{B\dot{B}}{}_{\beta }\Gamma ^{A}{}_{B\delta }+\gamma ^{A% \dot{A}}{}_{\beta }\Gamma ^{\dot{B}}{}_{\dot{A}\delta }\right) E_{A}\otimes E_{\dot{B}}$ $+\left( \gamma ^{B\dot{B}}{}_{\beta }\Gamma ^{\dot{A}}{}_{\dot{B}\delta }+\gamma ^{A\dot{A}}{}_{\beta }\Gamma ^{B}{}_{A\delta }\right) E_{\dot{A}% }\otimes E_{B}$ Rename these again to get basis tensors of the form $E_{C}\otimes E_{\dot{C}% } $ and transpose. $A\rightarrow C,\dot{B}\rightarrow \dot{C}$ in the first term $\dot{A}\rightarrow \dot{C},B\rightarrow C$ in the second term. $D_{e_{\delta }}e_{\beta }$ $=\left( \gamma ^{B\dot{C}}{}_{\beta }\Gamma ^{C}{}_{B\delta }+\gamma ^{C% \dot{A}}{}_{\beta }\Gamma ^{\dot{C}}{}_{\dot{A}\delta }\right) E_{C}\otimes E_{\dot{C}}$ $+\left( \gamma ^{C\dot{B}}{}_{\beta }\Gamma ^{\dot{C}}{}_{\dot{B}\delta }+\gamma ^{A\dot{C}}{}_{\beta }\Gamma ^{C}{}_{A\delta }\right) E_{\dot{C}% }\otimes E_{C}$ Rename the dummy indexes inside the parentheses: $D_{e_{\delta }}e_{\beta }$ $=\left( \gamma ^{K\dot{C}}{}_{\beta }\Gamma ^{C}{}_{K\delta }+\gamma ^{C% \dot{K}}{}_{\beta }\Gamma ^{\dot{C}}{}_{\dot{K}\delta }\right) E_{C}\otimes E_{\dot{C}}$ $+\left( \gamma ^{C\dot{K}}{}_{\beta }\Gamma ^{\dot{C}}{}_{\dot{K}\delta }+\gamma ^{K\dot{C}}{}_{\beta }\Gamma ^{C}{}_{K\delta }\right) E_{\dot{C}% }\otimes E_{C}$ and notice that the terms inside the parentheses are now identical so we get% \[ D_{e_{\delta }}e_{\beta }=\left( \gamma ^{K\dot{C}}{}_{\beta }\Gamma ^{C}{}_{K\delta }+\gamma ^{C\dot{K}}{}_{\beta }\Gamma ^{\dot{C}}{}_{\dot{K}% \delta }\right) \left( E_{C}\otimes E_{\dot{C}}+E_{\dot{C}}\otimes E_{C}\right) \]% or% \[ D_{e_{\delta }}e_{\beta }=\left( \gamma ^{K\dot{C}}{}_{\beta }\Gamma ^{C}{}_{K\delta }+\gamma ^{C\dot{K}}{}_{\beta }\Gamma ^{\dot{C}}{}_{\dot{K}% \delta }\right) e_{C\dot{C}} \]% and, using the inverse expansion,% \[ e_{C\dot{C}}=\gamma _{C\dot{C}}{}^{\alpha }e_{\alpha }. \]% $D_{e_{\delta }}e_{\beta }=\left( \gamma ^{K\dot{C}}{}_{\beta }\Gamma ^{C}{}_{K\delta }+\gamma ^{C\dot{K}}{}_{\beta }\Gamma ^{\dot{C}}{}_{\dot{K}% \delta }\right) \gamma _{C\dot{C}}{}^{\alpha }e_{\alpha }$ so that% \[ \Gamma ^{\alpha }{}_{\beta \delta }e_{\alpha }=\left( \gamma ^{K\dot{C}% }{}_{\beta }\Gamma ^{C}{}_{K\delta }+\gamma ^{C\dot{K}}{}_{\beta }\Gamma ^{% \dot{C}}{}_{\dot{K}\delta }\right) \gamma _{C\dot{C}}{}^{\alpha }e_{\alpha } \]% and we finally can identify% \[ \fbox{$\Gamma ^{\alpha }{}_{\beta \delta }=\left( \gamma ^{K\dot{C}% }{}_{\beta }\Gamma ^{C}{}_{K\delta }+\gamma ^{C\dot{K}}{}_{\beta }\Gamma ^{% \dot{C}}{}_{\dot{K}\delta }\right) \gamma _{C\dot{C}}{}^{\alpha }$} \]% \vspace{0.5in} \end{document} %%%%%%%%%%%%%%%%%%%%%%% End /document/hx2sol.tex %%%%%%%%%%%%%%%%%%%%%%