General Relativity Homework 5

Chasse_neige

Problem 1

Please calculate the connection and the Ricci tensor for the metric

$${\mathrm{d}}^{}{\tau }^2=B(t,r){{\mathrm{d}}^{}t}^2-A(t,r){{\mathrm{d}}^{}r}^2-r^2{\mathrm{d}}^{}{\mathrm{\Omega }}^2$$

For this time-relevant metric

$$g_{\mu \nu }=\left(\begin{array}{c}-B,A,r^2,r^2{\sin }^2\theta \end{array}\right)$$

First we derive the Christoffel symbol

$${\mathrm{\Gamma }}^{\lambda }{}_{\mu \nu }=g^{\lambda \eta }(g_{\eta \mu ,\nu }+g_{\eta \nu ,\mu }-g_{\mu \nu ,\eta })$$

Calculate the components, we can find all nonzero terms are

$$\begin{array}{c}{\mathrm{\Gamma }}^t{}_{tr}={\mathrm{\Gamma }}^t{}_{rt}=\frac{B^{\prime }}{2B}\\ {\mathrm{\Gamma }}^r{}_{tt}=\frac{B^{\prime }}{2A}\\ {\mathrm{\Gamma }}^r{}_{rr}=\frac{A^{\prime }}{2A}\\ {\mathrm{\Gamma }}^t{}_{tt}=\frac{\dot{B}}{2B}\\ {\mathrm{\Gamma }}^t{}_{rr}=\frac{\dot{A}}{2B}\end{array}$$$$\begin{array}{c}{\mathrm{\Gamma }}^r{}_{tr}={\mathrm{\Gamma }}^r{}_{rt}=\frac{\dot{A}}{2A}\\ {\mathrm{\Gamma }}^r{}_{\theta \theta }=-\frac{r}{A}\\ {\mathrm{\Gamma }}^r{}_{\varphi \varphi }=-\frac{r{\sin }^2\theta }{A}\\ {\mathrm{\Gamma }}^{\theta }{}_{r\theta }={\mathrm{\Gamma }}^{\theta }{}_{\theta r}=\frac{1}{r}\\ {\mathrm{\Gamma }}^{\theta }{}_{\varphi \varphi }=-\sin \theta \cos \theta \\ {\mathrm{\Gamma }}^{\varphi }{}_{r\varphi }={\mathrm{\Gamma }}^{\varphi }{}_{\varphi r}=\frac{1}{r}\\ {\mathrm{\Gamma }}^{\varphi }{}_{\theta \varphi }={\mathrm{\Gamma }}^{\varphi }{}_{\varphi \theta }=\cot \theta \end{array}$$

where prime means derivative of the space direction while dot means derivative of the time direction.

Calculate Ricci tensors using the definition

$$\begin{array}{c}{R}_{\mu \kappa }=g^{\lambda \nu }{R}_{\lambda \mu \nu \kappa }=R^{\lambda }{}_{\mu \lambda \kappa }\\ ={\mathrm{\Gamma }}^{\lambda }{}_{\mu \lambda ,\kappa }-{\mathrm{\Gamma }}^{\lambda }{}_{\mu \kappa ,\lambda }+{\mathrm{\Gamma }}^{\eta }{}_{\mu \lambda }{\mathrm{\Gamma }}^{\lambda }{}_{\eta \kappa }-{\mathrm{\Gamma }}^{\eta }{}_{\mu \kappa }{\mathrm{\Gamma }}^{\lambda }{}_{\eta \lambda }\end{array}$$

And we can get all nonzero components

$$\begin{array}{c}{R}_{tt}=\frac{\ddot{A}}{2A}-\frac{{\dot{A}}^2}{4A^2}-\frac{\dot{A}\dot{B}}{4AB}-\frac{B^{″}}{2A}+\frac{B^{\prime }}{4A}(\frac{A^{\prime }}{A}+\frac{B^{\prime }}{B})-\frac{B^{\prime }}{rA}\\ R_{tr}=-\frac{\dot{A}}{rA}\\ R_{rr}=-\frac{\ddot{A}}{2B}+\frac{\dot{A}}{4B}(\frac{\dot{A}}{A}+\frac{\dot{B}}{B})+\frac{B^{″}}{2B}-\frac{B^{\prime }}{4B}(\frac{A^{\prime }}{A}+\frac{B^{\prime }}{B})-\frac{A^{\prime }}{rA}\\ R_{\theta \theta }=-1+\frac{1}{A}-\frac{r}{2A}(\frac{A^{\prime }}{A}-\frac{B^{\prime }}{B})\\ R_{\varphi \varphi }={\sin }^2\theta R_{\theta \theta }\end{array}$$

Problem 2

Please derive the energy-momentum tensor for electromagnetic field.

We use the variation of the action to get the form for the energy-momentum tensor

$$T^{\mu \nu }=\frac{2}{\sqrt{-g}}\frac{\delta S}{\delta g_{\mu \nu }}$$

Plugging in the action of the EM field

$$S=-\frac{1}{4}\int {\mathrm{d}}^{4}x\sqrt{-g}{F}^{\mu \nu }{F}_{\mu \nu }$$

and do the variation

$$\begin{array}{c}\delta S=-\frac{1}{4}\int {\mathrm{d}}^{4}x{F}_{\alpha \beta }{F}_{\mu \nu }\delta (\sqrt{-g}{g}^{\mu \alpha }{g}^{\nu \beta })\\ =-\frac{1}{4}\int {\mathrm{d}}^{4}x{F}_{\alpha \beta }{F}_{\mu \nu }(g^{\mu \alpha }{g}^{\nu \beta }\cdot \frac{1}{2}\sqrt{-g}{g}^{\rho \sigma }\delta g_{\rho \sigma }-\sqrt{-g}{g}^{\nu \beta }{g}^{\sigma \alpha }{g}^{\mu \rho }\delta g_{\rho \sigma }-\sqrt{-g}{g}^{\mu \alpha }{g}^{\nu \rho }{g}^{\sigma \beta }\delta g_{\rho \sigma })\end{array}$$

So the energy-momentum tensor density is

$$\begin{array}{c}{T}^{\rho \sigma }=\frac{2}{\sqrt{-g}}\cdot (-\frac{1}{4}{F}_{\alpha \beta }{F}_{\mu \nu }(g^{\mu \alpha }{g}^{\nu \beta }\cdot \frac{1}{2}\sqrt{-g}g{g}^{\rho \sigma }-\sqrt{-g}{g}^{\nu \beta }{g}^{\sigma \alpha }{g}^{\mu \rho }-\sqrt{-g}{g}^{\mu \alpha }{g}^{\nu \rho }{g}^{\sigma \beta })\\ =F^{\rho \beta }{F}^{\sigma }{}_{\beta }-\frac{1}{4}{g}^{\rho \sigma }{F}^{\mu \nu }{F}_{\mu \nu }\end{array}$$