\documentclass{article} \def\ZZ{\hbox{\bf Z}} \def\RR{\hbox{\bf R}} \def\CC{\hbox{\bf C}} \def\QQ{\hbox{\bf Q}} \def\HH{\hbox{\bf H}} \def\Ga{\Gamma} \def\ga{\gamma} \def\al{\alpha} \def\De{\Delta} \def\de{\delta} \def\La{\Lambda} \def\la{\lambda} \def\eps{\varepsilon} \def\ka{\kappa} \def\oM{\overline{M}} \def\oX{\overline{X}} \def\onabla{\overline{\nabla}} \def\om{\omega} \def\Om{\Omega} \def\sup{\rm sup} \def\grad{\rm grad} \def\pa{\partial} \def\bpa{\bar{\partial}} \def\spec{\hbox{spec}} \def\Dom{\hbox{Dom}} \def\bb{\bar{b}} \def\bB{\bar{B}} \def\bM{\overline{M}} \def\bx{\bar{x}} \def\bw{\bar{w}} \def\bxi{\bar{\xi}} \def\bz{\bar{z}} \def\dist{{\rm dist}} \def\bpar{\bar{\partial}} \def\dimg{\dim_{\Gamma}} \def\codimg{\hbox{codim}_{\Gamma}} \def\cO{{\cal O}} \def\cF{{\cal F}} \def\Re{\hbox{Re}\,} \def\Im{\hbox{Im}\,} \def\Ker{\hbox{Ker}\,} \begin{document} \centerline {\bf Riemannian Geometry and General Relativity} \centerline{\bf (Differential Geometry-2, MTH 3412)} \centerline{\bf Home Assignement 4. Solutions of Selected Problems.} \centerline{\bf Professor M.Shubin} \centerline{\bf Winter 1997} \medskip\noindent {\bf 1.} Prove that in 4-dimensional case the cyclic relation $$R_{0123}+R_{0231}+R_{0312}=0\leqno (1)$$ is independent of the antisymmetry and symmetry relations $$R_{\mu\nu\rho\sigma}=-R_{\nu\mu\rho\sigma},\quad R_{\mu\nu\rho\sigma}=-R_{\mu\nu\sigma\rho}, \quad R_{\mu\nu\rho\sigma}=R_{\rho\sigma\mu\nu}\;,\leqno (2)$$ i.e. there exists a tensor of type (0,4) in $\RR^4$ such that it satisfies (2) but not (1). \medskip {\bf Solution.} We will restrict ourselves to tensors at a point or, equivalently, tensors whose components are constants. In fixed coordinates such a tensor $R_{\mu\nu\rho\sigma}$ is an arbitrary set of 256 numbers. Let us try to construct a tensor $R_{\mu\nu\rho\sigma}$ such that the symmetry relations (2) are satisfied and $$R_{0123}=R_{0231}=R_{0312}=1\;,$$ therefore (1) is not true. To see that this is possible it is sufficient to see that the symmetry group generated by the relations (2) applied to $R_{0123}, R_{0231}, R_{0312}$, obviously leads to three non-intersecting sets of components, so extension of $R_{\mu\nu\rho\sigma}$ to arbitrary $\mu, \nu, \rho, \sigma$ with the conditions (2) will not lead to a contradiction. \medskip\noindent {\bf 2.} Prove that in 2-dimensional case $$R_{\mu\nu\rho\sigma}={S\over 2}(g_{\mu\sigma}g_{\nu\rho}-g_{\mu\rho}g_{\nu\sigma})\;,$$ where $S=g^{\mu\nu}R_{\mu\nu}$ is the scalar curvature. \medskip {\bf Solution.} Both sides of the formula satisfy the symmetry relations (2) above. Therefore all components of each of them are expressed in terms of one (say $R_{0101}$). It follows that at each point they are proportional. To see the coincidence it is sufficient to check that their full contractions coincide i.e. $$g^{\nu\rho}g^{\mu\sigma}R_{\mu\nu\rho\sigma}= {S\over 2}g^{\nu\rho}g^{\mu\sigma}(g_{\mu\sigma}g_{\nu\rho}-g_{\mu\rho}g_{\nu\sigma} )\;.$$ Since the left-hand side is $S$ by definition of the scalar curvature, it is sufficient to check that $g^{\nu\rho}g^{\mu\sigma}(g_{\mu\sigma}g_{\nu\rho}-g_{\mu\rho}g_{\nu\sigma}) =2$, which is obvious because $g^{\nu\rho}g^{\mu\sigma}g_{\mu\sigma}g_{\nu\rho}=4$ and $g^{\nu\rho}g^{\mu\sigma}g_{\mu\rho}g_{\nu\sigma}=2$. \medskip\noindent {\bf 4.} Find the mass of a black hole if its horizon sphere has the size of an atom i.e. the radius about $10^{-8}$ cm. {\bf Solution.} $$m={c^2r\over 2G}\approx 7\cdot10^{16}\ \hbox{kg}\;. $$ \medskip\noindent {\bf 5.} Prove that the Schwarzschild solution indeed has a singularity (which can not be corrected by changing coordinates). \medskip {\bf Main ideas of the solution.} \medskip A. Let us say that a Lorentzian manifold is {\it geodesically complete} if every geodesic can be infinitely extended in both directions with respect to the natural parameter $s$ (interval or proper time). It is physically natural to consider such a manifold as a space/time without singularities because any observer moving there in a spaceship can do it forever by the ship's internal clock. Also if a manifold $M$ can be included as an open set into a geodesically complete manifold, then we will consider $M$ as having no true singularities, because the apparent singularities are caused by the fact that we consider only a part of a true space. So a genuine singularity of $M$ can be described as follows: $M$ should have a geodesic that ends at a finite value of its natural parameter and can not be extended even in a larger Lorentzian manifold (in which $M$ is isometrically imbedded). \medskip B. How can we detect a genuine singularity of $M$? The easiest way is to produce a geodesic $\ga=\ga(s)$ on $M$, defined for $s\in(0,s_0)$, and a scalar field $\cF$ on $M$ such that: \medskip (i) $\cF$ is expressed locally as a non-singular function of $g_{\mu\nu}$ and their derivatives; (ii) $\cF(\ga(s))$ is unbounded as $s\to s_0$. \medskip This would obviously imply that $M$ has a genuine singularity (``at the end of $\ga$"). \medskip C. Now let us consider the Schwarzschild metric $$ds^2=(1-{2m\over r})dt^2-(1-{2m\over r})^{-1}dr^2-r^2d\theta^2-r^2\sin^2\theta d\phi^2\;. \leqno (3)$$ How can we produce the geodesic $\ga$ and the scalar field $\cF$ for this metric? It is natural to take the geodesic $\ga$ which starts as a vertical (radial) fall of the test particle (or spaceship) from a distant point to the origin ($r=0$). After crossing the Schwarzschild radius $2m$ we should switch to other coordinates, e.g. replace $t,r$ by $\tau, \rho$ as described in Sect. 19 of the Dirac's book. There $r=0$ corresponds to $\rho=\tau$ and it is easy to see that our spaceship (moving along $\ga$) will reach $\rho=\tau$ in a finite proper time. \medskip D. Now let us try to produce an appropriate scalar field $\cF$. It is natural to try to do this starting from the curvature tensor. Scalar curvature is a natural scalar field but unfortunately it is identically 0 for the Schwarzschild metric because even the Ricci tensor vanishes identically. But there are other scalar quantities which can be constructed from the curvature tensor. Let us consider the curvature tensor $R^{\mu\nu}_{\ \ \ \rho\sigma}$ (with 2 upper suffixes). It is antisymmetric with respect to $\mu,\nu$ and also with respect to $\rho,\sigma$. Therefore for every point it defines a linear map on antisymmetric tensors of type (2,0). (This map is in fact symmetric due to the relation $R_{\mu\nu\rho\sigma}=R_{\rho\sigma\mu\nu}$.) Its eigenvalues or coefficients of the characteristic polynomial are scalar fields. For example the trace of this map is minus the scalar curvature. One of the scalar fields obtained in such a way is the ``Hilbert-Schmidt norm" of the curvature tensor which is $$\cF=R^{\mu\nu\rho\sigma}R_{\mu\nu\rho\sigma}\;.\leqno (4)$$ \medskip E. Let us calculate the scalar field $\cF$ from (4) for the Schwarzschild metric (3) starting at the points with $r>2m$ (i.e. outside of the Schwarzschild sphere). A standard (though tedious) calculation shows that $$\cF={48m^2\over r^6}\;.$$ Now note that $\cF$ is an analytic function of $r$ for all $r>0$. It follows that it is given by the same formula even in the coordinates $\tau,\rho$ with $$r=\Bigl[{3\over 2}\sqrt{2m}(\rho-\tau)\Bigr]^{2/3}\;.$$ (See formula (19.8) in Dirac.) It follows that we indeed meet a genuine singularity at $r=0$ (or $\rho=\tau$). \medskip\noindent {\bf 7.} Consider a satellite rotating around a black hole of the given mass $m$ (or given Schwarzschild radius $2m$) along a circular orbit of a radius $R>2m$ (outside the horizon sphere). Find the period of the rotation by a distant observer clock and also by the clock in the satellite. What happens with these periods as $R\to 2m$? \medskip {\bf Solution.} As usual denote $x^0=t,\ x^1=r,\ x^2=\theta,\ x^3=\phi$, and $v^\mu=dx^\mu/ds$. Then $$g_{\mu\nu}v^\mu v^\nu=1\leqno (5)$$ and $${dv^\mu\over ds}+\Ga^\mu_{\nu\sigma}v^\mu v^\sigma=0\;.\leqno (6)$$ For a given circular orbit we can choose coordinates so that $\theta=\pi/2$. Also $r\equiv R$. It follows that $v^1=v^2=0$. Symmetry implies that $v^0$ and $v^3$ are constants. Denote $v^0=a,\ v^3=b$. Equation (6) with $\mu=1$ takes the form $${dv^1\over ds}+\Ga^1_{00}(v^0)^2+2\Ga^1_{03}v^0v^3+\Ga^0_{33}(v^3)^2=0\;.$$ An easy calculation shows that $$\Ga^1_{00}={m(r-2m)\over r^3},\quad \Ga^1_{03}=0, \quad \Ga^1_{33}=-(r-2m)\sin^2\theta\;,$$ so we obtain $${m(R-2m)\over R^3}a^2-(R-2m)b^2=0$$ or $${ma^2\over R^3}=b^2\;.$$ On the other hand (5) gives in our situation $$(1-{2m\over R})a^2-R^2b^2=1\;.$$ Combining these 2 relations we find $a$ and $b$: $$b={1\over R}\Bigl({m\over R-3m}\Bigr)^{1/2}\;,\quad a=b\Bigl({R^3\over m}\Bigr)^{1/2}\;.$$ It follows that by the satellite clock the period will be $$T_{proper}={2\pi\over b}=2\pi R\Bigl({R-3m\over m}\Bigr)^{1/2}\;.$$ Note that this period becomes $0$ when $R=3m$ which means that the satellite should move with the speed of light to remain on the circular orbit. Therefore the circular orbits with $R<3m$ are impossible. (It follows that with the initial condition $r_0\in (2m,3m)$ and any tangential initial velocity the spaceship will fall into the black hole; to escape the attraction of the black hole, it should move radially away from it.) On the other hand by the distant observer clock the period will be $$T_{distant}=T_{proper}{dt\over ds}=v^0T_{proper}=aT_{proper}= T_{proper}\cdot b\Bigl({R^3\over m}\Bigr)^{1/2}=2\pi\Bigl({R^3\over m}\Bigr)^{1/2}\;.$$ Note that the distant observer will not notice that something special happens when the radius becomes equal to $3m$! And $T_{distant}$ as a function of $R$ is exactly the same as in the classics (i.e. it obeys the 3rd Kepler law: $T^2$ is proportional to $R^3$). \medskip\noindent {\bf 9.} Let $G$ be a compact Lie group with a biinvariant Riemannian metric, $X,Y,Z$ left-invariant vector fields on $G$. Prove that $\nabla_XY={1\over 2}[X,Y]$ and $R(X,Y)Z=-{1\over 4}[[X,Y],Z]$. \medskip {\bf Solution.} The fact that one-parametric subgroups and their left translations are geodesics implies that $\nabla_Z Z=0$ for any left-invariant vector field $Z$ on $G$. Applying this to $Z=X+Y$ we obtain $\nabla_XY+\nabla_YX=0$. On the other hand $\nabla_XY-\nabla_YX=[X,Y]$ due to the symmetry of the Levi-Civita connection. It follows that $\nabla_XY={1\over 2}[X,Y]$. Now $$R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z= {1\over 4}[X,[Y,Z]]-{1\over 4}[Y,[X,Z]]-{1\over 2}[[X,Y],Z]]=$$ $${1\over 4}[X,[Y,Z]]+{1\over 4}[Y,[Z,X]]-{1\over 2}[[X,Y],Z]]= -{1\over 4}[Z,[X,Y]]-{1\over 2}[[X,Y],Z]]=-{1\over 4}[[X,Y],Z]\;.$$ \medskip\noindent {\bf 10.} Let $G$ be a compact Lie group with a biinvariant Riemannian metric, $X,Y,Z,W$ left-invariant vector fields on $G$. Prove that $([X,Y],Z)=(X,[Y,Z])$ and \noindent $(R(X,Y)Z,W)={1\over4}([X,Y],[Z,W])$. \medskip {\bf Solution.} Note first that $(X,Z)=const$, therefore $$0=Y(X,Z)=(\nabla_YX,Z)+(X,\nabla_YZ)={1\over 2}\Bigl(([Y,X],Z)+(X,[Y,Z])\Bigr)\;,$$ hence $([X,Y],Z)=(X,[Y,Z])$. Now $$(R(X,Y)Z,W)=-{1\over 4}([[X,Y],Z],W)=-{1\over 4}([X,Y],[Z,W])\;.$$ \end{document}