Schwarzschild black hole
The spherically symmetric solution of Einstein's equations in vacuum for the spacetime metric has the form \cite{Schw} \begin{align}\label{Schw} ds^{2}=h(r)\,dt^2-h^{-1}(r)\,dr^2-r^2 d\Omega^{2}, &\qquad\mbox{where}\quad h(r)=1-\frac{r_g}{r};\quad r_{g}=\frac{2GM}{c^{2}};\\ d\Omega^{2}=d\theta^{2}+\sin^{2}\theta\, d\varphi^{2}&\;\text{-- metric of unit sphere.}\nonumber \end{align} The Birkhoff's theorem (1923) \cite{Birkhoff,Jebsen} states, that this solution is unique up to coordinate transformations. The quantity $r_g$ is called the Schwarzschild radius, or gravitational radius, $M$ is the mass of the central body or black hole.
Contents
Simple problems
Problem 1.
Find the interval of local time (proper time of stationary observer) at a point $(r,\theta,\varphi)$ in terms of coordinate time $t$, and show that $t$ is the proper time of an observer at infinity. What happens when $r\to r_{g}$?
The proper time of the stationary observer is $d\tau=ds|_{dr=d\theta=d\varphi=0}$: \[d\tau=\sqrt{g_{00}}dt=\sqrt{1-\frac{r_{g}}{r}}dt.\] At $r\to \infty$ it coincides with $dt$, so the coordinate time $t$ can be interpreted as the proper time of a "remote" observer. At $r\to r_g$ the local time flows slower and asymptotically stops. If one of two twins were to live some time at $r\approx r_g$, he will return to his remote twin having aged less (thought he might have acquired some grey hair due to constant fear of tumbling over the horizon).
Problem 2.
What is the physical distance between two points with coordinates $(r_{1},\theta,\varphi)$ and $(r_{2},\theta,\varphi)$? Between $(r,\theta,\varphi_{1})$ and $(r,\theta,\varphi_{2})$? How do these distances behave in the limit $r_{1},r\to r_{g}$?
\[l_{r1-r2}=\int\limits_{r_1}^{r_2} \frac{dr}{\sqrt{1-\frac{r_g}{r}}};\quad l_{\varphi1-\varphi_2}=2\pi r|\varphi_1-\varphi_2|; \quad l_{\theta1-\theta_2}=2\pi r|\theta_1-\theta_2|.\]
Problem 3.
What would be the answers to the previous two questions for $r<r_g$ and why*? Why the Schwarzschild metric cannot be imagined as a system of "welded" rigid rods in $r<r_g$, as it can be in the external region?
This is question is given not to be answered but to make think on the answer. Correct questions and correct answers can be given in terms of a proper coordinate frame, that is regular both in $r>r_g$ and in $r<r_g$. Still, one can say something meaningful as is.
At $r<r_g$ we have $g_{00}<0$ and $g_{11}>0$, thus the $t$ coordinate is spatial and $r$ coordinate is temporal (!): \[ds^{2}=|h|^{-1}(r)dr^{2}-|h|(r)dt^{2} -r^{2}d\Omega^{2}.\] The metric therefore is nonstationary in this region, depending on the temporal coordinate $r$, but homogeneous, as there is no dependence on spatial coordinates. Then for an observer "at rest" with respect to this coordinate system we would have $dt=d\theta=d\varphi=0$, and thus \[d\tau^2=ds^2=-\frac{dr^{2}}{1-r_{g}/r}= \frac{r dr^{2}}{r_{g}-r}>0, \quad\Rightarrow\quad d\tau=\frac{\sqrt{r}dr}{\sqrt{r_{g}-r}}.\] An observer at rest with respect to the old coordinate system $dr=d\theta=d\varphi=0$, though, does not exist, as it would be $ds^{2}<0$ for him, which corresponds to spacelike geodesics (i.e. particles traveling faster than light). The last of the two questions cannot be answered without additional assumptions, because time $t$, which is the spatial coordinate now, in the two points is not given. The physical distance at $d\theta=d\varphi=dr=0$ is defined as \[dl^{2}=|h(r)|dt^2.\] It evidently depends on time $r$. This very fact that Schwarzschild metric is nonstationary at $r<r_g$, and that a stationary one does not exist in this region, leads to the absence of stationary observers and thus to the impossibility to imagine it "welded" of a system of stiff rods.
*This was actually not a very simple problem
Problem 4.
Calculate the acceleration of a test particle with zero velocity.
If a particle is at rest, its 4-velocity is $u^{\mu}=g_{00}^{-1/2}\delta^{\mu}_{0}$, where the factor is determined from the normalizing condition \[1=g_{\mu\nu}u^{\mu}u^{\mu}=g_{00}(u^{0})^{2}.\] Then the $4$-acceleration can be found from the geodesic equation: \begin{align*} a^{0}\equiv&\frac{du^0}{ds}=-{\Gamma^{0}}_{00}u^{0}u^{0}\sim {\Gamma^{0}}_{00}\sim \Gamma_{0,\,00}\sim (\partial_{0}g_{00}+\partial_{0}g_{00} -\partial_{0}g_{00})=0;\\ a^{1}\equiv&\frac{du^1}{ds} =-{\Gamma^{1}}_{00}u^{0}u^{0} =-g^{11}\Gamma_{1,\,00}\;g_{00}^{-1} =-(g_{00}g_{11})^{-1}\Gamma_{1,\,00}=\Gamma_{1,00}= \\ &=\tfrac{1}{2} (\partial_{0}g_{10}+\partial_{0}g_{10} -\partial_{1}g_{00}) =-\frac{1}{2}\frac{dg_{00}}{dr} =-\frac{h'}{2} =\frac{r_{g}}{2r^2}. \end{align*} The scalar acceleration $a$ is then equal to \[a^{2}=-g_{11}(a^{1})^{2}=\frac{(h')^{2}}{4h} =-\frac{r_{g}^{2}}{4 r^4} \Big(1-\frac{r_g}{r}\Big)^{-1}\] and tends to infinity when we approach the horizon.
Problem 5.
Show that Schwarzschild metric is a solution of Einstein's equation in vacuum.
Straightforward calculation of Christoffel symbols and Ricci tensor yields the vacuum Einstein equation $R_{\mu\nu}=0$.
Symmetries and integrals of motion
For background on Killing vectors see problems \ref{equ_oto-kill1}, \ref{equ_oto-kill2}, \ref{equ_oto-kill3} of chapter 2.
Problem 6.
What integral of motion arises due to existance of a timelike Killing vector? Express it through the physical velocity of the particle.
If a Killing vector field is timelike we can use the coordinate frame in which $K^\mu$ is the unitary vector of the time coordinate\footnote{See problem \ref{equ_oto-kill2} of chapter 2.} $K^{\mu}=(1,0,0,0)$, while the spacelike basis vectors are orthogonal to it. Then the integral of motion is energy in the corresponding (stationary) frame $p_{\mu}K^{\mu}=p_{0}\equiv\varepsilon/c$. Using the result (\ref{u0}), which holds for arbitrary gravitational field, we obtain the expression for energy, which is the integral of motion in a stationary metric \begin{equation}\label{EnergyStat} \varepsilon=mc^{2}u_{0}= mc^{2}\sqrt{g_{00}}\cdot\gamma= \frac{mc^{2}\sqrt{g_{00}}} {\sqrt{1-\frac{v^2}{c^2}}}.\end{equation}
Problem 7.
Derive the Killing vectors for a sphere in Cartesian coordinate system; in spherical coordinates.
Killing vectors correspond to infinitesimal transformations that leave the metric invariant. Considering a two-dimensional sphere embedded in the tree-dimensional space, we can see that its symmetries are rotations. Three of them are independent: the rotations in each of the three coordinate planes. Let us consider an infinitesimal rotation in the plane $XY$: $dx=yd\lambda,\; dy=-xd\lambda$. Thus the Killing vector is* \[K_{1}=x\partial_y-y\partial_x.\] Using the spherical coordinates \[x=\sin\theta\cos\varphi;\quad y=\sin\theta\sin\varphi;\quad z=\cos\theta,\] we obtain $\partial_{x}=-\frac{\sin\varphi}{\sin\theta}\partial_\varphi$, $\partial_y=\frac{\cos\varphi}{\sin\theta}\partial_\varphi$, and therefore $K_{1}=\partial_{\varphi}$. Considering rotations in plains $XZ$ and $YZ$ in the same way, in the end we obtain \[\left\{\begin{array}{l} K_{1}=x\partial_{y}-y\partial_{x}=\partial_{\varphi}\\ K_{2}=z\partial_{x}-x\partial_{z}= \cos\varphi\;\partial_{\theta}-\cot\theta\sin\varphi\;\partial_\varphi\\ K_{3}=z\partial_{y}-y\partial_{x}= \sin\varphi\;\partial_{\theta}+\cot\theta\cos\varphi\;\partial_\varphi\;. \end{array}\right.\]
*Hereafter we use the following notation: a 4-vector $A$ is $A=A^{\mu}\partial_{\mu}$, where $\partial_{\mu}\equiv\partial/\partial x_{\mu}$ is the "coordinate" basis, $A^{\mu}$ the coordinates of $A$ in this basis; it is not hard to verify that transformation laws for $A^\mu$ and $\partial_{\mu}$ are adjusted so that $A$ is a quantity that does not depend on a coordinate frame. This also enables us to conveniently recalculate $A^\mu$ when we change the basis. For more detail see e.g. the textbook by Carroll:
Carroll S., Spacetime and geometry: an introduction to General Relativity. AW, 2003, ISBN 0805387323
Problem 8.
Verify that in coordinates $(t,r,\theta,\varphi)$ vectors \[ \begin{array}{l} \Omega^{\mu}=(1,0,0,0),\\ R^{\mu}=(0,0,0,1),\\ S^{\mu}=(0,0,\cos\varphi,-\cot\theta\sin\varphi),\\ T^{\mu}=(0,0,-\sin\varphi,-\cot\theta\cos\varphi) \end{array}\] are the Killing vectors of the Schwarzschild metric.
Vectors $\Omega^\mu$ and $R^\mu$ are Killing vectors because the metric does not depend explicitly either on $t$ or on $\varphi$. The last two correspond to the spherical symmetry (see previous problem). We can also check directly that they obey the Killing equation by evaluating the Christoffel symbols for the Schwarzschild metric.
Problem 9.
Show that existence of Killing vectors $S^\mu$ and $T^\mu$ leads to motion of particles in a plane.
The integrals of motion corresponding to $S^\mu$ and $T^\mu$ are \begin{align*} S=u_{\mu}S^{\mu}=& \;u_{2}\cos\varphi-u_{3}\cot\theta\sin\varphi;\\ T=u_{\mu}T^{\mu}=& -u_{2}\sin\varphi-u_{3}\cot\theta\cos\varphi. \end{align*} Let us align the coordinate frame in such a way that the initial conditions are \[\theta|_{t=0}=\pi/2;\quad u_{2}|_{t=0}\equiv u_{\theta}|_{t=0}=0.\] Then $S$ and $T$ are zero: \[u_{2}\cos\varphi=u_{3}\cot\theta\sin\varphi;\qquad u_{2}\sin\varphi=-u_{3}\cot\theta\cos\varphi.\] Taking the square and adding the two equations, and also multiplying them, we obtain \[\left\{\begin{array}{l} u_{2}^{2}=u_{3}^{2}\cot^{2}\theta,\\ u_{2}^{2}\sin\varphi\cos\varphi= -u_{3}^{2}\cot^{2}\theta\sin\varphi\cos\varphi \end{array}\right.\quad\Rightarrow\quad u_{3}^{2}\sin\varphi\cos\varphi \cdot\cot^{2}\theta=0.\] Then either $\varphi=const$, which means that $u_{3}=u_{2}=0$ and the motion is radial, or $\cot^{2}\theta=0$ and the motion takes place in the plane $\theta=\pi/2$.
Problem 10.
Show that the particles' motion in the plane is stable.
Conservation of $u_{\mu}R^{\mu}$ means that $R\equiv u_{3}\equiv\varphi'=const$. It is not hard to derive from the expressions for $S$ and $T$ that \begin{align*} &S\sin\varphi+T\cos\varphi=R\cot\theta,\\ &S\cos\varphi-T\sin\varphi=u_2. \end{align*} Taking the square and adding up, we are led to \[u_2^2 =S^2+T^2+R^2-\frac{R^2}{\sin^2 \theta},\] then on differentiating we obtain \[u'_{2}=R^2 \frac{\cos\theta}{\sin^{2}\theta}.\] Let the trajectory deviate slightly from the plane $\theta=\pi/2$. Then $\theta=\pi/2+\delta\theta$ and \[(\delta\theta)''=u'_{2} =R^2 \frac{\cos\theta}{\sin^{2}\theta} \approx -R^2 \delta\theta,\] therefore $\theta$ oscillates around the stable point $\pi/2$.
Problem 11.
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Problem 12.
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Radial motion
Consider a particle's radial motion: $\dot{\varphi}=\dot{\theta}=0$. In this problem one is especially interested in asymptotes of all functions as $r\to r_{g}$.
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Problem 14.
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Problem 15.
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Problem 16.
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Problem 17.
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Blackness of black holes
A source radiates photons of frequency $\omega_i$, its radial coordinate at the time of emission is $r=r_{em}$. Find the frequency of photons registered by a detector situated at $r=r_{det}$ on the same radial line in different situations described below. By stationary observers here, we mean stationary in the static Schwarzschild metric; "radius" is the radial coordinate $r$.
Problem 18.
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Problem 19.
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Problem 20.
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Problem 21.
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Orbital motion, effective potential
Due to high symmetry of the Schwarzschild metric, a particle's worldline is completely determined by the normalizing condition $u^{\mu}u_{\mu}=\epsilon$, where $\epsilon=1$ for a massive particle and $\epsilon=0$ for a massless one, plus two conservation laws---of energy and angular momentum.
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Miscellaneous problems
Problem 27.
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Problem 30.
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Solving Einstein's equations for a spherically symmetric metric of general form in vacuum (energy-momentum tensor equals to zero), one can reduce the metric to \[ds^2=f(t)\Big(1-\frac{C}{r}\Big)dt^2 -\Big(1-\frac{C}{r}\Big)^{-1}dr^2-r^2 d\Omega^2,\] where $C$ is some integration constant, and $f(t)$ an arbitrary function of time $t$.
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Problem 32.
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Problem 33.
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Different coordinates, maximal extension
We saw that a particle's proper time of reaching the singularity is finite. However, the Schwarzschild metric has a (removable) coordinate singularity at $r=r_{g}$. In order to eliminate it and analyze the casual structure of the full solution, it is convenient to use other coordinate frames. Everywhere below we transform the coordinates $r$ and $t$, while leaving the angular part unchanged.
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