Difference between revisions of "Schwarzschild black hole"

Revision as of 15:22, 17 June 2012

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.

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 $r\to r_{g}$?

solution

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 $r_{1},r\to r_{g}$?

solution

\[l_{r1-r2}=\int\limits_{r_1}^{r_2} \frac{dr}{\sqrt{1-\frac{r_g}{r}}};\quad l_{\varphi1-\varphi_2}=2\pi r|\varphi_1-\varphi_2|; \quad l_{\theta1-\theta_2}=2\pi r|\theta_1-\theta_2|.\]

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?

solution

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.

solution

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.

problem formulation

solution

problem solution

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 $t$.

Problem 31.

problem formulation

solution

problem solution

Problem 32.

problem formulation

solution

problem solution

Problem 33.

problem formulation

solution

problem formulation

solution

problem solution

@@ Line 79: / Line 79: @@
 === Problem 4. ===
-problem formulation
+Calculate the acceleration of a test particle with zero velocity.
 <div class="NavFrame collapsed">
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
      <p style="text-align: left;">
-problem solution	</p>
+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.	</p>
    </div>
 </div>
 <div id="BlackHoleExtra1"></div>
 === Problem 5. ===
 problem formulation

Difference between revisions of "Schwarzschild black hole"

Revision as of 15:22, 17 June 2012

Contents

Simple problems

Problem 1.

Problem 2.

Problem 3.

Problem 4.

Problem 5.

Symmetries and integrals of motion

Problem 6.

Problem 7.

Problem 8.

Problem 9.

Problem 10.

Problem 11.

Problem 12.

Radial motion

Problem 13.

Problem 14.

Problem 15.

Problem 16.

Problem 17.

Blackness of black holes

Problem 18.

Problem 19.

Problem 20.

Problem 21.

Orbital motion, effective potential

Problem 22.

Problem 23.

Problem 24.

Problem 25.

Problem 26.

Miscellaneous problems

Problem 27.

Problem 28.

Problem 29.

Problem 30.

Problem 31.

Problem 32.

Problem 33.

Different coordinates, maximal extension

Problem 34.

Problem 35.

Problem 36.

Problem 37.

Problem 38.

Problem 39.

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