Kerr black hole - Revision history

Igor: /* Problem 4: some more simple algebra */

2013-12-08T14:38:07Z

‎Problem 4: some more simple algebra

Igor at 17:25, 13 October 2012

2012-10-13T17:25:09Z

Igor at 20:53, 24 July 2012

2012-07-24T20:53:02Z

Cosmo All at 21:52, 23 July 2012

2012-07-23T21:52:55Z

Igor: /* Particles' motion in the equatorial plane */

2012-07-23T11:54:55Z

‎Particles' motion in the equatorial plane

Igor: /* Problem 36. */

2012-07-23T11:50:22Z

‎Problem 36.

Igor: /* Problem 36. */

2012-07-23T11:43:22Z

‎Problem 36.

Igor: /* Problem 36. */

2012-07-23T11:42:17Z

‎Problem 36.

Igor: /* Problem 36. */

2012-07-23T11:39:38Z

‎Problem 36.

Igor: /* Problem 36. */

2012-07-23T11:37:43Z

‎Problem 36.

@@ Line 126: / Line 126: @@
      <p style="text-align: left;">
 Let us introduce notation
-\[\Sigma^{2}=(r^2+a^2)^{2}-a^{2}\Delta \sin^{2}\theta,\]
+\begin{align*}
 so that for the Kerr metric $g_{\varphi\varphi}=-\tfrac{\Sigma^2}{\rho^2}\sin^{2}\theta$. After some straightforward calculations then we obtain
 \begin{equation}\label{Kerr-ABCD}

← Older revision		Revision as of 17:25, 13 October 2012
Line 1:		Line 1:
	[[Category:Black Holes\|3]]		[[Category:Black Holes\|3]]
−	~~=Kerr black hole=~~

	Kerr solution$^{}$ is the solution of Einstein's equations in vacuum that describes a rotating black hole (or the metric outside of a rotating axially symmetric body) . In the Boyer-Lindquist coordinates$^{*}$ it takes the form		Kerr solution$^{}$ is the solution of Einstein's equations in vacuum that describes a rotating black hole (or the metric outside of a rotating axially symmetric body) . In the Boyer-Lindquist coordinates$^{*}$ it takes the form

@@ Line 1: / Line 1: @@
 [[Category:Black Holes|3]]
 Kerr solution$^{*}$ is the solution of Einstein's equations in vacuum that describes a rotating black hole (or the metric outside of a rotating axially symmetric body) . In the Boyer-Lindquist coordinates$^{**}$ it takes the form
 \begin{align}\label{Kerr}
@@ Line 20: / Line 22: @@
 $^{**}$ R.H. Boyer, R.W. Lindquist. Maximal Analytic Extension of the Kerr Metric. ''J. Math. Phys'' '''8''', 265–281 (1967).
 ==General axially symmetric metric==

← Older revision		Revision as of 21:52, 23 July 2012
Line 1:		Line 1:
		+	[[Category:Black Holes\|3]]
	Kerr solution$^{}$ is the solution of Einstein's equations in vacuum that describes a rotating black hole (or the metric outside of a rotating axially symmetric body) . In the Boyer-Lindquist coordinates$^{*}$ it takes the form		Kerr solution$^{}$ is the solution of Einstein's equations in vacuum that describes a rotating black hole (or the metric outside of a rotating axially symmetric body) . In the Boyer-Lindquist coordinates$^{*}$ it takes the form
	\begin{align}\label{Kerr}		\begin{align}\label{Kerr}

@@ Line 881: / Line 881: @@
 <div id="BlackHole81"></div>
-=== Problem 30. ===
+=== Problem 30: preparatory algebra ===
 Put down explicit expressions for the metric components and parameters $A,B,C,D,\omega$
 <div class="NavFrame collapsed">
@@ Line 910: / Line 910: @@
 <div id="BlackHole82"></div>
-=== Problem 31. ===
+=== Problem 31: zero energy particles ===
 What is the angular velocity of a particle with zero energy?
 <div class="NavFrame collapsed">
@@ Line 936: / Line 936: @@
 <div id="BlackHole83"></div>
+=== Problem 32: geodesics and effective potential ===
-=== Problem 32. ===
 Use the normalizing conditions for the $4$-velocity $u^{\mu}u_{\mu}=\epsilon^2$ and two conservation laws to derive geodesic equations for particles, determine the effective potential for radial motion.
 <div class="NavFrame collapsed">
@@ Line 1,010: / Line 1,009: @@
 <div id="BlackHole84"></div>
-=== Problem 33. ===
+=== Problem 33: principal null geodesics ===
 Integrate the equations of motion for null geodesics with $L=aE$, investigate the asymptotes close to the horizons, limits $a\to 0$ and $a\to \mu$.
 <div class="NavFrame collapsed">
@@ Line 1,049: / Line 1,048: @@
 <div id="BlackHole85"></div>
-=== Problem 34. ===
+=== Problem 34: innermost stable circular orbits, massless case ===
 Find the minimal radii of circular geodesics for massless particles, the corresponding values of integrals of motion and angular velocities. Show that of the three solutions one lies beyond the horizon, one describes motion in positive direction and one in negative direction. Explore the limiting cases of Schwarzschild $a\to0$ and extreme Kerr $a\to\mu$.
 <div class="NavFrame collapsed">
@@ Line 1,132: / Line 1,131: @@
 <div id="BlackHole86"></div>
+=== Problem 35: circular orbits for massive particles ===
-=== Problem 35. ===
 Find $L^2$ and $E^2$ as functions of radii for circular geodesics of the massive particles.
 <div class="NavFrame collapsed">
@@ Line 1,193: / Line 1,191: @@
 <div id="BlackHole87"></div>
-=== Problem 36. ===
+=== Problem 36: innermost stable circular orbits, massive case ===
 Derive equation for the minimal radius of a stable circular orbit; find the energy and angular momentum of a particle on it, the minimal radius in the limiting cases $a/\mu\to 0,1$.
 <div class="NavFrame collapsed">

@@ Line 1,207: / Line 1,207: @@
 In the limit $\alpha\to 0$ we get $\xi>6$, i.e. $r>3\mu=\frac{3}{2}r_{g}$, which is the familiar result for Schwarzschild.</p>
-[[File:BHfig-Kerr-XiMin.png|center|thumb|400px| " ''Radius'' of innermost stable circular orbit as function of $\alpha=a/\mu$: the upper curve for $L>0$, the lower one for $L<0$. The horizon is shown by the dashed line."]]
+[[File:BHfig-Kerr-XiMin.png|center|thumb|400px| " ''Radius'' of innermost stable circular orbit as function of $\alpha=a/\mu$: the upper curve for $L<0$, the lower one for $L>0$. The horizon is shown by the dashed line."]]
-<p style="text-align: left;">In the limit $a\to 1$ different signs lead to different inequalities, which are easier to solve numerically. From the curves we see, that for the "$+$" sign the stability condition holds when $\xi>9$, and for the "$-$" sign when $\xi>1$. In the general case then
+<p style="text-align: left;">In the limit $a\to 1$ different signs lead to different inequalities, which are easier to solve numerically. From the curves we see, that for the "$-$" sign (retrograde or counter-rotating orbits) the stability condition holds when $\xi>9$, and for the "$+$" sign (prograde or co-rotating orbits) when $\xi>1$. In the general case then
 \begin{align*}
 &''-'':\qquad \alpha\in(0,1)\;\Rightarrow\;
 		\xi_{min}\in(6,1).
 \end{align*}
-The binding energy on the minimal stable orbit is
+The binding energy on the minimal stable (prograde) orbit is
 \[E^2=\nu^2 u^3+1-u=u/3+1-u=1-\frac{2}{3\xi},\]
 and for $\alpha\approx 1$ for the most strongly bound particle with $\xi\approx 1$

@@ Line 1,206: / Line 1,206: @@
 \[\xi^2-6\xi\pm 8\alpha \sqrt{\xi}-3\alpha^2>0.\]
 In the limit $\alpha\to 0$ we get $\xi>6$, i.e. $r>3\mu=\frac{3}{2}r_{g}$, which is the familiar result for Schwarzschild.
-\begin{figure}[!hbt]\center
+[[File:BHfig-Kerr-XiMin.png|400px|''Radius'' of innermost stable circular orbit as function of $\alpha=a/\mu$: the upper curve for $L>0$, the lower one for $L<0$. The horizon is shown by the dashed line.]]
-\includegraphics*[width=0.65\textwidth]{BHfig-Kerr-XiMin}
-\parbox{0.7\textwidth}{\caption{''Radius'' of minimal stable circular orbit as function of $\alpha=a/\mu$: the upper curve for $L>0$, the lower one for $L<0$. The horizon is shown by the dashed line.}}
-\end{figure}
-In the limit $a\to 1$ different signs lead to different inequalities, which are easier to solve numerically. From the curves we see, that for the ``$-$'' sign the stability condition holds when $\xi>9$, and for the ``$-$'' sign when $\xi>1$. In the general case then
+In the limit $a\to 1$ different signs lead to different inequalities, which are easier to solve numerically. From the curves we see, that for the "$+$" sign the stability condition holds when $\xi>9$, and for the "$-$" sign when $\xi>1$. In the general case then
 \begin{align*}
 &''+'':\qquad \alpha\in(0,1)\;\Rightarrow\;