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                "title": "Realization of interaction in the dark sector",
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                        "*": "[[Category:Interactions in the Dark Sector|7]]\n\n\n\n__NOTOC__\n\n\n\n\n= Vacuum Decay into Cold Dark Matter =\n\nLet us consider the Einstein field equations\n\\[R_{\\mu\\nu}-frac12Rg_{\\mu\\nu}=8\\pi G\\left(T_{\\mu\\nu}+\\frac\\Lambda{8\\pi G}g_{\\mu\\nu}\\right).\\]\t\nAccording to the Bianchi identities, (i) vacuum decay is possible only from a previous existence of some sort of non-vanishing matter and/or radiation, and (ii) the presence of a time-varying cosmological term results in a coupling between $T_{\\mu\\nu}$ and $\\Lambda$. We will assume (unless stated otherwise) coupling only between vacuum and CDM particles, so that\n\t\\[u_\\mu,T_{;\\nu}^{(CDM)\\mu\\nu}=-u_\\mu\\left(\\frac{\\Lambda g^{\\mu\\nu}}{8\\pi G}\\right)_{;\\nu}= -u_\\mu\\left(\\rho_\\Lambda g^{\\mu\\nu}\\right)_{;\\nu}\\]\nwhere $T_{\\mu\\nu}^{(CDM)}=\\rho_{dm}u^\\mu u^\\nu$ is the energy-momentum tensor of the CDM matter and $\\rho_\\Lambda$ is the vacuum energy density. It immediately follows that\n\t\\[\\dot\\rho_{dm}+3H\\rho_{dm}=-\\dot\\rho_\\Lambda.\\]\nNote that although the vacuum is decaying, $w_\\Lambda=-1$ is still constant, the physical equation of state (EoS) of the vacuum $w_\\Lambda\\equiv=p_\\Lambda/\\rho_\\Lambda$ is still equal to constant $-1$, which follows from the definition  of the cosmological constant.\n\n(see  [http://arxiv.org/abs/astro-ph/0408495 Can vacuum decay in our Universe?], [http://arxiv.org/abs/astro-ph/0507372 Interpreting Cosmological Vacuum Decay])\n\n\n<div id=\"IDE_78\"></div>\n<div style=\"border: 1px solid #AAA; padding:5px;\">\n=== Problem 1 ===\nSince vacuum energy is constantly decaying into CDM, CDM will dilute in a smaller rate compared with the standard relation $\\rho_{dm}\\propto a^{-3}$. Thus we assume that $\\rho_{dm}=\\rho_{dm0}a^{-3+\\varepsilon}$, where $\\varepsilon$ is a small positive small constant. Find the dependence $\\rho_\\Lambda(a)$ in this model.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">Conservation equation for CDM yields\n\\[\\rho_\\Lambda=\\bar\\rho_{\\Lambda0}+\\frac{\\varepsilon\\rho_{dm0}}{3-\\varepsilon}a^{-3+\\varepsilon}.\\]</p>\n  </div>\n</div></div>\n\n\n\n<div id=\"IDE_79\"></div>\n<div style=\"border: 1px solid #AAA; padding:5px;\">\n=== Problem 2 ===\nSolve the previous problem for the case when vacuum energy is constantly decaying into radiation.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">By considering that radiation will dilute more slowly compared to its standard evolution, $\\rho_r\\propto a^{-4}$, and that such a deviation is characterized by a positive constant $\\alpha$ one finds\n\t\\[\\rho_r=\\rho_{r0}a^{-4+\\alpha}.\\]\nBy inserting this expression into radiation conservation equation\n\t\\[\\dot\\rho_r+4H\\rho_r=-\\dot\\rho_\\Lambda,\\]\none obtains\n\\[\\rho_\\Lambda=\\bar\\rho_{\\Lambda0}+\\frac{\\alpha\\rho_{r0}}{4-\\alpha}a^{-4+\\alpha}.\\]\t</p>\n  </div>\n</div></div>\n\n\n\n<div id=\"IDE_80\"></div>\n<div style=\"border: 1px solid #AAA; padding:5px;\">\n=== Problem 3 ===\nShow that existence of a radiation dominated stage is always guaranteed in scenarios, considered in the previous problem.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">The ratio between the vacuum and radiation energy densities reads\n\\[\\frac{\\rho_\\Lambda}{\\rho_r}=\\frac{\\bar\\rho_{\\Lambda0}}{\\rho_{r0}}a^{4-\\alpha}+\\frac\\alpha{4-\\alpha}.\\]\nThe first term is asymptotically vanishing at early times whereas the second one is smaller than unity. Therefore, a radiation dominated stage is always guaranteed in this kind of scenarios.</p>\n  </div>\n</div></div>\n\n\n\n<div id=\"IDE_81\"></div>\n<div style=\"border: 1px solid #AAA; padding:5px;\">\n=== Problem 4 ===\nFind how the new temperature law scales with redshift in the case of vacuum energy decaying into radiation.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">For an adiabatic vacuum decay the equilibrium relations are preserved, as happens with the Stefan law, $\\rho_r\\propto T^4$.  As a consequence, one may check that the $Ta^{1-\\alpha/4}$. This implies that the new temperature law scales with redshift as\n\\[T=T_0(1+z)^{1-\\alpha/4}.\\]</p>\n  </div>\n</div></div>\n\n\n\n== Vacuum decay into CDM particles ==\n\nSince the energy density of the cold dark matter is $\\rho_{dm}=nm$, there are two possibilities for storage of the energy received from the vacuum decay process:\n\n(i) the equation describing concentration, $n$, has a source term while the proper mass of CDM particles remains constant;\n\n(ii) the mass $m$ of the CDM particles is itself a time-dependent quantity, while the total number of CDM particles, $N=na^3$, remains constant.\n\nLet us consider both the possibilities.\n\n\n\n\n\n<div id=\"IDE_82\"></div>\n<div style=\"border: 1px solid #AAA; padding:5px;\">\n=== Problem 5 ===\nFind dependence of total particle number on the scale factor in the model considered in problem [[#IDE_78]].\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">Since $\\rho_{dm}=nm$, we find, using $\\rho_{dm}=\\rho_{dm0}a^{-3+\\varepsilon}$, that\n\t\\[n=n_0a^{-3+\\varepsilon}.\\]\nIn the considered case, there is necessarily a source term in the current of CDM particles, that is, $N^\\alpha_{;\\alpha}$, where $N^\\alpha=nu^\\alpha$. In terms of the concentration it can be written as\n\t\\[\\dot n +3Hn=\\psi.\\]\nInserting $n=n_0a^{-3+\\varepsilon}$, one obtains\n\t\\[\\psi=\\varepsilon\\frac{\\dot a}an\\equiv n\\Gamma.\\]\nWe have written particle source $\\psi$ in terms of a decay rate, $\\Gamma\\equiv\\dot N/N$.</p>\n  </div>\n</div></div>\n\n\n\n<div id=\"IDE_83\"></div>\n<div style=\"border: 1px solid #AAA; padding:5px;\">\n=== Problem 6 ===\nFind time dependence of CDM particle mass in the case when there is no creation of CDM particles in the model considered in problem [[#IDE_78]].\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">In this case the concentration of CDM particles satisfies the equation\n\t\\[\\dot n +3Hn=0,\\]\nwhose solution is $n=n_0a^{-3}$, which implies that $N(t)=const$. Naturally, if CDM particles are not being created, the unique possibility is an increasing in the proper mass of CDM particles. Actually, since $\\rho_{dm0}=nm$, expressions \\[\\rho_{dm}=\\rho_{dm0}a^{-3+\\varepsilon}\\] and \\[n=n_0a^{-3}\\] imply that the mass of the CDM particles scales as\n\t\\[m(t)=m_0a(t)^\\varepsilon,\\]\nwhere $m_0$ is the present day mass of CDM particles.</p>\n  </div>\n</div></div>\n\n\n\n<div id=\"IDE_84\"></div>\n<div style=\"border: 1px solid #AAA; padding:5px;\">\n=== Problem 7 ===\nConsider a model where the cosmological constant $\\Lambda$ depends on time as $\\Lambda=\\sigma H$. Let a flat Universe be filled by the time-dependent cosmological constant and a component with the state equation $p_\\gamma= (\\gamma-1)\\rho_\\gamma$. Find solutions of Friedman equations for this system [http://arxiv.org/abs/0711.2686].\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">System of equations describing dynamics of the model reads\n\\begin{align}\n\\nonumber\n\\rho_\\gamma+\\Lambda & = 3H^2,\\\\\n\\nonumber\n\\dot\\rho_\\gamma + 3H\\gamma\\rho_\\gamma & =-\\dot\\Lambda,\\\\\n\\nonumber\n\\Lambda & =\\sigma H.\n\\end{align}\nCombine this equations to obtain\n\t\\[2\\dot H+3\\gamma H^2-\\sigma\\gamma H=0.\\]\nFor $H>0$ (expanding Universe) the equation has the following solution\n\t\\[a(t)=C\\left[\\exp\\left(\\frac{\\sigma\\gamma t}2\\right)-1\\right]^{3\\gamma/2}.\\]\nHere $C$ is one of the two integration constants (the equation for scale factor is the second-order one). The second integration constant is determined from the condition $a(t=0)=0$. Using the obtained solution, one finds\n\\begin{align}\n\\nonumber\n\\rho_\\gamma & =\\frac{\\sigma^2}3\\left[1+\\left(\\frac C a\\right)^{3\\gamma/2}\\right]\\left(\\frac C a\\right)^{3\\gamma/2},\\\\\n\\nonumber\n\\Lambda & =\\frac{\\sigma^2}3\\left[1+\\left(\\frac C a\\right)^{3\\gamma/2}\\right].\n\\end{align}</p>\n  </div>\n</div></div>\n\n\n\n<div id=\"IDE_85\"></div>\n<div style=\"border: 1px solid #AAA; padding:5px;\">\n\n=== Problem 8 ===\nShow that the model considered in the previous problem correctly reproduces the scale factor evolution both in the radiation-dominated and non-relativistic matter (dust) dominated cases.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">In the radiation-dominated epoch $\\gamma=w+1=4/3$. Therefore\n\t\\[a(t)=C\\left[\\exp\\left(\\frac{2\\sigma t}3\\right)-1\\right]^{1/2}.\\]\nFor small times $\\sigma t\\ll1$ the formula reproduces correct dependence $a(t)\\propto t^{1/2}$. In the matter dominated case $\\gamma=w+1=1$ and\n\t\\[a(t)=C\\left[\\exp\\left(\\frac{2\\sigma t}3\\right)-1\\right]^{2/3}.\\]\nAt $\\sigma t\\ll1$ we recover the standard law of non-relativistic matter evolution $a(t)\\propto t^{2/3}$. Note that in the limit of long times $\\sigma t\\gg1$ the scale factor grows exponentially for all values of the parameter $\\gamma$.</p>\n  </div>\n</div></div>\n\n\n\n<div id=\"IDE_86\"></div>\n<div style=\"border: 1px solid #AAA; padding:5px;\">\n=== Problem 9 ===\nFind the dependencies $\\rho_\\gamma(a)$ and $\\Lambda(a)$ both in the radiation-dominated and non-relativistic matter dominated cases in the model considered in problem [[#IDE_84]].\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">Using the expressions for $\\rho_\\gamma(a)$ and $\\Lambda(a)$ obtained in the problem [[#IDE_84]], one finds that in the radiation-dominated epoch $(\\gamma=4/3)$\n\\begin{align}\n\\nonumber\n\\rho_\\gamma & =\\frac{\\sigma^2C^4}3\\frac1{a^4}+\\frac{\\sigma^2C^2}3\\frac1{a^2},\\\\\n\\nonumber\n\\Lambda & =\\frac{\\sigma^2}3+\\frac{\\sigma^2C^2}3\\frac1{a^2}.\n\\end{align}\nIn the limit $a\\to0$ ($t\\to0$)\n\\begin{align}\n\\nonumber\n\\rho_\\gamma & =\\frac{\\sigma^2C^4}3\\frac1{a^4}=\\frac{3}{4t^2},\\\\\n\\nonumber\n\\Lambda & =\\frac{\\sigma^2C^2}3\\frac1{a^2}=\\frac\\sigma{2t}.\n\\end{align}\nAt the matter-dominated epoch ($\\gamma=1$)\n\\begin{align}\n\\nonumber\n\\rho_\\gamma & =\\frac{\\sigma^2C^3}3\\frac1{a^3}+\\frac{\\sigma^2C^{3/2}}3\\frac1{a^{3/2}},\\\\\n\\nonumber\n\\Lambda & =\\frac{\\sigma^2}3+\\frac{\\sigma^2C^{3/2}}3\\frac1{a^{3/2}}.\n\\end{align}</p>\n  </div>\n</div></div>\n\n\n\n<div id=\"IDE_87\"></div>\n<div style=\"border: 1px solid #AAA; padding:5px;\">\n=== Problem 10 ===\nShow that for the $\\Lambda(t)$ models \\[T\\frac{dS}{dt}=-\\dot\\rho_\\Lambda a^3.\\]\n<!--<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\"></p>\n  </div>\n</div>--></div>\n\n\n\n\n= Time-dependent cosmological \"constant\" =\n\n\n\n\n<div id=\"IDE_87_2\"></div>\n<div style=\"border: 1px solid #AAA; padding:5px;\">\n=== Problem 11 ===\nConsider a two-component Universe filled by matter with the state equation $p=w\\rho$ and cosmological constant and rewrite the second Friedman equation in the following form\n      \\begin{equation}\\label{mainDE}\n\\frac{\\ddot{a}}{a} = \\frac{1}{2}\\left( 1 + 3w\\right)\n                     \\left( \\frac{\\dot{a}^2}{a^2} + \\frac{k}{a^2} \\right)\n                   + \\frac{1-3w}{6} \\Lambda .\n\\end{equation}\n<!--<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\"></p>\n  </div>\n</div>--></div>\n\n\n\n<div id=\"IDE_87_3\"></div>\n<div style=\"border: 1px solid #AAA; padding:5px;\">\n=== Problem 12 ===\nConsider a two-component Universe filled by matter with the state equation $p=w\\rho$ and cosmological constant with quadratic time dependence $\\Lambda(\\tau)=\\mathcal{A}\\tau^2$ and find the time dependence for of the scale factor.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">Substitute the dependence $\\Lambda(t)=Bt^2$ into the equation (\\ref{mainDE}) and multiply both sides by $a^2$ to obtain\n\\begin{equation}\n\\ddot{a} a = \\frac 12 (1+3w)\\left(\\dot{a}^2 + k\\right) +\\frac{1-3w}{6} a^2\\mathcal{A}t^2\\Lambda\n\\label{mainDE1}\n\\end{equation}\nSeek the solution in the form $a(\\tau)=a_0\\tau^\\beta.$ Substitute this dependence into (\\ref{mainDE1}) to show that the equality takes place only under the following condition\n$$\\mathcal{A} = \\frac{6\\beta}{1-3w}\\left(1 -\\frac 1\\beta +\\frac 32 w \\right)$$</p>\n  </div>\n</div></div>\n\n\n\n<div id=\"tauell\"></div>\n<div style=\"border: 1px solid #AAA; padding:5px;\">\n=== Problem 13 ===\nConsider a flat two-component Universe filled by matter with the state equation $p=w\\rho$ and cosmological constant with quadratic time dependence $\\Lambda(\\tau)=\\mathcal{A}\\tau^{\\ell}$. Obtain the differential equation for Hubble parameter in this model and classify it.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">Recall that the definition of the Hubble parameter reads\n\\begin{equation}\nH \\equiv \\frac{\\dot{a}}{a} = H_0 \\left( \\frac{da}{a \\, d\\tau} \\right) ,\n\\label{Hdefn}\n\\end{equation}\nand use \\eqref{mainDE} to obtain:\n\\begin{equation}\n\\frac{dH}{d\\tau} = \\left( \\frac{-3\\gamma}{2H_0} \\right) H^2 +\n                   \\left( \\frac{\\gamma}{2H_0} \\right) \\Lambda +\n                   \\left( 1-\\frac{3\\gamma}{2} \\right) \\frac{k}{H_0 \\, a^2},\n\\label{1stDE}\n\\end{equation}\nwhere $\\gamma = 1+w.$ This equation can presented in the following form\n\\begin{equation}\n\\frac{dH}{d\\tau} = {\\mathcal P}(\\tau) \\, H^2 + {\\mathcal Q}(\\tau) \\, H +\n                   {\\mathcal R}(\\tau) ,\n\\label{RiccatiDE}\n\\end{equation}\nwhere ${\\mathcal P} \\equiv - 3\\gamma/2$, ${\\mathcal Q} \\equiv 0$ and\n${\\mathcal R}(\\tau) \\equiv (\\gamma/2) \\Lambda(\\tau).$\n\nThe obtained equation \\eqref{RiccatiDE} is of Riccati type. For further convenience we transit to dimensionless variables $H \\to x$:\n\\begin{equation}\nH \\equiv -\\frac{1}{{\\mathcal P} x} \\, \\frac{dx}{d\\tau} =\n          \\left( \\frac{2H_0}{3\\gamma} \\right) \\frac{dx}{x \\, d\\tau} ,\n\\label{xDefn}\n\\end{equation}\nAccording to the condition of the problem one should consider the spatially flat Universe $k=0,$ so substitute into the equation \\eqref{1stDE} the following dependence $\\Lambda(\\tau)=A\\tau^{\\ell}:$\n\\begin{equation}\n\\tau^{\\ell} \\; \\frac{d\\,^2x}{d\\tau^2} - \\alpha \\, x = 0 ,\n\\label{mainDE-t}\n\\end{equation}\nwhere\n\\begin{equation}\n\\alpha \\equiv \\frac{3\\gamma^2 {\\mathcal A}}{4} .\n\\label{alphaDef1}\n\\end{equation}\nIt is easy now to show that the scale factor $a(\\tau)$ and the function $x(\\tau)$ are linked by the relation\n\\begin{equation}\na(\\tau) = [ x(\\tau) ]^{2/3\\gamma} .\n\\label{xTOa}\n\\end{equation}\nThe constant $\\alpha$ is determined by the relation \\eqref{alphaDef1}.</p>\n  </div>\n</div></div>\n\n\n\n<div id=\"IDE_87_5\"></div>\n<div style=\"border: 1px solid #AAA; padding:5px;\">\n=== Problem 14 ===\nFind solution of the equation obtained in the previous problem in the case ${\\ell =1}$. Analyze the obtained solution.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">In the considered case the equation \\eqref{mainDE-t} takes the form\n\\begin{equation}\n\\tau \\; \\frac{d\\,^2x}{d\\tau^2} - \\alpha \\, x = 0 ,\n\\label{mainDE-t-1}\n\\end{equation}\nThe constant $\\alpha$ is determined by the relation \\eqref{alphaDef1}:\n\\begin{equation}\n\\alpha = (3\\gamma/2)^2 \\lambda_0 \\tau_0 .\n\\label{alphaDefn-1}\n\\end{equation}\nChange the independent variable from $\\tau$ to $z \\equiv 2\\sqrt{-\\alpha\\tau}$ to obtain\n\\begin{equation}\nz^2 \\, \\frac{d\\,^2x}{dz^2} - z \\, \\frac{dx}{dz} + z^2 \\, x = 0 .\n\\label{mainDE-t-1b}\n\\end{equation}\nThis equation can be reduced to the Bessel one\n$x(z) = c_1 z J_1(z) + c_2 z Y_1 (z),$\nwhere $J_1(z)$ and $Y_1(z)$ are respectively Bessel and Neumann functions of first order.\n\nUse the relation (\\ref{xTOa}) to obtain the explicit dependence for the scale factor:\n\\begin{equation}\na(\\tau) = \\tau^{1/3\\gamma} \\left[ c_1 J_1(z) +\n                                  c_2 Y_1 (z) \\right]^{\\; 2/3\\gamma} ,\n\\label{aSoln-1}\n\\end{equation}\nwhere the factor $2\\sqrt{-\\alpha}$ enters the coefficients $c_1,c_2$.\nThe Hubble parameter can be determined by substitution $x(z)$ into the expression \\eqref{xDefn}:\n\\begin{equation}\nH(\\tau) = H_0 \\, \\sqrt{-\\lambda_0 \\left( \\frac{\\tau_0}{\\tau} \\right)} \\,\n                 \\left[ \\frac{c_1 J_0(z) + c_2 Y_0(z)}\n                             {c_1 J_1(z) + c_2 Y_1(z)} \\right] ,\n\\label{HSoln-1}\n\\end{equation}\nwhere $J_0(z)$ and $Y_0(z)$ are respectively Bessel and Neumann functions of zero order.\nNote that from the definition (\\ref{alphaDefn-1}) it follows that $z(\\tau),$\n $a(\\tau)$ and $H(\\tau)$, can be real-valued (for $\\tau >0 $) if\n $\\lambda_0 \\leq 0.$\nThough such possibility exists in theory, it contradicts the observational data.\nThus the case $\\ell=1$ is likely to be impossible in real Universe.</p>\n  </div>\n</div></div>\n\n\n\n<div id=\"IDE_87_6\"></div>\n<div style=\"border: 1px solid #AAA; padding:5px;\">\n=== Problem 15 ===\nSolve the equation obtained in the problem [[#tauell]] for ${\\ell =2}.$ Consider the following cases\n<br/>\na) $\\lambda_0 > -1/(3\\gamma\\tau_0)^2,$<br/>\nb) $\\lambda_0 = -1/(3\\gamma\\tau_0)^2,$<br/>\nc) $\\lambda_0 < -1/(3\\gamma\\tau_0)^2$\n<br/>\n(see the previous problem). Analyze the obtained solution.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">In the case ${\\ell =2}$ the equation (\\ref{mainDE-t}) takes on the form\n\\begin{equation}\n\\tau^2 \\; \\frac{d\\,^2x}{d\\tau^2} - \\alpha \\, x = 0 ,\n\\label{mainDE-t-2}\n\\end{equation}\nwhere $\\alpha$ is determined by the equation (\\ref{alphaDefn}):\n\\begin{equation}\n\\alpha = (3\\gamma/2)^2 \\lambda_0 \\tau_0^2 .\n\\label{alphaDefn-2}\n\\end{equation}\nThis equation belongs to Euler type. Make the transformation of variables $y \\equiv \\ln\\tau$ in the equation (\\ref{mainDE-t-2}) to obtain\n\\begin{equation}\n\\frac{d\\,^2x}{dy^2} - \\frac{dx}{dy} - \\alpha \\, x = 0 .\n\\label{EulerDE}\n\\end{equation}\n\n\n'''a)''' $\\lambda_0 > -1/(3\\gamma\\tau_0)^2.$\nAs the astronomical observations give evidence in favor of positive value of $\\lambda_0,$ thais model makes certain interest.\nSolution of the equation (\\ref{EulerDE}) for $x(y)$, where $x(\\tau)$, is easy to obtain.\nThe scale factor and the Hubble parameter can be found using the relations (\\ref{xTOa}) and (\\ref{xDefn}), resulting in the following\n\\begin{equation}\na(\\tau)  =  \\tau^{1/3\\gamma} \\, \\left( c_1 \\tau^{m_0} + c_2\n                                  \\tau^{-m_0} \\right)^{2/3\\gamma}\n\\label{aSoln-2a-2} \n\\end{equation}\n\\begin{equation}\nH(\\tau)  =  \\frac{2H_0}{3\\gamma} \\, \\left[ \\frac\n              {m_1 c_1 \\tau^{m_0} + m_2 c_2 \\tau^{m_0}}\n              {\\tau \\, ( c_1 \\tau^{m_0} + c_2 \\tau^{-m_0})} \\right] ,\n\\label{HSoln-2a-2}\n\\end{equation}\nwhere  $m_0 \\equiv \\frac{1}{2} \\sqrt{1+(3\\gamma\\tau_0)^2 \\lambda_0}$.\nIt is easy to see that $a(\\tau)$ diverges at $\\tau=0$ for $\\lambda_0>0$ if\n$c_2 \\ne 0$. Thus the expressions (\\ref{aSoln-2a-2}) and (\\ref{HSoln-2a-2}) take on the simplified form\n\\begin{equation}\na(\\tau)  =  \\left( \\frac{\\tau}{\\tau_0} \\right)^{2m_1/3\\gamma}\n\\label{aSoln-2a-3} \n\\end{equation}\n\n\\begin{equation}\nH(\\tau)  =  \\frac{2H_0}{3\\gamma} \\, \\left[ \\frac{m_1}{\\tau} \\right] ,\n\\label{HSoln-2a-3}\n\\end{equation}\nwhere $m_1 \\equiv 1/2+m_0,$ \u00e0 $\\tau_0 = 2m_1/3\\gamma.$\n<br/>\n'''b)''' $\\lambda_0 = -1/(3\\gamma\\tau_0)^2.$\nProceed in analogous way to obtain\n\\begin{equation}\na(\\tau)  =  \\tau^{1/3\\gamma} \\, (c_3 + c_4 \\ln \\tau)^{2/3\\gamma}\n\\label{aSoln-2b-1} \n\\end{equation}\n\\begin{equation}\nH(\\tau)  =  \\frac{2H_0}{3\\gamma} \\, \\left[ \\frac\n              {(c_3+2c_4) + c_4 \\ln\\tau}\n              {2\\tau ( c_3 + c_4 \\ln\\tau)} \\right] ,\n\\label{HSoln-2b-1}\n\\end{equation}\nwhere $c_3,c_4$ are arbitrary constants.\nIn order to make $a(\\tau)$ finite at $\\tau=0$ set $c_4=0$, then\n\\begin{equation}\na(\\tau)  =  (c_3 \\sqrt{\\tau})^{2/3\\gamma}\n\\label{aSoln-2b-2} \n\\end{equation}\n\\begin{equation}\nH(\\tau)  =  \\frac{H_0}{3\\gamma\\tau} .\n\\label{HSoln-2b-2}\n\\end{equation}\nSubstitute $H(\\tau_0)=H_0$ into (\\ref{HSoln-2b-2}) to find age of the Universe in the considered model\n\\begin{equation}\n\\tau_0 = 1/3\\gamma.\n\\label{Age-2b}\n\\end{equation}\n\nSubstitute the expression (\\ref{Age-2b}) into (\\ref{aSoln-2b-2})\nand $a(\\tau_0)=1$ to find $c_3 = \\sqrt{3\\gamma}$. Then\n\\begin{equation}\na(\\tau) = \\left( \\frac{\\tau}{\\tau_0} \\right)^{1/3\\gamma}.\n\\end{equation}\n<br/>\n'''c)'''  $\\lambda_0 < -1/(3\\gamma\\tau_0)^2$\n\\label{sec:deSitter}\nThe dependencies $a(\\tau)$ and $H(\\tau)$ now take on the following form\n\\begin{equation}\na(\\tau)  =  \\tau^{1/3\\gamma} \\, \\left[ c_5 \\sin (m_3 \\ln\\tau) +\n              c_6 \\cos (m_3 \\ln\\tau) \\right]^{2/3\\gamma}\n\\label{aSoln-2c-1} \n\\end{equation}\n\\begin{equation}\nH(\\tau)  =  \\frac{2H_0}{3\\gamma} \\, \\left\\{ \\frac\n              {(c_5-2m_3c_6)\\sin(m_3\\ln\\tau)+(c_6+2m_3c_5)\\cos(m_3\\ln\\tau)}\n              {2\\tau[c_5\\sin(m_3\\ln\\tau)+c_6\\cos(m_3\\ln\\tau)]} \\right\\} ,\n\\label{HSoln-2c-1}\n\\end{equation}\nwhere $m_3 \\equiv \\frac{1}{2} \\sqrt{-(3\\gamma\\tau_0)^2 \\lambda_0 - 1}$\nand $c_5,c_6$ are arbitrary constants. Proceeding as usual, express $c_5$ and $c_6$ in terms of $\\tau_0$:\n\\begin{equation}\\label{c5defn-2}\nc_5  =  \\frac{1}{\\sqrt{\\tau_0}} \\sin(m_3\\ln\\tau_0) + \\frac{1}{m_3}\n          \\left[ \\left( \\frac{3\\gamma}{2} \\right) \\sqrt{\\tau_0} -\n          \\frac{1}{2\\sqrt{\\tau_0}} \\right] \\cos(m_3\\ln\\tau_0) \n\\end{equation}\n\\begin{equation}\nc_6  =  \\frac{1}{\\sqrt{\\tau_0}} \\cos(m_3\\ln\\tau_0) - \\frac{1}{m_3}\n          \\left[ \\left( \\frac{3\\gamma}{2} \\right) \\sqrt{\\tau_0} -\n          \\frac{1}{2\\sqrt{\\tau_0}} \\right] \\sin(m_3\\ln\\tau_0) .\n\\label{c6defn-2}\n\\end{equation}\nUnlike the previous cases, it is now impossible to keep finite values $a(\\tau)$ at $\\tau=0$ turning one of the constants $c_5,c_6$ to zero.</p>\n  </div>\n</div></div>\n\n\n\n<div id=\"IDE_87_7\"></div>\n<div style=\"border: 1px solid #AAA; padding:5px;\">\n\n=== Problem 16 ===\nConsider a flat two-component Universe filled by matter with the state equation $p=w\\rho$ and cosmological constant with the following scale factor dependence\n\\begin{equation}\n\\Lambda = {\\cal B} \\, a^{-m}.\n\\label{Bam}\n\\end{equation}\nFind dependence of energy density of matter on the scale factor in this model.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">Use the conservation equation and the first Friedman equation to obtain\n \\begin{equation}\n\\frac{d}{da} \\left( \\rho a^{3\\gamma} \\right) = \\left( \\frac{m{\\cal B}}{8\\pi G}\n                                               \\right) a^{3\\gamma-(m+1)} .\n\\end{equation}\nIts integration leads to the required dependence of energy density of matter on the scale factor\n\\begin{equation}\n\\rho (a) = \\rho_0 a^{-3\\gamma} f(a) ,\n\\label{NewRho}\n\\end{equation}\nwhere as usual $a(t_0)=a_0=1,\\rho(a_0)=\\rho_0$, and the function $f(a)$ takes on the form\n\\begin{equation}\nf(a)  \\equiv  1 + \\kappa_0 \\times \\left\\{ \\begin{array}{ll}\n                    \\frac{ {\\textstyle m(a^{3\\gamma-m}-1)} }\n                         { {\\textstyle 3\\gamma-m} }\n                          & \\mbox{ if } m \\neq 3\\gamma \\\\\n                    3\\gamma \\ln(a)\n                          & \\mbox{ if } m=3\\gamma\n                    \\end{array} \\right.\n                    \\label{fDefn} \n\\end{equation}\n\\begin{equation}\n\\kappa_0  \\equiv  {\\cal B}/8\\pi G \\rho_0 .\n                    \\label{kappaDef1}\n\\end{equation}\nIf $m=0$ then $f(a)=1$ and the equation (\\ref{NewRho}) recovers the usual result $\\rho\\sim a^{-3}$ for non-relativistic matter ($\\gamma=1$) and $a^{-4}$ for radiation ($\\gamma=4/3$). The new parameter $\\kappa_0$ can be determined by the decay law (\\ref{Bam}) and astronomical observations implying ${\\cal B} = \\Lambda_0 = 3H_0^2\\lambda_0$. Substitute this result into (\\ref{kappaDef1}) to find:\n\\begin{equation}\n\\kappa_0 = \\lambda_0/\\Omega_0.\n\\label{kappaDefn}\n\\end{equation}</p>\n  </div>\n</div></div>\n\n\n\n<div id=\"IDE_87_8\"></div>\n<div style=\"border: 1px solid #AAA; padding:5px;\">\n\n=== Problem 17 ===\nFind dependence of deceleration parameter on the scale factor for the model of previous problem.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">Substitute (\\ref{Bam}) and (\\ref{NewRho}) into the Friedman equation to obtain\n\\begin{equation}\n\\frac{da}{d\\tau} = a \\left[ \\Omega_0 a^{-3\\gamma} f(a) -\n                   (\\Omega_0+\\lambda_0-1) a^{-2} +\n                   \\lambda_0 a^{-m} \\right]^{1/2} ,\n\\label{Expansion}\n\\end{equation}\nwhere the function $f(a)$ is determined in the previous problem (see the formula \\eqref{fDefn}).\nThe second Friedman equation can be rewritten in the form\n\\begin{equation}\n\\frac{d^{\\, 2}a}{d\\tau^2} = \\left( 1 - \\frac{3\\gamma}{2} \\right)\n                            \\Omega_0 a^{1-3\\gamma} f(a) +\n                            \\lambda_0 a^{1-m} .\n\\label{Deceleration}\n\\end{equation}\nSubstitute this derivatives into the definition of the deceleration parameter to obtain the required dependence\n\\begin{equation}\n   q\\equiv -\\frac{\\ddot{a}a}{\\dot{a}^2}=-\\frac{\\left( 1 - \\frac{3\\gamma}{2} \\right)\n                            \\Omega_0 a^{1-3\\gamma} f(a) +\n                            \\lambda_0 a^{1-m}}{a\\left[ \\Omega_0 a^{-3\\gamma} f(a) -\n                   (\\Omega_0+\\lambda_0-1) a^{-2} +\n                   \\lambda_0 a^{-m} \\right]}\\end{equation}</p>\n  </div>\n</div></div>"
                    }
                ]
            },
            "22": {
                "pageid": 22,
                "ns": 0,
                "title": "Schwarzschild black hole",
                "revisions": [
                    {
                        "contentformat": "text/x-wiki",
                        "contentmodel": "wikitext",
                        "*": "[[Category:Black Holes|2]]\n\nThe spherically symmetric solution of Einstein's equations in vacuum for the spacetime metric has the form$^{*}$\n\\begin{align}\\label{Schw}\nds^{2}=h(r)\\,dt^2-h^{-1}(r)\\,dr^2-r^2 d\\Omega^{2},\n\t&\\qquad\\mbox{where}\\quad\n\th(r)=1-\\frac{r_g}{r};\\quad r_{g}=\\frac{2GM}{c^{2}};\\\\\nd\\Omega^{2}=d\\theta^{2}+\\sin^{2}\\theta\\, d\\varphi^{2}&\\;\\text{is the metric of unit sphere.}\\nonumber\n\\end{align}\nThe Birkhoff's theorem$^{**}$ (1923) 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.\n\n$^{*}$ K. Schwarzschild, On the gravitational field of a mass point according to Einstein's theory, ''Sitzungsber. Preuss. Akad. Wiss. Phys. Math. Kl.'',  p.189 (1916) (there's a translation of the original paper at [http://arxiv.org/abs/physics/9905030v1 arXiv:physics/9905030v1]; please disregard the abstract/foreword, which is incorrect).\n\n$^{**}$ G.D. Birkhoff, Relativity and Modern Physics, p.253, Harvard University Press, Cambridge (1923);\nJ.T. Jebsen, \"Ark. Mat. Ast. Fys.\" (Stockholm) 15, nr.18 (1921), see also [http://arxiv.org/abs/physics/0508163 arXiv:physics/0508163v2].\n\n__TOC__\n\n==Simple problems==\n<div id=\"BlackHole15\"></div>\n=== Problem 1: local time ===\nFind 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 $r\\to r_{g}$?\n\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nThe proper time of the stationary observer is $d\\tau=ds|_{dr=d\\theta=d\\varphi=0}$:\n\\[d\\tau=\\sqrt{g_{00}}\\;dt=\\sqrt{1-\\frac{r_{g}}{r}}\\;dt.\\]\nAt $r\\to \\infty$ 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).\t</p>\n  </div>\n</div>\n\n<div id=\"BlackHole16\"></div>\n\n=== Problem 2: measuring distances ===\nWhat 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 $r_{1},r\\to r_{g}$?\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\n\\[l_{r1-r2}=\\int\\limits_{r_1}^{r_2}\n\t\\frac{dr}{\\sqrt{1-\\frac{r_g}{r}}};\\quad\n\tl_{\\varphi1-\\varphi_2}=2\\pi r|\\varphi_1-\\varphi_2|;\n\t\\quad\n\tl_{\\theta1-\\theta_2}=2\\pi r|\\theta_1-\\theta_2|.\\]\t</p>\n  </div>\n</div>\n\n<div id=\"BlackHole17\"></div>\n\n=== Problem 3: the inner region ===\nWhat 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?\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\n''This question is given not to be answered but to make one think of the answer. Correct questions and correct answers can be given in terms of a proper coordinate frame, which is regular both in $r>r_g$ and in $r<r_g$. Still, one can say something meaningful as is.''</p>\n\n<p style=\"text-align: left;\">\nAt $r<r_g$ we have $g_{00}<0$ and $g_{11}>0$, thus the $t$ coordinate is ''spatial'' and $r$ coordinate is ''temporal'' (!):\n\\[ds^{2}=|h|^{-1}(r)dr^{2}-|h|(r)dt^{2}\n\t-r^{2}d\\Omega^{2}.\\]\nThe metric therefore is ''nonstationary'' in this region, depending on the temporal coordinate $r$, but homogeneous, as there is no dependence on spatial coordinates.\n\nThen for an observer \"at rest\" with respect to this coordinate system we would have  $dt=d\\theta=d\\varphi=0$, and thus\n\\[d\\tau^2=ds^2=-\\frac{dr^{2}}{1-r_{g}/r}=\n\t\t\\frac{r dr^{2}}{r_{g}-r}>0,\n\t\\quad\\Rightarrow\\quad\n\t\t d\\tau=\\frac{\\sqrt{r}dr}{\\sqrt{r_{g}-r}}.\\]\n\nAn 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).\n\nThe 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.\n\nThe physical distance at $d\\theta=d\\varphi=dr=0$ is defined as \n\\[dl^{2}=|h(r)|dt^2.\\]\nIt evidently depends on time $r$.\n\nThis 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.\t</p>\n\n<p style=\"text-align: left;\">*This was actually not a very simple problem</p>\n  </div>\n</div>\n\n\n<div id=\"BlackHole18\"></div>\n\n=== Problem 4: acceleration ===\nCalculate the acceleration of a test particle with zero velocity.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nIf 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 \n\\[1=g_{\\mu\\nu}u^{\\mu}u^{\\mu}=g_{00}(u^{0})^{2}.\\]\nThen the $4$-acceleration can be found from the geodesic equation:\n\\begin{align*}\na^{0}\\equiv&\\frac{du^0}{ds}=-{\\Gamma^{0}}_{00}u^{0}u^{0}\\sim\n\t{\\Gamma^{0}}_{00}\\sim \\Gamma_{0,\\,00}\\sim\n\t(\\partial_{0}g_{00}+\\partial_{0}g_{00}\n\t-\\partial_{0}g_{00})=0;\\\\\na^{1}\\equiv&\\frac{du^1}{ds}\n\t=-{\\Gamma^{1}}_{00}u^{0}u^{0}\n\t=-g^{11}\\Gamma_{1,\\,00}\\;g_{00}^{-1}\n\t=-(g_{00}g_{11})^{-1}\\Gamma_{1,\\,00}=\\Gamma_{1,00}=\n\t\\\\ &=\\tfrac{1}{2}\n\t(\\partial_{0}g_{10}+\\partial_{0}g_{10}\n\t-\\partial_{1}g_{00})\n\t=-\\frac{1}{2}\\frac{dg_{00}}{dr}\n\t=-\\frac{h'}{2}\n\t=\\frac{r_{g}}{2r^2}.\n\\end{align*}\nThe scalar acceleration $a$ is then equal to\n\\[a^{2}=-g_{11}(a^{1})^{2}=\\frac{(h')^{2}}{4h}\n\t=-\\frac{r_{g}^{2}}{4 r^4}\n\t\t\\Big(1-\\frac{r_g}{r}\\Big)^{-1}\\]\nand tends to infinity when we approach the horizon.\t</p>\n  </div>\n</div>\n\n<div id=\"BlackHoleExtra1\"></div>\n\n=== Problem 5: Schwarzschild is a vacuum solution ===\nShow that Schwarzschild metric is a solution of Einstein's equation in vacuum.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nStraightforward calculation of Christoffel symbols and Ricci tensor yields the vacuum Einstein equation $R_{\\mu\\nu}=0$.\t</p>\n  </div>\n</div>\n\n==Symmetries and integrals of motion==\nFor background on Killing vectors see problems [[Equations of General Relativity#equ_oto-kill1|K1]], [[Equations of General Relativity#equ_oto-kill2|K2]], [[Equations of General Relativity#equ_oto-kill3|K3]] of chapter 2.\n<div id=\"BlackHole19\"></div>\n=== Problem 6: timelike Killing vector ===\nWhat integral of motion arises due to existence of a timelike Killing vector? Express it through the physical velocity of the particle.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nIf a Killing vector field is timelike we can use the coordinate frame in which $K^\\mu$ is the unitary vector of the time coordinate (see problem [[Equations of General Relativity#equ_oto-kill2|K2]] 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 $p_{\\mu}K^{\\mu}=p_{0}\\equiv\\varepsilon/c$.\n\nUsing the result [[Technical_warm-up#u0|u0]], which holds for arbitrary gravitational field, we obtain the expression for energy, which is the integral of motion in a stationary metric\n\\begin{equation}\\label{EnergyStat}\n\t\\varepsilon=mc^{2}u_{0}=\n\t\tmc^{2}\\sqrt{g_{00}}\\cdot\\gamma=\n\t\t\\frac{mc^{2}\\sqrt{g_{00}}}\n\t\t\t{\\sqrt{1-\\frac{v^2}{c^2}}}.\\end{equation}\t</p>\n  </div>\n</div>\n\n\n\n<div id=\"BlackHole20\"></div>\n\n=== Problem 7: Killing vectors of a sphere ===\nDerive the Killing vectors for a sphere in Cartesian coordinate system; in spherical coordinates.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nKilling 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.\n\nLet us consider an infinitesimal rotation in the plane $XY$: $dx=yd\\lambda,\\; dy=-xd\\lambda$. Thus the Killing vector is*\n\\[K_{1}=x\\partial_y-y\\partial_x.\\]\n\nUsing the spherical coordinates\n\\[x=\\sin\\theta\\cos\\varphi;\\quad y=\\sin\\theta\\sin\\varphi;\\quad z=\\cos\\theta,\\]\nwe obtain $\\partial_{x}=-\\frac{\\sin\\varphi}{\\sin\\theta}\\partial_\\varphi$, $\\partial_y=\\frac{\\cos\\varphi}{\\sin\\theta}\\partial_\\varphi$, and therefore $K_{1}=\\partial_{\\varphi}$.\n\nConsidering rotations in plains $XZ$ and $YZ$ in the same way, in the end we obtain\n\\[\\left\\{\\begin{array}{l}\nK_{1}=x\\partial_{y}-y\\partial_{x}=\\partial_{\\varphi}\\\\\nK_{2}=z\\partial_{x}-x\\partial_{z}=\n\t\\cos\\varphi\\;\\partial_{\\theta}-\\cot\\theta\\sin\\varphi\\;\\partial_\\varphi\\\\\nK_{3}=z\\partial_{y}-y\\partial_{x}=\n\t\\sin\\varphi\\;\\partial_{\\theta}+\\cot\\theta\\cos\\varphi\\;\\partial_\\varphi\\;.\n\\end{array}\\right.\\]\t</p>\n\n<p style=\"text-align: left;\">*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:</p>\n\n<p style=\"text-align: left;\">Carroll S., Spacetime and geometry: an introduction to General Relativity. AW, 2003, ISBN 0805387323</p>\n  </div>\n</div>\n\n<div id=\"BlackHole21\"></div>\n\n=== Problem 8: spherical symmetry of Schwarzshild ===\nVerify that in coordinates $(t,r,\\theta,\\varphi)$ vectors \n\\[ \\begin{array}{l}\n\t\\Omega^{\\mu}=(1,0,0,0),\\\\\n\tR^{\\mu}=(0,0,0,1),\\\\\n\tS^{\\mu}=(0,0,\\cos\\varphi,-\\cot\\theta\\sin\\varphi),\\\\\n\tT^{\\mu}=(0,0,-\\sin\\varphi,-\\cot\\theta\\cos\\varphi)\n\\end{array}\\]\nare the Killing vectors of the Schwarzschild metric.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nVectors $\\Omega^\\mu$ 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.\t</p>\n  </div>\n</div>\n\n<div id=\"BlackHole22\"></div>\n=== Problem 9: planar motion ===\nShow that existence of Killing vectors $S^\\mu$ and $T^\\mu$ leads to motion of particles in a plane.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nThe integrals of motion corresponding to $S^\\mu$ and $T^\\mu$ are\n\\begin{align*}\nS=u_{\\mu}S^{\\mu}=&\n\t \\;u_{2}\\cos\\varphi-u_{3}\\cot\\theta\\sin\\varphi;\\\\\nT=u_{\\mu}T^{\\mu}=&\n\t-u_{2}\\sin\\varphi-u_{3}\\cot\\theta\\cos\\varphi.\n\\end{align*}\nLet us align the coordinate frame in such a way that the initial conditions are\n\\[\\theta|_{t=0}=\\pi/2;\\quad\n\t u_{2}|_{t=0}\\equiv u_{\\theta}|_{t=0}=0.\\]\nThen $S$ and $T$ are zero:\n\\[u_{2}\\cos\\varphi=u_{3}\\cot\\theta\\sin\\varphi;\\qquad\nu_{2}\\sin\\varphi=-u_{3}\\cot\\theta\\cos\\varphi.\\]\nTaking the square and adding the two equations, and also multiplying them, we obtain\n\\[\\left\\{\\begin{array}{l}\nu_{2}^{2}=u_{3}^{2}\\cot^{2}\\theta,\\\\\nu_{2}^{2}\\sin\\varphi\\cos\\varphi=\n\t-u_{3}^{2}\\cot^{2}\\theta\\sin\\varphi\\cos\\varphi\n\\end{array}\\right.\\quad\\Rightarrow\\quad\n\tu_{3}^{2}\\sin\\varphi\\cos\\varphi\n\t\\cdot\\cot^{2}\\theta=0.\\]\nThen 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 $\\theta=\\pi/2$.\t</p>\n  </div>\n</div>\n\n<div id=\"BlackHole23\"></div>\n\n=== Problem 10: stability of planar motion ===\nShow that the particles' motion in the plane is stable.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nConservation 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\n\\begin{align*}\n&S\\sin\\varphi+T\\cos\\varphi=R\\cot\\theta,\\\\\n&S\\cos\\varphi-T\\sin\\varphi=u_2.\n\\end{align*}\nTaking the square and adding up, we are led to\n\\[u_2^2 =S^2+T^2+R^2-\\frac{R^2}{\\sin^2 \\theta},\\]\nthen on differentiating we obtain\n\\[u'_{2}=R^2 \\frac{\\cos\\theta}{\\sin^{2}\\theta}.\\]\nLet the trajectory deviate slightly from the plane $\\theta=\\pi/2$. Then $\\theta=\\pi/2+\\delta\\theta$ and \n\\[(\\delta\\theta)''=u'_{2}\n\t=R^2 \\frac{\\cos\\theta}{\\sin^{2}\\theta}\n\t\\approx -R^2 \\delta\\theta,\\]\ntherefore $\\theta$ oscillates around the stable point $\\pi/2$.\t</p>\n  </div>\n</div>\n\n<div id=\"BlackHole24\"></div>\n\n=== Problem 11: remaining integrals of motion ===\nWrite down explicitly the conserved quantities  $p_{\\mu}\\Omega^{\\mu}$ and $p_{\\mu}R^{\\mu}$ for movement in the plane $\\theta=\\pi/2$.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nThe first integral of motion is, up to a multiplier, the energy (see Eq. [[Technical_warm-up#EnergyStat|EnergyStat]]):\n\\begin{equation}\\label{SchwInt-E1}\n\t E=\\Omega_{\\mu}u^{\\mu}=\n\tg_{00}\\frac{dx^{0}}{d\\lambda}=h\\frac{dt}{d\\lambda}=\n\t\\left(1-\\frac{r_g}{r}\\right)\\frac{dt}{d\\lambda},\\end{equation}\nwhere $\\lambda$ is the geodesic parameter: for a massive particle we can always choose the natural parametrization  $d\\lambda=ds=\\gamma^{-1}\\sqrt{g_{00}}\\,dt$ (see Eq. [[Technical_warm-up#IntervalStaticCase|IntervalStaticCase]]) so that $E$ is energy per unit mass; recovering the multipliers, for true energy we obtain\n\\begin{equation}\\label{SchwInt-E2}\n\t\\varepsilon_{m}\\equiv mcE=\n\tmc^{2}\\sqrt{g_{00}}\\,\\gamma=\n\t mc^{2}\\sqrt{h}\\,\\gamma=\n\t mc^{2}\\sqrt{\\frac{1-r_{g}/r}{1-v^{2}/c^{2}}}.\\end{equation}\n\nThe second integral of motion is the angular momentum (per unit mass also, and the sign is chosen so that in the Newtonian limit we obtain the usual angular momentum)\n\\begin{equation}\\label{SchwInt-L1}\n\t L=-R_{\\mu}u^{\\mu}=-g_{33}\\frac{dx^{3}}{d\\lambda}=\n\tr^{2}\\sin^{2}\\theta\\frac{d\\varphi}{d\\lambda}=\n\tr^{2}\\frac{d\\varphi}{d\\lambda}.\\end{equation}\nFor a massive particle we choose $\\lambda=s$, then\n$d\\lambda=ds=\\gamma^{-1}\\sqrt{g_{00}}\\,dt$ and\n\\begin{equation}\\label{SchwInt-L2}\n\t l_{m}\\equiv mL=\n\t mr^{2}\\dot{\\varphi} \\sqrt{g_{00}}\\gamma=\n\tm r^{2}\\dot{\\varphi}\n\t\\sqrt{\\frac{1-r_{g}/r}{1-v^{2}/c^{2}}}.\\end{equation}\nNote that conservation of $l_m$ is a generalization to the relativistic case of the second Kepler's law on the sweeping of equal areas per unit time $r^{2}\\dot{\\varphi}=const$.\t</p>\n  </div>\n</div>\n\n<div id=\"BlackHole25\"></div>\n\n=== Problem 12: work and mass ===\nWhat is the work needed to pull a particle from the horizon to infinity? Will a black hole's mass change if we drop a particle with zero initial velocity from immediate proximity of the horizon?\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nAt $r\\gg r_g$ we'll have $\\varepsilon_{m}=mc^{2}$, while near the horizon $\\varepsilon_{m}\\to 0$. The difference is the work needed to pull the particle away from the horizon to infinity, and it equals the rest mass (energy) of the particle. No, it will not change, because the energy of the falling particle is zero.\t</p>\n  </div>\n</div>\n\n==Radial motion==\nConsider a particle's radial motion: $\\dot{\\varphi}=\\dot{\\theta}=0$. In this problem one is especially interested in asymptotes of all functions as $r\\to r_{g}$.\n\n<p style=\"text-align: left;\">Let us set $c=1$ here and henceforth measure time in the units of length, so that $x^{0}=t$, $\\beta=v$, etc., and introduce the notation\n\\[h(r)\\equiv g_{00}(r)=-\\frac{1}{g_{11}(r)}=\n\t1-\\frac{r_{g}}{r}\n\t\t\\underset{r\\to r_g+0}{\\longrightarrow}+0.\\]</p>\n\n<div id=\"BlackHole26\"></div>\n=== Problem 13: null geodesics ===\nDerive the equation for null geodesics (worldlines of massless particles).\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nEquation of null curve is $ds^2=0$. Substituting the metric with $d\\theta=d\\varphi=0$, we get\n\\[dt=\\pm dr\\,\\sqrt{-\\frac{g_{11}}{g_{00}}}\n\t=\\pm\\frac{dr}{h}.\\]\n\nOn integration, we obtain\n\\begin{equation}\\label{Schw-NullGeodesic}\n\t\\pm t=r+r_{g}\\ln|r-r_{g}|+const.\\end{equation}\nDue to symmetry it is also a geodesic. This can be verified by evaluating the Christoffel symbols for the Schwarzschild metric and writing down the geodesic equatoin in explicit form.\t</p>\n  </div>\n</div>\n\n<div id=\"BlackHole27\"></div>\n=== Problem 14: geodesic motion of massive particle ===\nUse energy conservation to derive $v(r)$, $\\dot{r}(r)=dr/dt$, $r(t)$ for a massive particle. Initial conditions: $g_{00}|_{\\dot{r}=0}=h_{0}$ (the limiting case $h_{0}\\to 1$ is especially interesting and simple).\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nIn the radial case the physical velocity $v$ is expressed through the $4$-velocity $u^{\\mu}=(u^{0},u^{1},0,0)$ as\n\\begin{equation}\\label{Schw-v(u)}\n\t\\beta^{2}=\\frac{dx^{\\alpha}dx_{\\alpha}}{d\\tau^{2}}=\n\t\\frac{-g_{11}(dx^{1})^{2}}{g_{00}(dx^{0})^{2}}=\n\th^{-2}\\Big(\\frac{u^1}{u^0}\\Big)^2=\n\t\t\\Big(\\frac{\\dot{r}}{h}\\Big)^2,\\end{equation}\nwhere dot denotes differentiation by coordinate time $t$, and thus the energy conservation law (see (\\ref{SchwInt-E2})) takes the form\n\\[const=\\Big(\\frac{\\varepsilon}{mc^2}\\Big)^{2}=\n\th\\gamma^{2}=\\frac{h}{1-\\dot{r}^{2}/h^2}.\\]\n\nInitial condition fixes the constant*:\n\\begin{equation}\\label{Schw-h0}\n\t\\dot{r}\\sim v|_{r_{0}}=0\\quad\n\t\\Rightarrow\\quad\n\t\t\\Big(\\frac{\\varepsilon}{mc^2}\\Big)^{2}\n\t\t=g_{00}(r_{0})=h(r_0)=1-\\frac{r_{g}}{r_{0}}\n\t\\equiv h_{0},\\end{equation}\nso for the physical velocity we get\n\\begin{equation}\\label{SchwRad-V}\n\t\\gamma^{2}=\\frac{h_{0}}{h(r)}\n\t\\underset{r\\to r_{g}}{\\longrightarrow}\\infty;\n\t\\quad \\frac{v}{c}\\equiv\\beta=\n\t\t\\sqrt{1-\\gamma^{-2}}=\n\t\t\\sqrt{1-\\frac{h}{h_{0}}}\n\t\t\\underset{r\\to r_{g}}{\\longrightarrow}1.\\end{equation}\nIt tends to infinity close to the horizon! At the same time, as \n$v^{2}=\\dot{r}^{2}/h^{2}$ from (\\ref{Schw-v(u)}), the coordinate velocity\n\\begin{equation}\\label{Schw-dotR}\n\\Big|\\frac{dr}{dt}\\Big|=vh=h \\sqrt{1-\\frac{h}{h_{0}}}=\n\t\\Big(1-\\frac{r_{g}}{r}\\Big)\n\t\\sqrt{1-\\frac{1-\\frac{r_{g}}{r}}{1-\\frac{r_{g}}{r_{0}}}}\n\t\\underset{r\\to r_{g}}{\\thicksim}h\\to 0\\end{equation}\ntends to zero.\n\nAfter integration we obtain $t(r)$:\n\\begin{equation}\\label{Schw-t(r)}\nconst\\pm t=\\int\\limits_{r}^{r_{0}}\n\t\\frac{dr}{vh}=\n\t\\int\\limits_{r}^{r_{0}}\\frac{rdr}{r-r_{g}}\n\t\\bigg[1-\\frac{1-r_{g}/r}{1-r_{g}/r_{0}}\\bigg]^{-1/2}\n\t\\underset{\\substack{r\\approx r_{g}\\\\r_{0}\\gg r_{g}}}\n\t{\\approx}\\int\\limits_{r}^{r_{0}}\n\t\\frac{r\\,dr}{r-r_{g}}=r+r_{g}\\ln|r-r_{g}|.\\end{equation}\nAs should be expected, the asymptote is the same as for the null geodesics (for $r_{0}\\gg r_{g}$).</p>\n\n<p style=\"text-align: left;\">*Note that $r_0$ has the meaning of a turning point only for finite motion; this, however, does not prevent us from using the same notation in the non-finite case. Then $r_0$ is determined from $\\varepsilon$ by the same formula and takes negative values, while $h_{0}>1$.</p>\n  </div>\n</div>\n\n<div id=\"BlackHole28\"></div>\n=== Problem 15: radial motion in terms of proper time ===\nShow that the equation of radial motion in terms of proper time of the particle is the same as in the non-relativistic Newtonian theory. Calculate the proper time of the fall from $r=r_0$ to the center. Derive the first correction in $r_{g}/r$ to the Newtonian result. Initial velocity is zero.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nThe particle's proper time interval $\\tau$ is $d\\tau=\\gamma^{-1} d\\hat{t}$, where $d\\hat{t}=\\sqrt{h}dt$ is local time. Then\n\\begin{equation}\\label{ShwRad-tau}\n\t\\tau=\\int dt \\gamma^{-1}\\sqrt{h}=\n\t\\int\\limits_{r}^{r_{0}}\n\t\\frac{dr}{\\gamma v\\sqrt{h}}=\n\t\\int\\limits_{r}^{r_{0}}\n\t\t\\frac{dr}{v \\sqrt{h_{0}}}=\n\t\\int\\limits_{r}^{r_{0}}\n\t\t\\frac{dr}{\\sqrt{h_{0}-h}}=\n\t\\int\\limits_{r}^{r_{0}}\n\t\t\\frac{dr}{\\sqrt{\\frac{r_{g}}{r_{0}}-\n\t\t\t\\frac{r_{g}}{r}}}.\\end{equation}\nOn the other hand, in the Newtonian approach from energy conservation we obtain in the same way that \n\\[(\\dot{r})^{2}=2(E-U)=2\\;\\Big(E+\\frac{GM}{r}\\Big)\n\t=\\frac{r_{g}}{r}-\\frac{r_{g}}{r_{0}},\\]\nwhich gives us the same equation of motion.\n\nWe see that the integral is regular at $r=r_{g}$, is finite and converges even at $r\\to 0$. It is not hard to evaluate:\n\\begin{equation}\\label{Schw-TauParametric}\n\t\\tau=\\frac{r_0}{2}\\sqrt{\\frac{r_0}{r_g}}\n\t(\\eta+\\sin\\eta),\\qquad\n\tr=\\frac{r_0}{2}(1+\\cos\\eta),\\qquad \\eta\\in(0,\\pi).\n\\end{equation}\nThe proper time of reaching the singularity $r=0$ is therefore\n\\[\\tau_{f}=\\frac{\\pi}{2}r_{0}\n\t\\sqrt{\\frac{r_{0}}{r_{g}}}.\\]\n\nThe integral (\\ref{Schw-t(r)}) for $t(r)$ can be rewritten as\n\\[t(r)=\\int \\frac{dr}{vh}=\\sqrt{h_{0}}\n\t\\int\\frac{dr}{\\sqrt{h_{0}-h}}\\cdot \\frac{1}{h}=\n\t\\underbrace{\\sqrt{1-\\frac{r_{g}}{r_{0}}}}\n\t_{\\text{correction 1}}\\,\n\t\\underbrace{\\int\\!\\frac{dr}{\\sqrt{\\frac{r_g}{r}-\\frac{r_g}{r_0}}}}_{\\tau(r)}\n\t\t\\cdot \\underbrace{\\frac{1}{1-\\frac{r_g}{r}}}\n\t_{\\text{correction 2}}.\\]\nNow we see that the first factor gives the correction taking into account time dilation in the initial point $r_0$, while the second correction accounts for local time dilation and proper time dilation due to acceleration.\n\nIf the second factor is close to unity, we can expand it in powers of $r_{g}/r$ and obtain\n\\[t=\\sqrt{1-\\frac{r_g}{r_0}}\\;\\;\n\t\\big[\\tau+\\Delta\\tau\\big],\n\t\\quad\\text{where}\\quad\n\t\\Delta\\tau=\\int \\frac{dr\\,r_{g}/r}\n\t{\\sqrt{\\frac{r_g}{r}-\\frac{r_g}{r_0}}}.\\]\nOn integrating, we can express $\\Delta\\tau$ through $\\eta$ (\\ref{Schw-TauParametric}), and thus derive the equation of motion in the form analogous to (\\ref{Schw-TauParametric}), but with the correction term\n\\begin{equation}\\label{Schw-TauParametric2}\n\t\\tau=\\frac{r_0}{2}\\sqrt{\\frac{r_0}{r_g}}\\;\n\t\\Big([1+\\tfrac{r_g}{r_0}]\\eta+\\sin\\eta\\Big),\\qquad\n\tr=\\frac{r_0}{2}(1+\\cos\\eta).\n\\end{equation}\nEvidently, this approximation works only for $r\\gg r_{g}$.\t</p>\n  </div>\n</div>\n\n<div id=\"BlackHole29\"></div>\n\n=== Problem 16: ultra-relativistic limit ===\nDerive the equations of radial motion in the ultra-relativistic limit.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nIn the relativistic limit we should assume $\\gamma\\gg1$ (near $r_{g}$ this always holds), thus $v\\to 1$, and from (\\ref{Schw-dotR}) we get\n\\[\\frac{dr}{dt}\\approx h\\quad\n\t\\Rightarrow\\quad t=\\int \\frac{dr}{h}.\\]\nThis is, as should be expected, the null geodesic equation.</p>\n  </div>\n</div>\n\n<div id=\"BlackHole30\"></div>\n\n=== Problem 17: communication from near the black hole ===\nA particle (observer) falling into a black hole is emitting photons, which are detected on the same radial line far away from the horizon (i.e. the photons travel from emitter to detector radially). Find  $r$, $v$ and $\\dot{r}$ as functions of the signal's detection time in the limit  $r\\to r_g$.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nThe geodesics describing light signals propagating away from the horizon are given by the equation (\\ref{Schw-NullGeodesic}) with the plus sign:\n\\begin{equation}\\label{Schw-tLambda}\n\t t=r+r_{g}\\ln|r-r_{g}|+t_{\\Lambda},\\end{equation}\nwhere parameter $t_\\Lambda$ parametrizes this family of geodesics and gives the difference in coordinate times of signals' registration at a given $r$. At $r\\gg r_g$ this is also the proper time of the remote detector. Then\n\\begin{equation}\\label{Schw-dtL(dr)}\n\t dt_{\\Lambda}=dt-\\frac{r_{g}dr}{r-r_{g}}=\n\tdt-\\frac{r_{g}dr}{rh}=\n\t-\\frac{dr}{vh}-\\frac{r_{g}dr}{rh}=\n\t-\\frac{dr}{h}\\Big(\\frac{1}{v}+\\frac{r_{g}}{r}\\Big).\\end{equation}\nAs both $v$ and $r_{g}/r$ tend to unity near the horizon, the integral diverges logarithmically, twice faster than the one for $t(r)$. The meaning of this number is the following: as $r\\to r_{g}$, the geodesics of massive and massless particles asymptotically coincide, and it takes the light as much time to \"escape\" the potential well as it takes the particle to fall in.\n\nAsymptotically we obtain\n\\begin{equation}\\label{Schw-r(tL)}\n\t-t_{\\Lambda}\\approx 2\\int\\limits^{r}\n\t\\frac{rdr}{r-r_{g}}\\approx 2r_{g}\\ln|r-r_{g}|\n\t\t\\approx 2t(r)\\quad\\Rightarrow\\quad\n\tr(t_{\\Lambda})-r_{g}\\sim\n\t\tr_{g} e^{-t_{\\lambda}/2r_{g}};\\end{equation}\nalso $\\gamma=h_{0}/h\\approx r_{g}/(r-r_{g})$,  $v(r)=\\sqrt{1-h/h_{0}}\\approx 1-(r-r_{g})/2r_{g}$, and $\\dot{r}\\approx h$, thus\n\\begin{equation}\\label{Schw-v(tL)}\n\t\\gamma\\sim e^{t_{\\Lambda}/2r_{g}};\\qquad\n\t\t(1-v)\\sim {\\textstyle\\frac{1}{2}}\n \t\t e^{-t_{\\Lambda}/2r_{g}};\n\t\t\\qquad \\dot{r}\\sim e^{-t_{\\Lambda}/2r_{g}}.\\end{equation}</p>\n  </div>\n</div>\n\n==Blackness of black holes==\nA 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 $r$.\n\n<div id=\"BlackHole31\"></div>\n=== Problem 18: stationary source and detector ===\nThe source and detector are stationary.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nThe integral of motion that conserves along the light ray due to the timelike Killing vector $\\Omega^{\\mu}=\\partial_{t}$ is energy (up to a factor):\n\\begin{equation}\\label{Schw-OmegaConst}\n\tconst\\equiv\\omega_{0}=\n\t g_{\\mu\\nu}\\Omega^{\\mu}k^{\\nu}=\n\tg_{00}k^{0}=hk^{0}.\\end{equation}\nFor a static observer $u^{\\mu}=(u^{0},0,0,0)$, while\n\\begin{equation}\\label{Schw-u0}\n\t1=u^{\\mu}u_{\\mu}=g_{\\mu\\nu}u^{\\mu}u^{\\nu}=\n\tg_{00}(u^{0})^{2}\\quad\\Rightarrow\\quad\n\tu^{0}=h^{-1/2}.\\end{equation}\nTherefore the frequency registered by this static observer is\n\\begin{equation}\\label{Schw-omegastat}\n\t\\omega_{stat}=g_{\\mu\\nu}u^{\\mu}_{stat}k^{\\nu}=\n\tg_{00}u^{0}_{stat}k^{0}=\\sqrt{g_{00}}k^{0}=\n\t\t\\sqrt{h}k^{0}=\\omega_{0}h^{-1/2}.\\end{equation}\nThen\n\\begin{equation}\\label{Schw-RedShift0}\n\t\\omega_{0}=\\omega_{stat}(r)\\sqrt{h(r)}=const,\\end{equation}\nso we obtain\n\\begin{equation}\\label{Schw-RedShift}\n\t\\omega_{det}=\\omega_{em}\n\t\\sqrt{\\frac{g_{00}(r_{em})}{g_{00}(r_{det})}}.\\end{equation}\nThis is gravitational redshift.</p>\n  </div>\n</div>\n\n<div id=\"BlackHole32\"></div>\n=== Problem 19: free-falling source ===\nThe source is falling freely without initial velocity from radius $r_0$; it flies by the stationary detector at the moment of emission.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nIt is educative to derive the Doppler effect from analogous considerations. Let us consider an emitter (detector) moving radially with physical velocity $v$ (see. (\\ref{Schw-v(u)})). From the normalizing condition\n\\begin{equation}\\label{Schw-rad-u0}\n\t1=u^{\\mu}u_{\\mu}=g_{00}(u^{0})^{2}+g_{11}(u^{1})^{2}=\n\tg_{00}(u^0)^{2}(1-\\beta^{2})=\\gamma^{-2} h (u^0)^{2},\\end{equation}\nthus\n\\begin{equation}\\label{Schw-rad-u^mu}\n\t u^{\\mu}=\\gamma\\Big(\n\t\\frac{1}{\\sqrt{g_{00}}},\n\t\t\\frac{\\beta}{\\sqrt{-g_{11}}},0,0\\Big)=\n\t\\gamma\\Big(\\frac{1}{\\sqrt{h}},\n\t\t\\beta\\sqrt{h},0,0\\Big).\\end{equation}\nFor light\n\\[0=k^{\\mu}k_{\\mu}=g_{00}(k^0)^{2}+g_{11}(k^{1})^{2}\n\t\\quad\\Rightarrow\\quad\n\tk^{1}=\\pm k^{0}\\sqrt{-g_{00}/g_{11}}=\\pm hk^{0},\\]\ntherefor the observer registers (or the detector emits) light with frequency \n\\[\\omega_{em}\\equiv\\omega_{\\beta}=\n\t g_{\\mu}u^{\\mu}k^{\\nu}=\n\tg_{00}u^{0}k^{0}+g_{11}u^{1}k^{1}=\n\th\\cdot \\frac{\\gamma}{\\sqrt{h}}k^{0}\\pm\n\t\\frac{1}{h}\\cdot\\gamma\\beta\\sqrt{h}\\cdot hk^{0}=\n\t\t\\sqrt{h}\\gamma k^{0}(1\\pm\\beta).\\]\nChoosing the sign \"$+$\" here, which corresponds to the emitter moving towards the horizon and light travaling outwards, we obtain the relativistic Dopper effect \n\\begin{equation}\\label{RelatDoppler-radial}\n\t\\omega_{\\beta}=\n\t\\frac{\\omega_{0}}{\\sqrt{h}}\\cdot\\gamma(1+\\beta)=\n\t\\omega_{stat}\\cdot\\gamma(1+\\beta).\\end{equation}\nTaking into account that in this case the emitter is moving and detector is at rest, so that $\\omega_{em}=\\omega_{\\beta}$, $\\omega_{det}=\\omega_{stat}$, we get\n\\begin{equation}\\label{Schw-RadialDoppler}\n\t\\omega_{det}\\equiv \\omega_{stat}=\n\t\\omega_{em}\\frac{\\gamma^{-1}}{1+\\beta}=\n\t\\omega_{em}\\sqrt{\\frac{1-\\beta}{1+\\beta}}.\\end{equation}\t</p>\n  </div>\n</div>\n\n<div id=\"BlackHole33\"></div>\n\n=== Problem 20: adding the two effects ===\nThe source is freely falling the same way, while the detector is stationary at $r_{det}>r_{em}$.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nIn this case we just need to take into account both the gravitational redshift (\\ref{Schw-RedShift}) and the Doppler effect (\\ref{Schw-RadialDoppler}). On substituting $\\gamma=\\sqrt{h_{0}/h_{em}}$, we get\n\\begin{equation}\\label{Schw-Rad-FullRedshift}\n\t\\omega_{det}=\\omega_{stat}(r_{em})\n\t\\sqrt{\\frac{h_{em}}{h_{det}}}=\n\t\\omega_{em}\\frac{\\gamma^{-1}}{1+\\beta}\n\t\\sqrt{\\frac{h_{em}}{h_{det}}}=\n\t\\frac{\\omega_{em}}{1+\\beta}\\cdot\n\t\\frac{h_{em}}{\\sqrt{h_{0}h_{det}}}\\end{equation}\nIn the limit $r_{em}\\equiv r\\sim r_{g}\\ll r_{det}$ we obtain $\\beta\\approx 1$, thus\n\\[\\omega_{det}\\approx\n\t\\omega_{em}\\cdot \\frac{h_{em}}{2}=\n\t\\omega_{em}\\cdot {\\textstyle\\frac{1}{2}}\n\t\\Big(1-\\frac{r_{g}}{r_{em}}\\Big)\\to 0.\\]</p>\n  </div>\n</div>\n\n<div id=\"BlackHole34\"></div>\n=== Problem 21: intensity ===\nThe source is falling freely and emitting continuously photons with constant frequency, the detector is stationary far away from the horizon $r_{det}\\gg r_{g}$. How does the detected light's intensity depend on $r_{em}$ at the moment of emission? On the proper time of detector?\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nRepeating the argument of [[Schwarzschild black hole#BlackHole30|this problem]], for the proper time interval between detection of consecutive infinitely close signals (\\ref{Schw-dtL(dr)}) we have\n\\[dt_{\\Lambda}=\n\t-\\frac{dr}{h}\\Big(\\frac{1}{v}+\\frac{r_{g}}{r}\\Big),\\]\nwhere $dr$ is the corresponding displacement of the emitter, related with its proper time interval as (see \\ref{ShwRad-tau})\n\\[dr=-\\gamma v h\\, d\\tau.\\]\nThen using $\\gamma=h^{-1/2}$ from (\\ref{SchwRad-V}), we get\n\\begin{equation}\\label{Schw-dtL(dTau)}\n\tdt_{\\Lambda}=\\gamma\n\t\\Big(1+v\\frac{r_{g}}{r}\\Big) d\\tau=\n\th^{-1/2}\\Big(1+v\\frac{r_{g}}{r}\\Big) d\\tau.\\end{equation}\nIntensity of the emitted and detected light is proportional to the photons' frequency and inversely proportional to the proper time interval during which a given number $N$ of photons are being emitted/detected. So, combining (\\ref{Schw-Rad-FullRedshift}) and (\\ref{Schw-dtL(dTau)}), we get\n\\begin{equation}\\label{SchwRad-I}\n\t\\frac{I_{det}}{I_{em}}=\n\t\\frac{\\omega_{det}}{\\omega_{em}}\\cdot\n\t\\frac{d\\tau}{dt_{\\Lambda}}=\n\t\t\\frac{1}{1+v}\n\t\\frac{h_{em}}{\\sqrt{h_{0}h_{det}}}\\cdot\n\t\t\\frac{\\sqrt{h_{em}}}{1+v\\frac{r_{g}}{r}}.\\end{equation}\nIn the limit $h_{0},h_{det}\\to 1$ and $r\\to r_{g}$, so that $v\\to1$, we have\n\\[\\frac{I_{det}}{I_{em}}\\approx\n\t\t\\frac{1}{4}h_{em}^{3/2}=\n\t\t\\frac{1}{4}\\Big(1-\\frac{r_{g}}{r}\\Big)^{3/2}.\\]\nSubstituting $r(t_\\Lambda)$ from (\\ref{Schw-r(tL)}), we finally obtain\n\\begin{equation}\\label{Schw-I(tL)}\n\t I_{det}(t_\\Lambda)\\approx I_{em}\\cdot\n\t\\exp\\Big\\{-\\frac{3\\;t_\\Lambda}{4\\;r_g}\\Big\\}.\\end{equation}\t</p>\n  </div>\n</div>\n\n==Orbital motion, effective potential==\nDue 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.\n\n<div id=\"BlackHole35\"></div>\n=== Problem 22: impact parameter ===\nShow that the ratio of specific energy to specific angular momentum of a particle equals to $r_{g}/b$, where $b$ is the impact parameter at infinity (for unbounded motion).\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nFor $r\\gg r_{g}$ we'll have $h\\approx 1$ so if we count the angle from $\\pi/2$ the impact parameter is $b=r\\cos\\varphi$. If we denote the length element of the trajectory by $dl$, then the ratio between the angular momentum and energy is\n\\[\\frac{L}{E}\n=\\frac{r^2\\cdot d\\varphi/d\\lambda}\n\t\t{h\\cdot dt/d\\lambda}\n\t=\\frac{r^2 d\\varphi}{hdt}\\approx\n\t\\frac{r^2 (dl\\cos\\varphi\\, /r)}{dl/v}=\n\tr\\cdot r\\cos\\varphi=vb.\\]\nIn the ultrarelativistic case\n\\[L=Eb.\\]</p>\n  </div>\n</div>\n\n<div id=\"BlackHole36\"></div>\n\n=== Problem 23: geodesic equations and effective potential ===\nDerive the geodesics' equations; bring the equation for $r(\\lambda)$ to the form\n\\[\\frac{1}{2}\\Big(\\frac{dr}{d\\lambda}\\Big)^{2}\n\t+V_{\\epsilon}(r)=\\varepsilon,\\]\nwhere $V_{\\epsilon}(r)$ is a function conventionally termed as effective potential.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\n\tThe first equation is $\\epsilon=u^{\\mu}u_{\\mu}$. For a massive particle $\\epsilon=1$ and we can also choose the parameter $\\lambda$ as the natural parameter of the geodesic $d\\lambda=ds$. For a massless particle $\\epsilon=0$ and the parameter is arbitrary. In the Schwarzschild metric therefore\n\\begin{equation}\\label{Schw-Orbit1}\n\th\\Big(\\frac{dt}{d\\lambda}\\Big)^{2}-\n\t\\frac{1}{h}\\Big(\\frac{dr}{d\\lambda}\\Big)^{2}-\n\tr^{2}\\Big(\\frac{d\\varphi}{d\\lambda}\\Big)^{2}=\n\t\\epsilon.\\end{equation}\nThe two integrals of motion (\\ref{SchwInt-E1},\\ref{SchwInt-L1}) are\n\\begin{equation}\\label{Schw-Orbit-Integrals}\n\tE=h\\frac{dt}{d\\lambda};\\qquad\n\tL=r^{2}\\frac{d\\varphi}{d\\lambda}.\\end{equation}\nSubstituting this into (\\ref{Schw-Orbit1}) and rearranging terms, we derive the equation for $r(\\lambda)$ in the form\n\\begin{equation}\\label{Schw-V}\n\t\\frac{1}{2}\\Big(\\frac{dr}{d\\lambda}\\Big)^{2}+\n\tV_{\\epsilon}(r)=\\varepsilon,\\qquad\\mbox{where}\\quad\n\tV(r)=\\frac{h}{2}\n\t\\Big(\\epsilon+\\frac{L^2}{r^2}\\Big),\n\t\\quad \\varepsilon=\\frac{E^2}{2}.\\end{equation}\nThis is an analogue of one-dimensional motion of a particle in a potential $V$, with the quantity $\\varepsilon$ playing the role of full energy (its analytic solution for the given $V$ is expressed through the integral from the square root of a third degree polynomial in the denominator, which is reduced to elliptic Jacobi functions). Note that $V(r=r_{g})=0$.\n\nRecall that for massive particles $E$ and $L$ with the chosen parametrization are the energy and angular momentum per unit mass (see (\\ref{SchwInt-E2},\\ref{SchwInt-L2})).</p>\n  </div>\n</div>\n\n<div id=\"BlackHole37\"></div>\n\n=== Problem 24: bound and unbound motion ===\nPlot and investigate the function $V(r)$. Find the radii of circular orbits and analyze their stability; find the regions of parameters $(E,L)$ corresponding to bound and unbound motion, fall into the black hole. Consider the cases of a) massless, b) massive particles.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nLet us first consider $\\epsilon=0$.</p>\n\n[[File:BHfig-SchwPhaL.png|center|thumb|400px|The effective potential for a massless particle $V(r/r_{g})$. The position of the maximum, that corresponds to the photon sphere, is at $r/r_{g}=\\tfrac{3}{2}$ and does not depend on $L$]]\n\n<p style=\"text-align: left;\"> For massless particles the parameter $\\lambda$ is arbitrary, so we will fix it by demanding that far from the horizon, where the metric is asymptotically Euclidean, holds $u^{0}\\equiv dt/d\\lambda=1$. Then, first, $E=1$, and second, at $r\\gg r_{g}$\n\\[0=u^{\\mu}u_{\\mu}=(u^{0})^{2}-u^{\\alpha}u_{\\alpha}\n\t\\quad\\Rightarrow\\quad\n\t\\frac{dl}{d\\lambda}=1.\\]\nHere $dl$ is the line element of the (asymptotically) flat space, and for such parametrization we obtain that $L=b$, where $b$ is the impact parameter of the ray at infinity:\n\\begin{equation}\\label{Schw-orb-NullInts}\n\tu^{0}|_{r\\to\\infty}=1,\\;\\Rightarrow\\quad\n\t E=1;\\quad L=b.\\end{equation}\nThen\n\\[V_{\\epsilon=0}=b^{2}\\frac{h}{2r^{2}}=\n\t\\frac{b^{2}}{2}\n\t\\Big(\\frac{1}{r^2}-\\frac{r_{g}}{r^3}\\Big),\n\t\\quad \\varepsilon=\\frac{1}{2}.\\]\nAt small $r/r_{g}$ the effective potential behaves as $\\sim (-r^{-3})$, while at large $r/r_{g}$ as $\\sim r^{2}$, and reaches its maximum at\n\\[r_{max}=\\frac{3}{2}r_{g},\\quad\n\tV_{max}=V(r_{max})=\n\t\\frac{2}{27}\\Big(\\frac{b}{r_g}\\Big)^2.\\]\nAs this is a maximum, the corresponding circular orbit on the so-called \"photon sphere\" is unstable. For $\\varepsilon>V_{max}$ all the trajectories from one side of it escape to infinity, and the ones from the other side fall on the center (more accurately, they fall on the horizon, as we do not yet consider the motion beyond this point). Rewriting the inequality in terms of $b$, we have\n\\[b<b_{m}\\equiv\\frac{3\\sqrt{3}}{2}\\;r_{g}.\\]\nAt larger impact parameters there is a \"turning point\" for motion from infinity, in which $dr/d\\lambda=0$. In the region $r<r_{\\max}$ these values of $b$, which do not have the meaning of the impact parameter in this case, correspond to finite orbits falling on the center.</p>\n    <p style=\"text-align: left;\">\nFor massive particles it is convenient to express $V$ in terms of dimensionless quantities $\\xi=r/r_{g}$ and $l=L/r_{g}$:\n\\[V_{\\epsilon=1}-\\frac{1}{2}=-\\frac{1}{2\\xi}+\n\t\\frac{l^2}{2\\xi^2}-\\frac{l^2}{2\\xi^3}.\\]\nThe first term is the Newtonian potential energy, the second one is the centrifugal energy, and only the third term is absent in the Newtonian theory and is unique to General Relativity. It changes the asymptote of $V$ at small $\\xi$: $V\\sim -\\xi^{-3}$ instead of the usual $V\\sim\\xi^{-2}$.</p>\n\n[[File:BHfig-SchwM1aL.png|center|thumb|400px|Effective potential for a massive particle  $V(r/r_{g})$ for $l=\\{0, 1.25, \\sqrt{3}, 1.85, 2, 2.25, 2.5, 3\\}\\;\\;\\;$. There are stable as well as unstable circular orbits. The limiting value $l=\\sqrt{3}$ defines the inflection point.]]\n[[File:BHfig-SchwM2aL.png|center|thumb|400px|Same figure zoomed in to show the shallow minima]]\n\n<p style=\"text-align: left;\"> The extrema of $V$ are found as the roots of quadratic equation\n\\[\\xi^2-2l^2 \\xi +3l^2=0\\quad\\Rightarrow\\quad\n\t\\xi_{\\pm}=l^{2}\\left\\{1\\pm\\sqrt{1-3/l^2}\\right\\}.\\]\nTwo extrema exist, maximum and minimum, if $l>\\sqrt{3}$, i.e. $L>L_{cr}$, where\n\\[L_{cr}\\equiv\\sqrt{12}\\,GM.\\]\nThe minimum $\\xi=\\xi_{+}>l^2$ corresponds to a stable circular orbit and non-circular finite motions dangling around it (they are not elliptic). The maximum $\\xi=\\xi_{-}<l^2$ corresponds to an unstable circular orbit. If $\\varepsilon>V(\\xi_{-})$, then the motion is infinite with the fall on the center (i.e. at least on the horizon $r=r_g$).\n\nThe radius of the stable circular orbit is minimal when the discriminant turns to zero: $l=\\sqrt{3}$,\n$\\xi_{+}=\\xi_{-}=l^{2}=3$, and thus \\[r_{circ}^{min}=3r_{g}=6GM.\\]\n\nFor $l<\\sqrt{3}$, i.e. $L<L_{cr}$, there are no extrema of  $V$ and a particle's motion, either finite ($E<1$) or infinite ($E\\geq 1$) always ends with the fall on the center.\n\nIn the limit $l\\gg 1$, which corresponds to $L\\gg r_{g}$,\n\\[\\xi_{-}\\approx3/2,\\;\\xi_{+}\\approx2l^2,\n\t\\quad\\Rightarrow\\quad r_{-}=\\frac{3}{2}r_{g};\\;\n\t\tr_{+}=\\frac{2L^{2}}{r_{g}}=\n\t\t\t\\frac{L^2}{GM},\\]\nso the inner unstable orbit tends to the photon sphere, while the outer stable orbit tends to the classical circular one.\n\t</p>\n  </div>\n</div>\n\n<div id=\"BlackHole38\"></div>\n\n=== Problem 25: gravitational cross-section ===\nDerive the gravitational capture cross-section for a massless particle; the first correction to it for a massive particle ultra-relativistic at infinity. Find the cross-section for a non-relativistic particle to the first order in $v^2/c^2$.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nThe gravitational capture happens, i.e. a particle moving from infinity falls on the center, if it passes above the effective potential barrier\n\\[\\varepsilon\\geq V_{\\max}.\\]\n(a) For a massless particle we have already found the height of the barrier, so the limiting condition is\n\\[\\frac{1}{2}=\\frac{2}{27}\\Big(\\frac{b}{r_g}\\Big)^2,\\]\nand for the capture cross-section we get\n\\[S_{\\gamma}=\\pi b^2 =\\frac{27\\pi}{4}r_{g}^{2}\n\t=27\\pi \\Big(\\frac{GM}{c^2}\\Big)^2.\\]\n\nFor a massive particle $V_{\\epsilon=1}$ reaches its maximum in the lesser of the two roots $\\xi_{\\pm}$:\n\\[\\xi_{\\max}^{-1}=\\frac{1}{3}\\big(1+\\sqrt{1-3/l^2}\\big).\\]\nOn substituting this into $V(\\xi)$, after some transformations we find the maximum:\n\\[V_{\\max}=\\frac{1}{27}\\Big(\n\tl^2 +9+\\frac{(l^2-3)^{3/2}}{l}\\Big).\\]\n\n(b) Ultrarelativistic case:\n\\[\\gamma\\gg1,\\quad E=\\gamma_{\\infty}\\gg1,\n\t\\quad\\varepsilon\n\t=\\tfrac{1}{2}\\gamma_{\\infty}^{2}\\gg 1,\\]\nthus $V_{\\max}\\gg 1$, which only can be when $l\\gg1$. In this limit, in the first order by $l^{-2}$\n\\[V_{\\max}=\\frac{2}{27}l^2 +\\frac{1}{6},\\]\nso the limiting condition of capture is \n\\[l^{2}=\\frac{27}{4}\\big(\n\t\\gamma_{\\infty}^{2}-\\tfrac{1}{3}\\big).\\]\nExpressing it through the impact parameter $b=L/Ev$, we are led to\n\\[S_{\\gamma\\gg1}\n\t=\\pi b^2=\\pi r_{g}^{2}\\frac{l^2}{v^2 E^2}\n\t=\\frac{27\\pi}{4}r_{g}^{2}\\Big(\n\t\t1+\\frac{2}{3\\gamma_{\\infty}^{2}}\\Big).\\]\n\n(c) Nonrelativistic case: $E\\approx (1+v^2 /2)$, $\\varepsilon\\approx \\tfrac{1}{2}(1+v^2)$, and the limiting condition of capture takes the form\n\\[1+v^2 =\\frac{1}{27}\\Big(\n\tl^2 +9+\\frac{(l^2-3)^{3/2}}{l}\\Big).\\]\nIn the zeroth order by $v^2$ its solution is $l=2$ (note that at $l=\\sqrt{3}$ there is no maximum of $V$), while in the first order\n\\[l^2 =4(1+2v^2).\\]\nIn terms of impact parameter then the cross-section is \n\\[S_{v\\ll 1}=\\pi r_{g}^{2}\\frac{l^2}{v^2 E^2}\n\t\\approx \\frac{4\\pi r_{g}^{2}}{v^2}(1+v^2).\\]</p>\n  </div>\n</div>\n\n<div id=\"BlackHole39\"></div>\n=== Problem 26: innermost stable circular orbit ===\nFind the minimal radius of stable circular orbit and its parameters. What is the maximum gravitational binding energy of a particle in the Schwarzschild spacetime?\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nThe radius of stable circular orbit is minimal when\n\\[\\xi_{\\min}=l_{min}^{2}=3+0.\\]\nThe value of effective potential at $\\xi=\\xi_{min}$ determines the particle's energy on the corresponding circular orbit (\\ref{Schw-V}):\n\\[\\frac{E^2}{2}=\\varepsilon=V|_{\\epsilon=1}\n\t=\\frac{1}{2}\\Big(1-\\frac{1}{\\xi}\\Big)\n\t\t\\Big(1+\\frac{l^{2}}{\\xi^2}\\Big)=\n\t\\frac{1-\\xi_{\\min}^{-2}}{2},\\]\nwhere in the last equality we plugged in the value $\\xi=\\xi_{min}=l_{min}^{2}$. Thus\n\\[E=\\sqrt{1-\\xi_{min}^{-2}}=\\sqrt{\\frac{8}{9}}\n\t=\\frac{2\\sqrt{2}}{3}.\\]\nThe binding energy then is (in the units of $mc^{2}$)\n\\[1-E=1-\\frac{2\\sqrt{2}}{3}\\approx 0,06.\\]\t</p>\n  </div>\n</div>\n\n==Miscellaneous problems==\n\n<div id=\"BlackHole40\"></div>\n=== Problem 27: gravitational lensing ===\nGravitational lensing is the effect of deflection of a light beam's (photon's) trajectory in the gravitational field. Derive the deflection of a photon's trajectory in Schwarzschild metric in the limit $L/r_{g}\\gg 1$. Show that it is twice the value for a massive particle with velocity close to $c$ in the Newtonian theory.\n\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nWe choose $\\lambda$ for massless particles as before and obtain\n\\[\\Big(\\frac{dr}{d\\lambda}\\Big)^{2}+\n\t\tb^{2}\\frac{h}{r^2}=1;\\qquad\n\t\\frac{d\\varphi}{d\\lambda}=b\\frac{1}{r^2}.\\]\nExcluding $\\lambda$, after integration we obtain $\\varphi(r)$. We are interested here in the variation of angle when $r$ goes from infinity to the minimal value, fro which the square root is zero. In the Newtonian theory it is $\\pi/2$, so looking for the first correction to this value, we write\n\\[\\Delta\\varphi\\big|_{\\pi/2}=\n\t\tb\\int\\limits_{r_{min}}^{\\infty}\n\t\\frac{dr/r^2}\t{\\sqrt{1-hb^{2}/r^2}}=\n\t\\left\\|\n\t x=\\frac{b}{r},\\;\\varepsilon=\\frac{r_{g}}{r}\\ll 1\n\t\\right\\|=\n\t\\int\\limits_{0}^{x_{max}}\n\t\t\\frac{dx}{\\sqrt{1-x^2+\\varepsilon x^3}}.\\]\n\nChanging variables as $x^{2}-\\varepsilon x^3=y^2$ and using the small parameter $\\varepsilon$, we can transform the integral to one that is easy to compute\n\\[\\Delta\\varphi\\big|_{\\pi/2}\\approx\\int\\limits_{0}^{1}\n\t\\frac{dy}{\\sqrt{1-y^2}}(1+\\varepsilon y)=\n\t\\frac{\\pi}{2}+\\varepsilon=\n\t\t\\frac{\\pi}{2}+\\frac{r_{g}}{b}.\\]\nThe second term is the needed correction, which is half the full correction $\\delta\\phi$ to the angle's variation when a particle moves from infinity and back to infinity:\n\\[\\delta\\varphi\\big|_{\\pi}=\\frac{2r_{g}}{b}=\n\t\\frac{4GM}{b c^2}.\\]\n\nNow let us calculate the same thing in the Newtonian theory for a fast particle. Integrals of motion are\n\\begin{align*}\n&L=b v_{\\infty}=r^{2}\\dot{\\varphi},\\\\\n&E=U+\\frac{\\dot{r}^2}{2}+\\frac{(r\\dot{\\varphi})^2}{2}\n\t=\\frac{\\dot{r}^2}{2}-\\frac{r_g}{2r}+\\frac{L^2}{2r^2}.\n\\end{align*}\nThen $\\varphi(r)$ can be written as \n\\[\\varphi=L\\int\\frac{dr/r^2}\n\t{\\sqrt{2E+\\frac{r_g}{r}-\\frac{L^2}{r^2}}},\\]\nand changing variables to $u=r^{-1}$, we obtain for the variation of angle from infinity to the turning point\n\\[\\varphi|_{\\pi/2}=\\int\\limits_{0}^{u_{max}}\\frac{du}\n\t{\\sqrt{b^{-2}-u^2+\\frac{r_g}{L^2}u}}.\\]\nThe last term under the root is a small correction. Let us make another change of variables $u=u'+\\varepsilon/2$, where $\\varepsilon=r_{g}/L^2 \\ll 1$. Then up to terms of higher order by $\\varepsilon$ the integral is reduced to\n\\[\\varphi|_{\\pi/2}=\\int\\limits_{-\\varepsilon/2}^{b^{-1}}\n\t\\frac{du'}{\\sqrt{b^{-2}-{u'}^2}}=\n\t\\int\\limits_{-b\\varepsilon/2}^{1}\n\t\\frac{d\\xi}{\\sqrt{1-\\xi^2}}\n\t=\\frac{\\pi}{2}+\\frac{b\\varepsilon}{2}.\\]\nThen for motion from infinity to infinity the deflection of the trajectory from a straight line is \n\\[\\delta\\varphi|_{\\pi}=b\\varepsilon=\n\t\\frac{b r_g}{L^2}\n\t=\\frac{r_g}{b}\\frac{1}{v_{\\infty}^2}\n\t\t\\to \\frac{r_g}{b}.\\]\nThis is exactly half the correct quantity given by GTR.</p>\n  </div>\n</div>\n\n<div id=\"BlackHole41\"></div>\n=== Problem 28: generalization of Newtonian potential ===\nShow that the $4$-acceleration of a stationary particle in the Schwarzschild metric can be presented in the form\n\\[a_{\\mu}=-\\partial_{\\mu}\\Phi,\\quad\n\t\\text{where}\\quad \\Phi=\\ln \\sqrt{g_{00}}\n\t\t=\\tfrac{1}{2}\\ln g_{00}\\]\nis some generalization of the Newtonian gravitational potential.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nFor a static particle $u^{\\mu}=(u^0,0,0,0)$, where $u^0$ is found from the normalizing condition\n\\[1=u^\\mu u_\\mu = g_{00}(u^{0})^2\\quad\n\t\\Rightarrow\\quad u^{0}=\\frac{1}{\\sqrt{g_{00}}}.\\]\nThen $4$-acceleration is\n\\[a^\\mu=\\frac{du^\\mu}{ds}\n\t=-\\Gamma^{\\mu}_{\\nu\\lambda}u^{\\nu}u^{\\lambda}\n\t=-\\Gamma^{\\mu}_{00}(u^0)^{2}\\]\nand\n\\[a_{\\mu}=-\\frac{1}{g_{00}}\\Gamma_{\\mu\\,00}\n\t=-\\frac{1}{2g_{00}}(-\\partial_{\\mu}g_{00})\n\t=\\frac{1}{2}\\frac{\\partial_{\\mu}g_{00}}{g_{00}}\n\t=\\partial_{\\mu}\\Phi,\\]\nwhere $\\Phi=\\ln\\sqrt{g_{00}}$.\t</p>\n  </div>\n</div>\n\n<div id=\"BlackHole42\"></div>\n=== Problem 29: coordinate-invariant reformulation ===\nLet us reformulate the problem in a coordinate-independent manner. Suppose we have an arbitrary stationary metric with timelike Killing vector $\\xi^\\mu$, and we denote the $4$-velocity of a stationary observer by $u^{\\mu}=\\xi^{\\mu}/V$. What is the $4$-force per unit mass that we need to apply to a test particle in order to make it stay stationary? Show in coordinate-independent way that the answer coincides with $\\partial_{\\mu}\\Phi$ (up to the sign), and rewrite $\\Phi$ in coordinate-independent form.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nFrom the normalizing condition  $V^2=\\xi^\\mu \\xi_\\mu$. Here we will use a different meaning of acceleration, natural in the context of GTR. On a geodesic the covariant acceleration of a particle $w^\\mu = u^\\mu \\nabla_{\\mu} u^\\nu$ is zero; if a particle moves not on a geodesic, this means that some (non-gravitational) force (per unit mass) is acting on it, equal to $w^{\\mu}=-a^\\mu$.\n\nFirst of all, let us note that, as $\\nabla g=0$, \n\\[\\xi^{\\mu}\\nabla_{\\mu}V\n\t=\\xi^{\\mu}\\xi^{\\nu}\n\t(\\nabla_{\\mu}\\xi_{\\nu}+\\nabla_{\\nu}\\xi_{\\mu})=0,\\]\nand also\n\\[2V\\nabla_{\\mu}V=\\nabla_{\\mu}V^2\n\t=\\nabla_{\\mu}\\xi^{\\nu}\\xi_{\\nu}\n\t=2\\xi^{\\nu}\\nabla_{\\mu}\\xi_{\\nu}.\\]\nThen using stationarity $u^\\mu=\\xi_\\mu /V$ and the Killing equation, we get\n\\[\t-a^\\mu = w^\\mu=\\frac{1}{V}\\xi^{\\nu}\\nabla_{\\nu}\n\t\t\\big(\\frac{1}{V}\\xi^\\mu\\big)\n\t=\\frac{1}{V^2}\\xi_{\\nu}\\nabla^{\\nu}\\xi^{\\mu}\n\t=-\\frac{1}{V^2}\\xi_{\\nu}\\nabla^{\\mu}\\xi^{\\nu}\n\t=-\\frac{1}{V}\\nabla^{\\mu}V=-\\nabla^{\\mu}\\ln V,\\]\ntherefore\n\\[a^{\\mu}=\\nabla^{\\mu}\\Phi,\\quad\\text{where}\n\t\\quad \\Phi=\\ln V=\\tfrac{1}{2}\\ln \\xi^{\\mu}\\xi_{\\nu}.\\]\n\nIn the weak field limit $g_{00}\\approx 1+\\tfrac{2\\phi}{c^2}$, so $\\Phi\\approx \\phi/c^2$. $\\phi$ is the Newtonian gravitational potential.\t</p>\n  </div>\n</div>\n\n<div id=\"BlackHole43\"></div>\n=== Problem 30: surface gravity ===\nSurface gravity $\\kappa$ of the Schwarzschild horizon can be defined as acceleration of a stationary particle at the horizon, measured in the proper time of a stationary observer at infinity. Find $\\kappa$.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nThe scalar acceleration near the horizon, measured in proper time, is \n\\begin{align*}\n\t a=&\\sqrt{|a^{\\mu}a_{\\mu}|}\n\t=\\sqrt{|g^{\\mu\\nu}a_{\\mu}a_{\\nu}|}\n\t=\\sqrt{|g^{11}|(a_1)^{1}}=\\sqrt{|g^{11}|(\\Phi')^2}=\\\\\n\t&=\\sqrt{g_{00}}\\;\\Phi'(r)\n\t=\\sqrt{g_{00}}\\;\\frac{g_{00}'}{g_{00}}\n\t=\\frac{g_{00}'}{2\\sqrt{g_{00}}}.\n\\end{align*}\nIt tends to infinity at the horizon. The time of remote observer $t$ is related to the proper time as $dt=d\\tau\\sqrt{g_{00}}$, therefore acceleration measured in time $t$ is\n\\[a_{\\infty}=\\sqrt{g_{00}}\\;a=\\frac{1}{2}g_{00}',\\]\nand it is finite. At $r=r_{g}$ it gives us the surface gravity\n\\[\\kappa=a_{\\infty}\\Big|_{r=r_g}=\\frac{1}{2r_{g}}\n\t=\\frac{c^4}{4MG}.\\]\nIn the last equality we restored the dimensional factor $c^2$.\t</p>\n  </div>\n</div>\n\nSolving 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\n\\[ds^2=f(t)\\Big(1-\\frac{C}{r}\\Big)dt^2\n\t-\\Big(1-\\frac{C}{r}\\Big)^{-1}dr^2-r^2 d\\Omega^2,\\]\nwhere $C$ is some integration constant, and $f(t)$ an arbitrary function of time $t$.\n\n<div id=\"BlackHole44\"></div>\n\n=== Problem 31: uniqueness in exterior region ===\nSuppose all the matter is distributed around the center of symmetry, and its energy-momentum tensor is spherically symmetric, so that the form of $g_{\\mu\\nu}$ written above is correct. Show that the solution in the exterior region is reduced to the Schwarzschild metric and find the relation between $C$ and the system's mass $M$.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nWe eliminate the factor $f$ in $g_{00}$ by coordinate transformation $\\sqrt{f(t)}dt=dt'$; other components of the metric do not change. In the weak field limit  $g_{00}\\approx(1+\\tfrac{2\\varphi}{c^2})$, where  $\\varphi=-GM/r$ is the Newtonian gravitational potential. Comparing with the asymptote of $g_{00}$ we get\n\\[C=r_{g}=\\frac{2GM}{c^2}.\\]</p>\n  </div>\n</div>\n\n<div id=\"BlackHole45\"></div>\n=== Problem 32: solution in a spherically symmetric void ===\nLet there be a spherically symmetric void $r<r_{0}$ in the spherically symmetric matter distribution. Show that spacetime in the void is flat.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nThe integration constant is determined by the demand of boundedness of $g_{\\mu\\nu}$ in the region: $C=0$. Then on coordinate transformation $f(t)dt^2=d\\tau^2$ we will obtain the flat Minkowskii metric\n\\[ds^2=d\\tau^2-dr^2-r^2 d\\Omega^2.\\]\t</p>\n  </div>\n</div>\n\n<div id=\"BlackHole46\"></div>\n=== Problem 33: shells ===\nLet the matter distribution be spherically symmetric and filling regions  $r<r_{0}$ and $r_{1}<r<r_{2}$ ($r_{0}<r_{1}$). Can one affirm, that the solution in the layer of empty space $r_{0}<r<r_{1}$ is also the Schwarzschild metric?\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nWith such matter distribution the solution in region $r>r_{2}$ is Schwarzschild, as the constant $C$ and function $f(t)$ are fixed in the usual way. In region $r_{0}<r<r_{1}$, however, we do not have the freedom to choose a new time coordinate, as it will be determined by the smooth sewing-up of the metric at the boundaries (in region $r_{1}<r<r_{2}$ the solution is quite different, of course). Therefore $f(t)\\neq 1$ in the inner region.\t</p>\n  </div>\n</div>\n\n<div id=\"BlackHoleEin1\"></div>\n=== Problem 34: Einstein equations for spherically symmetric case ===\nConsider a static, spherically symmetric spacetime, described by metric\n\\[ds^{2} =-f(r)dt^{2} +f^{-1} (r)dr^{2} +r^{2} d\\Omega ^{2}, \\]\nand write the Einstein's equations for it.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\n\\[(-f)-rf'(r)=-8\\pi GPr^{2},\\]\nwhere $P=T_{r}^{r} $ is the radial pressure of the matter source. </p>\n  </div>\n</div>\n\n==Different coordinates, maximal extension==\nWe 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.\n\n<div id=\"BlackHole47\"></div>\n=== Problem 35: Rindler metric ===\nMake coordinate transformation in the Schwarzschild metric near the horizon $(r-r_{g})\\ll r_{g}$ by using physical distance to the horizon as a new radial coordinate instead of $r$, and show that in the new coordinates it reduces near the horizon to the Rindler metric.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nThe physical distance to the horizon along the \"radius\" $r$ at $(r-r_{g})\\ll r_{g}$ is\n\\[l=\\int\\limits_{r_{g}}^{r}\\frac{dr}{\\sqrt{1-r_{g}/r}}=\n\t\\int\\limits_{r_{g}}^{r}\n\t\\frac{\\sqrt{r}dr}{\\sqrt{r-r_{g}}}\\approx\n\t\\sqrt{r_{g}}\\int\\limits_{0}^{r-r_{g}}\n\t\t\\frac{d\\xi}{\\sqrt{\\xi}}=2\\sqrt{r_{g}(r-r_{g})}.\\]\nSubstituting this into the metric, we obtain (restoring the $c$ factors by dimensionality)\n\\[ds^{2}=l^{2}d\\omega^{2}-dl^{2}-r^{2}(l)d\\Omega^{2},\n\t\\qquad\\mbox{where}\\quad\\omega=\\frac{ct}{2r_{g}}.\\]\nComparing with the [[Technical_warm-up#Rindler|Rindler metric]], we see that $l\\equiv\\rho=\\frac{c^2}{a}$, and $\\tau=t\\frac{l}{2r_g}$. Thus the static Schwarzschild metric near the horizon has the same form as Minkowskii metric for a uniformly accelerated observer with acceleration $a=\\frac{c^2}{l}$, which tends to infinity at the horizon. Then using the inverse transformation, from Rindler to Minkowskii, we can turn the Schwarzschild metric near the horizon into the flat one, thus showing explicitly that $r=r_g$ is just a removable coordinate singularity.</p>\n  </div>\n</div>\n\n<div id=\"BlackHole48\"></div>\n\n=== Problem 36: tortoise coordinate ===\nDerive the Schwarzschild metric in coordinates $t$ and  $r^\\star=r+r_{g}\\ln|r-r_g|$. How do the null geodesics falling to the center look like in $(t,r^\\star)$? What range of values of $r^\\star$ corresponds to the region $r>r_g$?\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nMaking the change of variables\n $(t,r)\\to(t,r^\\star)$, we get\n\\[ds^2=\\Big(1-\\frac{r_g}{r}\\Big)\n\t\\big(dt^2-(dr^\\star)^2\\big)-r^2 d\\Omega^2,\\]\nwhere $r$ should be understood as a function of $r^\\star$. This metric is conformally flat$^{*}$ (without its angular part), and with no singularities. This is achieved, however, by pulling the horizon to infinity: $r=r_g$ corresponds to $r^\\star =-\\infty$.</p>\n\n<p style=\"text-align: left;\">$^{*}$A metric $g_{\\mu\\nu}$ is said to be conformally flat if it differs from the Minkowskii metric $\\eta_{\\mu\\nu}$ by a factor $g_{\\mu\\nu}=\\omega^{2}(x)\\eta_{\\mu\\nu}$.</p>\n  </div>\n</div>\n\n<div id=\"BlackHole49\"></div>\n\n=== Problem 37: introducing null coordinates ===\nRewrite the metric in coordinates $r$ and $u=t-r^\\star$, find the equations of null geodesics and the value of $g=det(g_{\\mu\\nu})$ at $r=r_{g}$. Likewise in coordinates $r$ and $v=t+r^\\star$; in coordinates $(u,v)$. The coordinate frames $(v,r)$ and $(u,r)$ are called the ingoing and outgoing Eddington-Finkelstein coordinates.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nIn different coordinates\n\\begin{align*}\n\tds^2&=\\Big(1-\\frac{r_g}{r}\\Big)dv^2\n\t-\\big(dv\\,dr+dr\\,dv\\big)-r^2 d\\Omega^2=\\\\\n\t&=\\Big(1-\\frac{r_g}{r}\\Big)du^2\n\t+\\big(du\\,dr+dr\\,du\\big)-r^2 d\\Omega^2,\\\\\n\t&=\\frac{1}{2}\\Big(1-\\frac{r_g}{r}\\Big)\n\t\\big(du\\,dv+dv\\,du\\big)-r^2 d\\Omega^2.\n\\end{align*}\nThe null geodesics equations are $v=const$ for the ones falling to the center and $u=const$ for the ones escaping to infinity. Thus, by using the null coordinates $(u,v)$, we rectify the geodesics everywhere including the vicinity of the horizon. In coordinates $(r,v)$ we can analytically extend the solution beyond the horizon. The resulting spacetime region contains all the geodesics corresponding to particles which cross the horizon from the outside in, as well as all the geodesics in the outer region.\n\nIn coordinates $(r,u)$ we can likewise extend the solution beyond the horizon, and the resulting spacetime will contain, in addition to all the geodesics in the outer region, all the geodesics corresponding to particles crossing the horizon from the inside out. The two considered analytic extensions extend out metric to \"different sides\". This will be seen better in Kruskal coordinates.\n\nIn coordinates $(r,u)$ and $(r,v)$ the determinant of $g_{\\mu\\nu}$ is $g=-r^{4}\\sin^{2}\\theta$, and in coordninates $(u,v)$ we get $g=-\\frac{1}{4}(1-r_{g}/r)^{2}r^{4}\\sin^{2}\\theta$.</p>\n  </div>\n</div>\n\n<div id=\"BlackHole50\"></div>\n=== Problem 38: Kruskal-Sekeres metric ===\nRewrite the Schwarzschild metric in coordinates $(u',v')$ and in the Kruskal coordinates $(T,R)$ (Kruskal solution), defined as follows:\n\\[v'=e^{v/2r_g},\\quad u'=-e^{-u/2r_g};\\qquad\n\tT=\\frac{u'+v'}{2},\\quad R=\\frac{v'-u'}{2}.\\]\nWhat are the equations of null geodesics, surfaces $r=const$ and $t=const$, of the horizon $r=r_{g}$, singularity $r=0$, in the coordinates $(T,R)$? What is the range space of $(T,R)$? Which regions in the Schwarzschild coordinates do the regions $\\{\\text{I}:\\;R>|T|\\}$, $\\{\\text{II}:\\;T>|R|\\}$, $\\{\\text{III}:\\;R<-|T|\\}$ and $\\{\\text{IV}:\\;T<-|R|\\}$ correspond to? Which of them are casually connected and which are not? What is the geometry of the spacelike slice $T=const$ and how does it evolve with time $T$?\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nAfter substitution we obtain\n\\[ds^2=\\frac{2r_{g}^3}{r}e^{-r/r_{g}}\n\t(dv'\\,du'+du'\\,dv')-r^{2}d\\Omega^2\n\t=\\frac{4r_{g}^3}{r}e^{-r/r_{g}}\n\t(dT^2-dR^2)-r^{2}d\\Omega^2.\\]\nEquations of null geodegics: $u=const$ and $v=const$, or in terms of $(T,R)$: $T=\\pm R+const$. The surface  $r=const$ is mapped to a hyperbola $T^2-R^2=-2(r-r_g)e^{r/r_g}$; singularity to a hyperbola $T^2-R^2=2r_g$ with ''two'' branches also; the horizon to \\emph{two} straight lines $T=\\pm R$. A surface $t=const$ is mapped to straight line $T=R\\tanh\\frac{t}{r_g}$; at $t\\to\\pm\\infty$ it coincides with the horizon. Every point $(T,R)$ here represents a two-sphere.</p>\n\n[[File:BHfig-KruskalCE.png|center|thumb|400px|The Kuskal diagram in coordinates $(T,R)$ and the mapping onto it of the coordinate grid $(t,r)$. Singularity is two hyperbola branches, in the past and future, while the horizon is two straight lines $T=\\pm R$. Null geodesics are also straight lines $T\\pm R=const$]]\n\n<p style=\"text-align: left;\">The physical region of values $(R,T)$ is given by condition $r>0$ and the region between the two hyperbola branches $T=\\pm\\sqrt{2r_g+R^2}$, which represent the past and future singularities. Region I is the asymptotically flat space outside of the horizon (we are living here); region II is the analytical extension along timelike geodesics directed inside across the horizon (coordinates $(r,v)$ map regions I and II); region III is the analytic extension along the timelike geodesics directed away from the horizon (coordinates $(r,u)$ map regions I and III); region IV is \\emph{another} asymptotically flat spacetime. It is not casually connected to our spacetime region I.\n\nA three-dimensional slice $T=const$ of the entire spacetime for $|T|>\\sqrt{2r_g}$ is comprised of two unconnected asymptotically flat regions; for $|T|<\\sqrt{2r_g}$ there is a connection between them, the so-called wormhole, or the Einstein-Rosen bridge. This bridge, however, is spacelike: it is seen clearly from the Kruskal diagram that there are no timelike geodesics (or any other timelike curves either) that would correspond to particles coming from one asymptotically flat region into the other.</p>\n  </div>\n</div>\n\n<div id=\"BlackHole51\"></div>\n=== Problem 39: Penrose diagram ===\nPass to coordinates\n\\[v''=\\arctan\\frac{v'}{\\sqrt{r_g}},\\quad\n\tu''=\\arctan\\frac{u'}{\\sqrt{r_g}}\\]\nand draw the spacetime diagram of the Kruskal solution in them.\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nThe coordinates $(v'',u'')$ are needed in order to conformally map the entire Kruskal solution onto a bounded region of parameters $(v'',u'')$,\n\\[|v''|<\\frac{\\pi}{2},\\quad |u''|<\\frac{\\pi}{2},\n\t\\quad |v''+u''|<\\frac{\\pi}{2}.\\]\nThis is a square with two opposite angles cut out, rotated by $\\pi/4$. Conformality here means that null geodesics, and thus the casual structure, is left invariant under such transformation, and equations of the null geodesics are still $u''=\\pm v'' +const$.\n[[File:BHfig-KruskalPenroseC2.png|center|thumb|400px|Penrose diagram for the Kruskal solution, with the grid of null geodesics]]</p>\n  </div>\n</div>\n\n<div id=\"BlackHole52\"></div>\n=== Problem 40: more realistic collapse ===\nThe Kruskal solution describes an eternal black hole. Suppose, for simplicity,  that some black hole is formed as a result of radial collapse of a spherically symmetric shell of massless particles. What part of the Kruskal solution will be realized, and what will not be? What is the casual structure of the resulting spacetime?\n<div class=\"NavFrame collapsed\">\n  <div class=\"NavHead\">solution</div>\n  <div style=\"width:100%;\" class=\"NavContent\">\n    <p style=\"text-align: left;\">\nOutside of the collapsing shell the spacetime should be the Schwarzschild solution. As the particles are massless, their worldlines on the Kruskal diagram lie on the straight line $T-R=const$, and the exterior region contains a part of region I and a part of region II.</p>\n\n[[File:BHfig-PenroseRealStarC.png|center|thumb|200px|Penrose diagram for the collapse of a massless spherically symmetric shell]]\n\n<p style=\"text-align: left;\">The inner region is just a chunk of Minkowskii spacetime, up to the moment of singularity formation. Sewing together of the tho solutions along the spheres of equal circumference $2\\pi r$ gives us the full spacetime realized in the considered idealized case of collapse. We see that singularity and horizon are only in the future.</p>\n  </div>\n</div>"
                    }
                ]
            }
        }
    }
}