{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Near-horizon geometry of the extremal Kerr black hole\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "This notebook derives the near-horizon geometry of the extremal (i.e. maximally spinning) Kerr black hole. It is based on SageMath tools developed through the [SageManifolds project](https://sagemanifolds.obspm.fr/). \n", "\n", "This notebook requires a version of SageMath at least equal to 9.4:" ] }, { "cell_type": "code", "execution_count": 1, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "'SageMath version 9.5.beta3, Release Date: 2021-10-11'" ] }, "execution_count": 1, "metadata": {}, "output_type": "execute_result" } ], "source": [ "version()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "First we set up the notebook to display mathematical objects using LaTeX rendering:" ] }, { "cell_type": "code", "execution_count": 2, "metadata": {}, "outputs": [], "source": [ "%display latex" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "To speed up computations, we ask for running them in parallel on 8 threads:" ] }, { "cell_type": "code", "execution_count": 3, "metadata": {}, "outputs": [], "source": [ "Parallelism().set(nproc=8)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Spacetime manifold\n", "\n", "We declare the Kerr spacetime (or more precisely the part of it covered by Boyer-Lindquist coordinates) as a 4-dimensional Lorentzian manifold $\\mathcal{M}$:" ] }, { "cell_type": "code", "execution_count": 4, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "4-dimensional Lorentzian manifold M\n" ] } ], "source": [ "M = Manifold(4, 'M', latex_name=r'\\mathcal{M}', structure='Lorentzian')\n", "print(M)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We then introduce the standard **Boyer-Lindquist coordinates** $(t,r,\\theta,\\phi)$ as a chart BL (for *Boyer-Lindquist*) on $\\mathcal{M}$, via the method chart(). The argument of the latter is a string\n", "(delimited by r\"...\" because of the backslash symbols) expressing the coordinates names, their ranges (the default is $(-\\infty,+\\infty)$) and their LaTeX symbols:" ] }, { "cell_type": "code", "execution_count": 5, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Chart (M, (t, r, th, ph))\n" ] }, { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\mathcal{M},(t, r, {\\theta}, {\\phi})\\right)\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\mathcal{M},(t, r, {\\theta}, {\\phi})\\right)$$" ], "text/plain": [ "Chart (M, (t, r, th, ph))" ] }, "execution_count": 5, "metadata": {}, "output_type": "execute_result" } ], "source": [ "BL. = M.chart(r\"t r th:(0,pi):\\theta ph:(0,2*pi):periodic:\\phi\") \n", "print(BL); BL" ] }, { "cell_type": "code", "execution_count": 6, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(t, r\\right)\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(t, r\\right)$$" ], "text/plain": [ "(t, r)" ] }, "execution_count": 6, "metadata": {}, "output_type": "execute_result" } ], "source": [ "BL[0], BL[1]" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Metric tensor of the extremal Kerr spacetime\n", "\n", "The mass parameter $m$ of the Kerr spacetime is declared as a symbolic variable:" ] }, { "cell_type": "code", "execution_count": 7, "metadata": {}, "outputs": [], "source": [ "m = var('m', domain='real')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "For the extremal Kerr black hole, the spin parameter $a$ reaches its upper bound, namely $m$:" ] }, { "cell_type": "code", "execution_count": 8, "metadata": {}, "outputs": [], "source": [ "a = m" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We get the (yet undefined) spacetime metric:" ] }, { "cell_type": "code", "execution_count": 9, "metadata": {}, "outputs": [], "source": [ "g = M.metric()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "

The metric is set by its components in the coordinate frame associated with Boyer-Lindquist coordinates, which is the current manifold's default frame:

" ] }, { "cell_type": "code", "execution_count": 10, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}g = \\left( \\frac{2 \\, m r}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} - 1 \\right) \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} {\\phi} + \\left( \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}}{m^{2} - 2 \\, m r + r^{2}} \\right) \\mathrm{d} r\\otimes \\mathrm{d} r + \\left( m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} t + {\\left(\\frac{2 \\, m^{3} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} + m^{2} + r^{2}\\right)} \\sin\\left({\\theta}\\right)^{2} \\mathrm{d} {\\phi}\\otimes \\mathrm{d} {\\phi}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}g = \\left( \\frac{2 \\, m r}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} - 1 \\right) \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} {\\phi} + \\left( \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}}{m^{2} - 2 \\, m r + r^{2}} \\right) \\mathrm{d} r\\otimes \\mathrm{d} r + \\left( m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} t + {\\left(\\frac{2 \\, m^{3} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} + m^{2} + r^{2}\\right)} \\sin\\left({\\theta}\\right)^{2} \\mathrm{d} {\\phi}\\otimes \\mathrm{d} {\\phi}$$" ], "text/plain": [ "g = (2*m*r/(m^2*cos(th)^2 + r^2) - 1) dt⊗dt - 2*m^2*r*sin(th)^2/(m^2*cos(th)^2 + r^2) dt⊗dph + (m^2*cos(th)^2 + r^2)/(m^2 - 2*m*r + r^2) dr⊗dr + (m^2*cos(th)^2 + r^2) dth⊗dth - 2*m^2*r*sin(th)^2/(m^2*cos(th)^2 + r^2) dph⊗dt + (2*m^3*r*sin(th)^2/(m^2*cos(th)^2 + r^2) + m^2 + r^2)*sin(th)^2 dph⊗dph" ] }, "execution_count": 10, "metadata": {}, "output_type": "execute_result" } ], "source": [ "rho2 = r^2 + (a*cos(th))^2\n", "Delta = r^2 -2*m*r + a^2\n", "g[0,0] = -(1-2*m*r/rho2)\n", "g[0,3] = -2*a*m*r*sin(th)^2/rho2\n", "g[1,1], g[2,2] = rho2/Delta, rho2\n", "g[3,3] = (r^2+a^2+2*m*r*(a*sin(th))^2/rho2)*sin(th)^2\n", "g.display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "

A matrix view of the components with respect to the manifold's default vector frame:

" ] }, { "cell_type": "code", "execution_count": 11, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\begin{array}{rrrr}\n", "\\frac{2 \\, m r}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} - 1 & 0 & 0 & -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\\\\n", "0 & \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}}{m^{2} - 2 \\, m r + r^{2}} & 0 & 0 \\\\\n", "0 & 0 & m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2} & 0 \\\\\n", "-\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} & 0 & 0 & {\\left(\\frac{2 \\, m^{3} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} + m^{2} + r^{2}\\right)} \\sin\\left({\\theta}\\right)^{2}\n", "\\end{array}\\right)\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\begin{array}{rrrr}\n", "\\frac{2 \\, m r}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} - 1 & 0 & 0 & -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\\\\n", "0 & \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}}{m^{2} - 2 \\, m r + r^{2}} & 0 & 0 \\\\\n", "0 & 0 & m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2} & 0 \\\\\n", "-\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} & 0 & 0 & {\\left(\\frac{2 \\, m^{3} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} + m^{2} + r^{2}\\right)} \\sin\\left({\\theta}\\right)^{2}\n", "\\end{array}\\right)$$" ], "text/plain": [ "[ 2*m*r/(m^2*cos(th)^2 + r^2) - 1 0 0 -2*m^2*r*sin(th)^2/(m^2*cos(th)^2 + r^2)]\n", "[ 0 (m^2*cos(th)^2 + r^2)/(m^2 - 2*m*r + r^2) 0 0]\n", "[ 0 0 m^2*cos(th)^2 + r^2 0]\n", "[ -2*m^2*r*sin(th)^2/(m^2*cos(th)^2 + r^2) 0 0 (2*m^3*r*sin(th)^2/(m^2*cos(th)^2 + r^2) + m^2 + r^2)*sin(th)^2]" ] }, "execution_count": 11, "metadata": {}, "output_type": "execute_result" } ], "source": [ "g[:]" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "

The list of the non-vanishing components:

" ] }, { "cell_type": "code", "execution_count": 12, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{lcl} g_{ \\, t \\, t }^{ \\phantom{\\, t}\\phantom{\\, t} } & = & \\frac{2 \\, m r}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} - 1 \\\\ g_{ \\, t \\, {\\phi} }^{ \\phantom{\\, t}\\phantom{\\, {\\phi}} } & = & -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\\\ g_{ \\, r \\, r }^{ \\phantom{\\, r}\\phantom{\\, r} } & = & \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}}{m^{2} - 2 \\, m r + r^{2}} \\\\ g_{ \\, {\\theta} \\, {\\theta} }^{ \\phantom{\\, {\\theta}}\\phantom{\\, {\\theta}} } & = & m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2} \\\\ g_{ \\, {\\phi} \\, t }^{ \\phantom{\\, {\\phi}}\\phantom{\\, t} } & = & -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\\\ g_{ \\, {\\phi} \\, {\\phi} }^{ \\phantom{\\, {\\phi}}\\phantom{\\, {\\phi}} } & = & {\\left(\\frac{2 \\, m^{3} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} + m^{2} + r^{2}\\right)} \\sin\\left({\\theta}\\right)^{2} \\end{array}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{lcl} g_{ \\, t \\, t }^{ \\phantom{\\, t}\\phantom{\\, t} } & = & \\frac{2 \\, m r}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} - 1 \\\\ g_{ \\, t \\, {\\phi} }^{ \\phantom{\\, t}\\phantom{\\, {\\phi}} } & = & -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\\\ g_{ \\, r \\, r }^{ \\phantom{\\, r}\\phantom{\\, r} } & = & \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}}{m^{2} - 2 \\, m r + r^{2}} \\\\ g_{ \\, {\\theta} \\, {\\theta} }^{ \\phantom{\\, {\\theta}}\\phantom{\\, {\\theta}} } & = & m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2} \\\\ g_{ \\, {\\phi} \\, t }^{ \\phantom{\\, {\\phi}}\\phantom{\\, t} } & = & -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\\\ g_{ \\, {\\phi} \\, {\\phi} }^{ \\phantom{\\, {\\phi}}\\phantom{\\, {\\phi}} } & = & {\\left(\\frac{2 \\, m^{3} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} + m^{2} + r^{2}\\right)} \\sin\\left({\\theta}\\right)^{2} \\end{array}$$" ], "text/plain": [ "g_t,t = 2*m*r/(m^2*cos(th)^2 + r^2) - 1 \n", "g_t,ph = -2*m^2*r*sin(th)^2/(m^2*cos(th)^2 + r^2) \n", "g_r,r = (m^2*cos(th)^2 + r^2)/(m^2 - 2*m*r + r^2) \n", "g_th,th = m^2*cos(th)^2 + r^2 \n", "g_ph,t = -2*m^2*r*sin(th)^2/(m^2*cos(th)^2 + r^2) \n", "g_ph,ph = (2*m^3*r*sin(th)^2/(m^2*cos(th)^2 + r^2) + m^2 + r^2)*sin(th)^2 " ] }, "execution_count": 12, "metadata": {}, "output_type": "execute_result" } ], "source": [ "g.display_comp()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Let us check that we are dealing with a solution of the **vacuum Einstein equation**:" ] }, { "cell_type": "code", "execution_count": 13, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{Ric}\\left(g\\right) = 0\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{Ric}\\left(g\\right) = 0$$" ], "text/plain": [ "Ric(g) = 0" ] }, "execution_count": 13, "metadata": {}, "output_type": "execute_result" } ], "source": [ "g.ricci().display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Near-horizon coordinates\n", "\n", "Let us introduce the chart NH of the near-horizon coordinates $(T, R, \\theta, \\Phi)$:" ] }, { "cell_type": "code", "execution_count": 14, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Chart (M, (T, R, th, Ph))\n" ] }, { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\mathcal{M},(T, R, {\\theta}, {\\Phi})\\right)\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\mathcal{M},(T, R, {\\theta}, {\\Phi})\\right)$$" ], "text/plain": [ "Chart (M, (T, R, th, Ph))" ] }, "execution_count": 14, "metadata": {}, "output_type": "execute_result" } ], "source": [ "NH. = M.chart(r\"T R th:(0,pi):\\theta Ph:(0,2*pi):periodic:\\Phi\") \n", "print(NH)\n", "NH" ] }, { "cell_type": "code", "execution_count": 15, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\left(\\mathcal{M},(t, r, {\\theta}, {\\phi})\\right), \\left(\\mathcal{M},(T, R, {\\theta}, {\\Phi})\\right)\\right]\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\left(\\mathcal{M},(t, r, {\\theta}, {\\phi})\\right), \\left(\\mathcal{M},(T, R, {\\theta}, {\\Phi})\\right)\\right]$$" ], "text/plain": [ "[Chart (M, (t, r, th, ph)), Chart (M, (T, R, th, Ph))]" ] }, "execution_count": 15, "metadata": {}, "output_type": "execute_result" } ], "source": [ "M.atlas()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Following J. Bardeen and G. T. Horowitz, \n", "[Phys. Rev. D **60**, 104030 (1999)](https://doi.org/10.1103/PhysRevD.60.104030) ([arXiv:hep-th/9905099](https://arxiv.org/abs/hep-th/9905099)), the near-horizon coordinates $(T, R, \\theta, \\Phi)$ are related to the Boyer-Lindquist coordinates by\n", "$$T = \\epsilon \\frac{t}{2m}, \\quad R = \\frac{r-m}{\\epsilon m},\\quad \\theta = \\theta, \n", "\\quad\\Phi = \\phi - \\frac{t}{2m},$$\n", "where $\\epsilon$ is a constant parameter.\n", "The horizon of the extremal Kerr black hole is located at $r=m$, which corresponds to $R=0$.\n", "\n", "*NB:* the coordinates $T$ and $R$ introduced by Bardeen and Horowitz (BH), which are denoted $t$ and $r$ by them, are actually $T_{\\rm BH} = 2m T$ and $R_{\\rm BH} = m R$.\n", "\n", "We implement the above relations as a transition map from the chart BL to the chart NH:" ] }, { "cell_type": "code", "execution_count": 16, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left\\{\\begin{array}{lcl} T & = & \\frac{{\\epsilon} t}{2 \\, m} \\\\ R & = & -\\frac{m - r}{{\\epsilon} m} \\\\ {\\theta} & = & {\\theta} \\\\ {\\Phi} & = & {\\phi} - \\frac{t}{2 \\, m} \\end{array}\\right.\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left\\{\\begin{array}{lcl} T & = & \\frac{{\\epsilon} t}{2 \\, m} \\\\ R & = & -\\frac{m - r}{{\\epsilon} m} \\\\ {\\theta} & = & {\\theta} \\\\ {\\Phi} & = & {\\phi} - \\frac{t}{2 \\, m} \\end{array}\\right.$$" ], "text/plain": [ "T = 1/2*eps*t/m\n", "R = -(m - r)/(eps*m)\n", "th = th\n", "Ph = ph - 1/2*t/m" ] }, "execution_count": 16, "metadata": {}, "output_type": "execute_result" } ], "source": [ "eps = var('eps', latex_name=r'\\epsilon')\n", "BL_to_NH = BL.transition_map(NH, [eps*t/(2*m), (r-m)/(eps*m), th, ph - t/(2*m)])\n", "BL_to_NH.display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "The inverse relation is" ] }, { "cell_type": "code", "execution_count": 17, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left\\{\\begin{array}{lcl} t & = & \\frac{2 \\, T m}{{\\epsilon}} \\\\ r & = & {\\left(R {\\epsilon} + 1\\right)} m \\\\ {\\theta} & = & {\\theta} \\\\ {\\phi} & = & \\frac{{\\Phi} {\\epsilon} + T}{{\\epsilon}} \\end{array}\\right.\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left\\{\\begin{array}{lcl} t & = & \\frac{2 \\, T m}{{\\epsilon}} \\\\ r & = & {\\left(R {\\epsilon} + 1\\right)} m \\\\ {\\theta} & = & {\\theta} \\\\ {\\phi} & = & \\frac{{\\Phi} {\\epsilon} + T}{{\\epsilon}} \\end{array}\\right.$$" ], "text/plain": [ "t = 2*T*m/eps\n", "r = (R*eps + 1)*m\n", "th = th\n", "ph = (Ph*eps + T)/eps" ] }, "execution_count": 17, "metadata": {}, "output_type": "execute_result" } ], "source": [ "BL_to_NH.inverse().display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Note that we have" ] }, { "cell_type": "code", "execution_count": 18, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\begin{array}{rrrr}\n", "\\frac{{\\epsilon}}{2 \\, m} & 0 & 0 & 0 \\\\\n", "0 & \\frac{1}{{\\epsilon} m} & 0 & 0 \\\\\n", "0 & 0 & 1 & 0 \\\\\n", "-\\frac{1}{2 \\, m} & 0 & 0 & 1\n", "\\end{array}\\right)\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\begin{array}{rrrr}\n", "\\frac{{\\epsilon}}{2 \\, m} & 0 & 0 & 0 \\\\\n", "0 & \\frac{1}{{\\epsilon} m} & 0 & 0 \\\\\n", "0 & 0 & 1 & 0 \\\\\n", "-\\frac{1}{2 \\, m} & 0 & 0 & 1\n", "\\end{array}\\right)$$" ], "text/plain": [ "[1/2*eps/m 0 0 0]\n", "[ 0 1/(eps*m) 0 0]\n", "[ 0 0 1 0]\n", "[ -1/2/m 0 0 1]" ] }, "execution_count": 18, "metadata": {}, "output_type": "execute_result" } ], "source": [ "BL_to_NH.jacobian()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "and" ] }, { "cell_type": "code", "execution_count": 19, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{2 \\, m^{2}}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{2 \\, m^{2}}$$" ], "text/plain": [ "1/2/m^2" ] }, "execution_count": 19, "metadata": {}, "output_type": "execute_result" } ], "source": [ "BL_to_NH.jacobian_det()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "The metric components with respect the coordinates \n", "$(T, R, \\theta, \\Phi)$ are computed by passing the \n", "chart NH to the method display():" ] }, { "cell_type": "code", "execution_count": 20, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\mathcal{M},(t, r, {\\theta}, {\\phi})\\right)\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\mathcal{M},(t, r, {\\theta}, {\\phi})\\right)$$" ], "text/plain": [ "Chart (M, (t, r, th, ph))" ] }, "execution_count": 20, "metadata": {}, "output_type": "execute_result" } ], "source": [ "M.default_chart()" ] }, { "cell_type": "code", "execution_count": 21, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}g = \\left( \\frac{2 \\, m r}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} - 1 \\right) \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} {\\phi} + \\left( \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}}{m^{2} - 2 \\, m r + r^{2}} \\right) \\mathrm{d} r\\otimes \\mathrm{d} r + \\left( m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} t + {\\left(\\frac{2 \\, m^{3} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} + m^{2} + r^{2}\\right)} \\sin\\left({\\theta}\\right)^{2} \\mathrm{d} {\\phi}\\otimes \\mathrm{d} {\\phi}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}g = \\left( \\frac{2 \\, m r}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} - 1 \\right) \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} {\\phi} + \\left( \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}}{m^{2} - 2 \\, m r + r^{2}} \\right) \\mathrm{d} r\\otimes \\mathrm{d} r + \\left( m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} t + {\\left(\\frac{2 \\, m^{3} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} + m^{2} + r^{2}\\right)} \\sin\\left({\\theta}\\right)^{2} \\mathrm{d} {\\phi}\\otimes \\mathrm{d} {\\phi}$$" ], "text/plain": [ "g = (2*m*r/(m^2*cos(th)^2 + r^2) - 1) dt⊗dt - 2*m^2*r*sin(th)^2/(m^2*cos(th)^2 + r^2) dt⊗dph + (m^2*cos(th)^2 + r^2)/(m^2 - 2*m*r + r^2) dr⊗dr + (m^2*cos(th)^2 + r^2) dth⊗dth - 2*m^2*r*sin(th)^2/(m^2*cos(th)^2 + r^2) dph⊗dt + (2*m^3*r*sin(th)^2/(m^2*cos(th)^2 + r^2) + m^2 + r^2)*sin(th)^2 dph⊗dph" ] }, "execution_count": 21, "metadata": {}, "output_type": "execute_result" } ], "source": [ "g.display()" ] }, { "cell_type": "code", "execution_count": 22, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}g = \\left( -\\frac{R^{2} m^{2} \\cos\\left({\\theta}\\right)^{4} + {\\left(R^{4} {\\epsilon}^{2} + 4 \\, R^{3} {\\epsilon} + 6 \\, R^{2}\\right)} m^{2} \\cos\\left({\\theta}\\right)^{2} - {\\left(R^{4} {\\epsilon}^{2} + 4 \\, R^{3} {\\epsilon} + 3 \\, R^{2}\\right)} m^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\left( -\\frac{R^{2} {\\epsilon} m^{2} \\sin\\left({\\theta}\\right)^{4} - {\\left(R^{4} {\\epsilon}^{3} + 4 \\, R^{3} {\\epsilon}^{2} + 8 \\, R^{2} {\\epsilon} + 4 \\, R\\right)} m^{2} \\sin\\left({\\theta}\\right)^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} {\\Phi} + \\left( \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + {\\left(R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + 1\\right)} m^{2}}{R^{2}} \\right) \\mathrm{d} R\\otimes \\mathrm{d} R + \\left( m^{2} \\cos\\left({\\theta}\\right)^{2} + {\\left(R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + 1\\right)} m^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( -\\frac{R^{2} {\\epsilon} m^{2} \\sin\\left({\\theta}\\right)^{4} - {\\left(R^{4} {\\epsilon}^{3} + 4 \\, R^{3} {\\epsilon}^{2} + 8 \\, R^{2} {\\epsilon} + 4 \\, R\\right)} m^{2} \\sin\\left({\\theta}\\right)^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} T + \\left( -\\frac{R^{2} {\\epsilon}^{2} m^{2} \\sin\\left({\\theta}\\right)^{4} - {\\left(R^{4} {\\epsilon}^{4} + 4 \\, R^{3} {\\epsilon}^{3} + 8 \\, R^{2} {\\epsilon}^{2} + 8 \\, R {\\epsilon} + 4\\right)} m^{2} \\sin\\left({\\theta}\\right)^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} {\\Phi}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}g = \\left( -\\frac{R^{2} m^{2} \\cos\\left({\\theta}\\right)^{4} + {\\left(R^{4} {\\epsilon}^{2} + 4 \\, R^{3} {\\epsilon} + 6 \\, R^{2}\\right)} m^{2} \\cos\\left({\\theta}\\right)^{2} - {\\left(R^{4} {\\epsilon}^{2} + 4 \\, R^{3} {\\epsilon} + 3 \\, R^{2}\\right)} m^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\left( -\\frac{R^{2} {\\epsilon} m^{2} \\sin\\left({\\theta}\\right)^{4} - {\\left(R^{4} {\\epsilon}^{3} + 4 \\, R^{3} {\\epsilon}^{2} + 8 \\, R^{2} {\\epsilon} + 4 \\, R\\right)} m^{2} \\sin\\left({\\theta}\\right)^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} {\\Phi} + \\left( \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + {\\left(R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + 1\\right)} m^{2}}{R^{2}} \\right) \\mathrm{d} R\\otimes \\mathrm{d} R + \\left( m^{2} \\cos\\left({\\theta}\\right)^{2} + {\\left(R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + 1\\right)} m^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( -\\frac{R^{2} {\\epsilon} m^{2} \\sin\\left({\\theta}\\right)^{4} - {\\left(R^{4} {\\epsilon}^{3} + 4 \\, R^{3} {\\epsilon}^{2} + 8 \\, R^{2} {\\epsilon} + 4 \\, R\\right)} m^{2} \\sin\\left({\\theta}\\right)^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} T + \\left( -\\frac{R^{2} {\\epsilon}^{2} m^{2} \\sin\\left({\\theta}\\right)^{4} - {\\left(R^{4} {\\epsilon}^{4} + 4 \\, R^{3} {\\epsilon}^{3} + 8 \\, R^{2} {\\epsilon}^{2} + 8 \\, R {\\epsilon} + 4\\right)} m^{2} \\sin\\left({\\theta}\\right)^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} {\\Phi}$$" ], "text/plain": [ "g = -(R^2*m^2*cos(th)^4 + (R^4*eps^2 + 4*R^3*eps + 6*R^2)*m^2*cos(th)^2 - (R^4*eps^2 + 4*R^3*eps + 3*R^2)*m^2)/(R^2*eps^2 + 2*R*eps + cos(th)^2 + 1) dT⊗dT - (R^2*eps*m^2*sin(th)^4 - (R^4*eps^3 + 4*R^3*eps^2 + 8*R^2*eps + 4*R)*m^2*sin(th)^2)/(R^2*eps^2 + 2*R*eps + cos(th)^2 + 1) dT⊗dPh + (m^2*cos(th)^2 + (R^2*eps^2 + 2*R*eps + 1)*m^2)/R^2 dR⊗dR + (m^2*cos(th)^2 + (R^2*eps^2 + 2*R*eps + 1)*m^2) dth⊗dth - (R^2*eps*m^2*sin(th)^4 - (R^4*eps^3 + 4*R^3*eps^2 + 8*R^2*eps + 4*R)*m^2*sin(th)^2)/(R^2*eps^2 + 2*R*eps + cos(th)^2 + 1) dPh⊗dT - (R^2*eps^2*m^2*sin(th)^4 - (R^4*eps^4 + 4*R^3*eps^3 + 8*R^2*eps^2 + 8*R*eps + 4)*m^2*sin(th)^2)/(R^2*eps^2 + 2*R*eps + cos(th)^2 + 1) dPh⊗dPh" ] }, "execution_count": 22, "metadata": {}, "output_type": "execute_result" } ], "source": [ "g.display(NH)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "From now on, we use the near-horizon coordinates as the default ones on the spacetime manifold:" ] }, { "cell_type": "code", "execution_count": 23, "metadata": {}, "outputs": [], "source": [ "M.set_default_chart(NH)\n", "M.set_default_frame(NH.frame())" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Hence NH becomes the default argument of display():" ] }, { "cell_type": "code", "execution_count": 24, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}g = \\left( -\\frac{R^{2} m^{2} \\cos\\left({\\theta}\\right)^{4} + {\\left(R^{4} {\\epsilon}^{2} + 4 \\, R^{3} {\\epsilon} + 6 \\, R^{2}\\right)} m^{2} \\cos\\left({\\theta}\\right)^{2} - {\\left(R^{4} {\\epsilon}^{2} + 4 \\, R^{3} {\\epsilon} + 3 \\, R^{2}\\right)} m^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\left( -\\frac{R^{2} {\\epsilon} m^{2} \\sin\\left({\\theta}\\right)^{4} - {\\left(R^{4} {\\epsilon}^{3} + 4 \\, R^{3} {\\epsilon}^{2} + 8 \\, R^{2} {\\epsilon} + 4 \\, R\\right)} m^{2} \\sin\\left({\\theta}\\right)^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} {\\Phi} + \\left( \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + {\\left(R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + 1\\right)} m^{2}}{R^{2}} \\right) \\mathrm{d} R\\otimes \\mathrm{d} R + \\left( m^{2} \\cos\\left({\\theta}\\right)^{2} + {\\left(R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + 1\\right)} m^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( -\\frac{R^{2} {\\epsilon} m^{2} \\sin\\left({\\theta}\\right)^{4} - {\\left(R^{4} {\\epsilon}^{3} + 4 \\, R^{3} {\\epsilon}^{2} + 8 \\, R^{2} {\\epsilon} + 4 \\, R\\right)} m^{2} \\sin\\left({\\theta}\\right)^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} T + \\left( -\\frac{R^{2} {\\epsilon}^{2} m^{2} \\sin\\left({\\theta}\\right)^{4} - {\\left(R^{4} {\\epsilon}^{4} + 4 \\, R^{3} {\\epsilon}^{3} + 8 \\, R^{2} {\\epsilon}^{2} + 8 \\, R {\\epsilon} + 4\\right)} m^{2} \\sin\\left({\\theta}\\right)^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} {\\Phi}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}g = \\left( -\\frac{R^{2} m^{2} \\cos\\left({\\theta}\\right)^{4} + {\\left(R^{4} {\\epsilon}^{2} + 4 \\, R^{3} {\\epsilon} + 6 \\, R^{2}\\right)} m^{2} \\cos\\left({\\theta}\\right)^{2} - {\\left(R^{4} {\\epsilon}^{2} + 4 \\, R^{3} {\\epsilon} + 3 \\, R^{2}\\right)} m^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\left( -\\frac{R^{2} {\\epsilon} m^{2} \\sin\\left({\\theta}\\right)^{4} - {\\left(R^{4} {\\epsilon}^{3} + 4 \\, R^{3} {\\epsilon}^{2} + 8 \\, R^{2} {\\epsilon} + 4 \\, R\\right)} m^{2} \\sin\\left({\\theta}\\right)^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} {\\Phi} + \\left( \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + {\\left(R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + 1\\right)} m^{2}}{R^{2}} \\right) \\mathrm{d} R\\otimes \\mathrm{d} R + \\left( m^{2} \\cos\\left({\\theta}\\right)^{2} + {\\left(R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + 1\\right)} m^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( -\\frac{R^{2} {\\epsilon} m^{2} \\sin\\left({\\theta}\\right)^{4} - {\\left(R^{4} {\\epsilon}^{3} + 4 \\, R^{3} {\\epsilon}^{2} + 8 \\, R^{2} {\\epsilon} + 4 \\, R\\right)} m^{2} \\sin\\left({\\theta}\\right)^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} T + \\left( -\\frac{R^{2} {\\epsilon}^{2} m^{2} \\sin\\left({\\theta}\\right)^{4} - {\\left(R^{4} {\\epsilon}^{4} + 4 \\, R^{3} {\\epsilon}^{3} + 8 \\, R^{2} {\\epsilon}^{2} + 8 \\, R {\\epsilon} + 4\\right)} m^{2} \\sin\\left({\\theta}\\right)^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} {\\Phi}$$" ], "text/plain": [ "g = -(R^2*m^2*cos(th)^4 + (R^4*eps^2 + 4*R^3*eps + 6*R^2)*m^2*cos(th)^2 - (R^4*eps^2 + 4*R^3*eps + 3*R^2)*m^2)/(R^2*eps^2 + 2*R*eps + cos(th)^2 + 1) dT⊗dT - (R^2*eps*m^2*sin(th)^4 - (R^4*eps^3 + 4*R^3*eps^2 + 8*R^2*eps + 4*R)*m^2*sin(th)^2)/(R^2*eps^2 + 2*R*eps + cos(th)^2 + 1) dT⊗dPh + (m^2*cos(th)^2 + (R^2*eps^2 + 2*R*eps + 1)*m^2)/R^2 dR⊗dR + (m^2*cos(th)^2 + (R^2*eps^2 + 2*R*eps + 1)*m^2) dth⊗dth - (R^2*eps*m^2*sin(th)^4 - (R^4*eps^3 + 4*R^3*eps^2 + 8*R^2*eps + 4*R)*m^2*sin(th)^2)/(R^2*eps^2 + 2*R*eps + cos(th)^2 + 1) dPh⊗dT - (R^2*eps^2*m^2*sin(th)^4 - (R^4*eps^4 + 4*R^3*eps^3 + 8*R^2*eps^2 + 8*R*eps + 4)*m^2*sin(th)^2)/(R^2*eps^2 + 2*R*eps + cos(th)^2 + 1) dPh⊗dPh" ] }, "execution_count": 24, "metadata": {}, "output_type": "execute_result" } ], "source": [ "g.display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## The near-horizon metric $h$ as the limit $\\epsilon\\to 0$ of the Kerr metric $g$\n", "\n", "Let us define the *near-horizon metric* as the metric $h$ on $\\mathcal{M}$ that is the limit $\\epsilon\\to 0$ of the Kerr metric $g$. The limit is taken by asking for a series expansion of $g$ with respect to $\\epsilon$ up to the 0-th order (i.e. keeping only $\\epsilon^0$ terms). This is achieved via the method truncate:" ] }, { "cell_type": "code", "execution_count": 25, "metadata": {}, "outputs": [], "source": [ "g0 = g.truncate(eps, 0)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Since NH is the default chart and NH.frame() the default frame, we initialize the NH components of $h$ via" ] }, { "cell_type": "code", "execution_count": 26, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}h = \\left( -\\frac{R^{2} m^{2} \\cos\\left({\\theta}\\right)^{4}}{\\cos\\left({\\theta}\\right)^{2} + 1} - \\frac{6 \\, R^{2} m^{2} \\cos\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} + \\frac{3 \\, R^{2} m^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\left( \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} {\\Phi} + \\left( \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2}}{R^{2}} + \\frac{m^{2}}{R^{2}} \\right) \\mathrm{d} R\\otimes \\mathrm{d} R + \\left( m^{2} \\cos\\left({\\theta}\\right)^{2} + m^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} T + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} {\\Phi}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}h = \\left( -\\frac{R^{2} m^{2} \\cos\\left({\\theta}\\right)^{4}}{\\cos\\left({\\theta}\\right)^{2} + 1} - \\frac{6 \\, R^{2} m^{2} \\cos\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} + \\frac{3 \\, R^{2} m^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\left( \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} {\\Phi} + \\left( \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2}}{R^{2}} + \\frac{m^{2}}{R^{2}} \\right) \\mathrm{d} R\\otimes \\mathrm{d} R + \\left( m^{2} \\cos\\left({\\theta}\\right)^{2} + m^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} T + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} {\\Phi}$$" ], "text/plain": [ "h = (-R^2*m^2*cos(th)^4/(cos(th)^2 + 1) - 6*R^2*m^2*cos(th)^2/(cos(th)^2 + 1) + 3*R^2*m^2/(cos(th)^2 + 1)) dT⊗dT + 4*R*m^2*sin(th)^2/(cos(th)^2 + 1) dT⊗dPh + (m^2*cos(th)^2/R^2 + m^2/R^2) dR⊗dR + (m^2*cos(th)^2 + m^2) dth⊗dth + 4*R*m^2*sin(th)^2/(cos(th)^2 + 1) dPh⊗dT + 4*m^2*sin(th)^2/(cos(th)^2 + 1) dPh⊗dPh" ] }, "execution_count": 26, "metadata": {}, "output_type": "execute_result" } ], "source": [ "h = M.lorentzian_metric('h')\n", "h[:] = g0[:] \n", "h.display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Some simplifications are in order:" ] }, { "cell_type": "code", "execution_count": 27, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}h = \\left( -\\frac{R^{2} m^{2} \\sin\\left({\\theta}\\right)^{4} - 8 \\, R^{2} m^{2} \\sin\\left({\\theta}\\right)^{2} + 4 \\, R^{2} m^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\left( \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} {\\Phi} + \\left( -\\frac{m^{2} \\sin\\left({\\theta}\\right)^{2} - 2 \\, m^{2}}{R^{2}} \\right) \\mathrm{d} R\\otimes \\mathrm{d} R + \\left( -m^{2} \\sin\\left({\\theta}\\right)^{2} + 2 \\, m^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} T + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} {\\Phi}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}h = \\left( -\\frac{R^{2} m^{2} \\sin\\left({\\theta}\\right)^{4} - 8 \\, R^{2} m^{2} \\sin\\left({\\theta}\\right)^{2} + 4 \\, R^{2} m^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\left( \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} {\\Phi} + \\left( -\\frac{m^{2} \\sin\\left({\\theta}\\right)^{2} - 2 \\, m^{2}}{R^{2}} \\right) \\mathrm{d} R\\otimes \\mathrm{d} R + \\left( -m^{2} \\sin\\left({\\theta}\\right)^{2} + 2 \\, m^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} T + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} {\\Phi}$$" ], "text/plain": [ "h = -(R^2*m^2*sin(th)^4 - 8*R^2*m^2*sin(th)^2 + 4*R^2*m^2)/(cos(th)^2 + 1) dT⊗dT + 4*R*m^2*sin(th)^2/(cos(th)^2 + 1) dT⊗dPh - (m^2*sin(th)^2 - 2*m^2)/R^2 dR⊗dR + (-m^2*sin(th)^2 + 2*m^2) dth⊗dth + 4*R*m^2*sin(th)^2/(cos(th)^2 + 1) dPh⊗dT + 4*m^2*sin(th)^2/(cos(th)^2 + 1) dPh⊗dPh" ] }, "execution_count": 27, "metadata": {}, "output_type": "execute_result" } ], "source": [ "h.apply_map(lambda x: x.simplify_full())\n", "h.display()" ] }, { "cell_type": "code", "execution_count": 28, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{lcl} h_{ \\, T \\, T }^{ \\phantom{\\, T}\\phantom{\\, T} } & = & -\\frac{R^{2} m^{2} \\sin\\left({\\theta}\\right)^{4} - 8 \\, R^{2} m^{2} \\sin\\left({\\theta}\\right)^{2} + 4 \\, R^{2} m^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\\\ h_{ \\, T \\, {\\Phi} }^{ \\phantom{\\, T}\\phantom{\\, {\\Phi}} } & = & \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\\\ h_{ \\, R \\, R }^{ \\phantom{\\, R}\\phantom{\\, R} } & = & -\\frac{m^{2} \\sin\\left({\\theta}\\right)^{2} - 2 \\, m^{2}}{R^{2}} \\\\ h_{ \\, {\\theta} \\, {\\theta} }^{ \\phantom{\\, {\\theta}}\\phantom{\\, {\\theta}} } & = & -m^{2} \\sin\\left({\\theta}\\right)^{2} + 2 \\, m^{2} \\\\ h_{ \\, {\\Phi} \\, T }^{ \\phantom{\\, {\\Phi}}\\phantom{\\, T} } & = & \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\\\ h_{ \\, {\\Phi} \\, {\\Phi} }^{ \\phantom{\\, {\\Phi}}\\phantom{\\, {\\Phi}} } & = & \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\end{array}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{lcl} h_{ \\, T \\, T }^{ \\phantom{\\, T}\\phantom{\\, T} } & = & -\\frac{R^{2} m^{2} \\sin\\left({\\theta}\\right)^{4} - 8 \\, R^{2} m^{2} \\sin\\left({\\theta}\\right)^{2} + 4 \\, R^{2} m^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\\\ h_{ \\, T \\, {\\Phi} }^{ \\phantom{\\, T}\\phantom{\\, {\\Phi}} } & = & \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\\\ h_{ \\, R \\, R }^{ \\phantom{\\, R}\\phantom{\\, R} } & = & -\\frac{m^{2} \\sin\\left({\\theta}\\right)^{2} - 2 \\, m^{2}}{R^{2}} \\\\ h_{ \\, {\\theta} \\, {\\theta} }^{ \\phantom{\\, {\\theta}}\\phantom{\\, {\\theta}} } & = & -m^{2} \\sin\\left({\\theta}\\right)^{2} + 2 \\, m^{2} \\\\ h_{ \\, {\\Phi} \\, T }^{ \\phantom{\\, {\\Phi}}\\phantom{\\, T} } & = & \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\\\ h_{ \\, {\\Phi} \\, {\\Phi} }^{ \\phantom{\\, {\\Phi}}\\phantom{\\, {\\Phi}} } & = & \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\end{array}$$" ], "text/plain": [ "h_T,T = -(R^2*m^2*sin(th)^4 - 8*R^2*m^2*sin(th)^2 + 4*R^2*m^2)/(cos(th)^2 + 1) \n", "h_T,Ph = 4*R*m^2*sin(th)^2/(cos(th)^2 + 1) \n", "h_R,R = -(m^2*sin(th)^2 - 2*m^2)/R^2 \n", "h_th,th = -m^2*sin(th)^2 + 2*m^2 \n", "h_Ph,T = 4*R*m^2*sin(th)^2/(cos(th)^2 + 1) \n", "h_Ph,Ph = 4*m^2*sin(th)^2/(cos(th)^2 + 1) " ] }, "execution_count": 28, "metadata": {}, "output_type": "execute_result" } ], "source": [ "h.display_comp()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "As a check, we compare with Eq. (2.6) of [Bardeen & Horowitz, Phys. Rev. D **60**, 104030 (1999)](https://doi.org/10.1103/PhysRevD.60.104030), which involves the bilinear form\n", "$$q = \\frac{4m^2\\sin^2\\theta}{1+\\cos^2\\theta} \n", "\\left( \\mathrm{d}\\Phi + R \\mathrm{d} T \\right)^2$$\n", "We construct $q$ from the 1-forms $\\mathrm{d} T$ and\n", "$\\mathrm{d}\\Phi$:" ] }, { "cell_type": "code", "execution_count": 29, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{d} T\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{d} T$$" ], "text/plain": [ "1-form dT on the 4-dimensional Lorentzian manifold M" ] }, "execution_count": 29, "metadata": {}, "output_type": "execute_result" } ], "source": [ "dT = NH.coframe()[0]\n", "dT" ] }, { "cell_type": "code", "execution_count": 30, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{d} {\\Phi}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{d} {\\Phi}$$" ], "text/plain": [ "1-form dPh on the 4-dimensional Lorentzian manifold M" ] }, "execution_count": 30, "metadata": {}, "output_type": "execute_result" } ], "source": [ "dPh = NH.coframe()[3]\n", "dPh" ] }, { "cell_type": "code", "execution_count": 31, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left( \\frac{4 \\, R^{2} m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\left( \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} {\\Phi} + \\left( \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} T + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} {\\Phi}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left( \\frac{4 \\, R^{2} m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\left( \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} {\\Phi} + \\left( \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} T + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} {\\Phi}$$" ], "text/plain": [ "4*R^2*m^2*sin(th)^2/(cos(th)^2 + 1) dT⊗dT + 4*R*m^2*sin(th)^2/(cos(th)^2 + 1) dT⊗dPh + 4*R*m^2*sin(th)^2/(cos(th)^2 + 1) dPh⊗dT + 4*m^2*sin(th)^2/(cos(th)^2 + 1) dPh⊗dPh" ] }, "execution_count": 31, "metadata": {}, "output_type": "execute_result" } ], "source": [ "q = 4*m^2*sin(th)^2/(1+cos(th)^2) * (dPh + R*dT) * (dPh + R*dT)\n", "q.display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We evaluate then $h-q$ to compare with Eq. (2.6) of Bardeen & Horowitz:" ] }, { "cell_type": "code", "execution_count": 32, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left( R^{2} m^{2} \\sin\\left({\\theta}\\right)^{2} - 2 \\, R^{2} m^{2} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\left( -\\frac{m^{2} \\sin\\left({\\theta}\\right)^{2} - 2 \\, m^{2}}{R^{2}} \\right) \\mathrm{d} R\\otimes \\mathrm{d} R + \\left( -m^{2} \\sin\\left({\\theta}\\right)^{2} + 2 \\, m^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left( R^{2} m^{2} \\sin\\left({\\theta}\\right)^{2} - 2 \\, R^{2} m^{2} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\left( -\\frac{m^{2} \\sin\\left({\\theta}\\right)^{2} - 2 \\, m^{2}}{R^{2}} \\right) \\mathrm{d} R\\otimes \\mathrm{d} R + \\left( -m^{2} \\sin\\left({\\theta}\\right)^{2} + 2 \\, m^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta}$$" ], "text/plain": [ "(R^2*m^2*sin(th)^2 - 2*R^2*m^2) dT⊗dT - (m^2*sin(th)^2 - 2*m^2)/R^2 dR⊗dR + (-m^2*sin(th)^2 + 2*m^2) dth⊗dth" ] }, "execution_count": 32, "metadata": {}, "output_type": "execute_result" } ], "source": [ "(h - q).display()" ] }, { "cell_type": "code", "execution_count": 33, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left( \\frac{R^{2} m^{2} \\sin\\left({\\theta}\\right)^{2} - 2 \\, R^{2} m^{2}}{{\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\left( -\\frac{m^{2} \\sin\\left({\\theta}\\right)^{2} - 2 \\, m^{2}}{{\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} R^{2} m^{2}} \\right) \\mathrm{d} R\\otimes \\mathrm{d} R + \\left( -\\frac{m^{2} \\sin\\left({\\theta}\\right)^{2} - 2 \\, m^{2}}{{\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left( \\frac{R^{2} m^{2} \\sin\\left({\\theta}\\right)^{2} - 2 \\, R^{2} m^{2}}{{\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\left( -\\frac{m^{2} \\sin\\left({\\theta}\\right)^{2} - 2 \\, m^{2}}{{\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} R^{2} m^{2}} \\right) \\mathrm{d} R\\otimes \\mathrm{d} R + \\left( -\\frac{m^{2} \\sin\\left({\\theta}\\right)^{2} - 2 \\, m^{2}}{{\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta}$$" ], "text/plain": [ "(R^2*m^2*sin(th)^2 - 2*R^2*m^2)/((cos(th)^2 + 1)*m^2) dT⊗dT - (m^2*sin(th)^2 - 2*m^2)/((cos(th)^2 + 1)*R^2*m^2) dR⊗dR - (m^2*sin(th)^2 - 2*m^2)/((cos(th)^2 + 1)*m^2) dth⊗dth" ] }, "execution_count": 33, "metadata": {}, "output_type": "execute_result" } ], "source": [ "s = (h - q)/(m^2*(1 + cos(th)^2))\n", "s.display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Some simplifications are in order:" ] }, { "cell_type": "code", "execution_count": 34, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}-R^{2} \\mathrm{d} T\\otimes \\mathrm{d} T + \\frac{1}{R^{2}} \\mathrm{d} R\\otimes \\mathrm{d} R +\\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}-R^{2} \\mathrm{d} T\\otimes \\mathrm{d} T + \\frac{1}{R^{2}} \\mathrm{d} R\\otimes \\mathrm{d} R +\\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta}$$" ], "text/plain": [ "-R^2 dT⊗dT + R^(-2) dR⊗dR + dth⊗dth" ] }, "execution_count": 34, "metadata": {}, "output_type": "execute_result" } ], "source": [ "s.apply_map(lambda x: x.subs({cos(th)^2: 1 - sin(th)^2}))\n", "s.apply_map(lambda x: x.simplify_full())\n", "s.display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "The above result shows the complete agreement between $h = m^2(1 + \\cos^2\\theta) s + q$ and the metric given by Eq. (2.6) of Bardeen & Horowitz." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Non-degenerate character\n", "\n", "A priori, the limit process defining $h$ ensures only that $h$ is a symmetric bilinear form. In order to establish that it is a proper metric on $\\mathcal{M}$, there remains to show that $h$ is non-degenerate. Let us do it by computing the determinant of $h$'s components with respect to the NH coordinates:" ] }, { "cell_type": "code", "execution_count": 35, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{llcl} & \\mathcal{M} & \\longrightarrow & \\mathbb{R} \\\\ & \\left(t, r, {\\theta}, {\\phi}\\right) & \\longmapsto & -4 \\, m^{8} \\sin\\left({\\theta}\\right)^{6} + 16 \\, m^{8} \\sin\\left({\\theta}\\right)^{4} - 16 \\, m^{8} \\sin\\left({\\theta}\\right)^{2} \\\\ & \\left(T, R, {\\theta}, {\\Phi}\\right) & \\longmapsto & -4 \\, m^{8} \\sin\\left({\\theta}\\right)^{6} + 16 \\, m^{8} \\sin\\left({\\theta}\\right)^{4} - 16 \\, m^{8} \\sin\\left({\\theta}\\right)^{2} \\end{array}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\begin{array}{llcl} & \\mathcal{M} & \\longrightarrow & \\mathbb{R} \\\\ & \\left(t, r, {\\theta}, {\\phi}\\right) & \\longmapsto & -4 \\, m^{8} \\sin\\left({\\theta}\\right)^{6} + 16 \\, m^{8} \\sin\\left({\\theta}\\right)^{4} - 16 \\, m^{8} \\sin\\left({\\theta}\\right)^{2} \\\\ & \\left(T, R, {\\theta}, {\\Phi}\\right) & \\longmapsto & -4 \\, m^{8} \\sin\\left({\\theta}\\right)^{6} + 16 \\, m^{8} \\sin\\left({\\theta}\\right)^{4} - 16 \\, m^{8} \\sin\\left({\\theta}\\right)^{2} \\end{array}$$" ], "text/plain": [ "M → ℝ\n", "(t, r, th, ph) ↦ -4*m^8*sin(th)^6 + 16*m^8*sin(th)^4 - 16*m^8*sin(th)^2\n", "(T, R, th, Ph) ↦ -4*m^8*sin(th)^6 + 16*m^8*sin(th)^4 - 16*m^8*sin(th)^2" ] }, "execution_count": 35, "metadata": {}, "output_type": "execute_result" } ], "source": [ "deth = h.determinant(frame=NH.frame())\n", "deth.display()" ] }, { "cell_type": "code", "execution_count": 36, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}-4 \\, {\\left(\\sin\\left({\\theta}\\right)^{2} - 2\\right)}^{2} m^{8} \\sin\\left({\\theta}\\right)^{2}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}-4 \\, {\\left(\\sin\\left({\\theta}\\right)^{2} - 2\\right)}^{2} m^{8} \\sin\\left({\\theta}\\right)^{2}$$" ], "text/plain": [ "-4*(sin(th)^2 - 2)^2*m^8*sin(th)^2" ] }, "execution_count": 36, "metadata": {}, "output_type": "execute_result" } ], "source": [ "deth.expr().factor()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Hence the determinant vanishes only for $\\sin\\theta = 0$, which corresponds simply to the standard coordinate singularity of spherical-type coordinates on the rotation axis. Away from the rotation axis, we \n", "have $\\det h < 0$, which implies that $h$ is non-degenerate and furthermore, that it is \n", "has the signature $(-,+,+,+)$. Hence we conclude that that $h$ is a **regular Lorentzian metric** on $\\mathcal{M}$. " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We note that the metric $h$ is not asymptotically flat. " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Einstein equation for $h$" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "As a limit of solutions of the vaccum Einstein equation, $h$ is itself a solution of the **vacuum Einstein equation**, as we can check:" ] }, { "cell_type": "code", "execution_count": 37, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{Ric}\\left(h\\right) = 0\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{Ric}\\left(h\\right) = 0$$" ], "text/plain": [ "Ric(h) = 0" ] }, "execution_count": 37, "metadata": {}, "output_type": "execute_result" } ], "source": [ "h.ricci().display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Killing vectors of the near-horizon geometry" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Let us first consider the vector field $\\eta := \\frac{\\partial}{\\partial\\Phi}$:" ] }, { "cell_type": "code", "execution_count": 38, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\eta = \\frac{\\partial}{\\partial {\\Phi} }\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\eta = \\frac{\\partial}{\\partial {\\Phi} }$$" ], "text/plain": [ "eta = ∂/∂Ph" ] }, "execution_count": 38, "metadata": {}, "output_type": "execute_result" } ], "source": [ "eta = M.vector_field(0, 0, 0, 1, name='eta', latex_name=r'\\eta')\n", "eta.display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "It is a Killing vector of the near-horizon metric, since the Lie derivative of $h$ along $\\eta$ vanishes:" ] }, { "cell_type": "code", "execution_count": 39, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}0\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}0$$" ], "text/plain": [ "0" ] }, "execution_count": 39, "metadata": {}, "output_type": "execute_result" } ], "source": [ "h.lie_derivative(eta).display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "This is not surprising since the components of $h$ are independent from $\\Phi$." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Similarly, we can check that $\\xi_1 := \\frac{\\partial}{\\partial T}$ is a Killing vector of $h$, reflecting the independence of the components of $h$ from $T$:" ] }, { "cell_type": "code", "execution_count": 40, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\xi_{1} = \\frac{\\partial}{\\partial T }\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\xi_{1} = \\frac{\\partial}{\\partial T }$$" ], "text/plain": [ "xi1 = ∂/∂T" ] }, "execution_count": 40, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi1 = M.vector_field(1, 0, 0, 0, name='xi1', latex_name=r'\\xi_{1}')\n", "xi1.display()" ] }, { "cell_type": "code", "execution_count": 41, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}0\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}0$$" ], "text/plain": [ "0" ] }, "execution_count": 41, "metadata": {}, "output_type": "execute_result" } ], "source": [ "h.lie_derivative(xi1).display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "The above two Killing vectors correspond respectively to the **axisymmetry** and the **pseudo-stationarity** of the Kerr metric. A third symmetry, which is not present in the original Kerr metric, is the invariance under the **scaling** $(T,R)\\mapsto (\\alpha T, R/\\alpha)$, as it is clear on the metric components in Out[27]. The corresponding Killing vector is " ] }, { "cell_type": "code", "execution_count": 42, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\xi_{2} = T \\frac{\\partial}{\\partial T } -R \\frac{\\partial}{\\partial R }\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\xi_{2} = T \\frac{\\partial}{\\partial T } -R \\frac{\\partial}{\\partial R }$$" ], "text/plain": [ "xi2 = T ∂/∂T - R ∂/∂R" ] }, "execution_count": 42, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi2 = M.vector_field(T, -R, 0, 0, name='xi2', latex_name=r'\\xi_{2}')\n", "xi2.display()" ] }, { "cell_type": "code", "execution_count": 43, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}0\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}0$$" ], "text/plain": [ "0" ] }, "execution_count": 43, "metadata": {}, "output_type": "execute_result" } ], "source": [ "h.lie_derivative(xi2).display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Finally, a fourth Killing vector is" ] }, { "cell_type": "code", "execution_count": 44, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\xi_{3} = \\left( \\frac{1}{2} \\, T^{2} + \\frac{1}{2 \\, R^{2}} \\right) \\frac{\\partial}{\\partial T } -R T \\frac{\\partial}{\\partial R } -\\frac{1}{R} \\frac{\\partial}{\\partial {\\Phi} }\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\xi_{3} = \\left( \\frac{1}{2} \\, T^{2} + \\frac{1}{2 \\, R^{2}} \\right) \\frac{\\partial}{\\partial T } -R T \\frac{\\partial}{\\partial R } -\\frac{1}{R} \\frac{\\partial}{\\partial {\\Phi} }$$" ], "text/plain": [ "xi3 = (1/2*T^2 + 1/2/R^2) ∂/∂T - R*T ∂/∂R - 1/R ∂/∂Ph" ] }, "execution_count": 44, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi3 = M.vector_field((T^2 + 1/R^2)/2, -R*T, 0, -1/R,\n", " name='xi3', latex_name=r'\\xi_{3}')\n", "xi3.display()" ] }, { "cell_type": "code", "execution_count": 45, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}0\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}0$$" ], "text/plain": [ "0" ] }, "execution_count": 45, "metadata": {}, "output_type": "execute_result" } ], "source": [ "h.lie_derivative(xi3).display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We shall see below that this Killing vector is actually related to the Killing vector $\\frac{\\partial}{\\partial \\tau}$ associated to the so-called *global NHEK coordinates* by \n", "$$\\xi_3 = \\frac{\\partial}{\\partial \\tau} - \\frac{1}{2} \\frac{\\partial}{\\partial T}$$" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Determination of the symmetry group" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We have four independent Killing vectors, $\\eta$, $\\xi_1$, $\\xi_2$ and $\\xi_3$, which implies that the symmetry group of the near-horizon geometry is a 4-dimensional Lie group $G$. Let us determine $G$ by investigating the **structure constants** of the basis \n", "$(\\eta, \\xi_1, \\xi_2, \\xi_3)$ of the Lie algebra of $G$. First of all, we notice that \n", "$\\eta$ commutes with the other Killing vectors:" ] }, { "cell_type": "code", "execution_count": 46, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\eta,\\xi_{1}\\right] = 0\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\eta,\\xi_{1}\\right] = 0$$" ], "text/plain": [ "[eta,xi1] = 0" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\eta,\\xi_{2}\\right] = 0\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\eta,\\xi_{2}\\right] = 0$$" ], "text/plain": [ "[eta,xi2] = 0" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\eta,\\xi_{3}\\right] = 0\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\eta,\\xi_{3}\\right] = 0$$" ], "text/plain": [ "[eta,xi3] = 0" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "for xi in [xi1, xi2, xi3]:\n", " show(eta.bracket(xi).display())" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Since $\\eta$ generates the rotation group $\\mathrm{SO}(2)=\\mathrm{U}(1)$, we may write that $G = \\mathrm{U}(1)\\times G_3$, where $G_3$\n", "is a 3-dimensional Lie group, whose generators are $(\\xi_1, \\xi_2, \\xi_3)$. Let us determine the structure constants of these three vectors. We have" ] }, { "cell_type": "code", "execution_count": 47, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\xi_{1},\\xi_{2}\\right] = \\frac{\\partial}{\\partial T }\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\xi_{1},\\xi_{2}\\right] = \\frac{\\partial}{\\partial T }$$" ], "text/plain": [ "[xi1,xi2] = ∂/∂T" ] }, "execution_count": 47, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi1.bracket(xi2).display()" ] }, { "cell_type": "code", "execution_count": 48, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}$$" ], "text/plain": [ "True" ] }, "execution_count": 48, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi1.bracket(xi2) == xi1" ] }, { "cell_type": "code", "execution_count": 49, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\xi_{1},\\xi_{3}\\right] = T \\frac{\\partial}{\\partial T } -R \\frac{\\partial}{\\partial R }\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\xi_{1},\\xi_{3}\\right] = T \\frac{\\partial}{\\partial T } -R \\frac{\\partial}{\\partial R }$$" ], "text/plain": [ "[xi1,xi3] = T ∂/∂T - R ∂/∂R" ] }, "execution_count": 49, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi1.bracket(xi3).display()" ] }, { "cell_type": "code", "execution_count": 50, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}$$" ], "text/plain": [ "True" ] }, "execution_count": 50, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi1.bracket(xi3) == xi2" ] }, { "cell_type": "code", "execution_count": 51, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\xi_{2},\\xi_{3}\\right] = \\left( \\frac{R^{2} T^{2} + 1}{2 \\, R^{2}} \\right) \\frac{\\partial}{\\partial T } -R T \\frac{\\partial}{\\partial R } -\\frac{1}{R} \\frac{\\partial}{\\partial {\\Phi} }\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\xi_{2},\\xi_{3}\\right] = \\left( \\frac{R^{2} T^{2} + 1}{2 \\, R^{2}} \\right) \\frac{\\partial}{\\partial T } -R T \\frac{\\partial}{\\partial R } -\\frac{1}{R} \\frac{\\partial}{\\partial {\\Phi} }$$" ], "text/plain": [ "[xi2,xi3] = 1/2*(R^2*T^2 + 1)/R^2 ∂/∂T - R*T ∂/∂R - 1/R ∂/∂Ph" ] }, "execution_count": 51, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi2.bracket(xi3).display()" ] }, { "cell_type": "code", "execution_count": 52, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}$$" ], "text/plain": [ "True" ] }, "execution_count": 52, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi2.bracket(xi3) == xi3" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "To summarize, we have" ] }, { "cell_type": "code", "execution_count": 53, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}$$" ], "text/plain": [ "True" ] }, "execution_count": 53, "metadata": {}, "output_type": "execute_result" } ], "source": [ "all([xi1.bracket(xi2) == xi1,\n", " xi1.bracket(xi3) == xi2,\n", " xi2.bracket(xi3) == xi3])" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "To recognize a standard Lie algebra, let us perform a slight change of basis:" ] }, { "cell_type": "code", "execution_count": 54, "metadata": {}, "outputs": [], "source": [ "vE = -sqrt(2)*xi3\n", "vF = sqrt(2)*xi1\n", "vH = 2*xi2" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We have then the following commutation relations:" ] }, { "cell_type": "code", "execution_count": 55, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}$$" ], "text/plain": [ "True" ] }, "execution_count": 55, "metadata": {}, "output_type": "execute_result" } ], "source": [ "all([vE.bracket(vF) == vH,\n", " vH.bracket(vE) == 2*vE,\n", " vH.bracket(vF) == -2*vF])" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We recognize the Lie algebra $\\mathfrak{sl}(2, \\mathbb{R})$. Indeed, we have, using the representation of $\\mathfrak{sl}(2, \\mathbb{R})$ by traceless $2\\times 2$ matrices:" ] }, { "cell_type": "code", "execution_count": 56, "metadata": {}, "outputs": [], "source": [ "sl2 = lie_algebras.sl(QQ, 2, representation='matrix') # QQ instead of RR to deal with an exact field\n", "E,F,H = sl2.gens()" ] }, { "cell_type": "code", "execution_count": 57, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\begin{array}{rr}\n", "0 & 1 \\\\\n", "0 & 0\n", "\\end{array}\\right)\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\begin{array}{rr}\n", "0 & 1 \\\\\n", "0 & 0\n", "\\end{array}\\right)$$" ], "text/plain": [ "[0 1]\n", "[0 0]" ] }, "execution_count": 57, "metadata": {}, "output_type": "execute_result" } ], "source": [ "E" ] }, { "cell_type": "code", "execution_count": 58, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\begin{array}{rr}\n", "0 & 0 \\\\\n", "1 & 0\n", "\\end{array}\\right)\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\begin{array}{rr}\n", "0 & 0 \\\\\n", "1 & 0\n", "\\end{array}\\right)$$" ], "text/plain": [ "[0 0]\n", "[1 0]" ] }, "execution_count": 58, "metadata": {}, "output_type": "execute_result" } ], "source": [ "F" ] }, { "cell_type": "code", "execution_count": 59, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\begin{array}{rr}\n", "1 & 0 \\\\\n", "0 & -1\n", "\\end{array}\\right)\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\begin{array}{rr}\n", "1 & 0 \\\\\n", "0 & -1\n", "\\end{array}\\right)$$" ], "text/plain": [ "[ 1 0]\n", "[ 0 -1]" ] }, "execution_count": 59, "metadata": {}, "output_type": "execute_result" } ], "source": [ "H" ] }, { "cell_type": "code", "execution_count": 60, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}$$" ], "text/plain": [ "True" ] }, "execution_count": 60, "metadata": {}, "output_type": "execute_result" } ], "source": [ "all([E.bracket(F) == H,\n", " H.bracket(E) == 2*E,\n", " H.bracket(F) == -2*F])" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Hence, we have $$\\mathrm{Lie}(G_3) = \\mathfrak{sl}(2, \\mathbb{R}).$$\n", "\n", "At this stage, $G_3$ could be $\\mathrm{SL}(2, \\mathbb{R})$, $\\mathrm{PSL}(2, \\mathbb{R})=\\mathrm{SL}(2, \\mathbb{R})/\\mathbb{Z}_2$\n", "or $\\overline{\\mathrm{SL}(2, \\mathbb{R})}$ (the universal covering group of $\\mathrm{SL}(2, \\mathbb{R})$). \n", "It cannot be $\\mathrm{PSL}(2, \\mathbb{R})$ because, as it appears clearly on $h$'s components,\n", "the transformation $(T,R) \\mapsto (-T,-R)$ is an element of $G_3$ and, in $\\mathrm{PSL}(2,\\mathbb{R})$, this element would be identified with the identity\n", "due to the quotient by $\\mathbb{Z}_2 = \\{\\mathrm{Id}, -\\mathrm{Id}\\}$ ($\\mathrm{Id}$ being the\n", "identity).\n", "$G_3$ is actually $\\mathrm{SL}(2, \\mathbb{R})$.\n", "We conclude that the full isometry group of the near-horizon geometry is \n", "$$G = \\mathrm{U}(1) \\times \\mathrm{SL}(2, \\mathbb{R}).$$" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Expression of $h$ and the near-horizon Killing vectors in the Boyer-Lindquist basis\n", "\n", "The Boyer-Lindquist components of the near-horizon metric $h$ are obtained by providing the argument BL to the method display:" ] }, { "cell_type": "code", "execution_count": 61, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}h = \\left( -\\frac{{\\left(m^{2} - 2 \\, m r + r^{2}\\right)} \\sin\\left({\\theta}\\right)^{4} - 4 \\, {\\left(5 \\, m^{2} - 6 \\, m r + 2 \\, r^{2}\\right)} \\sin\\left({\\theta}\\right)^{2} + 4 \\, m^{2} - 8 \\, m r + 4 \\, r^{2}}{4 \\, {\\left(m^{2} \\cos\\left({\\theta}\\right)^{2} + m^{2}\\right)}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( -\\frac{2 \\, {\\left(2 \\, m - r\\right)} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} t\\otimes \\mathrm{d} {\\phi} + \\left( -\\frac{m^{2} \\sin\\left({\\theta}\\right)^{2} - 2 \\, m^{2}}{m^{2} - 2 \\, m r + r^{2}} \\right) \\mathrm{d} r\\otimes \\mathrm{d} r + \\left( -m^{2} \\sin\\left({\\theta}\\right)^{2} + 2 \\, m^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( -\\frac{2 \\, {\\left(2 \\, m - r\\right)} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} t + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} {\\phi}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}h = \\left( -\\frac{{\\left(m^{2} - 2 \\, m r + r^{2}\\right)} \\sin\\left({\\theta}\\right)^{4} - 4 \\, {\\left(5 \\, m^{2} - 6 \\, m r + 2 \\, r^{2}\\right)} \\sin\\left({\\theta}\\right)^{2} + 4 \\, m^{2} - 8 \\, m r + 4 \\, r^{2}}{4 \\, {\\left(m^{2} \\cos\\left({\\theta}\\right)^{2} + m^{2}\\right)}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( -\\frac{2 \\, {\\left(2 \\, m - r\\right)} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} t\\otimes \\mathrm{d} {\\phi} + \\left( -\\frac{m^{2} \\sin\\left({\\theta}\\right)^{2} - 2 \\, m^{2}}{m^{2} - 2 \\, m r + r^{2}} \\right) \\mathrm{d} r\\otimes \\mathrm{d} r + \\left( -m^{2} \\sin\\left({\\theta}\\right)^{2} + 2 \\, m^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( -\\frac{2 \\, {\\left(2 \\, m - r\\right)} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} t + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} {\\phi}$$" ], "text/plain": [ "h = -1/4*((m^2 - 2*m*r + r^2)*sin(th)^4 - 4*(5*m^2 - 6*m*r + 2*r^2)*sin(th)^2 + 4*m^2 - 8*m*r + 4*r^2)/(m^2*cos(th)^2 + m^2) dt⊗dt - 2*(2*m - r)*sin(th)^2/(cos(th)^2 + 1) dt⊗dph - (m^2*sin(th)^2 - 2*m^2)/(m^2 - 2*m*r + r^2) dr⊗dr + (-m^2*sin(th)^2 + 2*m^2) dth⊗dth - 2*(2*m - r)*sin(th)^2/(cos(th)^2 + 1) dph⊗dt + 4*m^2*sin(th)^2/(cos(th)^2 + 1) dph⊗dph" ] }, "execution_count": 61, "metadata": {}, "output_type": "execute_result" } ], "source": [ "h.display(BL)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We note that the Boyer-Lindquist components of $h$ do not depend on the parameter $\\epsilon$, despite the change of coordinates NH $\\to$ BL does. " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We have established above that\n", "$$h = m^2 (1 + \\cos^2\\theta) s + q$$\n", "Let us evaluate the Boyer-Lindquist expressions of $s$ and $q$:" ] }, { "cell_type": "code", "execution_count": 62, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left( -\\frac{m^{2} - 2 \\, m r + r^{2}}{4 \\, m^{4}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( \\frac{1}{m^{2} - 2 \\, m r + r^{2}} \\right) \\mathrm{d} r\\otimes \\mathrm{d} r +\\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left( -\\frac{m^{2} - 2 \\, m r + r^{2}}{4 \\, m^{4}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( \\frac{1}{m^{2} - 2 \\, m r + r^{2}} \\right) \\mathrm{d} r\\otimes \\mathrm{d} r +\\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta}$$" ], "text/plain": [ "-1/4*(m^2 - 2*m*r + r^2)/m^4 dt⊗dt + 1/(m^2 - 2*m*r + r^2) dr⊗dr + dth⊗dth" ] }, "execution_count": 62, "metadata": {}, "output_type": "execute_result" } ], "source": [ "s.display(BL)" ] }, { "cell_type": "code", "execution_count": 63, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}-\\frac{{\\left(m - r\\right)}^{2}}{4 \\, m^{4}} \\mathrm{d} t\\otimes \\mathrm{d} t + \\frac{1}{{\\left(m - r\\right)}^{2}} \\mathrm{d} r\\otimes \\mathrm{d} r +\\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}-\\frac{{\\left(m - r\\right)}^{2}}{4 \\, m^{4}} \\mathrm{d} t\\otimes \\mathrm{d} t + \\frac{1}{{\\left(m - r\\right)}^{2}} \\mathrm{d} r\\otimes \\mathrm{d} r +\\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta}$$" ], "text/plain": [ "-1/4*(m - r)^2/m^4 dt⊗dt + (m - r)^(-2) dr⊗dr + dth⊗dth" ] }, "execution_count": 63, "metadata": {}, "output_type": "execute_result" } ], "source": [ "s.apply_map(factor, frame=BL.frame(), chart=BL, keep_other_components=True)\n", "s.display(BL)" ] }, { "cell_type": "code", "execution_count": 64, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left( \\frac{{\\left(4 \\, m^{2} - 4 \\, m r + r^{2}\\right)} \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + m^{2}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( -\\frac{2 \\, {\\left(2 \\, m - r\\right)} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} t\\otimes \\mathrm{d} {\\phi} + \\left( -\\frac{2 \\, {\\left(2 \\, m - r\\right)} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} t + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} {\\phi}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left( \\frac{{\\left(4 \\, m^{2} - 4 \\, m r + r^{2}\\right)} \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + m^{2}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( -\\frac{2 \\, {\\left(2 \\, m - r\\right)} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} t\\otimes \\mathrm{d} {\\phi} + \\left( -\\frac{2 \\, {\\left(2 \\, m - r\\right)} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} t + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} {\\phi}$$" ], "text/plain": [ "(4*m^2 - 4*m*r + r^2)*sin(th)^2/(m^2*cos(th)^2 + m^2) dt⊗dt - 2*(2*m - r)*sin(th)^2/(cos(th)^2 + 1) dt⊗dph - 2*(2*m - r)*sin(th)^2/(cos(th)^2 + 1) dph⊗dt + 4*m^2*sin(th)^2/(cos(th)^2 + 1) dph⊗dph" ] }, "execution_count": 64, "metadata": {}, "output_type": "execute_result" } ], "source": [ "q.display(BL)" ] }, { "cell_type": "code", "execution_count": 65, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{{\\left(2 \\, m - r\\right)}^{2} \\sin\\left({\\theta}\\right)^{2}}{{\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}} \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( -\\frac{2 \\, {\\left(2 \\, m - r\\right)} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} t\\otimes \\mathrm{d} {\\phi} + \\left( -\\frac{2 \\, {\\left(2 \\, m - r\\right)} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} t + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} {\\phi}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{{\\left(2 \\, m - r\\right)}^{2} \\sin\\left({\\theta}\\right)^{2}}{{\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}} \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( -\\frac{2 \\, {\\left(2 \\, m - r\\right)} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} t\\otimes \\mathrm{d} {\\phi} + \\left( -\\frac{2 \\, {\\left(2 \\, m - r\\right)} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} t + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} {\\phi}$$" ], "text/plain": [ "(2*m - r)^2*sin(th)^2/((cos(th)^2 + 1)*m^2) dt⊗dt - 2*(2*m - r)*sin(th)^2/(cos(th)^2 + 1) dt⊗dph - 2*(2*m - r)*sin(th)^2/(cos(th)^2 + 1) dph⊗dt + 4*m^2*sin(th)^2/(cos(th)^2 + 1) dph⊗dph" ] }, "execution_count": 65, "metadata": {}, "output_type": "execute_result" } ], "source": [ "q.apply_map(factor, frame=BL.frame(), chart=BL, keep_other_components=True)\n", "q.display(BL)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Hence the identity\n", "$$q = \\frac{\\sin^2\\theta}{1 + \\cos^2\\theta} \\left( \\frac{r - 2m}{m}\\mathrm{d}t + 2 m\\, \\mathrm{d}\\phi \\right)^2$$" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Near-horizon Killing vectors in terms of Boyer-Lindquist coordinates" ] }, { "cell_type": "code", "execution_count": 66, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\eta = \\frac{\\partial}{\\partial {\\phi} }\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\eta = \\frac{\\partial}{\\partial {\\phi} }$$" ], "text/plain": [ "eta = ∂/∂ph" ] }, "execution_count": 66, "metadata": {}, "output_type": "execute_result" } ], "source": [ "eta.display(BL)" ] }, { "cell_type": "code", "execution_count": 67, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\xi_{1} = \\frac{2 \\, m}{{\\epsilon}} \\frac{\\partial}{\\partial t } + \\frac{1}{{\\epsilon}} \\frac{\\partial}{\\partial {\\phi} }\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\xi_{1} = \\frac{2 \\, m}{{\\epsilon}} \\frac{\\partial}{\\partial t } + \\frac{1}{{\\epsilon}} \\frac{\\partial}{\\partial {\\phi} }$$" ], "text/plain": [ "xi1 = 2*m/eps ∂/∂t + 1/eps ∂/∂ph" ] }, "execution_count": 67, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi1.display(BL)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Given that for an extremal Kerr black hole $\\Omega_H = 1/(2m)$, we recognize that \n", "$$\\xi_1 = \\frac{2m}{\\epsilon} \\chi$$\n", "where $\\chi := \\frac{\\partial}{\\partial t} + \\Omega_H \\frac{\\partial}{\\partial \\phi}$ is the\n", "Killing vector that is tangent to the horizon null generators on $\\mathcal{H}$, or equivalently that is the null normal of the Killing horizon $\\mathcal{H}$. " ] }, { "cell_type": "code", "execution_count": 68, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\xi_{2} = t \\frac{\\partial}{\\partial t } + \\left( m - r \\right) \\frac{\\partial}{\\partial r } + \\frac{t}{2 \\, m} \\frac{\\partial}{\\partial {\\phi} }\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\xi_{2} = t \\frac{\\partial}{\\partial t } + \\left( m - r \\right) \\frac{\\partial}{\\partial r } + \\frac{t}{2 \\, m} \\frac{\\partial}{\\partial {\\phi} }$$" ], "text/plain": [ "xi2 = t ∂/∂t + (m - r) ∂/∂r + 1/2*t/m ∂/∂ph" ] }, "execution_count": 68, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi2.display(BL)" ] }, { "cell_type": "code", "execution_count": 69, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\xi_{3} = \\left( \\frac{4 \\, {\\epsilon} m^{4} + {\\left({\\epsilon} m^{2} - 2 \\, {\\epsilon} m r + {\\epsilon} r^{2}\\right)} t^{2}}{4 \\, {\\left(m^{3} - 2 \\, m^{2} r + m r^{2}\\right)}} \\right) \\frac{\\partial}{\\partial t } + \\frac{{\\left({\\epsilon} m - {\\epsilon} r\\right)} t}{2 \\, m} \\frac{\\partial}{\\partial r } + \\left( \\frac{12 \\, {\\epsilon} m^{4} - 8 \\, {\\epsilon} m^{3} r + {\\left({\\epsilon} m^{2} - 2 \\, {\\epsilon} m r + {\\epsilon} r^{2}\\right)} t^{2}}{8 \\, {\\left(m^{4} - 2 \\, m^{3} r + m^{2} r^{2}\\right)}} \\right) \\frac{\\partial}{\\partial {\\phi} }\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\xi_{3} = \\left( \\frac{4 \\, {\\epsilon} m^{4} + {\\left({\\epsilon} m^{2} - 2 \\, {\\epsilon} m r + {\\epsilon} r^{2}\\right)} t^{2}}{4 \\, {\\left(m^{3} - 2 \\, m^{2} r + m r^{2}\\right)}} \\right) \\frac{\\partial}{\\partial t } + \\frac{{\\left({\\epsilon} m - {\\epsilon} r\\right)} t}{2 \\, m} \\frac{\\partial}{\\partial r } + \\left( \\frac{12 \\, {\\epsilon} m^{4} - 8 \\, {\\epsilon} m^{3} r + {\\left({\\epsilon} m^{2} - 2 \\, {\\epsilon} m r + {\\epsilon} r^{2}\\right)} t^{2}}{8 \\, {\\left(m^{4} - 2 \\, m^{3} r + m^{2} r^{2}\\right)}} \\right) \\frac{\\partial}{\\partial {\\phi} }$$" ], "text/plain": [ "xi3 = 1/4*(4*eps*m^4 + (eps*m^2 - 2*eps*m*r + eps*r^2)*t^2)/(m^3 - 2*m^2*r + m*r^2) ∂/∂t + 1/2*(eps*m - eps*r)*t/m ∂/∂r + 1/8*(12*eps*m^4 - 8*eps*m^3*r + (eps*m^2 - 2*eps*m*r + eps*r^2)*t^2)/(m^4 - 2*m^3*r + m^2*r^2) ∂/∂ph" ] }, "execution_count": 69, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi3.display(BL)" ] }, { "cell_type": "code", "execution_count": 70, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\xi_{3} = \\frac{{\\left(4 \\, m^{4} + m^{2} t^{2} - 2 \\, m r t^{2} + r^{2} t^{2}\\right)} {\\epsilon}}{4 \\, {\\left(m - r\\right)}^{2} m} \\frac{\\partial}{\\partial t } + \\frac{{\\epsilon} {\\left(m - r\\right)} t}{2 \\, m} \\frac{\\partial}{\\partial r } + \\frac{{\\left(12 \\, m^{4} - 8 \\, m^{3} r + m^{2} t^{2} - 2 \\, m r t^{2} + r^{2} t^{2}\\right)} {\\epsilon}}{8 \\, {\\left(m - r\\right)}^{2} m^{2}} \\frac{\\partial}{\\partial {\\phi} }\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\xi_{3} = \\frac{{\\left(4 \\, m^{4} + m^{2} t^{2} - 2 \\, m r t^{2} + r^{2} t^{2}\\right)} {\\epsilon}}{4 \\, {\\left(m - r\\right)}^{2} m} \\frac{\\partial}{\\partial t } + \\frac{{\\epsilon} {\\left(m - r\\right)} t}{2 \\, m} \\frac{\\partial}{\\partial r } + \\frac{{\\left(12 \\, m^{4} - 8 \\, m^{3} r + m^{2} t^{2} - 2 \\, m r t^{2} + r^{2} t^{2}\\right)} {\\epsilon}}{8 \\, {\\left(m - r\\right)}^{2} m^{2}} \\frac{\\partial}{\\partial {\\phi} }$$" ], "text/plain": [ "xi3 = 1/4*(4*m^4 + m^2*t^2 - 2*m*r*t^2 + r^2*t^2)*eps/((m - r)^2*m) ∂/∂t + 1/2*eps*(m - r)*t/m ∂/∂r + 1/8*(12*m^4 - 8*m^3*r + m^2*t^2 - 2*m*r*t^2 + r^2*t^2)*eps/((m - r)^2*m^2) ∂/∂ph" ] }, "execution_count": 70, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi3.apply_map(factor, frame=BL.frame(), chart=BL, keep_other_components=True)\n", "xi3.display(BL)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We note that the BL components of $\\xi_2$ are independent from $\\epsilon$, while those of $\\xi_1$ and $\\xi_3$ are, with $\\xi_1$ diverging when $\\epsilon\\to 0$ and $\\xi_3$ vanishing when $\\epsilon\\to 0$." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### The Killing operator of $g$ applied to the near-horizon Killing vectors" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "In this part, we set the default chart and frame back to the Boyer-Lindquist ones:" ] }, { "cell_type": "code", "execution_count": 71, "metadata": {}, "outputs": [], "source": [ "M.set_default_chart(BL)\n", "M.set_default_frame(BL.frame())" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "The Levi-Civita connection $\\nabla$ of $g$:" ] }, { "cell_type": "code", "execution_count": 72, "metadata": {}, "outputs": [], "source": [ "nabla = g.connection()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "The **Killing form** $\\mathrm{K}v$ of a vector field $v$ is defined by\n", "$$(\\mathrm{K}v)_{ab} := \\nabla_a v_b + \\nabla_b v_a,$$\n", "where $v_a := g_{ab} v^b$. We implement it in SageMath via the following function:" ] }, { "cell_type": "code", "execution_count": 73, "metadata": {}, "outputs": [], "source": [ "def killing_form(v):\n", " Kv = 2*nabla(v.down(g)).symmetrize()\n", " if v._name is not None:\n", " Kv.set_name('K' + v._name, latex_name=r'\\mathrm{K}' + str(latex(v)))\n", " return Kv" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "The Killing form of $\\eta$ is identically zero:" ] }, { "cell_type": "code", "execution_count": 74, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Field of symmetric bilinear forms Keta on the 4-dimensional Lorentzian manifold M\n" ] }, { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{K}\\eta = 0\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{K}\\eta = 0$$" ], "text/plain": [ "Keta = 0" ] }, "execution_count": 74, "metadata": {}, "output_type": "execute_result" } ], "source": [ "Keta = killing_form(eta)\n", "print(Keta)\n", "Keta.display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We recover the fact that $\\eta$ is a Killing vector of $g$. Similarly, we have " ] }, { "cell_type": "code", "execution_count": 75, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Field of symmetric bilinear forms Kxi1 on the 4-dimensional Lorentzian manifold M\n" ] }, { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{K}\\xi_{1} = 0\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{K}\\xi_{1} = 0$$" ], "text/plain": [ "Kxi1 = 0" ] }, "execution_count": 75, "metadata": {}, "output_type": "execute_result" } ], "source": [ "Kxi1 = killing_form(xi1)\n", "print(Kxi1)\n", "Kxi1.display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We recover the fact that $\\xi_1 = (2m/\\epsilon) \\chi$ is a Killing vector of $g$." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "But $\\xi_2$ and $\\xi_3$ are not Killing vectors of $g$:" ] }, { "cell_type": "code", "execution_count": 76, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Field of symmetric bilinear forms Kxi2 on the 4-dimensional Lorentzian manifold M\n" ] }, { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{K}\\xi_{2} = \\left( -\\frac{2 \\, {\\left({\\left(m^{4} - m^{3} r\\right)} \\cos\\left({\\theta}\\right)^{4} + m^{2} r^{2} - 2 \\, m r^{3} + r^{4} - {\\left(m^{4} - 2 \\, m^{2} r^{2} + m r^{3}\\right)} \\cos\\left({\\theta}\\right)^{2}\\right)}}{m^{4} \\cos\\left({\\theta}\\right)^{4} + 2 \\, m^{2} r^{2} \\cos\\left({\\theta}\\right)^{2} + r^{4}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( \\frac{{\\left(m^{6} - 2 \\, m^{5} r + m^{4} r^{2}\\right)} \\sin\\left({\\theta}\\right)^{6} + 2 \\, {\\left(m^{6} + m^{5} r - 2 \\, m^{4} r^{2} + m^{3} r^{3} - m^{2} r^{4}\\right)} \\sin\\left({\\theta}\\right)^{4} - {\\left(3 \\, m^{6} - 7 \\, m^{4} r^{2} + 8 \\, m^{3} r^{3} - 3 \\, m^{2} r^{4} - r^{6}\\right)} \\sin\\left({\\theta}\\right)^{2}}{2 \\, {\\left(m^{5} \\cos\\left({\\theta}\\right)^{4} + 2 \\, m^{3} r^{2} \\cos\\left({\\theta}\\right)^{2} + m r^{4}\\right)}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} {\\phi} + \\left( \\frac{2 \\, r}{m - r} \\right) \\mathrm{d} r\\otimes \\mathrm{d} r + \\left( 2 \\, m r - 2 \\, r^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( \\frac{{\\left(m^{6} - 2 \\, m^{5} r + m^{4} r^{2}\\right)} \\sin\\left({\\theta}\\right)^{6} + 2 \\, {\\left(m^{6} + m^{5} r - 2 \\, m^{4} r^{2} + m^{3} r^{3} - m^{2} r^{4}\\right)} \\sin\\left({\\theta}\\right)^{4} - {\\left(3 \\, m^{6} - 7 \\, m^{4} r^{2} + 8 \\, m^{3} r^{3} - 3 \\, m^{2} r^{4} - r^{6}\\right)} \\sin\\left({\\theta}\\right)^{2}}{2 \\, {\\left(m^{5} \\cos\\left({\\theta}\\right)^{4} + 2 \\, m^{3} r^{2} \\cos\\left({\\theta}\\right)^{2} + m r^{4}\\right)}} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} t + \\left( -\\frac{2 \\, {\\left({\\left(m^{4} r^{2} - m^{3} r^{3} - {\\left(m^{6} - m^{5} r\\right)} \\cos\\left({\\theta}\\right)^{2}\\right)} \\sin\\left({\\theta}\\right)^{4} - {\\left(m r^{5} - r^{6} + {\\left(m^{5} r - m^{4} r^{2}\\right)} \\cos\\left({\\theta}\\right)^{4} + 2 \\, {\\left(m^{3} r^{3} - m^{2} r^{4}\\right)} \\cos\\left({\\theta}\\right)^{2}\\right)} \\sin\\left({\\theta}\\right)^{2}\\right)}}{m^{4} \\cos\\left({\\theta}\\right)^{4} + 2 \\, m^{2} r^{2} \\cos\\left({\\theta}\\right)^{2} + r^{4}} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} {\\phi}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{K}\\xi_{2} = \\left( -\\frac{2 \\, {\\left({\\left(m^{4} - m^{3} r\\right)} \\cos\\left({\\theta}\\right)^{4} + m^{2} r^{2} - 2 \\, m r^{3} + r^{4} - {\\left(m^{4} - 2 \\, m^{2} r^{2} + m r^{3}\\right)} \\cos\\left({\\theta}\\right)^{2}\\right)}}{m^{4} \\cos\\left({\\theta}\\right)^{4} + 2 \\, m^{2} r^{2} \\cos\\left({\\theta}\\right)^{2} + r^{4}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( \\frac{{\\left(m^{6} - 2 \\, m^{5} r + m^{4} r^{2}\\right)} \\sin\\left({\\theta}\\right)^{6} + 2 \\, {\\left(m^{6} + m^{5} r - 2 \\, m^{4} r^{2} + m^{3} r^{3} - m^{2} r^{4}\\right)} \\sin\\left({\\theta}\\right)^{4} - {\\left(3 \\, m^{6} - 7 \\, m^{4} r^{2} + 8 \\, m^{3} r^{3} - 3 \\, m^{2} r^{4} - r^{6}\\right)} \\sin\\left({\\theta}\\right)^{2}}{2 \\, {\\left(m^{5} \\cos\\left({\\theta}\\right)^{4} + 2 \\, m^{3} r^{2} \\cos\\left({\\theta}\\right)^{2} + m r^{4}\\right)}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} {\\phi} + \\left( \\frac{2 \\, r}{m - r} \\right) \\mathrm{d} r\\otimes \\mathrm{d} r + \\left( 2 \\, m r - 2 \\, r^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( \\frac{{\\left(m^{6} - 2 \\, m^{5} r + m^{4} r^{2}\\right)} \\sin\\left({\\theta}\\right)^{6} + 2 \\, {\\left(m^{6} + m^{5} r - 2 \\, m^{4} r^{2} + m^{3} r^{3} - m^{2} r^{4}\\right)} \\sin\\left({\\theta}\\right)^{4} - {\\left(3 \\, m^{6} - 7 \\, m^{4} r^{2} + 8 \\, m^{3} r^{3} - 3 \\, m^{2} r^{4} - r^{6}\\right)} \\sin\\left({\\theta}\\right)^{2}}{2 \\, {\\left(m^{5} \\cos\\left({\\theta}\\right)^{4} + 2 \\, m^{3} r^{2} \\cos\\left({\\theta}\\right)^{2} + m r^{4}\\right)}} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} t + \\left( -\\frac{2 \\, {\\left({\\left(m^{4} r^{2} - m^{3} r^{3} - {\\left(m^{6} - m^{5} r\\right)} \\cos\\left({\\theta}\\right)^{2}\\right)} \\sin\\left({\\theta}\\right)^{4} - {\\left(m r^{5} - r^{6} + {\\left(m^{5} r - m^{4} r^{2}\\right)} \\cos\\left({\\theta}\\right)^{4} + 2 \\, {\\left(m^{3} r^{3} - m^{2} r^{4}\\right)} \\cos\\left({\\theta}\\right)^{2}\\right)} \\sin\\left({\\theta}\\right)^{2}\\right)}}{m^{4} \\cos\\left({\\theta}\\right)^{4} + 2 \\, m^{2} r^{2} \\cos\\left({\\theta}\\right)^{2} + r^{4}} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} {\\phi}$$" ], "text/plain": [ "Kxi2 = -2*((m^4 - m^3*r)*cos(th)^4 + m^2*r^2 - 2*m*r^3 + r^4 - (m^4 - 2*m^2*r^2 + m*r^3)*cos(th)^2)/(m^4*cos(th)^4 + 2*m^2*r^2*cos(th)^2 + r^4) dt⊗dt + 1/2*((m^6 - 2*m^5*r + m^4*r^2)*sin(th)^6 + 2*(m^6 + m^5*r - 2*m^4*r^2 + m^3*r^3 - m^2*r^4)*sin(th)^4 - (3*m^6 - 7*m^4*r^2 + 8*m^3*r^3 - 3*m^2*r^4 - r^6)*sin(th)^2)/(m^5*cos(th)^4 + 2*m^3*r^2*cos(th)^2 + m*r^4) dt⊗dph + 2*r/(m - r) dr⊗dr + (2*m*r - 2*r^2) dth⊗dth + 1/2*((m^6 - 2*m^5*r + m^4*r^2)*sin(th)^6 + 2*(m^6 + m^5*r - 2*m^4*r^2 + m^3*r^3 - m^2*r^4)*sin(th)^4 - (3*m^6 - 7*m^4*r^2 + 8*m^3*r^3 - 3*m^2*r^4 - r^6)*sin(th)^2)/(m^5*cos(th)^4 + 2*m^3*r^2*cos(th)^2 + m*r^4) dph⊗dt - 2*((m^4*r^2 - m^3*r^3 - (m^6 - m^5*r)*cos(th)^2)*sin(th)^4 - (m*r^5 - r^6 + (m^5*r - m^4*r^2)*cos(th)^4 + 2*(m^3*r^3 - m^2*r^4)*cos(th)^2)*sin(th)^2)/(m^4*cos(th)^4 + 2*m^2*r^2*cos(th)^2 + r^4) dph⊗dph" ] }, "execution_count": 76, "metadata": {}, "output_type": "execute_result" } ], "source": [ "Kxi2 = killing_form(xi2)\n", "print(Kxi2)\n", "Kxi2.display()" ] }, { "cell_type": "code", "execution_count": 77, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Field of symmetric bilinear forms Kxi3 on the 4-dimensional Lorentzian manifold M\n" ] }, { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{K}\\xi_{3} = \\left( -\\frac{{\\left({\\epsilon} m^{4} - {\\epsilon} m^{3} r\\right)} t \\cos\\left({\\theta}\\right)^{4} - {\\left({\\epsilon} m^{4} - 2 \\, {\\epsilon} m^{2} r^{2} + {\\epsilon} m r^{3}\\right)} t \\cos\\left({\\theta}\\right)^{2} + {\\left({\\epsilon} m^{2} r^{2} - 2 \\, {\\epsilon} m r^{3} + {\\epsilon} r^{4}\\right)} t}{m^{5} \\cos\\left({\\theta}\\right)^{4} + 2 \\, m^{3} r^{2} \\cos\\left({\\theta}\\right)^{2} + m r^{4}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( \\frac{{\\epsilon} m^{4} \\cos\\left({\\theta}\\right)^{4} + {\\epsilon} r^{4} - 2 \\, {\\left(2 \\, {\\epsilon} m^{4} - {\\epsilon} m^{2} r^{2}\\right)} \\cos\\left({\\theta}\\right)^{2}}{2 \\, {\\left(m^{2} r^{2} - m r^{3} + {\\left(m^{4} - m^{3} r\\right)} \\cos\\left({\\theta}\\right)^{2}\\right)}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} r + \\left( \\frac{{\\left({\\epsilon} m^{6} - 2 \\, {\\epsilon} m^{5} r + {\\epsilon} m^{4} r^{2}\\right)} t \\sin\\left({\\theta}\\right)^{6} + 2 \\, {\\left({\\epsilon} m^{6} + {\\epsilon} m^{5} r - 2 \\, {\\epsilon} m^{4} r^{2} + {\\epsilon} m^{3} r^{3} - {\\epsilon} m^{2} r^{4}\\right)} t \\sin\\left({\\theta}\\right)^{4} - {\\left(3 \\, {\\epsilon} m^{6} - 7 \\, {\\epsilon} m^{4} r^{2} + 8 \\, {\\epsilon} m^{3} r^{3} - 3 \\, {\\epsilon} m^{2} r^{4} - {\\epsilon} r^{6}\\right)} t \\sin\\left({\\theta}\\right)^{2}}{4 \\, {\\left(m^{6} \\cos\\left({\\theta}\\right)^{4} + 2 \\, m^{4} r^{2} \\cos\\left({\\theta}\\right)^{2} + m^{2} r^{4}\\right)}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} {\\phi} + \\left( \\frac{{\\epsilon} m^{4} \\cos\\left({\\theta}\\right)^{4} + {\\epsilon} r^{4} - 2 \\, {\\left(2 \\, {\\epsilon} m^{4} - {\\epsilon} m^{2} r^{2}\\right)} \\cos\\left({\\theta}\\right)^{2}}{2 \\, {\\left(m^{2} r^{2} - m r^{3} + {\\left(m^{4} - m^{3} r\\right)} \\cos\\left({\\theta}\\right)^{2}\\right)}} \\right) \\mathrm{d} r\\otimes \\mathrm{d} t + \\left( \\frac{{\\epsilon} r t}{m^{2} - m r} \\right) \\mathrm{d} r\\otimes \\mathrm{d} r + \\left( -\\frac{{\\left(2 \\, {\\epsilon} m^{4} - {\\epsilon} m^{3} r\\right)} \\sin\\left({\\theta}\\right)^{4} - {\\left(2 \\, {\\epsilon} m^{4} - {\\epsilon} m^{3} r - {\\epsilon} m r^{3}\\right)} \\sin\\left({\\theta}\\right)^{2}}{m r^{2} - r^{3} + {\\left(m^{3} - m^{2} r\\right)} \\cos\\left({\\theta}\\right)^{2}} \\right) \\mathrm{d} r\\otimes \\mathrm{d} {\\phi} + \\frac{{\\left({\\epsilon} m r - {\\epsilon} r^{2}\\right)} t}{m} \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( \\frac{{\\left({\\epsilon} m^{6} - 2 \\, {\\epsilon} m^{5} r + {\\epsilon} m^{4} r^{2}\\right)} t \\sin\\left({\\theta}\\right)^{6} + 2 \\, {\\left({\\epsilon} m^{6} + {\\epsilon} m^{5} r - 2 \\, {\\epsilon} m^{4} r^{2} + {\\epsilon} m^{3} r^{3} - {\\epsilon} m^{2} r^{4}\\right)} t \\sin\\left({\\theta}\\right)^{4} - {\\left(3 \\, {\\epsilon} m^{6} - 7 \\, {\\epsilon} m^{4} r^{2} + 8 \\, {\\epsilon} m^{3} r^{3} - 3 \\, {\\epsilon} m^{2} r^{4} - {\\epsilon} r^{6}\\right)} t \\sin\\left({\\theta}\\right)^{2}}{4 \\, {\\left(m^{6} \\cos\\left({\\theta}\\right)^{4} + 2 \\, m^{4} r^{2} \\cos\\left({\\theta}\\right)^{2} + m^{2} r^{4}\\right)}} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} t + \\left( -\\frac{{\\left(2 \\, {\\epsilon} m^{4} - {\\epsilon} m^{3} r\\right)} \\sin\\left({\\theta}\\right)^{4} - {\\left(2 \\, {\\epsilon} m^{4} - {\\epsilon} m^{3} r - {\\epsilon} m r^{3}\\right)} \\sin\\left({\\theta}\\right)^{2}}{m r^{2} - r^{3} + {\\left(m^{3} - m^{2} r\\right)} \\cos\\left({\\theta}\\right)^{2}} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} r + \\left( \\frac{{\\left({\\left({\\epsilon} m^{6} - {\\epsilon} m^{5} r\\right)} t \\cos\\left({\\theta}\\right)^{2} - {\\left({\\epsilon} m^{4} r^{2} - {\\epsilon} m^{3} r^{3}\\right)} t\\right)} \\sin\\left({\\theta}\\right)^{4} + {\\left({\\left({\\epsilon} m^{5} r - {\\epsilon} m^{4} r^{2}\\right)} t \\cos\\left({\\theta}\\right)^{4} + 2 \\, {\\left({\\epsilon} m^{3} r^{3} - {\\epsilon} m^{2} r^{4}\\right)} t \\cos\\left({\\theta}\\right)^{2} + {\\left({\\epsilon} m r^{5} - {\\epsilon} r^{6}\\right)} t\\right)} \\sin\\left({\\theta}\\right)^{2}}{m^{5} \\cos\\left({\\theta}\\right)^{4} + 2 \\, m^{3} r^{2} \\cos\\left({\\theta}\\right)^{2} + m r^{4}} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} {\\phi}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{K}\\xi_{3} = \\left( -\\frac{{\\left({\\epsilon} m^{4} - {\\epsilon} m^{3} r\\right)} t \\cos\\left({\\theta}\\right)^{4} - {\\left({\\epsilon} m^{4} - 2 \\, {\\epsilon} m^{2} r^{2} + {\\epsilon} m r^{3}\\right)} t \\cos\\left({\\theta}\\right)^{2} + {\\left({\\epsilon} m^{2} r^{2} - 2 \\, {\\epsilon} m r^{3} + {\\epsilon} r^{4}\\right)} t}{m^{5} \\cos\\left({\\theta}\\right)^{4} + 2 \\, m^{3} r^{2} \\cos\\left({\\theta}\\right)^{2} + m r^{4}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( \\frac{{\\epsilon} m^{4} \\cos\\left({\\theta}\\right)^{4} + {\\epsilon} r^{4} - 2 \\, {\\left(2 \\, {\\epsilon} m^{4} - {\\epsilon} m^{2} r^{2}\\right)} \\cos\\left({\\theta}\\right)^{2}}{2 \\, {\\left(m^{2} r^{2} - m r^{3} + {\\left(m^{4} - m^{3} r\\right)} \\cos\\left({\\theta}\\right)^{2}\\right)}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} r + \\left( \\frac{{\\left({\\epsilon} m^{6} - 2 \\, {\\epsilon} m^{5} r + {\\epsilon} m^{4} r^{2}\\right)} t \\sin\\left({\\theta}\\right)^{6} + 2 \\, {\\left({\\epsilon} m^{6} + {\\epsilon} m^{5} r - 2 \\, {\\epsilon} m^{4} r^{2} + {\\epsilon} m^{3} r^{3} - {\\epsilon} m^{2} r^{4}\\right)} t \\sin\\left({\\theta}\\right)^{4} - {\\left(3 \\, {\\epsilon} m^{6} - 7 \\, {\\epsilon} m^{4} r^{2} + 8 \\, {\\epsilon} m^{3} r^{3} - 3 \\, {\\epsilon} m^{2} r^{4} - {\\epsilon} r^{6}\\right)} t \\sin\\left({\\theta}\\right)^{2}}{4 \\, {\\left(m^{6} \\cos\\left({\\theta}\\right)^{4} + 2 \\, m^{4} r^{2} \\cos\\left({\\theta}\\right)^{2} + m^{2} r^{4}\\right)}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} {\\phi} + \\left( \\frac{{\\epsilon} m^{4} \\cos\\left({\\theta}\\right)^{4} + {\\epsilon} r^{4} - 2 \\, {\\left(2 \\, {\\epsilon} m^{4} - {\\epsilon} m^{2} r^{2}\\right)} \\cos\\left({\\theta}\\right)^{2}}{2 \\, {\\left(m^{2} r^{2} - m r^{3} + {\\left(m^{4} - m^{3} r\\right)} \\cos\\left({\\theta}\\right)^{2}\\right)}} \\right) \\mathrm{d} r\\otimes \\mathrm{d} t + \\left( \\frac{{\\epsilon} r t}{m^{2} - m r} \\right) \\mathrm{d} r\\otimes \\mathrm{d} r + \\left( -\\frac{{\\left(2 \\, {\\epsilon} m^{4} - {\\epsilon} m^{3} r\\right)} \\sin\\left({\\theta}\\right)^{4} - {\\left(2 \\, {\\epsilon} m^{4} - {\\epsilon} m^{3} r - {\\epsilon} m r^{3}\\right)} \\sin\\left({\\theta}\\right)^{2}}{m r^{2} - r^{3} + {\\left(m^{3} - m^{2} r\\right)} \\cos\\left({\\theta}\\right)^{2}} \\right) \\mathrm{d} r\\otimes \\mathrm{d} {\\phi} + \\frac{{\\left({\\epsilon} m r - {\\epsilon} r^{2}\\right)} t}{m} \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( \\frac{{\\left({\\epsilon} m^{6} - 2 \\, {\\epsilon} m^{5} r + {\\epsilon} m^{4} r^{2}\\right)} t \\sin\\left({\\theta}\\right)^{6} + 2 \\, {\\left({\\epsilon} m^{6} + {\\epsilon} m^{5} r - 2 \\, {\\epsilon} m^{4} r^{2} + {\\epsilon} m^{3} r^{3} - {\\epsilon} m^{2} r^{4}\\right)} t \\sin\\left({\\theta}\\right)^{4} - {\\left(3 \\, {\\epsilon} m^{6} - 7 \\, {\\epsilon} m^{4} r^{2} + 8 \\, {\\epsilon} m^{3} r^{3} - 3 \\, {\\epsilon} m^{2} r^{4} - {\\epsilon} r^{6}\\right)} t \\sin\\left({\\theta}\\right)^{2}}{4 \\, {\\left(m^{6} \\cos\\left({\\theta}\\right)^{4} + 2 \\, m^{4} r^{2} \\cos\\left({\\theta}\\right)^{2} + m^{2} r^{4}\\right)}} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} t + \\left( -\\frac{{\\left(2 \\, {\\epsilon} m^{4} - {\\epsilon} m^{3} r\\right)} \\sin\\left({\\theta}\\right)^{4} - {\\left(2 \\, {\\epsilon} m^{4} - {\\epsilon} m^{3} r - {\\epsilon} m r^{3}\\right)} \\sin\\left({\\theta}\\right)^{2}}{m r^{2} - r^{3} + {\\left(m^{3} - m^{2} r\\right)} \\cos\\left({\\theta}\\right)^{2}} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} r + \\left( \\frac{{\\left({\\left({\\epsilon} m^{6} - {\\epsilon} m^{5} r\\right)} t \\cos\\left({\\theta}\\right)^{2} - {\\left({\\epsilon} m^{4} r^{2} - {\\epsilon} m^{3} r^{3}\\right)} t\\right)} \\sin\\left({\\theta}\\right)^{4} + {\\left({\\left({\\epsilon} m^{5} r - {\\epsilon} m^{4} r^{2}\\right)} t \\cos\\left({\\theta}\\right)^{4} + 2 \\, {\\left({\\epsilon} m^{3} r^{3} - {\\epsilon} m^{2} r^{4}\\right)} t \\cos\\left({\\theta}\\right)^{2} + {\\left({\\epsilon} m r^{5} - {\\epsilon} r^{6}\\right)} t\\right)} \\sin\\left({\\theta}\\right)^{2}}{m^{5} \\cos\\left({\\theta}\\right)^{4} + 2 \\, m^{3} r^{2} \\cos\\left({\\theta}\\right)^{2} + m r^{4}} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} {\\phi}$$" ], "text/plain": [ "Kxi3 = -((eps*m^4 - eps*m^3*r)*t*cos(th)^4 - (eps*m^4 - 2*eps*m^2*r^2 + eps*m*r^3)*t*cos(th)^2 + (eps*m^2*r^2 - 2*eps*m*r^3 + eps*r^4)*t)/(m^5*cos(th)^4 + 2*m^3*r^2*cos(th)^2 + m*r^4) dt⊗dt + 1/2*(eps*m^4*cos(th)^4 + eps*r^4 - 2*(2*eps*m^4 - eps*m^2*r^2)*cos(th)^2)/(m^2*r^2 - m*r^3 + (m^4 - m^3*r)*cos(th)^2) dt⊗dr + 1/4*((eps*m^6 - 2*eps*m^5*r + eps*m^4*r^2)*t*sin(th)^6 + 2*(eps*m^6 + eps*m^5*r - 2*eps*m^4*r^2 + eps*m^3*r^3 - eps*m^2*r^4)*t*sin(th)^4 - (3*eps*m^6 - 7*eps*m^4*r^2 + 8*eps*m^3*r^3 - 3*eps*m^2*r^4 - eps*r^6)*t*sin(th)^2)/(m^6*cos(th)^4 + 2*m^4*r^2*cos(th)^2 + m^2*r^4) dt⊗dph + 1/2*(eps*m^4*cos(th)^4 + eps*r^4 - 2*(2*eps*m^4 - eps*m^2*r^2)*cos(th)^2)/(m^2*r^2 - m*r^3 + (m^4 - m^3*r)*cos(th)^2) dr⊗dt + eps*r*t/(m^2 - m*r) dr⊗dr - ((2*eps*m^4 - eps*m^3*r)*sin(th)^4 - (2*eps*m^4 - eps*m^3*r - eps*m*r^3)*sin(th)^2)/(m*r^2 - r^3 + (m^3 - m^2*r)*cos(th)^2) dr⊗dph + (eps*m*r - eps*r^2)*t/m dth⊗dth + 1/4*((eps*m^6 - 2*eps*m^5*r + eps*m^4*r^2)*t*sin(th)^6 + 2*(eps*m^6 + eps*m^5*r - 2*eps*m^4*r^2 + eps*m^3*r^3 - eps*m^2*r^4)*t*sin(th)^4 - (3*eps*m^6 - 7*eps*m^4*r^2 + 8*eps*m^3*r^3 - 3*eps*m^2*r^4 - eps*r^6)*t*sin(th)^2)/(m^6*cos(th)^4 + 2*m^4*r^2*cos(th)^2 + m^2*r^4) dph⊗dt - ((2*eps*m^4 - eps*m^3*r)*sin(th)^4 - (2*eps*m^4 - eps*m^3*r - eps*m*r^3)*sin(th)^2)/(m*r^2 - r^3 + (m^3 - m^2*r)*cos(th)^2) dph⊗dr + (((eps*m^6 - eps*m^5*r)*t*cos(th)^2 - (eps*m^4*r^2 - eps*m^3*r^3)*t)*sin(th)^4 + ((eps*m^5*r - eps*m^4*r^2)*t*cos(th)^4 + 2*(eps*m^3*r^3 - eps*m^2*r^4)*t*cos(th)^2 + (eps*m*r^5 - eps*r^6)*t)*sin(th)^2)/(m^5*cos(th)^4 + 2*m^3*r^2*cos(th)^2 + m*r^4) dph⊗dph" ] }, "execution_count": 77, "metadata": {}, "output_type": "execute_result" } ], "source": [ "Kxi3 = killing_form(xi3)\n", "print(Kxi3)\n", "Kxi3.display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Global NHEK coordinates" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Let us introduce the global NHEK coordinates $(\\tau, y, \\theta,\\psi)$:" ] }, { "cell_type": "code", "execution_count": 78, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Chart (M, (ta, y, th, ps))\n" ] }, { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\mathcal{M},({\\tau}, y, {\\theta}, {\\psi})\\right)\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\mathcal{M},({\\tau}, y, {\\theta}, {\\psi})\\right)$$" ], "text/plain": [ "Chart (M, (ta, y, th, ps))" ] }, "execution_count": 78, "metadata": {}, "output_type": "execute_result" } ], "source": [ "GNH. = M.chart(r\"ta:\\tau y th:(0,pi):\\theta ps:(0,2*pi):periodic:\\psi\") \n", "print(GNH)\n", "GNH" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "They are related to the near-horizon coordinates $(T,R,\\theta,\\Phi)$ via Eqs. (2.7)-(2.8) of J. Bardeen and G. T. Horowitz, \n", "[Phys. Rev. D **60**, 104030 (1999)](https://doi.org/10.1103/PhysRevD.60.104030):" ] }, { "cell_type": "code", "execution_count": 79, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left\\{\\begin{array}{lcl} T & = & \\frac{\\sqrt{y^{2} + 1} \\sin\\left({\\tau}\\right)}{\\sqrt{y^{2} + 1} \\cos\\left({\\tau}\\right) + y} \\\\ R & = & \\sqrt{y^{2} + 1} \\cos\\left({\\tau}\\right) + y \\\\ {\\theta} & = & {\\theta} \\\\ {\\Phi} & = & {\\psi} + \\log\\left(\\frac{y \\sin\\left({\\tau}\\right) + \\cos\\left({\\tau}\\right)}{\\sqrt{y^{2} + 1} \\sin\\left({\\tau}\\right) + 1}\\right) \\end{array}\\right.\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left\\{\\begin{array}{lcl} T & = & \\frac{\\sqrt{y^{2} + 1} \\sin\\left({\\tau}\\right)}{\\sqrt{y^{2} + 1} \\cos\\left({\\tau}\\right) + y} \\\\ R & = & \\sqrt{y^{2} + 1} \\cos\\left({\\tau}\\right) + y \\\\ {\\theta} & = & {\\theta} \\\\ {\\Phi} & = & {\\psi} + \\log\\left(\\frac{y \\sin\\left({\\tau}\\right) + \\cos\\left({\\tau}\\right)}{\\sqrt{y^{2} + 1} \\sin\\left({\\tau}\\right) + 1}\\right) \\end{array}\\right.$$" ], "text/plain": [ "T = sqrt(y^2 + 1)*sin(ta)/(sqrt(y^2 + 1)*cos(ta) + y)\n", "R = sqrt(y^2 + 1)*cos(ta) + y\n", "th = th\n", "Ph = ps + log((y*sin(ta) + cos(ta))/(sqrt(y^2 + 1)*sin(ta) + 1))" ] }, "execution_count": 79, "metadata": {}, "output_type": "execute_result" } ], "source": [ "GNH_to_NH = GNH.transition_map(NH, [sqrt(1+y^2)*sin(ta)/(y + sqrt(1+y^2)*cos(ta)),\n", " y + sqrt(1+y^2)*cos(ta),\n", " th, \n", " ps + ln((cos(ta) + y*sin(ta))/(1 + sqrt(1+y^2)*sin(ta)))])\n", "GNH_to_NH.display()" ] }, { "cell_type": "code", "execution_count": 80, "metadata": {}, "outputs": [], "source": [ "assume(R^2*T^2 - R^2 - 1<0)\n", "assume(R>0)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "The inverse transformation is " ] }, { "cell_type": "code", "execution_count": 81, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Check of the inverse coordinate transformation:\n", " ta == arctan2(2*y^2*cos(ta)*sin(ta) + 2*sqrt(y^2 + 1)*y*sin(ta) + 2*cos(ta)*sin(ta), 2*((cos(ta)^4 + 3*cos(ta)^2)*y^4 + cos(ta)^4 + (2*cos(ta)^4 + 3*cos(ta)^2)*y^2 + ((3*cos(ta)^3 + cos(ta))*y^3 + 3*y*cos(ta)^3)*sqrt(y^2 + 1))/((cos(ta)^2 + 1)*y^2 + 2*sqrt(y^2 + 1)*y*cos(ta) + cos(ta)^2)) **failed**\n", " y == y *passed*\n", " th == th *passed*\n", " ps == ps + 1/2*log((cos(ta)^6 + 15*cos(ta)^4 + 15*cos(ta)^2 + 1)*y^6 + cos(ta)^6 + 3*(cos(ta)^6 + 10*cos(ta)^4 + 5*cos(ta)^2)*y^4 + 3*(cos(ta)^6 + 5*cos(ta)^4)*y^2 + 2*((3*cos(ta)^5 + 10*cos(ta)^3 + 3*cos(ta))*y^5 + 3*y*cos(ta)^5 + 2*(3*cos(ta)^5 + 5*cos(ta)^3)*y^3)*sqrt(y^2 + 1)) - log(y^2*cos(ta) + sqrt(y^2 + 1)*y + cos(ta)) + 1/2*log(y^2 + 1) - 1/2*log(abs((cos(ta)^4 + 6*cos(ta)^2 + 1)*y^4 + cos(ta)^4 + 2*(cos(ta)^4 + 3*cos(ta)^2)*y^2 + 4*((cos(ta)^3 + cos(ta))*y^3 + y*cos(ta)^3)*sqrt(y^2 + 1))) **failed**\n", " T == -2*R^2*T*abs(R^2*T^2 - R^2 - 1)/(R^4*T^4 - R^4 - 2*R^2*T^2 + (R^2*T^2 - R^2 - 1)*abs(R^2*T^2 - R^2 - 1) + 1) **failed**\n", " R == 1/2*(R^2*T^2 + R^2 + abs(R^2*T^2 - R^2 - 1) - 1)/R **failed**\n", " th == th *passed*\n", " Ph == Ph + log(-(R*T + 1)*abs(R^2*T^2 - R^2 - 1)/(R^2*T^2 - R*T*abs(R^2*T^2 - R^2 - 1) - R^2 - 1)) **failed**\n", "NB: a failed report can reflect a mere lack of simplification.\n" ] } ], "source": [ "GNH_to_NH.set_inverse(atan2(2*T*R^2, ((1 - T^2)*R^2 + 1)), \n", " ((1 + T^2)*R^2 - 1)/(2*R),\n", " th,\n", " Ph - ln(((1 - T*R)^2 + R^2)/sqrt(((1 + T^2)*R^2 - 1)^2 + 4*R^2)) )" ] }, { "cell_type": "code", "execution_count": 82, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left\\{\\begin{array}{lcl} {\\tau} & = & \\arctan\\left(2 \\, R^{2} T, -{\\left(T^{2} - 1\\right)} R^{2} + 1\\right) \\\\ y & = & \\frac{{\\left(T^{2} + 1\\right)} R^{2} - 1}{2 \\, R} \\\\ {\\theta} & = & {\\theta} \\\\ {\\psi} & = & {\\Phi} - \\log\\left(\\frac{{\\left(R T - 1\\right)}^{2} + R^{2}}{\\sqrt{{\\left({\\left(T^{2} + 1\\right)} R^{2} - 1\\right)}^{2} + 4 \\, R^{2}}}\\right) \\end{array}\\right.\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left\\{\\begin{array}{lcl} {\\tau} & = & \\arctan\\left(2 \\, R^{2} T, -{\\left(T^{2} - 1\\right)} R^{2} + 1\\right) \\\\ y & = & \\frac{{\\left(T^{2} + 1\\right)} R^{2} - 1}{2 \\, R} \\\\ {\\theta} & = & {\\theta} \\\\ {\\psi} & = & {\\Phi} - \\log\\left(\\frac{{\\left(R T - 1\\right)}^{2} + R^{2}}{\\sqrt{{\\left({\\left(T^{2} + 1\\right)} R^{2} - 1\\right)}^{2} + 4 \\, R^{2}}}\\right) \\end{array}\\right.$$" ], "text/plain": [ "ta = arctan2(2*R^2*T, -(T^2 - 1)*R^2 + 1)\n", "y = 1/2*((T^2 + 1)*R^2 - 1)/R\n", "th = th\n", "ps = Ph - log(((R*T - 1)^2 + R^2)/sqrt(((T^2 + 1)*R^2 - 1)^2 + 4*R^2))" ] }, "execution_count": 82, "metadata": {}, "output_type": "execute_result" } ], "source": [ "GNH_to_NH.inverse().display()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Plot of the near-horizon coordinates in terms of the global NHEK ones" ] }, { "cell_type": "code", "execution_count": 83, "metadata": {}, "outputs": [ { "data": { "image/png": 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\n", "text/plain": [ "Graphics object consisting of 31 graphics primitives" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "graph = NH.plot(GNH, ambient_coords=(y, ta), fixed_coords={th: pi/2, Ph: 0},\n", " ranges={T: (-10, 10), R: (0.03, 12)}, color={T: 'red', R: 'grey'}, \n", " number_values={T: 17, R: 13}, plot_points=800) + \\\n", " NH.plot(GNH, ambient_coords=(y, ta), fixed_coords={th: pi/2, R: 0.03, Ph: 0},\n", " ranges={T: (-60, 60)}, color={T: 'blue'}, thickness=2,\n", " number_values={T: 33}, plot_points=400)\n", "show(graph, xmin=-5, xmax=6)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Expression of the near horizon metric $h$ in terms of the global NHEK coordinates" ] }, { "cell_type": "code", "execution_count": 84, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}h = \\left( -\\frac{m^{2} \\cos\\left({\\theta}\\right)^{4} + 2 \\, m^{2} \\cos\\left({\\theta}\\right)^{2} + {\\left(m^{2} \\cos\\left({\\theta}\\right)^{4} + 6 \\, m^{2} \\cos\\left({\\theta}\\right)^{2} - 3 \\, m^{2}\\right)} y^{2} + m^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\tau}\\otimes \\mathrm{d} {\\tau} + \\left( \\frac{4 \\, m^{2} y \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\tau}\\otimes \\mathrm{d} {\\psi} + \\left( \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + m^{2}}{y^{2} + 1} \\right) \\mathrm{d} y\\otimes \\mathrm{d} y + \\left( -m^{2} \\sin\\left({\\theta}\\right)^{2} + 2 \\, m^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( \\frac{4 \\, m^{2} y \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\psi}\\otimes \\mathrm{d} {\\tau} + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\psi}\\otimes \\mathrm{d} {\\psi}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}h = \\left( -\\frac{m^{2} \\cos\\left({\\theta}\\right)^{4} + 2 \\, m^{2} \\cos\\left({\\theta}\\right)^{2} + {\\left(m^{2} \\cos\\left({\\theta}\\right)^{4} + 6 \\, m^{2} \\cos\\left({\\theta}\\right)^{2} - 3 \\, m^{2}\\right)} y^{2} + m^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\tau}\\otimes \\mathrm{d} {\\tau} + \\left( \\frac{4 \\, m^{2} y \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\tau}\\otimes \\mathrm{d} {\\psi} + \\left( \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + m^{2}}{y^{2} + 1} \\right) \\mathrm{d} y\\otimes \\mathrm{d} y + \\left( -m^{2} \\sin\\left({\\theta}\\right)^{2} + 2 \\, m^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( \\frac{4 \\, m^{2} y \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\psi}\\otimes \\mathrm{d} {\\tau} + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\psi}\\otimes \\mathrm{d} {\\psi}$$" ], "text/plain": [ "h = -(m^2*cos(th)^4 + 2*m^2*cos(th)^2 + (m^2*cos(th)^4 + 6*m^2*cos(th)^2 - 3*m^2)*y^2 + m^2)/(cos(th)^2 + 1) dta⊗dta + 4*m^2*y*sin(th)^2/(cos(th)^2 + 1) dta⊗dps + (m^2*cos(th)^2 + m^2)/(y^2 + 1) dy⊗dy + (-m^2*sin(th)^2 + 2*m^2) dth⊗dth + 4*m^2*y*sin(th)^2/(cos(th)^2 + 1) dps⊗dta + 4*m^2*sin(th)^2/(cos(th)^2 + 1) dps⊗dps" ] }, "execution_count": 84, "metadata": {}, "output_type": "execute_result" } ], "source": [ "h.display(GNH)" ] }, { "cell_type": "code", "execution_count": 85, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}h = \\left( -\\frac{{\\left(y^{2} \\cos\\left({\\theta}\\right)^{4} + 6 \\, y^{2} \\cos\\left({\\theta}\\right)^{2} + \\cos\\left({\\theta}\\right)^{4} - 3 \\, y^{2} + 2 \\, \\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\tau}\\otimes \\mathrm{d} {\\tau} + \\left( \\frac{4 \\, m^{2} y \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\tau}\\otimes \\mathrm{d} {\\psi} + \\left( \\frac{{\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}}{y^{2} + 1} \\right) \\mathrm{d} y\\otimes \\mathrm{d} y -{\\left(\\sin\\left({\\theta}\\right)^{2} - 2\\right)} m^{2} \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( \\frac{4 \\, m^{2} y \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\psi}\\otimes \\mathrm{d} {\\tau} + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\psi}\\otimes \\mathrm{d} {\\psi}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}h = \\left( -\\frac{{\\left(y^{2} \\cos\\left({\\theta}\\right)^{4} + 6 \\, y^{2} \\cos\\left({\\theta}\\right)^{2} + \\cos\\left({\\theta}\\right)^{4} - 3 \\, y^{2} + 2 \\, \\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\tau}\\otimes \\mathrm{d} {\\tau} + \\left( \\frac{4 \\, m^{2} y \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\tau}\\otimes \\mathrm{d} {\\psi} + \\left( \\frac{{\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}}{y^{2} + 1} \\right) \\mathrm{d} y\\otimes \\mathrm{d} y -{\\left(\\sin\\left({\\theta}\\right)^{2} - 2\\right)} m^{2} \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( \\frac{4 \\, m^{2} y \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\psi}\\otimes \\mathrm{d} {\\tau} + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\psi}\\otimes \\mathrm{d} {\\psi}$$" ], "text/plain": [ "h = -(y^2*cos(th)^4 + 6*y^2*cos(th)^2 + cos(th)^4 - 3*y^2 + 2*cos(th)^2 + 1)*m^2/(cos(th)^2 + 1) dta⊗dta + 4*m^2*y*sin(th)^2/(cos(th)^2 + 1) dta⊗dps + (cos(th)^2 + 1)*m^2/(y^2 + 1) dy⊗dy - (sin(th)^2 - 2)*m^2 dth⊗dth + 4*m^2*y*sin(th)^2/(cos(th)^2 + 1) dps⊗dta + 4*m^2*sin(th)^2/(cos(th)^2 + 1) dps⊗dps" ] }, "execution_count": 85, "metadata": {}, "output_type": "execute_result" } ], "source": [ "h.apply_map(factor, frame=GNH.frame(), chart=GNH, keep_other_components=True)\n", "h.display(GNH)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Let us check that we recover Eq. (2.9) of Bardeen & Horowitz, \n", "[Phys. Rev. D **60**, 104030 (1999)](https://doi.org/10.1103/PhysRevD.60.104030). This is actually obvious for all the terms, except for $h_{\\tau\\tau}$. For the latter, the computation" ] }, { "cell_type": "code", "execution_count": 86, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{4 \\, m^{2} y^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{4 \\, m^{2} y^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1}$$" ], "text/plain": [ "4*m^2*y^2*sin(th)^2/(cos(th)^2 + 1)" ] }, "execution_count": 86, "metadata": {}, "output_type": "execute_result" } ], "source": [ "h00 = h[GNH.frame(),0,0,GNH].expr()\n", "(h00 + m^2*(1 + cos(th)^2)*(1 + y^2)).simplify_full()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "shows that\n", "$$h_{\\tau\\tau} = m^2 \\left[ - (1 + \\cos^2\\theta)(1 + y^2) + 4 \\frac{y^2\\sin^2\\theta}{1 + \\cos^2\\theta} \\right],$$\n", "which yields a full agreement with Eq. (2.9) of Bardeen & Horowitz, up to an overall factor $2 m^2$." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Expression of the $h$-Killing vector $\\frac{\\partial}{\\partial\\tau}$ in terms of $\\xi_1$ and $\\xi_3$" ] }, { "cell_type": "code", "execution_count": 87, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{\\partial}{\\partial {\\tau} } = \\left( \\frac{R^{2} T^{2} + R^{2} + 1}{2 \\, R^{2}} \\right) \\frac{\\partial}{\\partial T } -R T \\frac{\\partial}{\\partial R } -\\frac{1}{R} \\frac{\\partial}{\\partial {\\Phi} }\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{\\partial}{\\partial {\\tau} } = \\left( \\frac{R^{2} T^{2} + R^{2} + 1}{2 \\, R^{2}} \\right) \\frac{\\partial}{\\partial T } -R T \\frac{\\partial}{\\partial R } -\\frac{1}{R} \\frac{\\partial}{\\partial {\\Phi} }$$" ], "text/plain": [ "∂/∂ta = 1/2*(R^2*T^2 + R^2 + 1)/R^2 ∂/∂T - R*T ∂/∂R - 1/R ∂/∂Ph" ] }, "execution_count": 87, "metadata": {}, "output_type": "execute_result" } ], "source": [ "GNH.frame()[0].display(NH)" ] }, { "cell_type": "code", "execution_count": 88, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}\$" ], "text/latex": [ "$$\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}$$" ], "text/plain": [ "True" ] }, "execution_count": 88, "metadata": {}, "output_type": "execute_result" } ], "source": [ "GNH.frame()[0] == xi3 + xi1/2" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Hence we have:\n", "$$\\frac{\\partial}{\\partial\\tau} = \\xi_3 + \\frac{1}{2} \\xi_1$$." ] } ], "metadata": { "kernelspec": { "display_name": "SageMath 9.5.beta3", "language": "sage", "name": "sagemath" }, "language": "python", "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.8.10" } }, "nbformat": 4, "nbformat_minor": 1 }