{ "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 10.3.rc3, Release Date: 2024-03-10'" ] }, "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": [ "\$\\displaystyle \\left(\\mathcal{M},(t, r, {\\theta}, {\\phi})\\right)\$" ], "text/latex": [ "$\\displaystyle \\left(\\mathcal{M},(t, r, {\\theta}, {\\phi})\\right)$" ], "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": [ "\$\\displaystyle t :\\ \\left( -\\infty, +\\infty \\right) ;\\quad r :\\ \\left( -\\infty, +\\infty \\right) ;\\quad {\\theta} :\\ \\left( 0 , \\pi \\right) ;\\quad {\\phi} :\\ \\left[ 0 , 2 \\, \\pi \\right] \\text{(periodic)}\$" ], "text/latex": [ "$\\displaystyle t :\\ \\left( -\\infty, +\\infty \\right) ;\\quad r :\\ \\left( -\\infty, +\\infty \\right) ;\\quad {\\theta} :\\ \\left( 0 , \\pi \\right) ;\\quad {\\phi} :\\ \\left[ 0 , 2 \\, \\pi \\right] \\text{(periodic)}$" ], "text/plain": [ "t: (-oo, +oo); r: (-oo, +oo); th: (0, pi); ph: [0, 2*pi] (periodic)" ] }, "execution_count": 6, "metadata": {}, "output_type": "execute_result" } ], "source": [ "BL.coord_range()" ] }, { "cell_type": "code", "execution_count": 7, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle t\$" ], "text/latex": [ "$\\displaystyle t$" ], "text/plain": [ "t" ] }, "execution_count": 7, "metadata": {}, "output_type": "execute_result" } ], "source": [ "BL[0]" ] }, { "cell_type": "code", "execution_count": 8, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle {\\theta}\$" ], "text/latex": [ "$\\displaystyle {\\theta}$" ], "text/plain": [ "th" ] }, "execution_count": 8, "metadata": {}, "output_type": "execute_result" } ], "source": [ "BL[2]" ] }, { "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": 9, "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": 10, "metadata": {}, "outputs": [], "source": [ "a = m" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We get the (yet undefined) spacetime metric:" ] }, { "cell_type": "code", "execution_count": 11, "metadata": {}, "outputs": [], "source": [ "g = M.metric()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "The metric is set by providing its nonzero components in the coordinate frame associated with Boyer-Lindquist coordinates, which is the current manifold's default frame:" ] }, { "cell_type": "code", "execution_count": 12, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle 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": [ "$\\displaystyle g = \\left( \\frac{2 \\, m r}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} - 1 \\right) \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} {\\phi} + \\left( \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}}{m^{2} - 2 \\, m r + r^{2}} \\right) \\mathrm{d} r\\otimes \\mathrm{d} r + \\left( m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} t + {\\left(\\frac{2 \\, m^{3} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} + m^{2} + r^{2}\\right)} \\sin\\left({\\theta}\\right)^{2} \\mathrm{d} {\\phi}\\otimes \\mathrm{d} {\\phi}$" ], "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": 12, "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": 13, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\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": [ "$\\displaystyle \\left(\\begin{array}{rrrr}\n", "\\frac{2 \\, m r}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} - 1 & 0 & 0 & -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\\\\n", "0 & \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}}{m^{2} - 2 \\, m r + r^{2}} & 0 & 0 \\\\\n", "0 & 0 & m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2} & 0 \\\\\n", "-\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} & 0 & 0 & {\\left(\\frac{2 \\, m^{3} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} + m^{2} + r^{2}\\right)} \\sin\\left({\\theta}\\right)^{2}\n", "\\end{array}\\right)$" ], "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": 13, "metadata": {}, "output_type": "execute_result" } ], "source": [ "g[:]" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "

The list of the non-vanishing components:

" ] }, { "cell_type": "code", "execution_count": 14, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\\(\\displaystyle \\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": [ "$\\displaystyle \\begin{array}{lcl} g_{ \\, t \\, t }^{ \\phantom{\\, t}\\phantom{\\, t} } & = & \\frac{2 \\, m r}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} - 1 \\\\ g_{ \\, t \\, {\\phi} }^{ \\phantom{\\, t}\\phantom{\\, {\\phi}} } & = & -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\\\ g_{ \\, r \\, r }^{ \\phantom{\\, r}\\phantom{\\, r} } & = & \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}}{m^{2} - 2 \\, m r + r^{2}} \\\\ g_{ \\, {\\theta} \\, {\\theta} }^{ \\phantom{\\, {\\theta}}\\phantom{\\, {\\theta}} } & = & m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2} \\\\ g_{ \\, {\\phi} \\, t }^{ \\phantom{\\, {\\phi}}\\phantom{\\, t} } & = & -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\\\ g_{ \\, {\\phi} \\, {\\phi} }^{ \\phantom{\\, {\\phi}}\\phantom{\\, {\\phi}} } & = & {\\left(\\frac{2 \\, m^{3} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} + m^{2} + r^{2}\\right)} \\sin\\left({\\theta}\\right)^{2} \\end{array}$" ], "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": 14, "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": 15, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\mathrm{Ric}\\left(g\\right) = 0\$" ], "text/latex": [ "$\\displaystyle \\mathrm{Ric}\\left(g\\right) = 0$" ], "text/plain": [ "Ric(g) = 0" ] }, "execution_count": 15, "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": 16, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Chart (M, (T, R, th, Ph))\n" ] }, { "data": { "text/html": [ "\$\\displaystyle \\left(\\mathcal{M},(T, R, {\\theta}, {\\Phi})\\right)\$" ], "text/latex": [ "$\\displaystyle \\left(\\mathcal{M},(T, R, {\\theta}, {\\Phi})\\right)$" ], "text/plain": [ "Chart (M, (T, R, th, Ph))" ] }, "execution_count": 16, "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": 17, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\left[\\left(\\mathcal{M},(t, r, {\\theta}, {\\phi})\\right), \\left(\\mathcal{M},(T, R, {\\theta}, {\\Phi})\\right)\\right]\$" ], "text/latex": [ "$\\displaystyle \\left[\\left(\\mathcal{M},(t, r, {\\theta}, {\\phi})\\right), \\left(\\mathcal{M},(T, R, {\\theta}, {\\Phi})\\right)\\right]$" ], "text/plain": [ "[Chart (M, (t, r, th, ph)), Chart (M, (T, R, th, Ph))]" ] }, "execution_count": 17, "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": 18, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\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": [ "$\\displaystyle \\left\\{\\begin{array}{lcl} T & = & \\frac{{\\epsilon} t}{2 \\, m} \\\\ R & = & -\\frac{m - r}{{\\epsilon} m} \\\\ {\\theta} & = & {\\theta} \\\\ {\\Phi} & = & {\\phi} - \\frac{t}{2 \\, m} \\end{array}\\right.$" ], "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": 18, "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": 19, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\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": [ "$\\displaystyle \\left\\{\\begin{array}{lcl} t & = & \\frac{2 \\, T m}{{\\epsilon}} \\\\ r & = & {\\left(R {\\epsilon} + 1\\right)} m \\\\ {\\theta} & = & {\\theta} \\\\ {\\phi} & = & \\frac{{\\Phi} {\\epsilon} + T}{{\\epsilon}} \\end{array}\\right.$" ], "text/plain": [ "t = 2*T*m/eps\n", "r = (R*eps + 1)*m\n", "th = th\n", "ph = (Ph*eps + T)/eps" ] }, "execution_count": 19, "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": 20, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\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": [ "$\\displaystyle \\left(\\begin{array}{rrrr}\n", "\\frac{{\\epsilon}}{2 \\, m} & 0 & 0 & 0 \\\\\n", "0 & \\frac{1}{{\\epsilon} m} & 0 & 0 \\\\\n", "0 & 0 & 1 & 0 \\\\\n", "-\\frac{1}{2 \\, m} & 0 & 0 & 1\n", "\\end{array}\\right)$" ], "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": 20, "metadata": {}, "output_type": "execute_result" } ], "source": [ "BL_to_NH.jacobian()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "and" ] }, { "cell_type": "code", "execution_count": 21, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\frac{1}{2 \\, m^{2}}\$" ], "text/latex": [ "$\\displaystyle \\frac{1}{2 \\, m^{2}}$" ], "text/plain": [ "1/2/m^2" ] }, "execution_count": 21, "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": 22, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\left(\\mathcal{M},(t, r, {\\theta}, {\\phi})\\right)\$" ], "text/latex": [ "$\\displaystyle \\left(\\mathcal{M},(t, r, {\\theta}, {\\phi})\\right)$" ], "text/plain": [ "Chart (M, (t, r, th, ph))" ] }, "execution_count": 22, "metadata": {}, "output_type": "execute_result" } ], "source": [ "M.default_chart()" ] }, { "cell_type": "code", "execution_count": 23, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle 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": [ "$\\displaystyle g = \\left( \\frac{2 \\, m r}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} - 1 \\right) \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} {\\phi} + \\left( \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}}{m^{2} - 2 \\, m r + r^{2}} \\right) \\mathrm{d} r\\otimes \\mathrm{d} r + \\left( m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( -\\frac{2 \\, m^{2} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} t + {\\left(\\frac{2 \\, m^{3} r \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + r^{2}} + m^{2} + r^{2}\\right)} \\sin\\left({\\theta}\\right)^{2} \\mathrm{d} {\\phi}\\otimes \\mathrm{d} {\\phi}$" ], "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": 23, "metadata": {}, "output_type": "execute_result" } ], "source": [ "g.display() # display in the default chart, i.e. BL" ] }, { "cell_type": "code", "execution_count": 24, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\\(\\displaystyle 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": [ "$\\displaystyle g = \\left( -\\frac{R^{2} m^{2} \\cos\\left({\\theta}\\right)^{4} + {\\left(R^{4} {\\epsilon}^{2} + 4 \\, R^{3} {\\epsilon} + 6 \\, R^{2}\\right)} m^{2} \\cos\\left({\\theta}\\right)^{2} - {\\left(R^{4} {\\epsilon}^{2} + 4 \\, R^{3} {\\epsilon} + 3 \\, R^{2}\\right)} m^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\left( -\\frac{R^{2} {\\epsilon} m^{2} \\sin\\left({\\theta}\\right)^{4} - {\\left(R^{4} {\\epsilon}^{3} + 4 \\, R^{3} {\\epsilon}^{2} + 8 \\, R^{2} {\\epsilon} + 4 \\, R\\right)} m^{2} \\sin\\left({\\theta}\\right)^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} {\\Phi} + \\left( \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + {\\left(R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + 1\\right)} m^{2}}{R^{2}} \\right) \\mathrm{d} R\\otimes \\mathrm{d} R + \\left( m^{2} \\cos\\left({\\theta}\\right)^{2} + {\\left(R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + 1\\right)} m^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( -\\frac{R^{2} {\\epsilon} m^{2} \\sin\\left({\\theta}\\right)^{4} - {\\left(R^{4} {\\epsilon}^{3} + 4 \\, R^{3} {\\epsilon}^{2} + 8 \\, R^{2} {\\epsilon} + 4 \\, R\\right)} m^{2} \\sin\\left({\\theta}\\right)^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} T + \\left( -\\frac{R^{2} {\\epsilon}^{2} m^{2} \\sin\\left({\\theta}\\right)^{4} - {\\left(R^{4} {\\epsilon}^{4} + 4 \\, R^{3} {\\epsilon}^{3} + 8 \\, R^{2} {\\epsilon}^{2} + 8 \\, R {\\epsilon} + 4\\right)} m^{2} \\sin\\left({\\theta}\\right)^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} {\\Phi}$" ], "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(NH) # triggers the computation of the components w.r.t. NH chart" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "From now on, we set the near-horizon coordinates as the default ones on the spacetime manifold:" ] }, { "cell_type": "code", "execution_count": 25, "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": 26, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\\(\\displaystyle 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": [ "$\\displaystyle g = \\left( -\\frac{R^{2} m^{2} \\cos\\left({\\theta}\\right)^{4} + {\\left(R^{4} {\\epsilon}^{2} + 4 \\, R^{3} {\\epsilon} + 6 \\, R^{2}\\right)} m^{2} \\cos\\left({\\theta}\\right)^{2} - {\\left(R^{4} {\\epsilon}^{2} + 4 \\, R^{3} {\\epsilon} + 3 \\, R^{2}\\right)} m^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\left( -\\frac{R^{2} {\\epsilon} m^{2} \\sin\\left({\\theta}\\right)^{4} - {\\left(R^{4} {\\epsilon}^{3} + 4 \\, R^{3} {\\epsilon}^{2} + 8 \\, R^{2} {\\epsilon} + 4 \\, R\\right)} m^{2} \\sin\\left({\\theta}\\right)^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} {\\Phi} + \\left( \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + {\\left(R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + 1\\right)} m^{2}}{R^{2}} \\right) \\mathrm{d} R\\otimes \\mathrm{d} R + \\left( m^{2} \\cos\\left({\\theta}\\right)^{2} + {\\left(R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + 1\\right)} m^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( -\\frac{R^{2} {\\epsilon} m^{2} \\sin\\left({\\theta}\\right)^{4} - {\\left(R^{4} {\\epsilon}^{3} + 4 \\, R^{3} {\\epsilon}^{2} + 8 \\, R^{2} {\\epsilon} + 4 \\, R\\right)} m^{2} \\sin\\left({\\theta}\\right)^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} T + \\left( -\\frac{R^{2} {\\epsilon}^{2} m^{2} \\sin\\left({\\theta}\\right)^{4} - {\\left(R^{4} {\\epsilon}^{4} + 4 \\, R^{3} {\\epsilon}^{3} + 8 \\, R^{2} {\\epsilon}^{2} + 8 \\, R {\\epsilon} + 4\\right)} m^{2} \\sin\\left({\\theta}\\right)^{2}}{R^{2} {\\epsilon}^{2} + 2 \\, R {\\epsilon} + \\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} {\\Phi}$" ], "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": 26, "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": 27, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\\(\\displaystyle 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": [ "$\\displaystyle h = \\left( -\\frac{R^{2} m^{2} \\cos\\left({\\theta}\\right)^{4}}{\\cos\\left({\\theta}\\right)^{2} + 1} - \\frac{6 \\, R^{2} m^{2} \\cos\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} + \\frac{3 \\, R^{2} m^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\left( \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} {\\Phi} + \\left( \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2}}{R^{2}} + \\frac{m^{2}}{R^{2}} \\right) \\mathrm{d} R\\otimes \\mathrm{d} R + \\left( m^{2} \\cos\\left({\\theta}\\right)^{2} + m^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} T + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} {\\Phi}$" ], "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": 27, "metadata": {}, "output_type": "execute_result" } ], "source": [ "h = M.lorentzian_metric('h')\n", "h.set( g.truncate(eps, 0) )\n", "h.display()" ] }, { "cell_type": "code", "execution_count": 28, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle h = \\left( -\\frac{{\\left(\\cos\\left({\\theta}\\right)^{4} + 6 \\, \\cos\\left({\\theta}\\right)^{2} - 3\\right)} 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} + \\frac{{\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}}{R^{2}} \\mathrm{d} R\\otimes \\mathrm{d} R + {\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2} \\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": [ "$\\displaystyle h = \\left( -\\frac{{\\left(\\cos\\left({\\theta}\\right)^{4} + 6 \\, \\cos\\left({\\theta}\\right)^{2} - 3\\right)} R^{2} m^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\left( \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} {\\Phi} + \\frac{{\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}}{R^{2}} \\mathrm{d} R\\otimes \\mathrm{d} R + {\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2} \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} T + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} {\\Phi}$" ], "text/plain": [ "h = -(cos(th)^4 + 6*cos(th)^2 - 3)*R^2*m^2/(cos(th)^2 + 1) dT⊗dT + 4*R*m^2*sin(th)^2/(cos(th)^2 + 1) dT⊗dPh + (cos(th)^2 + 1)*m^2/R^2 dR⊗dR + (cos(th)^2 + 1)*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": 28, "metadata": {}, "output_type": "execute_result" } ], "source": [ "h.apply_map(factor)\n", "h.display()" ] }, { "cell_type": "code", "execution_count": 29, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\\(\\displaystyle \\begin{array}{lcl} h_{ \\, T \\, T }^{ \\phantom{\\, T}\\phantom{\\, T} } & = & -\\frac{{\\left(\\cos\\left({\\theta}\\right)^{4} + 6 \\, \\cos\\left({\\theta}\\right)^{2} - 3\\right)} 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{{\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}}{R^{2}} \\\\ h_{ \\, {\\theta} \\, {\\theta} }^{ \\phantom{\\, {\\theta}}\\phantom{\\, {\\theta}} } & = & {\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} 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": [ "$\\displaystyle \\begin{array}{lcl} h_{ \\, T \\, T }^{ \\phantom{\\, T}\\phantom{\\, T} } & = & -\\frac{{\\left(\\cos\\left({\\theta}\\right)^{4} + 6 \\, \\cos\\left({\\theta}\\right)^{2} - 3\\right)} R^{2} m^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\\\ h_{ \\, T \\, {\\Phi} }^{ \\phantom{\\, T}\\phantom{\\, {\\Phi}} } & = & \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\\\ h_{ \\, R \\, R }^{ \\phantom{\\, R}\\phantom{\\, R} } & = & \\frac{{\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}}{R^{2}} \\\\ h_{ \\, {\\theta} \\, {\\theta} }^{ \\phantom{\\, {\\theta}}\\phantom{\\, {\\theta}} } & = & {\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2} \\\\ h_{ \\, {\\Phi} \\, T }^{ \\phantom{\\, {\\Phi}}\\phantom{\\, T} } & = & \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\\\ h_{ \\, {\\Phi} \\, {\\Phi} }^{ \\phantom{\\, {\\Phi}}\\phantom{\\, {\\Phi}} } & = & \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\end{array}$" ], "text/plain": [ "h_T,T = -(cos(th)^4 + 6*cos(th)^2 - 3)*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 = (cos(th)^2 + 1)*m^2/R^2 \n", "h_th,th = (cos(th)^2 + 1)*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": 29, "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": 30, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\mathrm{d} T\$" ], "text/latex": [ "$\\displaystyle \\mathrm{d} T$" ], "text/plain": [ "1-form dT on the 4-dimensional Lorentzian manifold M" ] }, "execution_count": 30, "metadata": {}, "output_type": "execute_result" } ], "source": [ "dT = NH.coframe()[0]\n", "dT" ] }, { "cell_type": "code", "execution_count": 31, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\mathrm{d} {\\Phi}\$" ], "text/latex": [ "$\\displaystyle \\mathrm{d} {\\Phi}$" ], "text/plain": [ "1-form dPh on the 4-dimensional Lorentzian manifold M" ] }, "execution_count": 31, "metadata": {}, "output_type": "execute_result" } ], "source": [ "dPh = NH.coframe()[3]\n", "dPh" ] }, { "cell_type": "code", "execution_count": 32, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\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": [ "$\\displaystyle \\left( \\frac{4 \\, R^{2} m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\left( \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} T\\otimes \\mathrm{d} {\\Phi} + \\left( \\frac{4 \\, R m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} T + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\Phi}\\otimes \\mathrm{d} {\\Phi}$" ], "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": 32, "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": 33, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\left( -R^{2} m^{2} \\cos\\left({\\theta}\\right)^{2} - R^{2} m^{2} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\frac{{\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}}{R^{2}} \\mathrm{d} R\\otimes \\mathrm{d} R + {\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2} \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta}\$" ], "text/latex": [ "$\\displaystyle \\left( -R^{2} m^{2} \\cos\\left({\\theta}\\right)^{2} - R^{2} m^{2} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\frac{{\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}}{R^{2}} \\mathrm{d} R\\otimes \\mathrm{d} R + {\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2} \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta}$" ], "text/plain": [ "(-R^2*m^2*cos(th)^2 - R^2*m^2) dT⊗dT + (cos(th)^2 + 1)*m^2/R^2 dR⊗dR + (cos(th)^2 + 1)*m^2 dth⊗dth" ] }, "execution_count": 33, "metadata": {}, "output_type": "execute_result" } ], "source": [ "(h - q).display()" ] }, { "cell_type": "code", "execution_count": 34, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\left( -\\frac{R^{2} m^{2} \\cos\\left({\\theta}\\right)^{2} + R^{2} m^{2}}{{\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}} \\right) \\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": [ "$\\displaystyle \\left( -\\frac{R^{2} m^{2} \\cos\\left({\\theta}\\right)^{2} + R^{2} m^{2}}{{\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}} \\right) \\mathrm{d} T\\otimes \\mathrm{d} T + \\frac{1}{R^{2}} \\mathrm{d} R\\otimes \\mathrm{d} R +\\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta}$" ], "text/plain": [ "-(R^2*m^2*cos(th)^2 + R^2*m^2)/((cos(th)^2 + 1)*m^2) dT⊗dT + R^(-2) dR⊗dR + dth⊗dth" ] }, "execution_count": 34, "metadata": {}, "output_type": "execute_result" } ], "source": [ "s = (h - q)/(m^2*(1 + cos(th)^2))\n", "s.display()" ] }, { "cell_type": "code", "execution_count": 35, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle -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": [ "$\\displaystyle -R^{2} \\mathrm{d} T\\otimes \\mathrm{d} T + \\frac{1}{R^{2}} \\mathrm{d} R\\otimes \\mathrm{d} R +\\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta}$" ], "text/plain": [ "-R^2 dT⊗dT + R^(-2) dR⊗dR + dth⊗dth" ] }, "execution_count": 35, "metadata": {}, "output_type": "execute_result" } ], "source": [ "s.apply_map(factor)\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": 36, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\begin{array}{llcl} & \\mathcal{M} & \\longrightarrow & \\mathbb{R} \\\\ & \\left(t, r, {\\theta}, {\\phi}\\right) & \\longmapsto & 4 \\, m^{8} \\cos\\left({\\theta}\\right)^{6} + 4 \\, m^{8} \\cos\\left({\\theta}\\right)^{4} - 4 \\, m^{8} \\cos\\left({\\theta}\\right)^{2} - 4 \\, m^{8} \\\\ & \\left(T, R, {\\theta}, {\\Phi}\\right) & \\longmapsto & 4 \\, m^{8} \\cos\\left({\\theta}\\right)^{6} + 4 \\, m^{8} \\cos\\left({\\theta}\\right)^{4} - 4 \\, m^{8} \\cos\\left({\\theta}\\right)^{2} - 4 \\, m^{8} \\end{array}\$" ], "text/latex": [ "$\\displaystyle \\begin{array}{llcl} & \\mathcal{M} & \\longrightarrow & \\mathbb{R} \\\\ & \\left(t, r, {\\theta}, {\\phi}\\right) & \\longmapsto & 4 \\, m^{8} \\cos\\left({\\theta}\\right)^{6} + 4 \\, m^{8} \\cos\\left({\\theta}\\right)^{4} - 4 \\, m^{8} \\cos\\left({\\theta}\\right)^{2} - 4 \\, m^{8} \\\\ & \\left(T, R, {\\theta}, {\\Phi}\\right) & \\longmapsto & 4 \\, m^{8} \\cos\\left({\\theta}\\right)^{6} + 4 \\, m^{8} \\cos\\left({\\theta}\\right)^{4} - 4 \\, m^{8} \\cos\\left({\\theta}\\right)^{2} - 4 \\, m^{8} \\end{array}$" ], "text/plain": [ "M → ℝ\n", "(t, r, th, ph) ↦ 4*m^8*cos(th)^6 + 4*m^8*cos(th)^4 - 4*m^8*cos(th)^2 - 4*m^8\n", "(T, R, th, Ph) ↦ 4*m^8*cos(th)^6 + 4*m^8*cos(th)^4 - 4*m^8*cos(th)^2 - 4*m^8" ] }, "execution_count": 36, "metadata": {}, "output_type": "execute_result" } ], "source": [ "deth = h.determinant(frame=NH.frame())\n", "deth.display()" ] }, { "cell_type": "code", "execution_count": 37, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle 4 \\, {\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)}^{2} m^{8} {\\left(\\cos\\left({\\theta}\\right) + 1\\right)} {\\left(\\cos\\left({\\theta}\\right) - 1\\right)}\$" ], "text/latex": [ "$\\displaystyle 4 \\, {\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)}^{2} m^{8} {\\left(\\cos\\left({\\theta}\\right) + 1\\right)} {\\left(\\cos\\left({\\theta}\\right) - 1\\right)}$" ], "text/plain": [ "4*(cos(th)^2 + 1)^2*m^8*(cos(th) + 1)*(cos(th) - 1)" ] }, "execution_count": 37, "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 $h$ 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": 38, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\mathrm{Ric}\\left(h\\right) = 0\$" ], "text/latex": [ "$\\displaystyle \\mathrm{Ric}\\left(h\\right) = 0$" ], "text/plain": [ "Ric(h) = 0" ] }, "execution_count": 38, "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": 39, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\eta = \\frac{\\partial}{\\partial {\\Phi} }\$" ], "text/latex": [ "$\\displaystyle \\eta = \\frac{\\partial}{\\partial {\\Phi} }$" ], "text/plain": [ "eta = ∂/∂Ph" ] }, "execution_count": 39, "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": 40, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle 0\$" ], "text/latex": [ "$\\displaystyle 0$" ], "text/plain": [ "0" ] }, "execution_count": 40, "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": 41, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\xi_{1} = \\frac{\\partial}{\\partial T }\$" ], "text/latex": [ "$\\displaystyle \\xi_{1} = \\frac{\\partial}{\\partial T }$" ], "text/plain": [ "xi1 = ∂/∂T" ] }, "execution_count": 41, "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": 42, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle 0\$" ], "text/latex": [ "$\\displaystyle 0$" ], "text/plain": [ "0" ] }, "execution_count": 42, "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 **stationarity** of the Kerr metric. A third symmetry, which is not present in the original extremal 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": 43, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\xi_{2} = T \\frac{\\partial}{\\partial T } -R \\frac{\\partial}{\\partial R }\$" ], "text/latex": [ "$\\displaystyle \\xi_{2} = T \\frac{\\partial}{\\partial T } -R \\frac{\\partial}{\\partial R }$" ], "text/plain": [ "xi2 = T ∂/∂T - R ∂/∂R" ] }, "execution_count": 43, "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": 44, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle 0\$" ], "text/latex": [ "$\\displaystyle 0$" ], "text/plain": [ "0" ] }, "execution_count": 44, "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": 45, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\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": [ "$\\displaystyle \\xi_{3} = \\left( \\frac{1}{2} \\, T^{2} + \\frac{1}{2 \\, R^{2}} \\right) \\frac{\\partial}{\\partial T } -R T \\frac{\\partial}{\\partial R } -\\frac{1}{R} \\frac{\\partial}{\\partial {\\Phi} }$" ], "text/plain": [ "xi3 = (1/2*T^2 + 1/2/R^2) ∂/∂T - R*T ∂/∂R - 1/R ∂/∂Ph" ] }, "execution_count": 45, "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": 46, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle 0\$" ], "text/latex": [ "$\\displaystyle 0$" ], "text/plain": [ "0" ] }, "execution_count": 46, "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": 47, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\left[\\eta,\\xi_{1}\\right] = 0\$" ], "text/latex": [ "$\\displaystyle \\left[\\eta,\\xi_{1}\\right] = 0$" ], "text/plain": [ "[eta,xi1] = 0" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/html": [ "\$\\displaystyle \\left[\\eta,\\xi_{2}\\right] = 0\$" ], "text/latex": [ "$\\displaystyle \\left[\\eta,\\xi_{2}\\right] = 0$" ], "text/plain": [ "[eta,xi2] = 0" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/html": [ "\$\\displaystyle \\left[\\eta,\\xi_{3}\\right] = 0\$" ], "text/latex": [ "$\\displaystyle \\left[\\eta,\\xi_{3}\\right] = 0$" ], "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": 48, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\left[\\xi_{1},\\xi_{2}\\right] = \\frac{\\partial}{\\partial T }\$" ], "text/latex": [ "$\\displaystyle \\left[\\xi_{1},\\xi_{2}\\right] = \\frac{\\partial}{\\partial T }$" ], "text/plain": [ "[xi1,xi2] = ∂/∂T" ] }, "execution_count": 48, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi1.bracket(xi2).display()" ] }, { "cell_type": "code", "execution_count": 49, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\mathrm{True}\$" ], "text/latex": [ "$\\displaystyle \\mathrm{True}$" ], "text/plain": [ "True" ] }, "execution_count": 49, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi1.bracket(xi2) == xi1" ] }, { "cell_type": "code", "execution_count": 50, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\left[\\xi_{1},\\xi_{3}\\right] = T \\frac{\\partial}{\\partial T } -R \\frac{\\partial}{\\partial R }\$" ], "text/latex": [ "$\\displaystyle \\left[\\xi_{1},\\xi_{3}\\right] = T \\frac{\\partial}{\\partial T } -R \\frac{\\partial}{\\partial R }$" ], "text/plain": [ "[xi1,xi3] = T ∂/∂T - R ∂/∂R" ] }, "execution_count": 50, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi1.bracket(xi3).display()" ] }, { "cell_type": "code", "execution_count": 51, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\mathrm{True}\$" ], "text/latex": [ "$\\displaystyle \\mathrm{True}$" ], "text/plain": [ "True" ] }, "execution_count": 51, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi1.bracket(xi3) == xi2" ] }, { "cell_type": "code", "execution_count": 52, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\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": [ "$\\displaystyle \\left[\\xi_{2},\\xi_{3}\\right] = \\left( \\frac{R^{2} T^{2} + 1}{2 \\, R^{2}} \\right) \\frac{\\partial}{\\partial T } -R T \\frac{\\partial}{\\partial R } -\\frac{1}{R} \\frac{\\partial}{\\partial {\\Phi} }$" ], "text/plain": [ "[xi2,xi3] = 1/2*(R^2*T^2 + 1)/R^2 ∂/∂T - R*T ∂/∂R - 1/R ∂/∂Ph" ] }, "execution_count": 52, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi2.bracket(xi3).display()" ] }, { "cell_type": "code", "execution_count": 53, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\mathrm{True}\$" ], "text/latex": [ "$\\displaystyle \\mathrm{True}$" ], "text/plain": [ "True" ] }, "execution_count": 53, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi2.bracket(xi3) == xi3" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "To summarize, we have" ] }, { "cell_type": "code", "execution_count": 54, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\mathrm{True}\$" ], "text/latex": [ "$\\displaystyle \\mathrm{True}$" ], "text/plain": [ "True" ] }, "execution_count": 54, "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": 55, "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": 56, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\mathrm{True}\$" ], "text/latex": [ "$\\displaystyle \\mathrm{True}$" ], "text/plain": [ "True" ] }, "execution_count": 56, "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": 57, "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": 58, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\left(\\begin{array}{rr}\n", "0 & 1 \\\\\n", "0 & 0\n", "\\end{array}\\right)\$" ], "text/latex": [ "$\\displaystyle \\left(\\begin{array}{rr}\n", "0 & 1 \\\\\n", "0 & 0\n", "\\end{array}\\right)$" ], "text/plain": [ "[0 1]\n", "[0 0]" ] }, "execution_count": 58, "metadata": {}, "output_type": "execute_result" } ], "source": [ "E" ] }, { "cell_type": "code", "execution_count": 59, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\left(\\begin{array}{rr}\n", "0 & 0 \\\\\n", "1 & 0\n", "\\end{array}\\right)\$" ], "text/latex": [ "$\\displaystyle \\left(\\begin{array}{rr}\n", "0 & 0 \\\\\n", "1 & 0\n", "\\end{array}\\right)$" ], "text/plain": [ "[0 0]\n", "[1 0]" ] }, "execution_count": 59, "metadata": {}, "output_type": "execute_result" } ], "source": [ "F" ] }, { "cell_type": "code", "execution_count": 60, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\left(\\begin{array}{rr}\n", "1 & 0 \\\\\n", "0 & -1\n", "\\end{array}\\right)\$" ], "text/latex": [ "$\\displaystyle \\left(\\begin{array}{rr}\n", "1 & 0 \\\\\n", "0 & -1\n", "\\end{array}\\right)$" ], "text/plain": [ "[ 1 0]\n", "[ 0 -1]" ] }, "execution_count": 60, "metadata": {}, "output_type": "execute_result" } ], "source": [ "H" ] }, { "cell_type": "code", "execution_count": 61, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\mathrm{True}\$" ], "text/latex": [ "$\\displaystyle \\mathrm{True}$" ], "text/plain": [ "True" ] }, "execution_count": 61, "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": 62, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\\(\\displaystyle h = \\left( -\\frac{{\\left(m^{2} - 2 \\, m r + r^{2}\\right)} \\cos\\left({\\theta}\\right)^{4} + 2 \\, {\\left(9 \\, m^{2} - 10 \\, m r + 3 \\, r^{2}\\right)} \\cos\\left({\\theta}\\right)^{2} - 15 \\, m^{2} + 14 \\, m r - 3 \\, 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} \\cos\\left({\\theta}\\right)^{2} + m^{2}}{m^{2} - 2 \\, m r + r^{2}} \\right) \\mathrm{d} r\\otimes \\mathrm{d} r + {\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2} \\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": [ "$\\displaystyle h = \\left( -\\frac{{\\left(m^{2} - 2 \\, m r + r^{2}\\right)} \\cos\\left({\\theta}\\right)^{4} + 2 \\, {\\left(9 \\, m^{2} - 10 \\, m r + 3 \\, r^{2}\\right)} \\cos\\left({\\theta}\\right)^{2} - 15 \\, m^{2} + 14 \\, m r - 3 \\, r^{2}}{4 \\, {\\left(m^{2} \\cos\\left({\\theta}\\right)^{2} + m^{2}\\right)}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( -\\frac{2 \\, {\\left(2 \\, m - r\\right)} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} t\\otimes \\mathrm{d} {\\phi} + \\left( \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + m^{2}}{m^{2} - 2 \\, m r + r^{2}} \\right) \\mathrm{d} r\\otimes \\mathrm{d} r + {\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2} \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( -\\frac{2 \\, {\\left(2 \\, m - r\\right)} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} t + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} {\\phi}$" ], "text/plain": [ "h = -1/4*((m^2 - 2*m*r + r^2)*cos(th)^4 + 2*(9*m^2 - 10*m*r + 3*r^2)*cos(th)^2 - 15*m^2 + 14*m*r - 3*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*cos(th)^2 + m^2)/(m^2 - 2*m*r + r^2) dr⊗dr + (cos(th)^2 + 1)*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": 62, "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": 63, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\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": [ "$\\displaystyle \\left( -\\frac{m^{2} - 2 \\, m r + r^{2}}{4 \\, m^{4}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( \\frac{1}{m^{2} - 2 \\, m r + r^{2}} \\right) \\mathrm{d} r\\otimes \\mathrm{d} r +\\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta}$" ], "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": 63, "metadata": {}, "output_type": "execute_result" } ], "source": [ "s.display(BL)" ] }, { "cell_type": "code", "execution_count": 64, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle -\\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": [ "$\\displaystyle -\\frac{{\\left(m - r\\right)}^{2}}{4 \\, m^{4}} \\mathrm{d} t\\otimes \\mathrm{d} t + \\frac{1}{{\\left(m - r\\right)}^{2}} \\mathrm{d} r\\otimes \\mathrm{d} r +\\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta}$" ], "text/plain": [ "-1/4*(m - r)^2/m^4 dt⊗dt + (m - r)^(-2) dr⊗dr + dth⊗dth" ] }, "execution_count": 64, "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": 65, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\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": [ "$\\displaystyle \\left( \\frac{{\\left(4 \\, m^{2} - 4 \\, m r + r^{2}\\right)} \\sin\\left({\\theta}\\right)^{2}}{m^{2} \\cos\\left({\\theta}\\right)^{2} + m^{2}} \\right) \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( -\\frac{2 \\, {\\left(2 \\, m - r\\right)} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} t\\otimes \\mathrm{d} {\\phi} + \\left( -\\frac{2 \\, {\\left(2 \\, m - r\\right)} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} t + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} {\\phi}$" ], "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": 65, "metadata": {}, "output_type": "execute_result" } ], "source": [ "q.display(BL)" ] }, { "cell_type": "code", "execution_count": 66, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\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": [ "$\\displaystyle \\frac{{\\left(2 \\, m - r\\right)}^{2} \\sin\\left({\\theta}\\right)^{2}}{{\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}} \\mathrm{d} t\\otimes \\mathrm{d} t + \\left( -\\frac{2 \\, {\\left(2 \\, m - r\\right)} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} t\\otimes \\mathrm{d} {\\phi} + \\left( -\\frac{2 \\, {\\left(2 \\, m - r\\right)} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} t + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\phi}\\otimes \\mathrm{d} {\\phi}$" ], "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": 66, "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": 67, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\eta = \\frac{\\partial}{\\partial {\\phi} }\$" ], "text/latex": [ "$\\displaystyle \\eta = \\frac{\\partial}{\\partial {\\phi} }$" ], "text/plain": [ "eta = ∂/∂ph" ] }, "execution_count": 67, "metadata": {}, "output_type": "execute_result" } ], "source": [ "eta.display(BL)" ] }, { "cell_type": "code", "execution_count": 68, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\xi_{1} = \\frac{2 \\, m}{{\\epsilon}} \\frac{\\partial}{\\partial t } + \\frac{1}{{\\epsilon}} \\frac{\\partial}{\\partial {\\phi} }\$" ], "text/latex": [ "$\\displaystyle \\xi_{1} = \\frac{2 \\, m}{{\\epsilon}} \\frac{\\partial}{\\partial t } + \\frac{1}{{\\epsilon}} \\frac{\\partial}{\\partial {\\phi} }$" ], "text/plain": [ "xi1 = 2*m/eps ∂/∂t + 1/eps ∂/∂ph" ] }, "execution_count": 68, "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": 69, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\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": [ "$\\displaystyle \\xi_{2} = t \\frac{\\partial}{\\partial t } + \\left( m - r \\right) \\frac{\\partial}{\\partial r } + \\frac{t}{2 \\, m} \\frac{\\partial}{\\partial {\\phi} }$" ], "text/plain": [ "xi2 = t ∂/∂t + (m - r) ∂/∂r + 1/2*t/m ∂/∂ph" ] }, "execution_count": 69, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi2.display(BL)" ] }, { "cell_type": "code", "execution_count": 70, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\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": [ "$\\displaystyle \\xi_{3} = \\left( \\frac{4 \\, {\\epsilon} m^{4} + {\\left({\\epsilon} m^{2} - 2 \\, {\\epsilon} m r + {\\epsilon} r^{2}\\right)} t^{2}}{4 \\, {\\left(m^{3} - 2 \\, m^{2} r + m r^{2}\\right)}} \\right) \\frac{\\partial}{\\partial t } + \\frac{{\\left({\\epsilon} m - {\\epsilon} r\\right)} t}{2 \\, m} \\frac{\\partial}{\\partial r } + \\left( \\frac{12 \\, {\\epsilon} m^{4} - 8 \\, {\\epsilon} m^{3} r + {\\left({\\epsilon} m^{2} - 2 \\, {\\epsilon} m r + {\\epsilon} r^{2}\\right)} t^{2}}{8 \\, {\\left(m^{4} - 2 \\, m^{3} r + m^{2} r^{2}\\right)}} \\right) \\frac{\\partial}{\\partial {\\phi} }$" ], "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": 70, "metadata": {}, "output_type": "execute_result" } ], "source": [ "xi3.display(BL)" ] }, { "cell_type": "code", "execution_count": 71, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\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": [ "$\\displaystyle \\xi_{3} = \\frac{{\\left(4 \\, m^{4} + m^{2} t^{2} - 2 \\, m r t^{2} + r^{2} t^{2}\\right)} {\\epsilon}}{4 \\, {\\left(m - r\\right)}^{2} m} \\frac{\\partial}{\\partial t } + \\frac{{\\epsilon} {\\left(m - r\\right)} t}{2 \\, m} \\frac{\\partial}{\\partial r } + \\frac{{\\left(12 \\, m^{4} - 8 \\, m^{3} r + m^{2} t^{2} - 2 \\, m r t^{2} + r^{2} t^{2}\\right)} {\\epsilon}}{8 \\, {\\left(m - r\\right)}^{2} m^{2}} \\frac{\\partial}{\\partial {\\phi} }$" ], "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": 71, "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": 72, "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": 73, "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": 74, "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": 75, "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": [ "\$\\displaystyle \\mathrm{K}\\eta = 0\$" ], "text/latex": [ "$\\displaystyle \\mathrm{K}\\eta = 0$" ], "text/plain": [ "Keta = 0" ] }, "execution_count": 75, "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": 76, "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": [ "\$\\displaystyle \\mathrm{K}\\xi_{1} = 0\$" ], "text/latex": [ "$\\displaystyle \\mathrm{K}\\xi_{1} = 0$" ], "text/plain": [ "Kxi1 = 0" ] }, "execution_count": 76, "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": 77, "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": [ "\\(\\displaystyle \\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": [ "$\\displaystyle \\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": 77, "metadata": {}, "output_type": "execute_result" } ], "source": [ "Kxi2 = killing_form(xi2)\n", "print(Kxi2)\n", "Kxi2.display()" ] }, { "cell_type": "code", "execution_count": 78, "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": [ "\\(\\displaystyle \\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": [ "$\\displaystyle \\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": 78, "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": 79, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Chart (M, (ta, y, th, ps))\n" ] }, { "data": { "text/html": [ "\$\\displaystyle \\left(\\mathcal{M},({\\tau}, y, {\\theta}, {\\psi})\\right)\$" ], "text/latex": [ "$\\displaystyle \\left(\\mathcal{M},({\\tau}, y, {\\theta}, {\\psi})\\right)$" ], "text/plain": [ "Chart (M, (ta, y, th, ps))" ] }, "execution_count": 79, "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": 80, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\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": [ "$\\displaystyle \\left\\{\\begin{array}{lcl} T & = & \\frac{\\sqrt{y^{2} + 1} \\sin\\left({\\tau}\\right)}{\\sqrt{y^{2} + 1} \\cos\\left({\\tau}\\right) + y} \\\\ R & = & \\sqrt{y^{2} + 1} \\cos\\left({\\tau}\\right) + y \\\\ {\\theta} & = & {\\theta} \\\\ {\\Phi} & = & {\\psi} + \\log\\left(\\frac{y \\sin\\left({\\tau}\\right) + \\cos\\left({\\tau}\\right)}{\\sqrt{y^{2} + 1} \\sin\\left({\\tau}\\right) + 1}\\right) \\end{array}\\right.$" ], "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": 80, "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": 81, "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": 82, "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": 83, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\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": [ "$\\displaystyle \\left\\{\\begin{array}{lcl} {\\tau} & = & \\arctan\\left(2 \\, R^{2} T, -{\\left(T^{2} - 1\\right)} R^{2} + 1\\right) \\\\ y & = & \\frac{{\\left(T^{2} + 1\\right)} R^{2} - 1}{2 \\, R} \\\\ {\\theta} & = & {\\theta} \\\\ {\\psi} & = & {\\Phi} - \\log\\left(\\frac{{\\left(R T - 1\\right)}^{2} + R^{2}}{\\sqrt{{\\left({\\left(T^{2} + 1\\right)} R^{2} - 1\\right)}^{2} + 4 \\, R^{2}}}\\right) \\end{array}\\right.$" ], "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": 83, "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": 84, "metadata": {}, "outputs": [ { "data": { "image/png": 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", "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": 85, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\\(\\displaystyle 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} \\cos\\left({\\theta}\\right)^{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": [ "$\\displaystyle h = \\left( -\\frac{m^{2} \\cos\\left({\\theta}\\right)^{4} + 2 \\, m^{2} \\cos\\left({\\theta}\\right)^{2} + {\\left(m^{2} \\cos\\left({\\theta}\\right)^{4} + 6 \\, m^{2} \\cos\\left({\\theta}\\right)^{2} - 3 \\, m^{2}\\right)} y^{2} + m^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\tau}\\otimes \\mathrm{d} {\\tau} + \\left( \\frac{4 \\, m^{2} y \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\tau}\\otimes \\mathrm{d} {\\psi} + \\left( \\frac{m^{2} \\cos\\left({\\theta}\\right)^{2} + m^{2}}{y^{2} + 1} \\right) \\mathrm{d} y\\otimes \\mathrm{d} y + \\left( m^{2} \\cos\\left({\\theta}\\right)^{2} + m^{2} \\right) \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( \\frac{4 \\, m^{2} y \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\psi}\\otimes \\mathrm{d} {\\tau} + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\psi}\\otimes \\mathrm{d} {\\psi}$" ], "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*cos(th)^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.display(GNH)" ] }, { "cell_type": "code", "execution_count": 86, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\\(\\displaystyle 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(\\cos\\left({\\theta}\\right)^{2} + 1\\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": [ "$\\displaystyle h = \\left( -\\frac{{\\left(y^{2} \\cos\\left({\\theta}\\right)^{4} + 6 \\, y^{2} \\cos\\left({\\theta}\\right)^{2} + \\cos\\left({\\theta}\\right)^{4} - 3 \\, y^{2} + 2 \\, \\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\tau}\\otimes \\mathrm{d} {\\tau} + \\left( \\frac{4 \\, m^{2} y \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\tau}\\otimes \\mathrm{d} {\\psi} + \\left( \\frac{{\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2}}{y^{2} + 1} \\right) \\mathrm{d} y\\otimes \\mathrm{d} y + {\\left(\\cos\\left({\\theta}\\right)^{2} + 1\\right)} m^{2} \\mathrm{d} {\\theta}\\otimes \\mathrm{d} {\\theta} + \\left( \\frac{4 \\, m^{2} y \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\psi}\\otimes \\mathrm{d} {\\tau} + \\left( \\frac{4 \\, m^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1} \\right) \\mathrm{d} {\\psi}\\otimes \\mathrm{d} {\\psi}$" ], "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 + (cos(th)^2 + 1)*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": 86, "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": 87, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\frac{4 \\, m^{2} y^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1}\$" ], "text/latex": [ "$\\displaystyle \\frac{4 \\, m^{2} y^{2} \\sin\\left({\\theta}\\right)^{2}}{\\cos\\left({\\theta}\\right)^{2} + 1}$" ], "text/plain": [ "4*m^2*y^2*sin(th)^2/(cos(th)^2 + 1)" ] }, "execution_count": 87, "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": 88, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\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": [ "$\\displaystyle \\frac{\\partial}{\\partial {\\tau} } = \\left( \\frac{R^{2} T^{2} + R^{2} + 1}{2 \\, R^{2}} \\right) \\frac{\\partial}{\\partial T } -R T \\frac{\\partial}{\\partial R } -\\frac{1}{R} \\frac{\\partial}{\\partial {\\Phi} }$" ], "text/plain": [ "∂/∂ta = 1/2*(R^2*T^2 + R^2 + 1)/R^2 ∂/∂T - R*T ∂/∂R - 1/R ∂/∂Ph" ] }, "execution_count": 88, "metadata": {}, "output_type": "execute_result" } ], "source": [ "GNH.frame()[0].display(NH)" ] }, { "cell_type": "code", "execution_count": 89, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\$\\displaystyle \\mathrm{True}\$" ], "text/latex": [ "$\\displaystyle \\mathrm{True}$" ], "text/plain": [ "True" ] }, "execution_count": 89, "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 10.3.rc3", "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": "3.10.12" } }, "nbformat": 4, "nbformat_minor": 4 }