PDA

View Full Version : Equivalence of Lagrangian and ADM Hamiltonian


David Norton
Nov25-04, 03:37 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>I\'ve got a little problem that I\'d appreciate some help on. I\'m trying\nto demonstrate that the equations of motion produced by the ADM\nHamiltonian are the same as those produced by the Lagrangian for GR.\nSpecifically, I take the Lagrangian for GR to be\n\n\\begin{equation}\nL = \\int d^3x N\\sqrt{g}( R + (trK)^2 - K_{ij}K^{ij} )\n\\end{equation}\n\nwhere \\$R\\$ is the scalar curvature of a three manifold, \\$K_{ij}\\$ is the\nextrinsic curvature of the three manifold embedded in spacetime, and\n\\$trK=g^{ij}K_{ij}\\$ is the trace of the extrinsic curvature. Producing\nan expression for \\$\\dot{g}_{ij}\\$ is no problem, but I\'m stuck when\ntrying to get \\$\\dot{\\pi}^{ij}\\$. What I want to do is to show that the\nequation for \\$\\dot{\\pi}^{ij}\\$ that I get from varying the Lagrangian\nwith respect to \\$g_{ij}\\$ is the same as Eq. (21.115) in Misner, Thorne,\nand Wheeler, thereby proving that the Lagrangian and Hamiltonian\npictures predict the same dynamics. For the benefit of those who don\'t\nhave a copy of MTW to hand, I\'ll write out the equation here (apologies\nfor the \\LaTeX):\n\n\\begin{align}\n\\dot{\\pi}^{ij}\n&= -N\\sqrt{g}(R^{ij} - \\frac{1}{2}g^{ij}R) +\n\\frac{N}{2\\sqrt{g}}g^{ij}(\\pi^{mn}\\pi_{mn} - \\frac{1}{2}(tr\\pi)^2)\n\\nonumber \\\\\n& - \\frac{2N}{\\sqrt{g}}(\\pi^{im}\\pi_m^j - \\frac{1}{2}\\pi^{ij}tr\\pi)\n+ \\sqrt{g}(N^{|ij} - g^{ij}N^{|m}_{|m}) \\nonumber \\\\\n& +(\\pi^{ij}N^m)_{|m} - N^i_{|m}\\pi^{mj} - N^j_{|m}\\pi^{mi}\n\\end{align}\n\nNow, I can produce everything here apart from the last three terms. I\nsuspect that these terms arise through variation of \\$K_{ij}\\$, but can\'t\nimmediately see how to show this. Can anyone help point me in the right\ndirection?\n\nThanks in advance.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>I've got a little problem that I'd appreciate some help on. I'm trying
to demonstrate that the equations of motion produced by the ADM
Hamiltonian are the same as those produced by the Lagrangian for GR.
Specifically, I take the Lagrangian for GR to be

\begin{equation}L = \int d^{3x} N\sqrt{g}( R + (trK)^2 - K_{ij}K^{ij} )\end{equation}

where $R$ is the scalar curvature of a three manifold, $K_{ij}$ is the
extrinsic curvature of the three manifold embedded in spacetime, and
$trK=g^{ij}K_{ij}$ is the trace of the extrinsic curvature. Producing
an expression for $\dot{g}_{ij}$ is no problem, but I'm stuck when
trying to get $\dot{\pi}^{ij}$. What I want to do is to show that the
equation for $\dot{\pi}^{ij}$ that I get from varying the Lagrangian
with respect to $g_{ij}$ is the same as Eq. (21.115) in Misner, Thorne,
and Wheeler, thereby proving that the Lagrangian and Hamiltonian
pictures predict the same dynamics. For the benefit of those who don't
have a copy of MTW to hand, I'll write out the equation here (apologies
for the \LaTeX):\begin{align}\dot{\pi}^{ij}&= -N\sqrt{g}(R^{ij} - \frac{1}{2}g^{ij}R) +\frac{N}{2\sqrt{g}}g^{ij}(\pi^{mn}\pi_{mn} - \frac{1}{2}(tr\pi)^2)\nonumber \\& - \frac{2N}{\sqrt{g}}(\pi^{im}\pi_m^j - \frac{1}{2}\pi^{ij}tr\pi)+ \sqrt{g}(N^{|ij} - g^{ij}N^{|m}_{|m}) \nonumber \\& +(\pi^{ij}N^m)_{|m} - N^{i_}{|m}\pi^{mj} - N^{j_}{|m}\pi^{mi}\end{align}

Now, I can produce everything here apart from the last three terms. I
suspect that these terms arise through variation of $K_{ij}$, but can't
immediately see how to show this. Can anyone help point me in the right
direction?

Thanks in advance.

Philip McSweeney
Nov27-04, 06:37 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>% Note to moderators: Surprisingly, I couldn\'t locate the derivation of\nthis particular equation anywhere online so I think it may be helpful\nto include some TeX here just to illustrate a couple of things. I know\nthis is frowned upon, but he started it. :-)\n\n\\documentclass{article}\n\n\n\\usepackage{am smath}\n\\usepackage{amsfonts}\n\n\n\\begin{docume nt}\n\nSure. The first thing to note, however, is that the Lagrangian that\nyou\'ve given above is incorrect if you\'re working from MTW. It should\nbe\n\n\\begin{equation}\nL = \\int d^3x N\\sqrt{g}\n( R + K_{ij}K^{ij} - (\\text{tr}K)^2 ),\n\\end{equation}\n\n\\noindent so make sure that you\'re working through this problem with\nthis correct form for the Lagrangian. If you vary this with respect to\n\\$g_{ij}\\$ and can produce everything but the final three terms in\nEq. (21.115), eventually you\'ll see a term in the variation of the\nfollowing form\n\n\\begin{equation}\n\\delta A = \\int d^3x 2N\\pi^{ij}\\delta K_{ij}\n= \\int d^3x \\pi^{ij}\\delta(\\nabla_aN_b + \\nabla_bN_a).\n\\end{equation}\n\n\\noindent (Pay no attention to the fact that I\'ve denoted it by\n\\$\\delta A\\$.) You may have to rewrite some extrinsic curvature terms in\norder to get the \\$\\pi^{ij}\\$ in this. You are indeed correct when you\nsay that the interesting stuff comes from the variation of the\nextrinsic curvature. The temptation is to do something silly like\nsetting\n\n\\begin{equation}\n\\nabla_aN_b = \\nabla_a(g_{bm}N^m)\n\\end{equation}\n\n\\noinden t and then varying this equation with respect to \\$g_{ab}\\$ to\nproduce something like\n\n\\begin{equation}\n\\nabla_aN_b = N^m\\nabla_a\\delta g_{bm} +\n(\\nabla_aN^m)\\delta g_{bm}.\n\\end{equation}\n\n\\noindent This is incorrect! The way to do it is to write out\n\n\\begin{equation}\n\\nabla_a N_b\n= \\partial_aN_b - \\Gamma^m_{ab}N_m\n\\end{equation}\n\n\\noindent and then vary this equation. The two important things to\nnote are (i) in this picture we treat the lapse, shift, metric, and\ntime-derivative of the metric as being variables that are varied\nindependently of one another, so \\$\\delta N_m = 0\\$ and (ii) the\nvariation of the Levi-Civita connection is given by\n\n\\begin{equation}\n\\delta \\Gamma^m_{ab}\n= \\frac{1}{2}g^{mn}\n(\\nabla_a\\delta g_{bn} + \\nabla_b\\delta g_{na}\n- \\nabla_n\\delta g_{ab} ).\n\\end{equation}\n\n\\noindent (You can show this easily by considering permutations of\nindices.) Now vary Eq. (5) according to this rule, substitute the\nresults back into Eq. (2), integrate by parts, and Bob\'s your uncle -\nyou\'ve got the desired final three terms in Eq. (21.115) of MTW.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>% Note to moderators: Surprisingly, I couldn't locate the derivation of
this particular equation anywhere online so I think it may be helpful
to include some TeX here just to illustrate a couple of things. I know
this is frowned upon, but he started it. :-)

\documentclass{article}\usepackage{amsmath}\usepac kage{amsfonts}\begin{document}Sure. The first thing to note, however, is that the Lagrangian thatyou've given above is incorrect if you're working from MTW. It should
be

\begin{equation}L = \int d^{3x} N\sqrt{g}( R + K_{ij}K^{ij} - (\text{tr}K)^2 ),\end{equation}\noindent so make sure that you're working through this problem with
this correct form for the Lagrangian. If you vary this with respect to
$g_{ij}$ and can produce everything but the final three terms in
Eq. (21.115), eventually you'll see a term in the variation of the
following form

\begin{equation}\delta A = \int d^{3x} 2N\pi^{ij}\delta K_{ij}= \int d^{3x} \pi^{ij}\delta(\nabla_aN_b + \nabla_bN_a).\end{equation}\noindent (Pay no attention to the fact that I've denoted it by$\delta A$.) You may have to rewrite some extrinsic curvature terms in
order to get the $\pi^{ij}$ in this. You are indeed correct when you
say that the interesting stuff comes from the variation of the
extrinsic curvature. The temptation is to do something silly like
setting

\begin{equation}\nabla_aN_b = \nabla_a(g_{bm}N^m)\end{equation}\noindent and then varying this equation with respect to $g_{ab}$ to
produce something like

\begin{equation}\nabla_aN_b = N^m\nabla_a\delta g_{bm} +(\nabla_aN^m)\delta g_{bm}.\end{equation}\noindent This is incorrect! The way to do it is to write out

\begin{equation}\nabla_a N_b= \partial_aN_b - \Gamma^m_{ab}N_m\end{equation}\noindent and then vary this equation. The two important things to
note are (i) in this picture we treat the lapse, shift, metric, and
time-derivative of the metric as being variables that are varied
independently of one another, so $\delta N_m = 0$ and (ii) the
variation of the Levi-Civita connection is given by

\begin{equation}\delta \Gamma^m_{ab}= \frac{1}{2}g^{mn}(\nabla_a\delta g_{bn} + \nabla_b\delta g_{na}- \nabla_n\delta g_{ab} ).\end{equation}\noindent (You can show this easily by considering permutations of
indices.) Now vary Eq. (5) according to this rule, substitute the
results back into Eq. (2), integrate by parts, and Bob's your uncle -
you've got the desired final three terms in Eq. (21.115) of MTW.