View Full Version : Re: Scharzchild Coordinates for Interior of Event Horizon
tessel@um.bot
Apr7-04, 08:33 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resize=yes,status=no,wi dth=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>On Thu, 1 Apr 2004, Daryl McCullough wrote:\n\n> But that\'s exactly what I can\'t do, because the integrand blows\n> up in the middle of the range (at r=2M).\n\nYes. You have here two -distinct- charts which are defined on\n-nonoverlapping domains-:\n\n1. (right) exterior region:\n\nds^2 = -(1-2m/r) dt^2 + dr^2/(1-2m/r) + r^2 (du^2 + sin(u)^2 dv^2),\n\n-infty < t < infty, 2m < r < infty, 0 < u < pi, -pi < v < pi\n\n888888\n/\\ /\\\n/ \\ /**\\ scri^+\n/ \\/****\\\n\\ /\\****/\n\\ / \\**/\n\\/ \\/ scri^-\n888888\n\n\n2. (future) interior region:\n\nds^2 = -(1-2m/r) dt^2 + dr^2/(1-2m/r) + r^2 (du^2 + sin(u)^2 dv^2),\n\n-infty < t < infty, 0 < r < 2m, 0 < u < pi, -pi < v < pi\n\nwhere I suggested renaming the coordinats as follows:\n\nds^2 = -dt^2/(2m/t -1) + (2m/t-1) dz^2 + t^2 (du^2 + sin(u)^2 dv^2),\n\n0 < t < 2m, -infty < z < infty, 0 < u < pi, -pi < v < pi\n\n888888\n/\\****/\\\n/ \\**/ \\ scri^+\n/ \\/ \\\n\\ /\\ /\n\\ / \\ /\n\\/ \\/ scri^-\n888888\n\nwhere now increasing t <--> decreasing proper time of infalling observer,\nand where the spacelike curvature singularity is located at t=0; note too\nthat here the coordinate vector @/@z is a Killing vector field,\ncorresponding to axial translation. (See also the pictures in MTW\nsketching E^3 embeddings of two-dimensional spacelike slices through the\nSchwarzschild vacuum!) Thus, as Charles Torre already noted, in the\ninterior chart, the value of your "t" coordinate (my "z" coordinate) at\nwhich a given timelike geodesic strikes the singularity is determined by\nthe initial conditions.\n\nThe fact that the domains of the two charts are -disjoint- means that to\ncontinue any curve from the exterior chart into the interior chart, you\nmust introduce a -third- chart defined (at least) on 2m-e < r < 2m + e,\nwhere e > 0 is some small constant, solve the geodesic equation in all\nthree charts, and find a piecewise definition of the world line in\nquestion.\n\n(Once you know how to continue -one- geodesic, you know how to continue\nall of them. But saying how to continue -one- geodesic is essentially\nchoosing -one- interior chart from an infinite family related by\n"transition diffeos" t -> t + c defined on\n\n-infty < t < infty, 0 < r < 2m, 0 < u < pi, -pi < v < pi\n\nwhere I refer to the original, misnamed coordinates. Does this help, or\ndoes it only add to your confusion? If the latter, sorry I mentioned it.)\n\nIt is considerably easier to just introduce just -one- chart covering the\nentire history of your observer, and solve the geodesic equation in that\nchart. The Painleve chart (1921) is the easiest to use, but some\ntextbooks also do this using the LeMaitre chart, a comoving chart--- see\nfor example Dirac\'s textbook. For Painleve, see the exercise on\nPainleve-Novikov charts in my previous post, which are suitable for\nobservers falling in freely and radially from rest at r = A, where A > 2m\nis a constant. Taking the limit A -> infty gives back the Painleve chart\nanalysis of observers falling in freely and radially from rest at spatial\ninfinity.\n\n(Amusing trivia question: how are the prominent mathematicians Painleve\nand Poincare related to the prominent contemporary politicians by those\nnames?)\n\n> My feeling is that this indicates that there is something fishy about\n> trying to equate values of t on one side of the event horizon with\n> values of t on the other side. You basically have one region (the\n> interior region) with Schwarzschild coordinates (r_interior,t_interior)\n> and another region with Schwarzschild coordinates (r_exterior,\n> t_exterior), and the only connection between them is that at the event\n> horizon\n\nYou have here very nearly arrived at what I just said!:\n\n"The fact that the domains of the two charts are -disjoint- means that to\ncontinue any curve from one into the other, you must introduce a -third-\nchart [straddling the horizon], solve the geodesic equation in all three\ncharts, and find a piecewise definition of the world line in question."\n\nI am pretty sure the entire confusion here derives from inadequate\nappreciation of the definition of a smooth manifold in terms of\noverlapping coordinate charts (with "transition diffeos" on the overlaps),\nso you might want to review that, e.g. in an early chapter of\n\nauthor = {William M. Boothby},\ntitle = {An Introduction to Differentiable Manifolds and {R}iemannian\nGeometry},\nedition = {Second},\nseries = {Pure and Applied Mathematics},\nvolume = 120,\npublisher = {Academic Press},\nyear = 1986}\n\n(see also the exercises at the end of the chapter in question).\n\nThe exact point under discussion here is also discussed at length -with\ngood illustrations- somewhere in MTW.\n\nHTH,\n\n"T. Essel" (hiding somewhere in cyberspace)\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Thu, 1 Apr 2004, Daryl McCullough wrote:
> But that's exactly what I can't do, because the integrand blows
> up in the middle of the range (at r=2M).
Yes. You have here two -distinct- charts which are defined on
-nonoverlapping domains-:
1. (right) exterior region:
ds^2 = -(1-2m/r) dt^2 + dr^2/(1-2m/r) + r^2 (du^2 + sin(u)^2 dv^2),
-infty < t < infty, 2m < r < infty, < u < \pi, -\pi < v < \pi
888888
/\ /\
/ \ /**\ scri^+
/ \/****\
\ /\****/
\ / \**/
\/ \/ scri^-
888888
2. (future) interior region:
ds^2 = -(1-2m/r) dt^2 + dr^2/(1-2m/r) + r^2 (du^2 + sin(u)^2 dv^2),
-infty < t < infty, < r < 2m, < u < \pi, -\pi < v < \pi
where I suggested renaming the coordinats as follows:
ds^2 = -dt^2/(2m/t -1) + (2m/t-1) dz^2 + t^2 (du^2 + sin(u)^2 dv^2),
< t < 2m, -infty < z < infty, < u < \pi, -\pi < v < \pi
888888
/\****/\
/ \**/ \ scri^+
/ \/ \
\ /\ /
\ / \ /
\/ \/ scri^-
888888
where now increasing t <--> decreasing proper time of infalling observer,
and where the spacelike curvature singularity is located at t=0; note too
that here the coordinate vector @/@z is a Killing vector field,
corresponding to axial translation. (See also the pictures in MTW
sketching E^3 embeddings of two-dimensional spacelike slices through the
Schwarzschild vacuum!) Thus, as Charles Torre already noted, in the
interior chart, the value of your "t" coordinate (my "z" coordinate) at
which a given timelike geodesic strikes the singularity is determined by
the initial conditions.
The fact that the domains of the two charts are -disjoint- means that to
continue any curve from the exterior chart into the interior chart, you
must introduce a -third- chart defined (at least) on 2m-e < r < 2m + e,
where e > is some small constant, solve the geodesic equation in all
three charts, and find a piecewise definition of the world line in
question.
(Once you know how to continue -one- geodesic, you know how to continue
all of them. But saying how to continue -one- geodesic is essentially
choosing -one- interior chart from an infinite family related by
"transition diffeos" t -> t + c defined on
-infty < t < infty, < r < 2m, < u < \pi, -\pi < v < \pi
where I refer to the original, misnamed coordinates. Does this help, or
does it only add to your confusion? If the latter, sorry I mentioned it.)
It is considerably easier to just introduce just -one- chart covering the
entire history of your observer, and solve the geodesic equation in that
chart. The Painleve chart (1921) is the easiest to use, but some
textbooks also do this using the LeMaitre chart, a comoving chart--- see
for example Dirac's textbook. For Painleve, see the exercise on
Painleve-Novikov charts in my previous post, which are suitable for
observers falling in freely and radially from rest at r = A, where A > 2m
is a constant. Taking the limit A -> infty gives back the Painleve chart
analysis of observers falling in freely and radially from rest at spatial
infinity.
(Amusing trivia question: how are the prominent mathematicians Painleve
and Poincare related to the prominent contemporary politicians by those
names?)
> My feeling is that this indicates that there is something fishy about
> trying to equate values of t on one side of the event horizon with
> values of t on the other side. You basically have one region (the
> interior region) with Schwarzschild coordinates (r_{interior},t_{interior})
> and another region with Schwarzschild coordinates (r_{exterior},
> t_{exterior}), and the only connection between them is that at the event
> horizon
You have here very nearly arrived at what I just said!:
"The fact that the domains of the two charts are -disjoint- means that to
continue any curve from one into the other, you must introduce a -third-
chart [straddling the horizon], solve the geodesic equation in all three
charts, and find a piecewise definition of the world line in question."
I am pretty sure the entire confusion here derives from inadequate
appreciation of the definition of a smooth manifold in terms of
overlapping coordinate charts (with "transition diffeos" on the overlaps),
so you might want to review that, e.g. in an early chapter of
author = {William M. Boothby},
title = {An Introduction to Differentiable Manifolds and {R}iemannian
Geometry},
edition = {Second},
series = {Pure and Applied Mathematics},
volume = 120,
publisher = {Academic Press},
year = 1986}
(see also the exercises at the end of the chapter in question).
The exact point under discussion here is also discussed at length -with
good illustrations- somewhere in MTW.
HTH,
"T. Essel" (hiding somewhere in cyberspace)
Bruce Pew
Apr7-04, 08:44 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resize=yes,status=no,wi dth=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>daryl@atc-nycorp.com (Daryl McCullough) wrote in message news:<c4hjf50msv@drn.newsguy.com>...\n> Charles Torre says...\n>\n> >To get a formula for the final t value for geodesics as a function\n> >of the initial conditions you will have to solve the geodesic equation.\n>\n> Yes, that\'s what my question was about: how to solve the geodesic\n> equation. In my original post, I *solved* the geodesic equation,\n> except for an arbitrary undetermined constant. I wanted to know\n> how to evaluate that constant.\n>\n> Basically, the geodesic equation gives you a differential equation\n> relating the Schwarzschild coordinate t to the Schwarzschild coordinate\n> r:\n>\n> dt = - f(r) dr\n>\n> In my original post, I gave the explicit form of f(r). The important\n> fact about f(r) is that it has a pole at r=2M. I don\'t know how to\n> integrate f(r) past a region where it contains a pole. (I can do it\n> using an analytic continuation, but then t picks up an imaginary\n> component as you pass the pole. That doesn\'t seem to make any physical\n> sense.)\n>\n> If you let t0 be the value of the Schwarzschild coordinate t at which\n> the observer starts falling (from r > 2M), and let t1 be the value of\n> the Schwarzschild coordinate t at which the observer reaches r=0, then\n> you can solve for t(r) (the value of t at which the observer reaches\n> radius r) for any r as follows:\n>\n> for r > 2M\n> t(r) = t0 + Integral from r to r0 of |f(r)| dr\n>\n> for r < 2M\n> t(r) = t1 + Integral from 0 to r of |f(r)| dr\n>\n> I don\'t see how to relate t0 to t1.\n\nIf you use Steve Carlip\'s definition of the Schwarzschild observer\n\n"The term ``Schwarzschild observer\'\' is, in fact, used as a technical\nterm in the literature, where it means---without any exception I have\never seen---``a single observer at rest in Schwarzschild\ncoordinates.\'\'"\n\nhttp://groups.google.com/groups?hl=en&lr=&ie=UTF-8&frame=right&rnum=31&thl=878612473,878520962,878435850,878504939,878285 746,877818641,877577114,878544382,878519996,878647 639,878624232,877816977&seekm=406d6837%240%243054%2461fed72c%40news.rcn.co m#link35\n\nthere are no shells inside the event horizon. The correlation between\ndt [at infinity] and dTau\n\n(1-2M/r)dt/dTau = 1\n\nThe correlation between radial r-coordinate and dt\n\ndr/dt = -(1-2M/r)(2M/r)^1/2\n\nSolving for dr/dTau\n\ndr/dTau = -(2M/r)^1/2\n\ndTau = - (r^1/2 dr)/(2M^1/2)\n\nIntegrate this and\n\nTau_2 - Tau_1 = (1/3)(2/M)^1/2 (r^3/2_1 - r^3/2_2)\n\nThe elapsed wristwatch time for any choice of r_1 and r_2.\n\ndTau = (1-2M/r)\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>daryl@atc-nycorp.com (Daryl McCullough) wrote in message news:<c4hjf50msv@drn.newsguy.com>...
> Charles Torre says...
>
> >To get a formula for the final t value for geodesics as a function
> >of the initial conditions you will have to solve the geodesic equation.
>
> Yes, that's what my question was about: how to solve the geodesic
> equation. In my original post, I *solved* the geodesic equation,
> except for an arbitrary undetermined constant. I wanted to know
> how to evaluate that constant.
>
> Basically, the geodesic equation gives you a differential equation
> relating the Schwarzschild coordinate t to the Schwarzschild coordinate
> r:
>
> dt = - f(r) dr
>
> In my original post, I gave the explicit form of f(r). The important
> fact about f(r) is that it has a pole at r=2M. I don't know how to
> integrate f(r) past a region where it contains a pole. (I can do it
> using an analytic continuation, but then t picks up an imaginary
> component as you pass the pole. That doesn't seem to make any physical
> sense.)
>
> If you let t0 be the value of the Schwarzschild coordinate t at which
> the observer starts falling (from r > 2M), and let t1 be the value of
> the Schwarzschild coordinate t at which the observer reaches r=0, then
> you can solve for t(r) (the value of t at which the observer reaches
> radius r) for any r as follows:
>
> for r > 2M
> t(r) = t0 + Integral from r to r0 of |f(r)| dr
>
> for r < 2M
> t(r) = t1 + Integral from to r of |f(r)| dr
>
> I don't see how to relate t0 to t1.
If you use Steve Carlip's definition of the Schwarzschild observer
"The term ``Schwarzschild observer'' is, in fact, used as a technical
term in the literature, where it means---without any exception I have
ever seen---``a single observer at rest in Schwarzschild
coordinates.''"
http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&frame=right&rnum=31&thl=878612473,878520962,878435850,878504939,878285 746,877818641,877577114,878544382,878519996,878647 639,878624232,877816977&seekm=406d6837%240%243054%2461fed72c%40news.rcn.co m#link35
there are no shells inside the event horizon. The correlation between
dt [at infinity] and dTau
(1-2M/r)dt/dTau = 1
The correlation between radial r-coordinate and dt
dr/dt = -(1-2M/r)(2M/r)^1/2
Solving for dr/dTau
dr/dTau = -(2M/r)^1/2
dTau = - (r^1/2 dr)/(2M^1/2)
Integrate this and
\Tau_2 - \Tau_1 = (1/3)(2/M)^1/2 (r^3/2_1 - r^3/2_2)
The elapsed wristwatch time for any choice of r_1 and r_2.
dTau = (1-2M/r)
Daryl McCullough
Apr7-04, 08:55 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resize=yes,status=no,wi dth=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>John Anderson says...\n\n>There\'s a coordinate singularity at the horizon in\n>Schwarzschildcoordinates. There\'s no real singularity. Therefore, it\'s\n>wrong to use Schwarzschild coordinates inside the horizon.\n>They aren\'t valid inside the horizon.\n\nThey aren\'t singular *inside* the horizon---they are only singular\n*at* the horizon. So there is nothing wrong in principle with using\nSchwarzschild coordinates everywhere except for a patch of spacetime\nnear the horizon.\n\n>Use Kruskal coordinates which, are not singular at the horizon,\n>to relate what\'s going on inside and outside.\n\nI actually looked at Kruskal coordinates, and I still\nfind it mysterious.\n\nFor a freefalling particle that falls radially from infinity (so that\nfar from the event horizon the 4-velocity is given by U^t = 1, U^r = 0),\nthe relationship between r and t is given by\n\nt = t0 + [r^{3/2}/(3/2) + r^{1/2}/(1/2) + log((1 + sqrt(r))/(1 - sqrt(r)))]\n\nwhere t0 = the value of t for which the infalling particle hits the\nsingularity at r=0, (and where I\'ve chosen units so that 2M = 1)\n\nThe problem is that the argument to the log becomes *negative* outside\nthe horizon. That corresponds to a *complex* value of t. For r > 1,\nthe above expression can be rewritten as\n\nt = t0 + i pi +\n[r^{3/2}/(3/2) + r^{1/2}/(1/2) + log((1 + sqrt(r))/(sqrt(r) - 1))]\n\n(where I used log(-1) = i pi).\n\nSo it seems that for t to be real outside the horizon, we\nhave to choose t0 to be complex. In particular, t0 must\nbe of the form A + pi i, where A is real.\n\nThis sounds nonsensical (or maybe not---coordinates\nare just labels, rather than observables, so there may be nothing\nwrong with having a complex value of t). So I\'ve been wanting\nsomeone to either explain how to get rid of that imaginary\ncomponent, or else reassure me that it\'s harmless.\n\nAnyway, anticipating John Anderson\'s suggestion, I looked up\nKruskal coordinates which are well-behaved at the horizon. The\nrelationship between Kruskal coordinates and the Schwarzschild\ncoordinate t is (again in units where 2M = 1):\n(See\nhttp://scienceworld.wolfram.com/physics/SchwarzschildBlackHoleKruskalCoordinates.html)\n\n t = 2 arctanh(v/u)\n\nWhere u is the spacelike coordinate, and v is the timelike coordinate.\n\nIn Kruskal coordinates, "outside" the event horizon corresponds to\nu > v, and "inside" the event horizon corresponds to u < v.\nA fact about arctanh is this: arctanh(x) is only real for |x| < 1.\nFor |x| > 1, we can use\n\narctanh(x) = i pi/2 + arctanh(1/x)\n\nThis means that inside the event horizon (where v > u), the above\nequation for t gives:\n\nt = i pi + arctanh(u/v)\n\nSo the Kruskal coordinates *also* seem to suggest that t becomes\ncomplex inside the event horizon.\n\nI\'m guessing that the resolution is that the value of t is arbitrary\nin Schwarzschild coordinates, and that the zero for t can be chosen\n*independently* on each size of the event horizon so as to make everything\nreal. But this means that there is really no meaningful relationship\nbetween values of t on one side and values of t on the other side.\n\n--\nDaryl McCullough\nIthaca, NY\n\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>John Anderson says...
>There's a coordinate singularity at the horizon in
>Schwarzschildcoordinates. There's no real singularity. Therefore, it's
>wrong to use Schwarzschild coordinates inside the horizon.
>They aren't valid inside the horizon.
They aren't singular *inside* the horizon---they are only singular
*at* the horizon. So there is nothing wrong in principle with using
Schwarzschild coordinates everywhere except for a patch of spacetime
near the horizon.
>Use Kruskal coordinates which, are not singular at the horizon,
>to relate what's going on inside and outside.
I actually looked at Kruskal coordinates, and I still
find it mysterious.
For a freefalling particle that falls radially from infinity (so that
far from the event horizon the 4-velocity is given by U^t = 1, U^r = 0),
the relationship between r and t is given by
t = t0 + [r^{3/2}/(3/2) + r^{1/2}/(1/2) +[/itex] log((1 + \sqrt(r))/(1 - \sqrt(r)))]
where t0 = the value of t for which the infalling particle hits the
singularity at r=0, (and where I've chosen units so that 2M = 1)
The problem is that the argument to the log becomes *negative* outside
the horizon. That corresponds to a *complex* value of t. For r > 1,
the above expression can be rewritten as
t = t0 + i \pi +
[r^{3/2}/(3/2) + r^{1/2}/(1/2) + log((1 [itex]+ \sqrt(r))/(\sqrt(r) - 1))]
(where I used log(-1) = i \pi).
So it seems that for t to be real outside the horizon, we
have to choose t0 to be complex. In particular, t0 must
be of the form A + \pi i, where A is real.
This sounds nonsensical (or maybe not---coordinates
are just labels, rather than observables, so there may be nothing
wrong with having a complex value of t). So I've been wanting
someone to either explain how to get rid of that imaginary
component, or else reassure me that it's harmless.
Anyway, anticipating John Anderson's suggestion, I looked up
Kruskal coordinates which are well-behaved at the horizon. The
relationship between Kruskal coordinates and the Schwarzschild
coordinate t is (again in units where 2M = 1):
(See
http://scienceworld.wolfram.com/physics/SchwarzschildBlackHoleKruskalCoordinates.html)
t = 2 arctanh(v/u)
Where u is the spacelike coordinate, and v is the timelike coordinate.
In Kruskal coordinates, "outside" the event horizon corresponds to
u > v, and "inside" the event horizon corresponds to u < v.
A fact about arctanh is this: arctanh(x) is only real for |x| < 1.
For |x| > 1, we can use
arctanh(x) = i \pi/2 + arctanh(1/x)
This means that inside the event horizon (where v > u), the above
equation for t gives:
t = i \pi + arctanh(u/v)
So the Kruskal coordinates *also* seem to suggest that t becomes
complex inside the event horizon.
I'm guessing that the resolution is that the value of t is arbitrary
in Schwarzschild coordinates, and that the zero for t can be chosen
*independently* on each size of the event horizon so as to make everything
real. But this means that there is really no meaningful relationship
between values of t on one side and values of t on the other side.
--
Daryl McCullough
Ithaca, NY
<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>daryl@atc-nycorp.com (Daryl McCullough) wrote in message news:<c4p71701qe8@drn.newsguy.com>...\n> John Anderson says...\n>\n> >There\'s a coordinate singularity at the horizon in\n> >Schwarzschildcoordinates. There\'s no real singularity. Therefore, it\'s\n> >wrong to use Schwarzschild coordinates inside the horizon.\n> >They aren\'t valid inside the horizon.\n>\n> They aren\'t singular *inside* the horizon---they are only singular\n> *at* the horizon. So there is nothing wrong in principle with using\n> Schwarzschild coordinates everywhere except for a patch of spacetime\n> near the horizon.\n>\n> >Use Kruskal coordinates which, are not singular at the horizon,\n> >to relate what\'s going on inside and outside.\n>\n> I actually looked at Kruskal coordinates, and I still\n> find it mysterious.\n>\n> For a freefalling particle that falls radially from infinity (so that\n> far from the event horizon the 4-velocity is given by U^t = 1, U^r = 0),\n> the relationship between r and t is given by\n>\n> t = t0 + [r^{3/2}/(3/2) + r^{1/2}/(1/2) + log((1 + sqrt(r))/(1 - sqrt(r)))]\n>\n> where t0 = the value of t for which the infalling particle hits the\n> singularity at r=0, (and where I\'ve chosen units so that 2M = 1)\n>\n> The problem is that the argument to the log becomes *negative* outside\n> the horizon. That corresponds to a *complex* value of t. For r > 1,\n> the above expression can be rewritten as\n>\n> t = t0 + i pi +\n> [r^{3/2}/(3/2) + r^{1/2}/(1/2) + log((1 + sqrt(r))/(sqrt(r) - 1))]\n>\n> (where I used log(-1) = i pi).\n>\n> So it seems that for t to be real outside the horizon, we\n> have to choose t0 to be complex. In particular, t0 must\n> be of the form A + pi i, where A is real.\n>\n> This sounds nonsensical (or maybe not---coordinates\n> are just labels, rather than observables, so there may be nothing\n> wrong with having a complex value of t). So I\'ve been wanting\n> someone to either explain how to get rid of that imaginary\n> component, or else reassure me that it\'s harmless.\n>\n> Anyway, anticipating John Anderson\'s suggestion, I looked up\n> Kruskal coordinates which are well-behaved at the horizon. The\n> relationship between Kruskal coordinates and the Schwarzschild\n> coordinate t is (again in units where 2M = 1):\n> (See\n> http://scienceworld.wolfram.com/physics/SchwarzschildBlackHoleKruskalCoordinates.html)\n>\ n> t = 2 arctanh(v/u)\n> Where u is the spacelike coordinate, and v is the timelike coordinate.\n\nActually that is not complete. It is\nt = 2 arctanh[(v/u)^(+/-1)]\nThe sign is chosen depending on which side of the horizon you are\nconsidering. Compare equations 12.2.1 and 12.2.2 at\nhttp://www.geocities.com/zcphysicsms/chap12.htm#BM12_2\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>daryl@atc-nycorp.com (Daryl McCullough) wrote in message news:<c4p71701qe8@drn.newsguy.com>...
> John Anderson says...
>
> >There's a coordinate singularity at the horizon in
> >Schwarzschildcoordinates. There's no real singularity. Therefore, it's
> >wrong to use Schwarzschild coordinates inside the horizon.
> >They aren't valid inside the horizon.
>
> They aren't singular *inside* the horizon---they are only singular
> *at* the horizon. So there is nothing wrong in principle with using
> Schwarzschild coordinates everywhere except for a patch of spacetime
> near the horizon.
>
> >Use Kruskal coordinates which, are not singular at the horizon,
> >to relate what's going on inside and outside.
>
> I actually looked at Kruskal coordinates, and I still
> find it mysterious.
>
> For a freefalling particle that falls radially from infinity (so that
> far from the event horizon the 4-velocity is given by U^t = 1, U^r = 0),
> the relationship between r and t is given by
>
> t = t0 + [r^{3/2}/(3/2) + r^{1/2}/(1/2) + log((1 + \sqrt(r))/(1 - \sqrt(r)))]
>
> where t0 = the value of t for which the infalling particle hits the
> singularity at r=0, (and where I've chosen units so that 2M = 1)
>
> The problem is that the argument to the log becomes *negative* outside
> the horizon. That corresponds to a *complex* value of t. For r > 1,
> the above expression can be rewritten as
>
> t = t0 + i \pi +
> [r^{3/2}/(3/2) + r^{1/2}/(1/2) + log((1 + \sqrt(r))/(\sqrt(r) - 1))]
>
> (where I used log(-1) = i \pi).
>
> So it seems that for t to be real outside the horizon, we
> have to choose t0 to be complex. In particular, t0 must
> be of the form A + \pi i, where A is real.
>
> This sounds nonsensical (or maybe not---coordinates
> are just labels, rather than observables, so there may be nothing
> wrong with having a complex value of t). So I've been wanting
> someone to either explain how to get rid of that imaginary
> component, or else reassure me that it's harmless.
>
> Anyway, anticipating John Anderson's suggestion, I looked up
> Kruskal coordinates which are well-behaved at the horizon. The
> relationship between Kruskal coordinates and the Schwarzschild
> coordinate t is (again in units where 2M = 1):
> (See
> http://scienceworld.wolfram.com/physics/SchwarzschildBlackHoleKruskalCoordinates.html)
>
> t = 2 arctanh(v/u)
> Where u is the spacelike coordinate, and v is the timelike coordinate.
Actually that is not complete. It is
t = 2 arctanh[(v/u)^(+/-1)]
The sign is chosen depending on which side of the horizon you are
considering. Compare equations 12.2.1 and 12.2.2 at
http://www.geocities.com/zcphysicsms/chap12.htm#BM12_2
Bill Hobba
Apr7-04, 09:28 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>\n\n"Daryl McCullough" <daryl@atc-nycorp.com> wrote in message\nnews:c4p71701qe8@drn.newsguy.com...\n> John Anderson says...\n>\n> >There\'s a coordinate singularity at the horizon in\n> >Schwarzschildcoordinates. There\'s no real singularity. Therefore, it\'s\n> >wrong to use Schwarzschild coordinates inside the horizon.\n> >They aren\'t valid inside the horizon.\n>\n> They aren\'t singular *inside* the horizon---they are only singular\n> *at* the horizon. So there is nothing wrong in principle with using\n> Schwarzschild coordinates everywhere except for a patch of spacetime\n> near the horizon.\n>\n> >Use Kruskal coordinates which, are not singular at the horizon,\n> >to relate what\'s going on inside and outside.\n>\n> I actually looked at Kruskal coordinates, and I still\n> find it mysterious.\n\nNothing mysterious at all. See page 148 Wald. Quite similar to the trick\nwith the Rindler space-time metric where its singularity can be transformed\naway and it revealed to really be flat space-time. That is all that is\nreally happening with Kruskal coordinates. However, as Wald discusses, the\nphysically of the regions revealed by this transformation are another\nmatter - some are probably not physical.\n\nThanks\nBill\n\n\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Daryl McCullough" <daryl@atc-nycorp.com> wrote in message
news:c4p71701qe8@drn.newsguy.com...
> John Anderson says...
>
> >There's a coordinate singularity at the horizon in
> >Schwarzschildcoordinates. There's no real singularity. Therefore, it's
> >wrong to use Schwarzschild coordinates inside the horizon.
> >They aren't valid inside the horizon.
>
> They aren't singular *inside* the horizon---they are only singular
> *at* the horizon. So there is nothing wrong in principle with using
> Schwarzschild coordinates everywhere except for a patch of spacetime
> near the horizon.
>
> >Use Kruskal coordinates which, are not singular at the horizon,
> >to relate what's going on inside and outside.
>
> I actually looked at Kruskal coordinates, and I still
> find it mysterious.
Nothing mysterious at all. See page 148 Wald. Quite similar to the trick
with the Rindler space-time metric where its singularity can be transformed
away and it revealed to really be flat space-time. That is all that is
really happening with Kruskal coordinates. However, as Wald discusses, the
physically of the regions revealed by this transformation are another
matter - some are probably not physical.
Thanks
Bill
Daryl McCullough
Apr7-04, 03:11 PM
<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>David says...\n\nI wrote (relating Kruskal coordinates to Schwarzschild coordinates)\n\n>> t = 2 arctanh(v/u)\n>> Where u is the spacelike coordinate, and v is the timelike coordinate.\n>\n>Actually that is not complete. It is\n> t = 2 arctanh[(v/u)^(+/-1)]\n>The sign is chosen depending on which side of the horizon you are\n>considering.\n\nI\'m sorry for belaboring this point, but I\'m having a hard time\nmaking myself understood. The question is *why* do you choose a\n+ sign on one side of the horizon and a - sign on the other? It\nturns out that that\'s equivalent to shifting t by a constant amount.\n\nIn general, the relationship between Kruskal coordinates and\nSchwarzschild coordinates is determined only up to a constant. The\nmore general expression is this:\n\nfor r < 2M (where |u| < |v|)\n\nt = A + 2 arctanh(u/v)\n\nfor r > 2M (where |u| > |v|)\n\nt = B + 2 arctanh(v/u)\n\nThen the question is: how do you relate the constants A and B?\n\nYour solution amounts to the choice A = B. My solution is what\nyou would get if you analytically continued the solution for r < 2M\ninto the region r > 2M, which amounts to the choice\n\nA = B + i pi\n\nI agree that my solution is weird, but in contrast, your solution\nseems a little arbitrary. It seems to me that any choice of A and B\nare equally acceptable. I don\'t see how continuity or any other\nconstraint will tell you what the value of A - B is.\n\n--\nDaryl McCullough\nIthaca, NY\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>David says...
I wrote (relating Kruskal coordinates to Schwarzschild coordinates)
>> t = 2 arctanh(v/u)
>> Where u is the spacelike coordinate, and v is the timelike coordinate.
>
>Actually that is not complete. It is
> t = 2 arctanh[(v/u)^(+/-1)]
>The sign is chosen depending on which side of the horizon you are
>considering.
I'm sorry for belaboring this point, but I'm having a hard time
making myself understood. The question is *why* do you choose a
+ sign on one side of the horizon and a - sign on the other? It
turns out that that's equivalent to shifting t by a constant amount.
In general, the relationship between Kruskal coordinates and
Schwarzschild coordinates is determined only up to a constant. The
more general expression is this:
for r < 2M (where |u| < |v|)
t = A + 2[/itex] arctanh(u/v)
for r > 2M (where [itex]|u| > |v|)
t = B + 2 arctanh(v/u)
Then the question is: how do you relate the constants A and B?
Your solution amounts to the choice A = B. My solution is what
you would get if you analytically continued the solution for r < 2M
into the region r > 2M, which amounts to the choice
A = B + i \pi
I agree that my solution is weird, but in contrast, your solution
seems a little arbitrary. It seems to me that any choice of A and B
are equally acceptable. I don't see how continuity or any other
constraint will tell you what the value of A - B is.
--
Daryl McCullough
Ithaca, NY
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