Metrics, geometries, and coordinate transformations [Archive]

John Baez

Apr7-04, 09:06 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><ebunn\\@lfa221051.richmond.edu> wrote in message\nnews:c3ssmt$fm5$1\\@lfa222122.richmond.ed u...\n\n> It seems to me quite unwise to say that different coordinate systems\n> give different metrics. The metric is a tensor, and a tensor\n> is something that exists independent of what coordinates one chooses.\n> So personally, I wouldn\'t talk about "the Rindler metric" as\n> something distinct from the Minkowski metric; I\'d talk about\n> Rindler coordinates.\n>\n> Am I alone in this?\n\nNo, this is what all mathematicians do, and many physicists - but\nnot *all* physicists.\n\nAnd even if we follow Ted Bunn\'s wise advice, we can easily get ourselves\nconfused if we\'re not careful. We agree that changing coordinates doesn\'t\nchange the metric on a manifold. This is what some people call a\n"passive" coordinate transformation. But, applying a diffeomorphism to\nthe metric *does* change it - and this is what they call an "active"\ncoordinate transformation.\n\nSince there can be situations where what Herr Professor Schmidt regards\nas a "passive" coordinate transformation is regarded by Herr Professor\nSchultz as an "active" one, the situation is ripe for confusion. Schmidt\nwill say that the metric doesn\'t change, while Schultz will insist it does!\n\nSo, we gotta be careful.\n\nIf you want to talk about a metric modulo diffeomorphisms, call it\na "geometry". Both active and passive coordinate transformations leave\na geometry unchanged.\n\nIn GR, two metrics giving the same geometry give the same physics.\n\nIn short, we\'ve got:\n\nformulas for metrics in terms of coordinates -\nchanged by both active and passive coordinate transformations\n\nmetrics -\nchanged by active coordinate transformations but not passive ones\n\ngeometries -\nunchanged by both active and passive coordinate transformations\n\nI leave as a puzzle to figure out what the 4th possibility is like,\nand whether people actually talk about this one!\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><ebunn@lfa221051.richmond.edu> wrote in message
news:c3ssmt$fm5$1@lfa222122.richmond.edu...

> It seems to me quite unwise to say that different coordinate systems
> give different metrics. The metric is a tensor, and a tensor
> is something that exists independent of what coordinates one chooses.
> So personally, I wouldn't talk about "the Rindler metric" as
> something distinct from the Minkowski metric; I'd talk about
> Rindler coordinates.
>
> Am I alone in this?

No, this is what all mathematicians do, and many physicists - but
not *all* physicists.

And even if we follow Ted Bunn's wise advice, we can easily get ourselves
confused if we're not careful. We agree that changing coordinates doesn't
change the metric on a manifold. This is what some people call a
"passive" coordinate transformation. But, applying a diffeomorphism to
the metric *does* change it - and this is what they call an "active"
coordinate transformation.

Since there can be situations where what Herr Professor Schmidt regards
as a "passive" coordinate transformation is regarded by Herr Professor
Schultz as an "active" one, the situation is ripe for confusion. Schmidt
will say that the metric doesn't change, while Schultz will insist it does!

So, we gotta be careful.

If you want to talk about a metric modulo diffeomorphisms, call it
a "geometry". Both active and passive coordinate transformations leave
a geometry unchanged.

In GR, two metrics giving the same geometry give the same physics.

In short, we've got:

formulas for metrics in terms of coordinates -
changed by both active and passive coordinate transformations

metrics -
changed by active coordinate transformations but not passive ones

geometries -
unchanged by both active and passive coordinate transformations

I leave as a puzzle to figure out what the 4th possibility is like,
and whether people actually talk about this one!

Arkadiusz Jadczyk

Apr8-04, 06:31 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>On Tue, 6 Apr 2004 17:54:43 +0000 (UTC), baez@galaxy.ucr.edu (John Baez)\nwrote:\n\n>If you want to talk about a metric modulo diffeomorphisms, call it\n>a "geometry". Both active and passive coordinate transformations leave\n>a geometry unchanged.\n>\n>In GR, two metrics giving the same geometry give the same physics.\n\nPerhaps with one little comments:\n\nSpinors are sometimes considered to be a part of GR. Dirac operator is,\nby some, considered as quite important, perhaps even more important than\nthe metric itself (Alain Connes) Its spectrum may depend on the spin\nstructure, and there may be inequivalent structures for two equivalent\nmetrics.\n\nBut that is also a side remark, not that relevant to the original\nquestion....\n\nark\n--\n\nArkadiusz Jadczyk\nhttp://www.cassiopaea.org/quantum_future/homepage.htm\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>On Tue, 6 Apr 2004 17:54:43 +0000 (UTC), baez@galaxy.ucr.edu (John Baez)
wrote:

>If you want to talk about a metric modulo diffeomorphisms, call it
>a "geometry". Both active and passive coordinate transformations leave
>a geometry unchanged.
>
>In GR, two metrics giving the same geometry give the same physics.

Perhaps with one little comments:

Spinors are sometimes considered to be a part of GR. Dirac operator is,
by some, considered as quite important, perhaps even more important than
the metric itself (Alain Connes) Its spectrum may depend on the spin
structure, and there may be inequivalent structures for two equivalent
metrics.

But that is also a side remark, not that relevant to the original
question....

ark
--

Arkadiusz Jadczyk
http://www.cassiopaea.org/quantum_future/homepage.htm

--

Danny Ross Lunsford

Apr11-04, 11:44 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\nArkadiusz Jadczyk wrote:\n\n>>If you want to talk about a metric modulo diffeomorphisms, call it\n>>a "geometry". Both active and passive coordinate transformations leave\n>>a geometry unchanged.\n>>\n>>In GR, two metrics giving the same geometry give the same physics.\n>\n> Perhaps with one little comments:\n>\n> Spinors are sometimes considered to be a part of GR. Dirac operator is,\n> by some, considered as quite important, perhaps even more important than\n> the metric itself (Alain Connes) Its spectrum may depend on the spin\n> structure, and there may be inequivalent structures for two equivalent\n> metrics.\n>\n> But that is also a side remark, not that relevant to the original\n> question....\n\nNo no, this is the whole point! You can always cook up a metric given a\nframe. Likewise given some coordinate system one can set up frames.\n(Just minutes ago I was trying to imagine some tensors written out\nexplicitly in terms of the frame instead of g. And then I was sort of\nstruck by the thought that both GR and field theory are about these\nframes, this is the one thing they really have in common.)\n\nThe real problem with this is - the metric is not gravity, it\'s just the\nmetric. The particular odd form of the Christoffel connection is\ngravity. It doesn\'t need spinors for expression.\n\n-drl\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>Arkadiusz Jadczyk wrote:

>>If you want to talk about a metric modulo diffeomorphisms, call it
>>a "geometry". Both active and passive coordinate transformations leave
>>a geometry unchanged.
>>
>>In GR, two metrics giving the same geometry give the same physics.
>
> Perhaps with one little comments:
>
> Spinors are sometimes considered to be a part of GR. Dirac operator is,
> by some, considered as quite important, perhaps even more important than
> the metric itself (Alain Connes) Its spectrum may depend on the spin
> structure, and there may be inequivalent structures for two equivalent
> metrics.
>
> But that is also a side remark, not that relevant to the original
> question....

No no, this is the whole point! You can always cook up a metric given a
frame. Likewise given some coordinate system one can set up frames.
(Just minutes ago I was trying to imagine some tensors written out
explicitly in terms of the frame instead of g. And then I was sort of
struck by the thought that both GR and field theory are about these
frames, this is the one thing they really have in common.)

The real problem with this is - the metric is not gravity, it's just the
metric. The particular odd form of the Christoffel connection is
gravity. It doesn't need spinors for expression.

-drl

tessel@um.bot

Apr13-04, 05:55 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>On Wed, 7 Apr 2004, [ISO-8859-1] Morris Carré wrote:\n\n> This reminds me of a fact I heard about in this newsgroup - that the\n> equivalence of two solutions to Einstein\'s equations was in all\n> generality an undecidable problem. Is my memory correct ? Can I read\n> your portrait of Schmidt and Schultz as a representation of the same\n> fact ?\n\nGoogle for old posts to this group with keywords "Karlhede" and "Cartan",\nwhere you will find elaboration of the following point:\n\nCartan\'s very general procedure for recognizing locally isometric\nmanifolds (which was streamlined by Karlhede for the special case of\nLorentzian geometry) ultimately reduces the problem to an algebraic one.\nThere is no algorithm for solving this algebraic problem in every case.\nNonetheless, in practice this potential defect seems not to have arisen.\n\nFor Cartan\'s procedure, see\n\nauthor = {Peter J. Olver},\ntitle = {Equivalence, Invariants, and Symmetry},\npublisher = {Cambridge University Press},\nyear = 1995}\n\nFor Karlhede\'s procedure, see\n\nauthor = {A. Karlhede and M. A. H. MacCallum},\ntitle = {On Determing the Isometry Group of a Riemannian Space},\njournal = {Gen. Rel. Grav.},\nvolume = 14,\nyear = 1982,\npages = {673--682}}\n\nauthor = {D. Pollney and J. E. F. Skea and R. A. D\'Inverno},\ntitle = {Classifying geometries in general relativity. {I.} {S}tandard\nforms for symmetric spinors},\njournal = {Class. Quant. Grav.},\nvolume = 17,\nnumber = 3,\nyear = 2000,\npages = {643-663}}\n\n(and succeeding issues for next two parts).\n\nHTH,\n\n"T. Essel"\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>On Wed, 7 Apr 2004, [ISO-8859-1] Morris Carré wrote:

> This reminds me of a fact I heard about in this newsgroup - that the
> equivalence of two solutions to Einstein's equations was in all
> generality an undecidable problem. Is my memory correct ? Can I read
> your portrait of Schmidt and Schultz as a representation of the same
> fact ?

Google for old posts to this group with keywords "Karlhede" and "Cartan",
where you will find elaboration of the following point:

Cartan's very general procedure for recognizing locally isometric
manifolds (which was streamlined by Karlhede for the special case of
Lorentzian geometry) ultimately reduces the problem to an algebraic one.
There is no algorithm for solving this algebraic problem in every case.
Nonetheless, in practice this potential defect seems not to have arisen.

For Cartan's procedure, see

author = {Peter J. Olver},
title = {Equivalence, Invariants, and Symmetry},
publisher = {Cambridge University Press},
year = 1995}

For Karlhede's procedure, see

author = {A. Karlhede and M. A. H. MacCallum},
title = {On Determing the Isometry Group of a Riemannian Space},
journal = {Gen. Rel. Grav.},
volume = 14,
year = 1982,
pages = {673--682}}

author = {D. Pollney and J. E. F. Skea and R. A. D'Inverno},
title = {Classifying geometries in general relativity. {I.} {S}tandard
forms for symmetric spinors},
journal = {Class. Quant. Grav.},
volume = 17,
number = 3,
year = 2000,
pages = {643-663}}

(and succeeding issues for next two parts).

HTH,

"T. Essel"