0. Law of Thermodynamics [Archive]

Florian Dufey

Apr28-04, 02:25 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>The 0. law of thermodynamics is stated as follows:\nI have three simple systems A({x}), B({y}), C({z}).\nIf A and B are in thermal equilibrium and B and C are in equilibrium\nsimultaneously, then A and C are also in equilibrium (Transitivity of\nthermal equilibrium).\nThe assumtion of equilibrium leads to 3 equations,\nF_1({x},{y})=0,\nF_2({y},{z})=0,\nF_3( {x},{z})=0.\nThe assumtion of transitivity means that any two of these equations\nalso include the third.\nIt is stated that from these assumptions one can show that there are\nfunctions t_1({x}), t_2({y}), t_3({z}), so that in equilibrium\nt_1=t_2=t_3. Does anybody know a formal proof for the last statement?\nThanks,\nFlorian\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>The . law of thermodynamics is stated as follows:
I have three simple systems A({x}), B({y}), C({z}).
If A and B are in thermal equilibrium and B and C are in equilibrium
simultaneously, then A and C are also in equilibrium (Transitivity of
thermal equilibrium).
The assumtion of equilibrium leads to 3 equations,
F_1({x},{y})=0,F_2({y},{z})=0,F_3({x},{z})=0.
The assumtion of transitivity means that any two of these equations
also include the third.
It is stated that from these assumptions one can show that there are
functions t_1({x}), t_2({y}), t_3({z}), so that in equilibrium
t_1=t_2=t_3. Does anybody know a formal proof for the last statement?
Thanks,
Florian

Rob Woodside

Apr30-04, 02:03 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>The zeroth law is required by people who don\'t know what an\nequivalence relation is. An equivalence relation, like equality, is\nreflexive: A ~ A; symmetric: A ~ B <=> B ~ A; and transitive: A ~ B ^\nB ~ C => A ~ C.\n\nOne requirement of thermal equilibrium for a system is a uniform\ntemperature. Thermal equilibrium is an equivalence relation between\nsystems and thus by definition requires the "zeroth law".\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>The zeroth law is required by people who don't know what an
equivalence relation is. An equivalence relation, like equality, is
reflexive: A ~ A; symmetric: A ~ B <=> B ~ A; and transitive: A ~ B ^B ~ C => A ~ C.

One requirement of thermal equilibrium for a system is a uniform
temperature. Thermal equilibrium is an equivalence relation between
systems and thus by definition requires the "zeroth law".

David Williams

May1-04, 07:52 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>-> One requirement of thermal equilibrium for a system is a uniform\n-> temperature. Thermal equilibrium is an equivalence relation between\n-> systems and thus by definition requires the "zeroth law".\n\nNot in a gravitational field. Radiation gains energy if it moves\n"downhill", and vice versa, so an object at the top of a hill is at\nthermal equilibrium with a hotter object at the bottom of the hill.\n\ndow\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>-> One requirement of thermal equilibrium for a system is a uniform
-> temperature. Thermal equilibrium is an equivalence relation between
-> systems and thus by definition requires the "zeroth law".

Not in a gravitational field. Radiation gains energy if it moves
"downhill", and vice versa, so an object at the top of a hill is at
thermal equilibrium with a hotter object at the bottom of the hill.

dow

Rob Woodside

May2-04, 04:52 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.williams@bayman.org (David Williams) wrote in message news:<1087764485.1572.1087741897@bayman.org>...\n> -> One requirement of thermal equilibrium for a system is a uniform\n> -> temperature. Thermal equilibrium is an equivalence relation between\n> -> systems and thus by definition requires the "zeroth law".\n>\n> Not in a gravitational field. Radiation gains energy if it moves\n> "downhill", and vice versa, so an object at the top of a hill is at\n> thermal equilibrium with a hotter object at the bottom of the hill.\n>\n> dow\n\nNot what in a gravitational field???- No zero law?, No equivalence\nrelation?, No thermal equilibrium?, No uniform temperature? I suspect\nthe problem is with thermal equilibrium. For a box of blackbody\nradiation sitting in a uniform gravitaional field, locally one has\nthermal equilibrium, but not over large vertical distances. Another\nrequirement for themal equilibrium is mechanical equilibrium and in\nparticular no pressure gradiants. So any static column of fluid in a\nuniform gravitational field must have hydrostatic pressure gradiants\nand cannot be in thermal equilibrium. Perhaps gravity is telling us\nthat thermal equilibrium is a local concept only.\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.williams@bayman.org (David Williams) wrote in message news:<1087764485.1572.1087741897@bayman.org>...
> -> One requirement of thermal equilibrium for a system is a uniform
> -> temperature. Thermal equilibrium is an equivalence relation between
> -> systems and thus by definition requires the "zeroth law".
>
> Not in a gravitational field. Radiation gains energy if it moves
> "downhill", and vice versa, so an object at the top of a hill is at
> thermal equilibrium with a hotter object at the bottom of the hill.
>
> dow

Not what in a gravitational field???- No zero law?, No equivalence
relation?, No thermal equilibrium?, No uniform temperature? I suspect
the problem is with thermal equilibrium. For a box of blackbody
radiation sitting in a uniform gravitaional field, locally one has
thermal equilibrium, but not over large vertical distances. Another
requirement for themal equilibrium is mechanical equilibrium and in
particular no pressure gradiants. So any static column of fluid in a
uniform gravitational field must have hydrostatic pressure gradiants
and cannot be in thermal equilibrium. Perhaps gravity is telling us
that thermal equilibrium is a local concept only.

Peter Tobias

May3-04, 04:52 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>\nDavid Williams:\n> Radiation gains energy if it moves\n> "downhill", and vice versa, so an object at the top of a hill is at\n> thermal equilibrium with a hotter object at the bottom of the hill.\n\nSounds convincing. Have you calculated how the temperature profile in a\ngravitational field should look like when a system is in equilibrium, or\ncould you give a link to a calculation?\n\nRegards,\n\nPeter\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 Williams:
> Radiation gains energy if it moves
> "downhill", and vice versa, so an object at the top of a hill is at
> thermal equilibrium with a hotter object at the bottom of the hill.

Sounds convincing. Have you calculated how the temperature profile in a
gravitational field should look like when a system is in equilibrium, or
could you give a link to a calculation?

Regards,

Peter

J. J. Lodder

May3-04, 01:32 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>\nRob Woodside <rwmw@telus.net> wrote:\n\n> david.williams@bayman.org (David Williams) wrote in message news:<1087764485.1\n572.1087741897@bayman.org>...\ n> > -> One requirement of thermal equilibrium for a system is a uniform\n> > -> temperature. Thermal equilibrium is an equivalence relation between\n> > -> systems and thus by definition requires the "zeroth law".\n> >\n> > Not in a gravitational field. Radiation gains energy if it moves\n> > "downhill", and vice versa, so an object at the top of a hill is at\n> > thermal equilibrium with a hotter object at the bottom of the hill.\n> >\n> > dow\n>\n> Not what in a gravitational field???- No zero law?, No equivalence\n> relation?, No thermal equilibrium?, No uniform temperature? I suspect\n> the problem is with thermal equilibrium. For a box of blackbody\n> radiation sitting in a uniform gravitaional field, locally one has\n> thermal equilibrium, but not over large vertical distances. Another\n> requirement for themal equilibrium is mechanical equilibrium and in\n> particular no pressure gradiants. So any static column of fluid in a\n> uniform gravitational field must have hydrostatic pressure gradiants\n> and cannot be in thermal equilibrium. Perhaps gravity is telling us\n> that thermal equilibrium is a local concept only.\n\nCertainly not.\nAs Tolman already discovered the temperature in a gravitational field\n-in thermal equilibrium- must depend (in the expected way)\non the gravitational potential. (1920-ies, wuld have to look up)\n\nYou can derive that by considering red/blue shifting of blackbody\nradiation for example. Out of a hole in the top of a blackbody cavity\ncomes radiation at a lower temperature than out of a hole in the bottom.\n\nA more fundamental derivation is possible using the thermodynamic\ndefinition of temperature by means of Carnot cycles.\nSince the bottom reservoir is at a higher temperature\nthan the higher onea carnot cycle between the two thermal reservoirs\ncan be made to do work.\nHowever, for this work to be made\nheat must be transferred from the low to the high reservoir,\nagainst the gravitational potential difference.\nThis takes work to be done.\n\nThe generalized condition for thermal equilibrium\nis that the temperature difference between the high and the low\nmust be just right to generate the work needed to overcome\nthe potential difference for the heat.\nOut comes the same result as you get from the red-shift argument.\n\nBest,\n\nJan\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>Rob Woodside <rwmw@telus.net> wrote:

> david.williams@bayman.org (David Williams) wrote in message news:<1087764485.1
572.1087741897@bayman.org>...
> > -> One requirement of thermal equilibrium for a system is a uniform
> > -> temperature. Thermal equilibrium is an equivalence relation between
> > -> systems and thus by definition requires the "zeroth law".
> >
> > Not in a gravitational field. Radiation gains energy if it moves
> > "downhill", and vice versa, so an object at the top of a hill is at
> > thermal equilibrium with a hotter object at the bottom of the hill.
> >
> > dow
>
> Not what in a gravitational field???- No zero law?, No equivalence
> relation?, No thermal equilibrium?, No uniform temperature? I suspect
> the problem is with thermal equilibrium. For a box of blackbody
> radiation sitting in a uniform gravitaional field, locally one has
> thermal equilibrium, but not over large vertical distances. Another
> requirement for themal equilibrium is mechanical equilibrium and in
> particular no pressure gradiants. So any static column of fluid in a
> uniform gravitational field must have hydrostatic pressure gradiants
> and cannot be in thermal equilibrium. Perhaps gravity is telling us
> that thermal equilibrium is a local concept only.

Certainly not.
As Tolman already discovered the temperature in a gravitational field
-in thermal equilibrium- must depend (in the expected way)
on the gravitational potential. (1920-ies, wuld have to look up)

You can derive that by considering red/blue shifting of blackbody
radiation for example. Out of a hole in the top of a blackbody cavity
comes radiation at a lower temperature than out of a hole in the bottom.

A more fundamental derivation is possible using the thermodynamic
definition of temperature by means of Carnot cycles.
Since the bottom reservoir is at a higher temperature
than the higher onea carnot cycle between the two thermal reservoirs
can be made to do work.
However, for this work to be made
heat must be transferred from the low to the high reservoir,
against the gravitational potential difference.
This takes work to be done.

The generalized condition for thermal equilibrium
is that the temperature difference between the high and the low
must be just right to generate the work needed to overcome
the potential difference for the heat.
Out comes the same result as you get from the red-shift argument.

Best,

Jan

Peter Tobias

May3-04, 02:05 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>Rob Woodside:\n> I suspect\n> the problem is with thermal equilibrium. For a box of blackbody\n> radiation sitting in a uniform gravitaional field, locally one has\n> thermal equilibrium, but not over large vertical distances.\n\nI understand David\'s remark differently: There can be a thermal\nequilibrium between a high and a low object, even if miles apart, but\nin this equilibrium the temperature of the high object is slightly\nlower than the temperature of the low object.\n\n> Another\n> requirement for themal equilibrium is mechanical equilibrium and in\n> particular no pressure gradiants.\n\nI don\'t believe that statement. Except for the miniscule temperature\ngradient due to gravity\'s effect on time that David mentioned, any\nself-establishing temperature gradient in an enclosed system would\ncontradict the second law of thermodynamics, and it hasn\'t been\nobserved in any experiment, even with observed and steady pressure\ngradients.\n\nRegards,\n\nPeter\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>Rob Woodside:
> I suspect
> the problem is with thermal equilibrium. For a box of blackbody
> radiation sitting in a uniform gravitaional field, locally one has
> thermal equilibrium, but not over large vertical distances.

I understand David's remark differently: There can be a thermal
equilibrium between a high and a low object, even if miles apart, but
in this equilibrium the temperature of the high object is slightly
lower than the temperature of the low object.

> Another
> requirement for themal equilibrium is mechanical equilibrium and in
> particular no pressure gradiants.

I don't believe that statement. Except for the miniscule temperature
gradient due to gravity's effect on time that David mentioned, any
self-establishing temperature gradient in an enclosed system would
contradict the second law of thermodynamics, and it hasn't been
observed in any experiment, even with observed and steady pressure
gradients.

Regards,

Peter

Ken S. Tucker

May3-04, 05:45 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>rwmw@telus.net (Rob Woodside) wrote in message news:<4d06b9d7.0405021106.52bcfdd9@posting.google. com>...\n>david.williams@bayman.org (David Williams) wrote in message news:<1087764485.1572.1087741897@bayman.org>...\n> > -> One requirement of thermal equilibrium for a system is a uniform\n>> -> temperature. Thermal equilibrium is an equivalence relation between\n>> -> systems and thus by definition requires the "zeroth law".\n>>\n>> Not in a gravitational field. Radiation gains energy if it moves\n>> "downhill", and vice versa, so an object at the top of a hill is at\n>> thermal equilibrium with a hotter object at the bottom of the hill.\n>> dow\n>\n>Not what in a gravitational field???- No zero law?, No equivalence\n>relation?, No thermal equilibrium?, No uniform temperature? I suspect\n>the problem is with thermal equilibrium. For a box of blackbody\n>radiation sitting in a uniform gravitaional field, locally one has\n>thermal equilibrium, but not over large vertical distances. Another\n>requirement for themal equilibrium is mechanical equilibrium and in\n>particular no pressure gradiants. So any static column of fluid in a\n>uniform gravitational field must have hydrostatic pressure gradiants\n>and cannot be in thermal equilibrium. Perhaps gravity is telling us\n>that thermal equilibrium is a local concept only.\n\nYeah, I understand *thermal equilbrium* requres the\nexchange of radiant energy, even as a molecular\nrelation. Using the gravitational frequency shift David\nWilliams suggests, as heat conveys "goes down hill",\nit should be *blue-shifted*.\nKen S. Tucker\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>rwmw@telus.net (Rob Woodside) wrote in message news:<4d06b9d7.0405021106.52bcfdd9@posting.google.com>...
>david.williams@bayman.org (David Williams) wrote in message news:<1087764485.1572.1087741897@bayman.org>...
>> -> One requirement of thermal equilibrium for a system is a uniform
>> -> temperature. Thermal equilibrium is an equivalence relation between
>> -> systems and thus by definition requires the "zeroth law".
>>
>> Not in a gravitational field. Radiation gains energy if it moves
>> "downhill", and vice versa, so an object at the top of a hill is at
>> thermal equilibrium with a hotter object at the bottom of the hill.
>> dow
>
>Not what in a gravitational field???- No zero law?, No equivalence
>relation?, No thermal equilibrium?, No uniform temperature? I suspect
>the problem is with thermal equilibrium. For a box of blackbody
>radiation sitting in a uniform gravitaional field, locally one has
>thermal equilibrium, but not over large vertical distances. Another
>requirement for themal equilibrium is mechanical equilibrium and in
>particular no pressure gradiants. So any static column of fluid in a
>uniform gravitational field must have hydrostatic pressure gradiants
>and cannot be in thermal equilibrium. Perhaps gravity is telling us
>that thermal equilibrium is a local concept only.

Yeah, I understand *thermal equilbrium* requres the
exchange of radiant energy, even as a molecular
relation. Using the gravitational frequency shift David
Williams suggests, as heat conveys "goes down hill",
it should be *blue-shifted*.
Ken S. Tucker

Florian Dufey

May3-04, 05:45 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>\n\nRob Woodside schrieb:\n> The zeroth law is required by people who don\'t know what an\n> equivalence relation is. An equivalence relation, like equality, is\n> reflexive: A ~ A; symmetric: A ~ B <=> B ~ A; and transitive: A ~ B ^\n> B ~ C => A ~ C.\n>\n> One requirement of thermal equilibrium for a system is a uniform\n> temperature. Thermal equilibrium is an equivalence relation between\n> systems and thus by definition requires the "zeroth law".\n>\nThe statement of the zeroth Law may be formulated as an equivalence\nrelation between thermal equilibrium states.\nBut how do you prove that there is a variable called Temperature?\nI ask this because I found some proofs in the literature which seem to\nbe wrong or incomplete.\nIs there something like a mathematical statement that every equivalence\nrelation leads to something like a Temperature?\n\n\n\n\n\n--\n\n---------------------------------------------------------------------\nFlorian Dufey\nPhysik Department T38\nTechnische Universitaet Muenchen\nJames-Franck-Strasse\nD-85747 Garching b. Muenchen\nGermany\n\nemail: florian.dufey@physik.tu-muenchen.de\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>Rob Woodside schrieb:
> The zeroth law is required by people who don't know what an
> equivalence relation is. An equivalence relation, like equality, is
> reflexive: A ~ A; symmetric: A ~ B <=> B ~ A; and transitive: A ~ B ^
> B ~ C => A ~ C.
>
> One requirement of thermal equilibrium for a system is a uniform
> temperature. Thermal equilibrium is an equivalence relation between
> systems and thus by definition requires the "zeroth law".
>
The statement of the zeroth Law may be formulated as an equivalence
relation between thermal equilibrium states.
But how do you prove that there is a variable called Temperature?
I ask this because I found some proofs in the literature which seem to
be wrong or incomplete.
Is there something like a mathematical statement that every equivalence
relation leads to something like a Temperature?

--

---------------------------------------------------------------------
Florian Dufey
Physik Department T38
Technische Universitaet Muenchen
James-Franck-Strasse
D-85747 Garching b. Muenchen
Germany

email: florian.dufey@physik.tu-muenchen.de
---------------------------------------------------------------------

J. J. Lodder

May3-04, 05:52 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>Reply-To: jjlxa25@xs4all.nl (J. J. Lodder)\nNNTP-Posting-Host: lfa222122.richmond.edu\nX-Trace: rumor.richmond.edu 1083625094 17997 141.166.222.122 (3 May 2004 22:58:14 GMT)\nX-Complaints-To: usenet@rumor.richmond.edu\nNNTP-Posting-Date: Mon, 3 May 2004 22:58:14 +0000 (UTC)\nUser-Agent: MacSOUP/2.6 (unregistered for 7405 days)\nXref: core-easynews sci.physics.research:55910\nX-Received-Date: Mon, 03 May 2004 16:01:16 MST (news.easynews.com)\n\nPeter Tobias <tobias@pa.msu.edu> wrote:\n\n> Rob Woodside:\n> > I suspect\n> > the problem is with thermal equilibrium. For a box of blackbody\n> > radiation sitting in a uniform gravitaional field, locally one has\n> > thermal equilibrium, but not over large vertical distances.\n>\n> I understand David\'s remark differently: There can be a thermal\n> equilibrium between a high and a low object, even if miles apart, but\n> in this equilibrium the temperature of the high object is slightly\n> lower than the temperature of the low object.\n>\n> > Another\n> > requirement for themal equilibrium is mechanical equilibrium and in\n> > particular no pressure gradiants.\n>\n> I don\'t believe that statement.\n\nCorrect. In a gravitational field mechanical equilibrium -requires-\nthe existence of pressure gradients.\nThink of the earth\'s atmosphere for example.\nBut even in a black-body cavity\nthere will be a gradient of the radiaton pressure.\nIn the way expected, just as for the earth\'s atmosphere:\nthe difference in radiation pressure across a small cube\nwill equal the weight of the radiaton in it.\n(even though photons don\'t interact)\nYou can derive this in the traditional way\nby assuming an infinitesimal \'kohlenstoffpartikel\'\nto be floating in the radiation gas, at neutral boyancy.\n\n> Except for the miniscule temperature\n> gradient due to gravity\'s effect on time that David mentioned, any\n> self-establishing temperature gradient in an enclosed system would\n> contradict the second law of thermodynamics, and it hasn\'t been\n> observed in any experiment, even with observed and steady pressure\n> gradients.\n\nOn the contrary: in a gravitational field the second law\nrequires the existence of a temperature gradient,\nas explained in my other posting.\n\nBest,\n\nJan\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>Reply-To: jjlxa25@xs4all.nl (J. J. Lodder)
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Peter Tobias <tobias@pa.msu.edu> wrote:

> Rob Woodside:
> > I suspect
> > the problem is with thermal equilibrium. For a box of blackbody
> > radiation sitting in a uniform gravitaional field, locally one has
> > thermal equilibrium, but not over large vertical distances.
>
> I understand David's remark differently: There can be a thermal
> equilibrium between a high and a low object, even if miles apart, but
> in this equilibrium the temperature of the high object is slightly
> lower than the temperature of the low object.
>
> > Another
> > requirement for themal equilibrium is mechanical equilibrium and in
> > particular no pressure gradiants.
>
> I don't believe that statement.

Correct. In a gravitational field mechanical equilibrium -requires-
the existence of pressure gradients.
Think of the earth's atmosphere for example.
But even in a black-body cavity
there will be a gradient of the radiaton pressure.
In the way expected, just as for the earth's atmosphere:
the difference in radiation pressure across a small cube
will equal the weight of the radiaton in it.
(even though photons don't interact)
You can derive this in the traditional way
by assuming an infinitesimal 'kohlenstoffpartikel'
to be floating in the radiation gas, at neutral boyancy.

> Except for the miniscule temperature
> gradient due to gravity's effect on time that David mentioned, any
> self-establishing temperature gradient in an enclosed system would
> contradict the second law of thermodynamics, and it hasn't been
> observed in any experiment, even with observed and steady pressure
> gradients.

On the contrary: in a gravitational field the second law
requires the existence of a temperature gradient,
as explained in my other posting.

Best,

Jan

Florian Dufey

May4-04, 03:06 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>J. J. Lodder schrieb:\n\n\n>\n> Certainly not.\n> As Tolman already discovered the temperature in a gravitational field\n> -in thermal equilibrium- must depend (in the expected way)\n> on the gravitational potential. (1920-ies, wuld have to look up)\n\nI doubt some that the statements made in this thread about the\ntemperature in gravitation fields are correct:\n\nIf I would sit on some arbitrary point (the reference point) in the\ngravitation field, the temperature of the radiation that I would receive\nof all the systems which are in thermal equilibrium with each other\nwould be the same, regardless of their distance. This temperature is\nthe temperature which I assign also to the system which emitted the\nradiation. It fulfills the 0. law. It only depends on the\ninternal variables of the systems (like P and V) and on its position in\nthe gravitation field relative to the point of reference.\n\nHowever, this temperature transforms as a scalar with respect to\ngeneral relativity under a change of reference point. That is the\ntemperature of a system on the bottom of the gravitation field measured\nat that point would be higher than its temperature measured at some\npoint far away from the center of gravity. I can well imagine that\nthese local temperatures conform the observation of Tolman.\n\nIt is also not necessary to assume mechanical equilibrium.\nWhy should the temperature of a pressure bottle of gas in thermal\nequilibrium with its surrounding depend on the pressure only because it\nis in a gravitation field?\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>J. J. Lodder schrieb:

>
> Certainly not.
> As Tolman already discovered the temperature in a gravitational field
> -in thermal equilibrium- must depend (in the expected way)
> on the gravitational potential. (1920-ies, wuld have to look up)

I doubt some that the statements made in this thread about the
temperature in gravitation fields are correct:

If I would sit on some arbitrary point (the reference point) in the
gravitation field, the temperature of the radiation that I would receive
of all the systems which are in thermal equilibrium with each other
would be the same, regardless of their distance. This temperature is
the temperature which I assign also to the system which emitted the
radiation. It fulfills the . law. It only depends on the
internal variables of the systems (like P and V) and on its position in
the gravitation field relative to the point of reference.

However, this temperature transforms as a scalar with respect to
general relativity under a change of reference point. That is the
temperature of a system on the bottom of the gravitation field measured
at that point would be higher than its temperature measured at some
point far away from the center of gravity. I can well imagine that
these local temperatures conform the observation of Tolman.

It is also not necessary to assume mechanical equilibrium.
Why should the temperature of a pressure bottle of gas in thermal
equilibrium with its surrounding depend on the pressure only because it
is in a gravitation field?

Rob Woodside

May4-04, 03:07 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>nospam@de-ster.demon.nl (J. J. Lodder) wrote in message news:<1gd8ea0.1mroxiu14nfjpfN@de-ster.xs4all.nl>...\nsnip\n> In a gravitational field mechanical equilibrium -requires-\n> the existence of pressure gradients.\nsnip\n> In a gravitational field the second law\n> requires the existence of a temperature gradient,\n> as explained in my other posting.\nsnip\nThanks Jan, Are you suggesting that the state functions for a system\nin thermal equilibrium may vary from place to place in the system as\nlong as no useful work can be done with the differences?\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>nospam@de-ster.demon.nl (J. J. Lodder) wrote in message news:<1gd8ea0.1mroxiu14nfjpfN@de-ster.xs4all.nl>...
snip
> In a gravitational field mechanical equilibrium -requires-
> the existence of pressure gradients.
snip
> In a gravitational field the second law
> requires the existence of a temperature gradient,
> as explained in my other posting.
snip
Thanks Jan, Are you suggesting that the state functions for a system
in thermal equilibrium may vary from place to place in the system as
long as no useful work can be done with the differences?

Alfred Einstead

May4-04, 03:07 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>nospam@de-ster.demon.nl (J. J. Lodder) wrote:\n> Certainly not.\n> As Tolman already discovered the temperature in a gravitational field\n> -in thermal equilibrium- must depend (in the expected way)\n> on the gravitational potential. (1920-ies, wuld have to look up)\n\nActually, the constraints between gravity and thermodynamics can\nalso be reversed to DERIVE Einstein\'s equations. This was done\nby Jacobsen; see gr-qc/9504004.\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>nospam@de-ster.demon.nl (J. J. Lodder) wrote:
> Certainly not.
> As Tolman already discovered the temperature in a gravitational field
> -in thermal equilibrium- must depend (in the expected way)
> on the gravitational potential. (1920-ies, wuld have to look up)

Actually, the constraints between gravity and thermodynamics can
also be reversed to DERIVE Einstein's equations. This was done
by Jacobsen; see http://www.arxiv.org/abs/gr-qc/9504004.

Ken S. Tucker

May6-04, 07:52 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>\nwhopkins@csd.uwm.edu (Alfred Einstead) wrote in message news:<e58d56ae.0405040903.50b8d228@posting.google. com>...\n>nospam@de-ster.demon.nl (J. J. Lodder) wrote:\n>> Certainly not.\n>> As Tolman already discovered the temperature in a gravitational field\n>> -in thermal equilibrium- must depend (in the expected way)\n>> on the gravitational potential. (1920-ies, wuld have to look up)\n>\n>Actually, the constraints between gravity and thermodynamics can\n>also be reversed to DERIVE Einstein\'s equations. This was done\n>by Jacobsen; see gr-qc/9504004.\n\nWell yeah that makes sense, the Einstein\'s Law\nG_uv = kT_uv holds for any form of energy.\n\n>From: Florian Dufey\n>If I would sit on some arbitrary point (the reference point) in the\n>gravitation field, the temperature of the radiation that I would receive\n>of all the systems which are in thermal equilibrium with each other\n>would be the same, regardless of their distance.\nok, more below....\n\n>This temperature is\n>the temperature which I assign also to the system which emitted the\n>radiation. It fulfills the 0. law. It only depends on the\n>internal variables of the systems (like P and V) and on its position in\n>the gravitation field relative to the point of reference.\n\nThat is highly interesting, if I understand correctly.\nIf a solid metal rod is hung vertically in a g-field,\nand the rod is in thermal equilbrium, then any observer\nat any location will receive the same radiant energy\nfrom any and all points of the rod.\nThe reason for that is the higher temperature at\nthe bottom of the rod, is exactly compensated by\nthe radiant energy I would receive from that point\nas that radiant energy is red shifted in accord\nwith my relative altitude. Likewise the radiant energy\nfrom the top of the rod, which is assumed to be\ncooler would be blue shifted if my altitude was lower\nthan the top of the rod.\n\nSo my question is this, would the temperature,\nof rod, measured by the frequency of it\'s radiant\nenergy (assuming thermal equilbrium) be the same\nat any point for all observers?\n\nPardon my simplicity, but Alfred Einstead above suggests\nreverse engeering the invariance of the rod\'s radiant energy\nto "DERIVE" Einstein\'s Law (G_uv = kT_uv).\n\nEnergy conservation in g-fields is difficult to\nlearn, it\'s nice to get an invariant.\n\nThanks in Advance.\nKen S. Tucker\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>whopkins@csd.uwm.edu (Alfred Einstead) wrote in message news:<e58d56ae.0405040903.50b8d228@posting.google.com>...
>nospam@de-ster.demon.nl (J. J. Lodder) wrote:
>> Certainly not.
>> As Tolman already discovered the temperature in a gravitational field
>> -in thermal equilibrium- must depend (in the expected way)
>> on the gravitational potential. (1920-ies, wuld have to look up)
>
>Actually, the constraints between gravity and thermodynamics can
>also be reversed to DERIVE Einstein's equations. This was done
>by Jacobsen; see http://www.arxiv.org/abs/gr-qc/9504004.

Well yeah that makes sense, the Einstein's Law
G_{uv} = kT_uv holds for any form of energy.

>From: Florian Dufey
>If I would sit on some arbitrary point (the reference point) in the
>gravitation field, the temperature of the radiation that I would receive
>of all the systems which are in thermal equilibrium with each other
>would be the same, regardless of their distance.
ok, more below....

>This temperature is
>the temperature which I assign also to the system which emitted the
>radiation. It fulfills the . law. It only depends on the
>internal variables of the systems (like P and V) and on its position in
>the gravitation field relative to the point of reference.

That is highly interesting, if I understand correctly.
If a solid metal rod is hung vertically in a g-field,
and the rod is in thermal equilbrium, then any observer
at any location will receive the same radiant energy
from any and all points of the rod.
The reason for that is the higher temperature at
the bottom of the rod, is exactly compensated by
the radiant energy I would receive from that point
as that radiant energy is red shifted in accord
with my relative altitude. Likewise the radiant energy
from the top of the rod, which is assumed to be
cooler would be blue shifted if my altitude was lower
than the top of the rod.

So my question is this, would the temperature,
of rod, measured by the frequency of it's radiant
energy (assuming thermal equilbrium) be the same
at any point for all observers?

Pardon my simplicity, but Alfred Einstead above suggests
reverse engeering the invariance of the rod's radiant energy
to "DERIVE" Einstein's Law (G_{uv} = kT_uv).

Energy conservation in g-fields is difficult to
learn, it's nice to get an invariant.

Thanks in Advance.
Ken S. Tucker

Alfred Einstead

May6-04, 07:52 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>\nFlorian Dufey <dufey@jupiter.t30.physik.tu-muenchen.de> wrote:\n> The assumtion of transitivity means that any two of these equations\n> also include the third.\n> It is stated that from these assumptions one can show that there are\n> functions t_1({x}), t_2({y}), t_3({z}), so that in equilibrium\n> t_1=t_2=t_3. Does anybody know a formal proof for the last statement?\n\nSince equilibrium is an equivalence relations, then equilibrium\nstates come in equivalence classes. One needs also to show (or\nassume) that if two systems from two different classes are\nin contact with one another, heat will flow only in one direction;\nfrom one to the other. Denote by A->B the relation that heat\nwill flow from A to B when A and B are in contact with each other.\n\nThus one has:\nIf A & B are not equivalent, then either A->B or B->A.\n\nOne needs to show that this property respects equivalence:\n\nIf A & B are equivalent, B->C, then A->C\nIf A & B are equivalent, C->B, then A->B.\n\nThe 2nd Law has to be used somewhere for this. For if A & B\nare equivalent, B->C, but C->A, then heat is flowing (through C)\nfrom B to A. This is supposedly forbidden by the 2nd Law.\n\nTransitivity follows as well:\nIf A->B, B->C then A->C\nby a similar argument. Thus, the equilibrium states are sorted\ninto a linear ordering.\n\nOne may need to make some kind of assumption on how many equilibrium\nstates are possible (finite or countable). Then, one shows that any\ntotal ordering satisfying the restriction can be mapped in an\norder-preserving way to the real interval [0,1].\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>Florian Dufey <dufey@jupiter.t30.physik.tu-muenchen.de> wrote:
> The assumtion of transitivity means that any two of these equations
> also include the third.
> It is stated that from these assumptions one can show that there are
> functions t_1({x}), t_2({y}), t_3({z}), so that in equilibrium
> t_1=t_2=t_3. Does anybody know a formal proof for the last statement?

Since equilibrium is an equivalence relations, then equilibrium
states come in equivalence classes. One needs also to show (or
assume) that if two systems from two different classes are
in contact with one another, heat will flow only in one direction;
from one to the other. Denote by A->B the relation that heat
will flow from A to B when A and B are in contact with each other.

Thus one has:
If A & B are not equivalent, then either A->B or B->A.

One needs to show that this property respects equivalence:

If A & B are equivalent, B->C, then A->C
If A & B are equivalent, C->B, then A->B.

The 2nd Law has to be used somewhere for this. For if A & B
are equivalent, B->C, but C->A, then heat is flowing (through C)
from B to A. This is supposedly forbidden by the 2nd Law.

Transitivity follows as well:
If A->B, B->C then A->C
by a similar argument. Thus, the equilibrium states are sorted
into a linear ordering.

One may need to make some kind of assumption on how many equilibrium
states are possible (finite or countable). Then, one shows that any
total ordering satisfying the restriction can be mapped in an
order-preserving way to the real interval [0,1].

Rob Woodside

May6-04, 07:52 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\nFlorian Dufey <dufey@jupiter.t30.physik.tu-muenchen.de> wrote in message news:<4095F2C7.4090402@jupiter.t30.physik.tu-muenchen.de>...\n> The statement of the zeroth Law may be formulated as an equivalence\n> relation between thermal equilibrium states.\n> But how do you prove that there is a variable called Temperature?\n\nI think you have to postulate it. "The potential that governs heat\nflow", "The lagrange multiplier that keeps the total energy constant"\nor "inversely as the rate of change of entropy with internal energy"\n\n> I ask this because I found some proofs in the literature which seem to\n> be wrong or incomplete.\n\nSadly there is a lot of this.\n\n> Is there something like a mathematical statement that every equivalence\n> relation leads to something like a Temperature?\n\nI doubt it. Notice that any intensive state function will obey a\n"zeroth law" due to the equivalence relation of thermal equilibrium.\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>Florian Dufey <dufey@jupiter.t30.physik.tu-muenchen.de> wrote in message news:<4095F2C7.4090402@jupiter.t30.physik.tu-muenchen.de>...
> The statement of the zeroth Law may be formulated as an equivalence
> relation between thermal equilibrium states.
> But how do you prove that there is a variable called Temperature?

I think you have to postulate it. "The potential that governs heat
flow", "The lagrange multiplier that keeps the total energy constant"
or "inversely as the rate of change of entropy with internal energy"

> I ask this because I found some proofs in the literature which seem to
> be wrong or incomplete.

Sadly there is a lot of this.

> Is there something like a mathematical statement that every equivalence
> relation leads to something like a Temperature?

I doubt it. Notice that any intensive state function will obey a
"zeroth law" due to the equivalence relation of thermal equilibrium.

J. J. Lodder

May6-04, 12:21 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>Florian Dufey <dufey@jupiter.t30.physik.tu-muenchen.de> wrote:\n\n> J. J. Lodder schrieb:\n\n> > Certainly not.\n> > As Tolman already discovered the temperature in a gravitational field\n> > -in thermal equilibrium- must depend (in the expected way)\n> > on the gravitational potential. (1920-ies, would have to look up)\n>\n> I doubt some that the statements made in this thread about the\n> temperature in gravitation fields are correct:\n>\n> If I would sit on some arbitrary point (the reference point) in the\n> gravitation field, the temperature of the radiation that I would receive\n> of all the systems which are in thermal equilibrium with each other\n> would be the same, regardless of their distance.\n\nRight.\n\n> This temperature is\n> the temperature which I assign also to the system which emitted the\n> radiation.\n\nYou can do that, and it would be consistent,\nbut it makes no sense physically.\nIt makes the temperature of a system\ndepend on your position relative to that system.\nWhat is needed physically is that temperature must be an intrinsic\nvariable of the body, independent of which observer looks at it.\nFor example,\nyou wouldn\'t want the temperature of the surface of the sun\nto depend on whether you observe it from Earth or from Mars.\n\n> It fulfills the 0. law.\n\nNot in the proper generalization to situations with a gravitational\nfield. Please wait for my posting explaining this to make it past\nmoderators.\n\n> It only depends on the\n> internal variables of the systems (like P and V) and on its position in\n> the gravitation field relative to the point of reference.\n>\n> However, this temperature transforms as a scalar with respect to\n> general relativity under a change of reference point. That is the\n> temperature of a system on the bottom of the gravitation field measured\n> at that point would be higher than its temperature measured at some\n> point far away from the center of gravity. I can well imagine that\n> these local temperatures conform the observation of Tolman.\n\nIndeed. And these local temperatures are the real temperatures\nin the physical sense.\nIf you stick thermometers into the column at various places,\nand read those thermometers from a distance with a telescope\nyou find Tolman\'s potential-dependent temperatures.\n\nIn practice the effect is of course quite small,\nneutron stars and black holes excepted,\n\nJan\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>Florian Dufey <dufey@jupiter.t30.physik.tu-muenchen.de> wrote:

> J. J. Lodder schrieb:

> > Certainly not.
> > As Tolman already discovered the temperature in a gravitational field
> > -in thermal equilibrium- must depend (in the expected way)
> > on the gravitational potential. (1920-ies, would have to look up)
>
> I doubt some that the statements made in this thread about the
> temperature in gravitation fields are correct:
>
> If I would sit on some arbitrary point (the reference point) in the
> gravitation field, the temperature of the radiation that I would receive
> of all the systems which are in thermal equilibrium with each other
> would be the same, regardless of their distance.

Right.

> This temperature is
> the temperature which I assign also to the system which emitted the
> radiation.

You can do that, and it would be consistent,
but it makes no sense physically.
It makes the temperature of a system
depend on your position relative to that system.
What is needed physically is that temperature must be an intrinsic
variable of the body, independent of which observer looks at it.
For example,
you wouldn't want the temperature of the surface of the sun
to depend on whether you observe it from Earth or from Mars.

> It fulfills the . law.

Not in the proper generalization to situations with a gravitational
field. Please wait for my posting explaining this to make it past
moderators.

> It only depends on the
> internal variables of the systems (like P and V) and on its position in
> the gravitation field relative to the point of reference.
>
> However, this temperature transforms as a scalar with respect to
> general relativity under a change of reference point. That is the
> temperature of a system on the bottom of the gravitation field measured
> at that point would be higher than its temperature measured at some
> point far away from the center of gravity. I can well imagine that
> these local temperatures conform the observation of Tolman.

Indeed. And these local temperatures are the real temperatures
in the physical sense.
If you stick thermometers into the column at various places,
and read those thermometers from a distance with a telescope
you find Tolman's potential-dependent temperatures.

In practice the effect is of course quite small,
neutron stars and black holes excepted,

Jan

Florian Dufey

May7-04, 06:41 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>\n\nJ. J. Lodder schrieb:\n> Florian Dufey <dufey@jupiter.t30.physik.tu-muenchen.de> wrote:\n\n>>This temperature is=20\n>>the temperature which I assign also to the system which emitted the\n>>radiation.\n>=20\n>=20\n> You can do that, and it would be consistent,\n> but it makes no sense physically.\n> It makes the temperature of a system\n> depend on your position relative to that system.\n> What is needed physically is that temperature must be an intrinsic\n> variable of the body, independent of which observer looks at it.\n> For example,\n> you wouldn\'t want the temperature of the surface of the sun\n> to depend on whether you observe it from Earth or from Mars.\n\nLet us remember, that we are talking of very strong gravitational fields =\n\nso that we have to use general relativity.\nLeaving high precision measurements aside, the temperature readings on=20\nMars and Earth would practically coincide.\n\n[snip]\n> If you stick thermometers into the column at various places,\n> and read those thermometers from a distance with a telescope\n> you find Tolman\'s potential-dependent temperatures.\n\nThis I do not doubt. But you have to admit, that an astronomer observing =\n\nneutron stars, won\'t poke thermometers into the star and watch their=20\nreadings with its telescope, but determine the temperature of the=20\nemitted radiation.\n\nI just returned from the library and found there a book by Tolman=20\nhimself, which was first published in 1934:\nRichard C. Tolman, Relativity, Thermodynamics and Cosmology (Clarendon=20\nPress, Oxford, 1949)\nHe derives in section =A7129 the general condition of thermal equilibrium=\n=20\nin a stationary gravitational field and finds that the quantity=20\nT=3DT_0/sqrt(g_{44}) has to be constant where T_0 is the local=20\ntemperature. Although he seemingly prefers to call T_0 the "temperature" =\n\nhe cites Einstein speaking of "wahre Temperatur" (true temperature)=20\nwhich he identifies with T and of "Ehrenfests Taschentemperatur"=20\n(temperature inside Ehrenfest\'s pocket) which he identifies with T_0=20\neven before Einstein had derived the general form of relativity.\nAll in all it seems to be a very nice book, especially when you compare=20\nit with modern treatments which skip equilibrium thermodynamics at all=20\nand directly try to write down equations for irreversible relativistical =\n\nfluid dynamics.\n\n>=20\n>=20\n>>It fulfills the 0. law.\n>=20\n>=20\n> Not in the proper generalization to situations with a gravitational\n> field. Please wait for my posting explaining this to make it past\n> moderators.\nI am eagerly awaiting it.\n\nCheers,\nFlorian\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>J. J. Lodder schrieb:
> Florian Dufey <dufey@jupiter.t30.physik.tu-muenchen.de> wrote:

>>This temperature is=20
>>the temperature which I assign also to the system which emitted the
>>radiation.
>=20>=20
> You can do that, and it would be consistent,
> but it makes no sense physically.
> It makes the temperature of a system
> depend on your position relative to that system.
> What is needed physically is that temperature must be an intrinsic
> variable of the body, independent of which observer looks at it.
> For example,
> you wouldn't want the temperature of the surface of the sun
> to depend on whether you observe it from Earth or from Mars.

Let us remember, that we are talking of very strong gravitational fields =

so that we have to use general relativity.
Leaving high precision measurements aside, the temperature readings on=20
Mars and Earth would practically coincide.

[snip]
> If you stick thermometers into the column at various places,
> and read those thermometers from a distance with a telescope
> you find Tolman's potential-dependent temperatures.

This I do not doubt. But you have to admit, that an astronomer observing =

neutron stars, won't poke thermometers into the star and watch their=20
readings with its telescope, but determine the temperature of the=20
emitted radiation.

I just returned from the library and found there a book by Tolman=20
himself, which was first published in 1934:
Richard C. Tolman, Relativity, Thermodynamics and Cosmology (Clarendon=20
Press, Oxford, 1949)
He derives in section =A7129 the general condition of thermal equilibrium=
=20
in a stationary gravitational field and finds that the quantity=20
T=3DT_0/\sqrt(g_{44}) has to be constant where T_0 is the local=20
temperature. Although he seemingly prefers to call T_0 the "temperature" =

he cites Einstein speaking of "wahre Temperatur" (true temperature)=20
which he identifies with T and of "Ehrenfests Taschentemperatur"=20
(temperature inside Ehrenfest's pocket) which he identifies with T_0=20
even before Einstein had derived the general form of relativity.
All in all it seems to be a very nice book, especially when you compare=20
it with modern treatments which skip equilibrium thermodynamics at all=20
and directly try to write down equations for irreversible relativistical =

fluid dynamics.

>=20>=20
>>It fulfills the . law.
>=20>=20
> Not in the proper generalization to situations with a gravitational
> field. Please wait for my posting explaining this to make it past
> moderators.
I am eagerly awaiting it.

Cheers,
Florian

Niels L. Ellegaard

May7-04, 06:41 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>Florian Dufey <dufey@jupiter.t30.physik.tu-muenchen.de> writes:\n\n> But how do you prove that there is a variable called Temperature? I\n> ask this because I found some proofs in the literature which seem to\n> be wrong or incomplete.\n\nIf you are looking for an axiomatic introduction to thermodynamics,\nthen I highly recommend\n\nThermodynamics and an Introduction to Thermostatistics\nby Herbert B. Callen\n\nIf I remember correctly you zeroth law of thermodynamics is replaced\nby the assumption that entropy is an extensive quantity. Suppose that\nhave two subsystems with the energies E1 and E2. Furthermore the\nsubsystems are described by the entropy functions S1(E1) and\nS2(E2). Now Callen\'s assumption tells us that the total entropy is\ngiven by\n\nS(E1,E2) = S1(E1) + S2(E2)\n\nYou can decribe the thermal contact by a constraint E1 + E2 = E. Given\nthis constraint you can look for the value of E1, that maximizes the\ntotal entropy. This value is the equilibrium value. It fulfills\n\nd(S1(E1))/dE1 = d(S2(E2))/dE2\n\nFrom this description of equilibrium it is trivial to derive your\nzeroth law.\n\nNow you can ask: Why is entropy extensive? In order to answer this\nquestion you need statistical mechanics. I pressume that it can only\nbe introduced in thermodynamics as an axiom or a zeroth law if you\nwill.\n\nNiels.\n\n--\nNiels L Ellegaard http://dirac.ruc.dk/~gnalle/\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>Florian Dufey <dufey@jupiter.t30.physik.tu-muenchen.de> writes:

> But how do you prove that there is a variable called Temperature? I
> ask this because I found some proofs in the literature which seem to
> be wrong or incomplete.

If you are looking for an axiomatic introduction to thermodynamics,
then I highly recommend

Thermodynamics and an Introduction to Thermostatistics
by Herbert B. Callen

If I remember correctly you zeroth law of thermodynamics is replaced
by the assumption that entropy is an extensive quantity. Suppose that
have two subsystems with the energies E1 and E2. Furthermore the
subsystems are described by the entropy functions S1(E1) and
S2(E2). Now Callen's assumption tells us that the total entropy is
given by

S(E1,E2) = S1(E1) + S2(E2)

You can decribe the thermal contact by a constraint E1 + E2 = E. Given
this constraint you can look for the value of E1, that maximizes the
total entropy. This value is the equilibrium value. It fulfills

d(S1(E1))/dE1 = d(S2(E2))/dE2

From this description of equilibrium it is trivial to derive your
zeroth law.

Now you can ask: Why is entropy extensive? In order to answer this
question you need statistical mechanics. I pressume that it can only
be introduced in thermodynamics as an axiom or a zeroth law if you
will.

Niels.

--
Niels L Ellegaard http://dirac.ruc.dk/~gnalle/

J. J. Lodder

May10-04, 05:02 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>\nKen S. Tucker <dynamics@vianet.on.ca> wrote:\n\n> That is highly interesting, if I understand correctly.\n> If a solid metal rod is hung vertically in a g-field,\n> and the rod is in thermal equilbrium, then any observer\n> at any location will receive the same radiant energy\n> from any and all points of the rod.\n> The reason for that is the higher temperature at\n> the bottom of the rod, is exactly compensated by\n> the radiant energy I would receive from that point\n> as that radiant energy is red shifted in accord\n> with my relative altitude. Likewise the radiant energy\n> from the top of the rod, which is assumed to be\n> cooler would be blue shifted if my altitude was lower\n> than the top of the rod.\n\nRight, and you can get that directly from the second law.\nIf there were a temperature difference,\nyou could run a carnot cycle on it,\nand gain work from a system in thermal equilibrium,\nwhich must be impossible.\nAssuming the rod to be immersed in a heat bath\nit would be a perpetuum mobile of the second kind.\n\n> So my question is this, would the temperature,\n> of rod, measured by the frequency of it\'s radiant\n> energy (assuming thermal equilbrium) be the same\n> at any point for all observers?\n\nNo. The higher you stand, the lower is the temperature you see.\nUnless you correct for the redshift caused by the potential difference,\nto know what the temperature really is.\nWhen you do, you see a higher temperature for the bottom of the rod.\n\nBy no less than 10^[-16}, relatively, for a one meter rod\nin the gravitational field of the earth,\nto be measured absolutely for verification.\nRelative methods, like attaching a thermocouple\nbetween top and bottom will not work for precisely the same reason.\n\nBest,\n\nJan\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>Ken S. Tucker <dynamics@vianet.on.ca> wrote:

> That is highly interesting, if I understand correctly.
> If a solid metal rod is hung vertically in a g-field,
> and the rod is in thermal equilbrium, then any observer
> at any location will receive the same radiant energy
> from any and all points of the rod.
> The reason for that is the higher temperature at
> the bottom of the rod, is exactly compensated by
> the radiant energy I would receive from that point
> as that radiant energy is red shifted in accord
> with my relative altitude. Likewise the radiant energy
> from the top of the rod, which is assumed to be
> cooler would be blue shifted if my altitude was lower
> than the top of the rod.

Right, and you can get that directly from the second law.
If there were a temperature difference,
you could run a carnot cycle on it,
and gain work from a system in thermal equilibrium,
which must be impossible.
Assuming the rod to be immersed in a heat bath
it would be a perpetuum mobile of the second kind.

> So my question is this, would the temperature,
> of rod, measured by the frequency of it's radiant
> energy (assuming thermal equilbrium) be the same
> at any point for all observers?

No. The higher you stand, the lower is the temperature you see.
Unless you correct for the redshift caused by the potential difference,
to know what the temperature really is.
When you do, you see a higher temperature for the bottom of the rod.

By no less than 10^[-16}, relatively, for a one meter rod
in the gravitational field of the earth,
to be measured absolutely for verification.
Relative methods, like attaching a thermocouple
between top and bottom will not work for precisely the same reason.

Best,

Jan

Florian Dufey

May12-04, 01:29 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>\n\nAlfred Einstead schrieb:\n\n> The 2nd Law has to be used somewhere for this. For if A & B\n> are equivalent, B->C, but C->A, then heat is flowing (through C)\n> from B to A. This is supposedly forbidden by the 2nd Law.\n>\nOk, that is quite interesting. Often you can hear the statement that the\n0. Law guarantees the existence of a variable "empirical temperature".\nBut actually it is insufficient for that purpose alone but has to be\nused in combination with the 2. Law.\n\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>Alfred Einstead schrieb:

> The 2nd Law has to be used somewhere for this. For if A & B
> are equivalent, B->C, but C->A, then heat is flowing (through C)
> from B to A. This is supposedly forbidden by the 2nd Law.
>
Ok, that is quite interesting. Often you can hear the statement that the
. Law guarantees the existence of a variable "empirical temperature".
But actually it is insufficient for that purpose alone but has to be
used in combination with the 2. Law.

--

J. J. Lodder

May14-04, 04:09 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>Florian Dufey <dufey@jupiter.t30.physik.tu-muenchen.de> wrote:\n\n> J. J. Lodder schrieb:\n> > Florian Dufey <dufey@jupiter.t30.physik.tu-muenchen.de> wrote:\n>\n> >>This temperature is=20\n> >>the temperature which I assign also to the system which emitted the\n> >>radiation.\n> >=20\n> >=20\n> > You can do that, and it would be consistent,\n> > but it makes no sense physically.\n> > It makes the temperature of a system\n> > depend on your position relative to that system.\n> > What is needed physically is that temperature must be an intrinsic\n> > variable of the body, independent of which observer looks at it.\n> > For example,\n> > you wouldn\'t want the temperature of the surface of the sun\n> > to depend on whether you observe it from Earth or from Mars.\n>\n> Let us remember, that we are talking of very strong gravitational fields =\n>\n> so that we have to use general relativity.\n> Leaving high precision measurements aside, the temperature readings on=20\n> Mars and Earth would practically coincide.\n\nPractice doesn\'t invalidate a matter of principle.\n\n> [snip]\n> > If you stick thermometers into the column at various places,\n> > and read those thermometers from a distance with a telescope\n> > you find Tolman\'s potential-dependent temperatures.\n>\n> This I do not doubt. But you have to admit, that an astronomer observing =\n>\n> neutron stars, won\'t poke thermometers into the star and watch their=20\n> readings with its telescope, but determine the temperature of the=20\n> emitted radiation.\n\nYes.\nAnd a competent astronomer would correct for known or estimated\ngravitational redshift (by Wien\'s law)\nto get the true surface temperature of a neutron star.\n(which will be nearly a blackbody radiator,\napart from rotation Doppler shifts)\n\n> I just returned from the library and found there a book by Tolman=20\n> himself, which was first published in 1934:\n> Richard C. Tolman, Relativity, Thermodynamics and Cosmology (Clarendon=20\n> Press, Oxford, 1949)\n> He derives in section =A7129 the general condition of thermal equilibrium=\n> =20\n> in a stationary gravitational field and finds that the quantity=20\n> T=3DT_0/sqrt(g_{44}) has to be constant where T_0 is the local=20\n> temperature. Although he seemingly prefers to call T_0 the "temperature" =\n>\n> he cites Einstein speaking of "wahre Temperatur" (true temperature)=20\n> which he identifies with T and of "Ehrenfests Taschentemperatur"=20\n> (temperature inside Ehrenfest\'s pocket) which he identifies with T_0=20\n> even before Einstein had derived the general form of relativity.\n> All in all it seems to be a very nice book, especially when you compare=20\n> it with modern treatments which skip equilibrium thermodynamics at all=20\n> and directly try to write down equations for irreversible relativistical =\n>\n> fluid dynamics.\n>\n> >=20\n> >=20\n> >>It fulfills the 0. law.\n> >=20\n> >=20\n> > Not in the proper generalization to situations with a gravitational\n> > field. Please wait for my posting explaining this to make it past\n> > moderators.\n> I am eagerly awaiting it.\n\nVanished in a black cyberhole, apparently. I reposted.\n\nAnd further comment: Einstein, Planck, and Ehrenfest argued about the\n\'correct\'\nLorentz transformation for temperature and other thermodynamic\nquantities.\nIt turned out (much later) that there is no \'correct\' way.\nThe Lorentz tyransformed temperature is a calculated quantity without\nphysical meaning,\nand more than one choice is possible.\n\nA temperature with physical meaning exists only in those cases\nwhere you can (in principle) determine the absolute temperature\nthrough its definition by heat exchange using carnot cycles.\nIn practice one may proceed by proving that the temperature scale\ndefined by a blackbody thermometry coincides with the absolute scale.\n\nBest,\n\nJan\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>Florian Dufey <dufey@jupiter.t30.physik.tu-muenchen.de> wrote:

> J. J. Lodder schrieb:
> > Florian Dufey <dufey@jupiter.t30.physik.tu-muenchen.de> wrote:
>
> >>This temperature is=20
> >>the temperature which I assign also to the system which emitted the
> >>radiation.
> >=20
> >=20
> > You can do that, and it would be consistent,
> > but it makes no sense physically.
> > It makes the temperature of a system
> > depend on your position relative to that system.
> > What is needed physically is that temperature must be an intrinsic
> > variable of the body, independent of which observer looks at it.
> > For example,
> > you wouldn't want the temperature of the surface of the sun
> > to depend on whether you observe it from Earth or from Mars.
>
> Let us remember, that we are talking of very strong gravitational fields =
>
> so that we have to use general relativity.
> Leaving high precision measurements aside, the temperature readings on=20
> Mars and Earth would practically coincide.

Practice doesn't invalidate a matter of principle.

> [snip]
> > If you stick thermometers into the column at various places,
> > and read those thermometers from a distance with a telescope
> > you find Tolman's potential-dependent temperatures.
>
> This I do not doubt. But you have to admit, that an astronomer observing =
>
> neutron stars, won't poke thermometers into the star and watch their=20
> readings with its telescope, but determine the temperature of the=20
> emitted radiation.

Yes.
And a competent astronomer would correct for known or estimated
gravitational redshift (by Wien's law)
to get the true surface temperature of a neutron star.
(which will be nearly a blackbody radiator,
apart from rotation Doppler shifts)

> I just returned from the library and found there a book by Tolman=20
> himself, which was first published in 1934:
> Richard C. Tolman, Relativity, Thermodynamics and Cosmology (Clarendon=20
> Press, Oxford, 1949)
> He derives in section =A7129 the general condition of thermal equilibrium=
> =20
> in a stationary gravitational field and finds that the quantity=20
> T=3DT_0/\sqrt(g_{44}) has to be constant where T_0 is the local=20
> temperature. Although he seemingly prefers to call T_0 the "temperature" =
>
> he cites Einstein speaking of "wahre Temperatur" (true temperature)=20
> which he identifies with T and of "Ehrenfests Taschentemperatur"=20
> (temperature inside Ehrenfest's pocket) which he identifies with T_0=20
> even before Einstein had derived the general form of relativity.
> All in all it seems to be a very nice book, especially when you compare=20
> it with modern treatments which skip equilibrium thermodynamics at all=20
> and directly try to write down equations for irreversible relativistical =
>
> fluid dynamics.
>
> >=20
> >=20
> >>It fulfills the . law.
> >=20
> >=20
> > Not in the proper generalization to situations with a gravitational
> > field. Please wait for my posting explaining this to make it past
> > moderators.
> I am eagerly awaiting it.

Vanished in a black cyberhole, apparently. I reposted.

And further comment: Einstein, Planck, and Ehrenfest argued about the
'correct'
Lorentz transformation for temperature and other thermodynamic
quantities.
It turned out (much later) that there is no 'correct' way.
The Lorentz tyransformed temperature is a calculated quantity without
physical meaning,
and more than one choice is possible.

A temperature with physical meaning exists only in those cases
where you can (in principle) determine the absolute temperature
through its definition by heat exchange using carnot cycles.
In practice one may proceed by proving that the temperature scale
defined by a blackbody thermometry coincides with the absolute scale.

Best,

Jan

J. J. Lodder

May14-04, 04:09 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>Peter Tobias <tobiasep@comcast.net> wrote:\n\n> David Williams:\n> > Radiation gains energy if it moves\n> > "downhill", and vice versa, so an object at the top of a hill is at\n> > thermal equilibrium with a hotter object at the bottom of the hill.\n>\n> Sounds convincing. Have you calculated how the temperature profile in a\n> gravitational field should look like when a system is in equilibrium, or\n> could you give a link to a calculation?\n\nYou can easily do it for yourself.\n(natural units, newtonian approximation, skipping some steps)\nEnergy = frequency = temperature = gravitational mass, so\n\ndE = E F dh = E d\\Phi\n\nIntegration and restoring plumbers units gives\n\nT_1 = T_2 exp(\\Phi_2 - \\Phi_1)/c^2,\n\nor, expanding, since \\Phi << c^2\n\n\\Delta T = -T \\Delta\\Phi/c^2\n\nVerify Carnot condition:\nFrom the temperature difference you can get work dW by moving dQ\nthrough a carnot cycle\n\ndW = dQ \\Delta T/T,\n\nwhich is just the mechanical energy needed\nto move dQ through the potential difference.\n\nThe confusion felt by some on the subject\nmay be caused by the misleading name of \'zero-th law\',\nwhich suggests it is a fundamental principle\nthat goes before everything else.\nActually it isn\'t, because absolute (aka thermodynamic) temperature\n-is defined- by the second law using carnot cycles.\n\nWhat the zero\'th law really says is\nthat for an extended body in thermal equilibrium\nno net work can be made by using a Carnot cycle\nto move heat from one part of the body to another.\n\nConversely, a body at constant thermodynamic temperature in a\ngravitational field would not be at max entropy,\nfor some work could be extracted from it by moving some heat downwards.\n\nBest,\n\nJan\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>Peter Tobias <tobiasep@comcast.net> wrote:

> David Williams:
> > Radiation gains energy if it moves
> > "downhill", and vice versa, so an object at the top of a hill is at
> > thermal equilibrium with a hotter object at the bottom of the hill.
>
> Sounds convincing. Have you calculated how the temperature profile in a
> gravitational field should look like when a system is in equilibrium, or
> could you give a link to a calculation?

You can easily do it for yourself.
(natural units, newtonian approximation, skipping some steps)
Energy = frequency = temperature = gravitational mass, so

dE = E F dh = E d\Phi

Integration and restoring plumbers units gives

T_1 = T_2 \exp(\Phi_2 - \Phi_1)/c^2,

or, expanding, since \Phi << c^2\Delta T = -T \Delta\Phi/c^2

Verify Carnot condition:
From the temperature difference you can get work dW by moving dQ
through a carnot cycle

dW = dQ \Delta T/T,

which is just the mechanical energy needed
to move dQ through the potential difference.

The confusion felt by some on the subject
may be caused by the misleading name of 'zero-th law',
which suggests it is a fundamental principle
that goes before everything else.
Actually it isn't, because absolute (aka thermodynamic) temperature
-is defined- by the second law using carnot cycles.

What the zero'th law really says is
that for an extended body in thermal equilibrium
no net work can be made by using a Carnot cycle
to move heat from one part of the body to another.

Conversely, a body at constant thermodynamic temperature in a
gravitational field would not be at max entropy,
for some work could be extracted from it by moving some heat downwards.

Best,

Jan

Florian Dufey

May17-04, 08:14 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>\n\n> That is highly interesting, if I understand correctly.\n> If a solid metal rod is hung vertically in a g-field,\n> and the rod is in thermal equilbrium, then any observer\n> at any location will receive the same radiant energy\n> from any and all points of the rod.\n> The reason for that is the higher temperature at\n> the bottom of the rod, is exactly compensated by\n> the radiant energy I would receive from that point\n> as that radiant energy is red shifted in accord\n> with my relative altitude. Likewise the radiant energy\n> from the top of the rod, which is assumed to be\n> cooler would be blue shifted if my altitude was lower\n> than the top of the rod.\n>\n> So my question is this, would the temperature,\n> of rod, measured by the frequency of it\'s radiant\n> energy (assuming thermal equilbrium) be the same\n> at any point for all observers?\n\n\nAs far as I understand it, yes.\n\n\n>\n> Pardon my simplicity, but Alfred Einstead above suggests\n> reverse engeering the invariance of the rod\'s radiant energy\n> to "DERIVE" Einstein\'s Law (G_uv = kT_uv).\n>\n> Energy conservation in g-fields is difficult to\n> learn, it\'s nice to get an invariant.\n>\nThat\'s an interesting point. But remember that thermodynamics should\nbetter be termed thermostatics. That is, we can define thermal\nequilibrium strictly only for systems which don\'t change with time.\nFor these systems energy will always be a conserved quantity as a\nconsequence of time translation invariance.\n\n\n\n\n\n> Thanks in Advance.\n> Ken S. Tucker\n\n\n--\n\n---------------------------------------------------------------------\nFlorian Dufey\nPhysik Department T38\nTechnische Universitaet Muenchen\nJames-Franck-Strasse\nD-85747 Garching b. Muenchen\nGermany\n\nemail: florian.dufey@physik.tu-muenchen.de\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>That is highly interesting, if I understand correctly.
> If a solid metal rod is hung vertically in a g-field,
> and the rod is in thermal equilbrium, then any observer
> at any location will receive the same radiant energy
> from any and all points of the rod.
> The reason for that is the higher temperature at
> the bottom of the rod, is exactly compensated by
> the radiant energy I would receive from that point
> as that radiant energy is red shifted in accord
> with my relative altitude. Likewise the radiant energy
> from the top of the rod, which is assumed to be
> cooler would be blue shifted if my altitude was lower
> than the top of the rod.
>
> So my question is this, would the temperature,
> of rod, measured by the frequency of it's radiant
> energy (assuming thermal equilbrium) be the same
> at any point for all observers?

As far as I understand it, yes.

>
> Pardon my simplicity, but Alfred Einstead above suggests
> reverse engeering the invariance of the rod's radiant energy
> to "DERIVE" Einstein's Law (G_{uv} = kT_uv).
>
> Energy conservation in g-fields is difficult to
> learn, it's nice to get an invariant.
>
That's an interesting point. But remember that thermodynamics should
better be termed thermostatics. That is, we can define thermal
equilibrium strictly only for systems which don't change with time.
For these systems energy will always be a conserved quantity as a
consequence of time translation invariance.

> Thanks in Advance.
> Ken S. Tucker

--

---------------------------------------------------------------------
Florian Dufey
Physik Department T38
Technische Universitaet Muenchen
James-Franck-Strasse
D-85747 Garching b. Muenchen
Germany

email: florian.dufey@physik.tu-muenchen.de
---------------------------------------------------------------------

Florian Dufey

May18-04, 03:44 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>J. J. Lodder schrieb:\n[snip]\n>\n> You can easily do it for yourself.\n> (natural units, newtonian approximation, skipping some steps)\n> Energy = frequency = temperature = gravitational mass, so\n\n\nI don\'t see any reason why temperature should be proportional to energy\nin GR.\n\n>\n> dE = E F dh = E d\\Phi\n>\n> Integration and restoring plumbers units gives\n>\n> T_1 = T_2 exp(\\Phi_2 - \\Phi_1)/c^2,\n>\n> or, expanding, since \\Phi << c^2\n>\n> \\Delta T = -T \\Delta\\Phi/c^2\n>\n> Verify Carnot condition:\n> From the temperature difference you can get work dW by moving dQ\n> through a carnot cycle\n>\n> dW = dQ \\Delta T/T,\n>\n> which is just the mechanical energy needed\n> to move dQ through the potential difference.\n>\n> The confusion felt by some on the subject\n> may be caused by the misleading name of \'zero-th law\',\n> which suggests it is a fundamental principle\n> that goes before everything else.\n> Actually it isn\'t, because absolute (aka thermodynamic) temperature\n> -is defined- by the second law using carnot cycles.\n\nI don\'t think that you are right here. As has been nicely laid out by\nAlfred Einstead in some other branch of this thread, the zeroth law\nguarantees that we can sort systems which are in thermal equilibrium\ninto equivalence classes and define a linear ordering of these classes.\nThese ordering is equivalent to the existence of empirical temperature\nscales.\nNone of these conditions seem to be violated in GR, so that there must\nbe an empirical temperature theta also for a neutron star. I think it is\nperfectly ok to take the temperature of the radiation that is received\nfrom every system in the star as measured at a common reference point.\nAs it is a common quantity of all the systems which are in equilibrium.\nIf the energy density of the radiation at the point where it is\ngenerated is E_i, the energy density as the reference point would be\ntheta_i= E_i exp((\\Phi_i)/c^2), or, more generally theta=E/sqrt(g_44).\n\n\nThe second law says in one of its famous formulations by Clausius that\nit is impossible to transfer heat from a colder to a hotter body\nwithout doing work. Your temperature scale would violate this principle.\n\nYou are right when you say that absolute temperature may be defined by\nthe use of Carnot cycles, but note that it also defines the\ntransformation law of temperature.\nI know that there are some difficulties with energy conservation in GR.\nI suppose that this is not a problem for static systems, as there exists\ntime translation.\nSo I may write W+Q_1-Q_2 for the Carnot cycle.\nThe efficiency is then as usual\neta=W/Q_1=1-Q_2/Q_1\nwhich has to be a function of the empirical temperatures theta_1 and\ntheta_2 alone, which has to be of the form eta=1=f(theta_2)/f(theta_1)\nIt is then easy to show that Q_1/theta_1= Q_2/theta_2, that is,\nf(theta)=const*theta. Hence, the empirical temperature scale we have\nchosen is also an absolute temperature scale. The constant itself will\nbe a function of the potential of the reference point chosen.\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>J. J. Lodder schrieb:
[snip]
>
> You can easily do it for yourself.
> (natural units, newtonian approximation, skipping some steps)
> Energy = frequency = temperature = gravitational mass, so

I don't see any reason why temperature should be proportional to energy
in GR.

>
> dE = E F dh = E d\Phi
>
> Integration and restoring plumbers units gives
>
> T_1 = T_2 \exp(\Phi_2 - \Phi_1)/c^2,
>
> or, expanding, since \Phi << c^2
>
> \Delta T = -T \Delta\Phi/c^2
>
> Verify Carnot condition:
> From the temperature difference you can get work dW by moving dQ
> through a carnot cycle
>
> dW = dQ \Delta T/T,
>
> which is just the mechanical energy needed
> to move dQ through the potential difference.
>
> The confusion felt by some on the subject
> may be caused by the misleading name of 'zero-th law',
> which suggests it is a fundamental principle
> that goes before everything else.
> Actually it isn't, because absolute (aka thermodynamic) temperature
> -is defined- by the second law using carnot cycles.

I don't think that you are right here. As has been nicely laid out by
Alfred Einstead in some other branch of this thread, the zeroth law
guarantees that we can sort systems which are in thermal equilibrium
into equivalence classes and define a linear ordering of these classes.
These ordering is equivalent to the existence of empirical temperature
scales.
None of these conditions seem to be violated in GR, so that there must
be an empirical temperature \theta also for a neutron star. I think it is
perfectly ok to take the temperature of the radiation that is received
from every system in the star as measured at a common reference point.
As it is a common quantity of all the systems which are in equilibrium.
If the energy density of the radiation at the point where it is
generated is E_i, the energy density as the reference point would be
\theta_i= E_i \exp((\Phi_i)/c^2), or, more generally \theta=E/\sqrt(g_{44}).

The second law says in one of its famous formulations by Clausius that
it is impossible to transfer heat from a colder to a hotter body
without doing work. Your temperature scale would violate this principle.

You are right when you say that absolute temperature may be defined by
the use of Carnot cycles, but note that it also defines the
transformation law of temperature.
I know that there are some difficulties with energy conservation in GR.
I suppose that this is not a problem for static systems, as there exists
time translation.
So I may write W+Q_1-Q_2 for the Carnot cycle.
The efficiency is then as usual
\eta=W/Q_1=1-Q_2/Q_1
which has to be a function of the empirical temperatures \theta_1 and
\theta_2 alone, which has to be of the form \eta=1=f(\theta_2)/f(\theta_1)
It is then easy to show that Q_1/\theta_1= Q_2/\theta_2, that is,
f(\theta)=const*\theta. Hence, the empirical temperature scale we have
chosen is also an absolute temperature scale. The constant itself will
be a function of the potential of the reference point chosen.

J. J. Lodder

May23-04, 03:14 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>Florian Dufey <dufey@jupiter.t30.physik.tu-muenchen.de> wrote:\n\n> J. J. Lodder schrieb:\n> [snip]\n> >\n> > You can easily do it for yourself.\n> > (natural units, newtonian approximation, skipping some steps)\n> > Energy = frequency = temperature = gravitational mass, so\n>\n>\n> I don\'t see any reason why temperature should be proportional to energy\n> in GR.\n\nEverywhere.\nTemperature is nothing but energy per degree of freedom,\nand best measured in the same unit.\n\n> > dE = E F dh = E d\\Phi\n> >\n> > Integration and restoring plumbers units gives\n> >\n> > T_1 = T_2 exp(\\Phi_2 - \\Phi_1)/c^2,\n> >\n> > or, expanding, since \\Phi << c^2\n> >\n> > \\Delta T = -T \\Delta\\Phi/c^2\n> >\n> > Verify Carnot condition:\n> > From the temperature difference you can get work dW by moving dQ\n> > through a carnot cycle\n> >\n> > dW = dQ \\Delta T/T,\n> >\n> > which is just the mechanical energy needed\n> > to move dQ through the potential difference.\n> >\n> > The confusion felt by some on the subject\n> > may be caused by the misleading name of \'zero-th law\',\n> > which suggests it is a fundamental principle\n> > that goes before everything else.\n> > Actually it isn\'t, because absolute (aka thermodynamic) temperature\n> > -is defined- by the second law using carnot cycles.\n>\n> I don\'t think that you are right here. As has been nicely laid out by\n> Alfred Einstead in some other branch of this thread, the zeroth law\n> guarantees that we can sort systems which are in thermal equilibrium\n> into equivalence classes and define a linear ordering of these classes.\n\nHe is mistaken in the case of a gravitational field.\nSee below.\n\n> These ordering is equivalent to the existence of empirical temperature\n> scales.\n> None of these conditions seem to be violated in GR, so that there must\n> be an empirical temperature theta also for a neutron star. I think it is\n> perfectly ok to take the temperature of the radiation that is received\n> from every system in the star as measured at a common reference point.\n> As it is a common quantity of all the systems which are in equilibrium.\n> If the energy density of the radiation at the point where it is\n> generated is E_i, the energy density as the reference point would be\n> theta_i= E_i exp((\\Phi_i)/c^2), or, more generally theta=E/sqrt(g_44).\n>\n>\n> The second law says in one of its famous formulations by Clausius that\n> it is impossible to transfer heat from a colder to a hotter body\n> without doing work. Your temperature scale would violate this principle.\n\nThat formulation too must be revised in the case of general relativity\nbeing taken into account.\nAnd not only in general relativity: any force that works on all forms of\nenergy equally would have such an effect.\nIt is just that no others are known.\nAdding E = mc^2 as an ad-hoc postulate to Newtonian physics\nhas the same consequences.\n\nThe opposite may even be the case: energy may be gained by transferring\nheat from a colder to a hotter body through a gravitational potential\ngradient, when the gravitational energy gain wins.\nAnd a forteriori: heat will flow spontaneously in an isothermal body\nto establish a temperature gradient matching the gravitational potential\nby the formula derived above.\n\nAnd no: it is not \'my\' temperature scale. It is -the- absolute\ntemperature. There is nothing to choose.\n\nTo see the thruth of these assertions consider the following\ngedankenexperiment.\nTake a long collumn, in thermal equilibrium in a gravitational field,\nand a small test body.\n(to make everything simple you can take both to be blackbody cavities,\nwith the traditional kohlenstoffpartikel inside to equibrilate the\nmodes)\nPut the small body at half height, connect it through small holes and a\nwaveguide, and wait for equilibrium.\nNow lift the testbody adiabatically to the top, and connect there.\nHeat will then flow from the testbody into the collumn.\nConversely, lower the testbody to the bottom.\nThen heat will flow the oppposite way.\nIt is therfore impossible to establish a temperature ordering\nbased on the direction of heat flow alone.\nAnd one can easily go on to construct carnot machinery on the same\nprinciples.\n\n> You are right when you say that absolute temperature may be defined by\n> the use of Carnot cycles, but note that it also defines the\n> transformation law of temperature.\n\nThere is no absolute temperature beyond that defined by using carnot\ncycles.\nAnd every other \'temperature\' must either be related to it,\nor be no more than a formal concept without physical relevance.\n\nAs I explained in my earlier posting\nthere is no problem at all with thermodynamics,\nif properly generalised to take account of the weight of heat and work,\nand of energy in general.\nThere are just some too narrow attemps at axiomatization\nwhich need to be reformulated.\n\nBest,\n\nJan\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>Florian Dufey <dufey@jupiter.t30.physik.tu-muenchen.de> wrote:

> J. J. Lodder schrieb:
> [snip]
> >
> > You can easily do it for yourself.
> > (natural units, newtonian approximation, skipping some steps)
> > Energy = frequency = temperature = gravitational mass, so
>
>
> I don't see any reason why temperature should be proportional to energy
> in GR.

Everywhere.
Temperature is nothing but energy per degree of freedom,
and best measured in the same unit.

> > dE = E F dh = E d\Phi
> >
> > Integration and restoring plumbers units gives
> >
> > T_1 = T_2 \exp(\Phi_2 - \Phi_1)/c^2,
> >
> > or, expanding, since \Phi << c^2
> >
> > \Delta T = -T \Delta\Phi/c^2
> >
> > Verify Carnot condition:
> > From the temperature difference you can get work dW by moving dQ
> > through a carnot cycle
> >
> > dW = dQ \Delta T/T,
> >
> > which is just the mechanical energy needed
> > to move dQ through the potential difference.
> >
> > The confusion felt by some on the subject
> > may be caused by the misleading name of 'zero-th law',
> > which suggests it is a fundamental principle
> > that goes before everything else.
> > Actually it isn't, because absolute (aka thermodynamic) temperature
> > -is defined- by the second law using carnot cycles.
>
> I don't think that you are right here. As has been nicely laid out by
> Alfred Einstead in some other branch of this thread, the zeroth law
> guarantees that we can sort systems which are in thermal equilibrium
> into equivalence classes and define a linear ordering of these classes.

He is mistaken in the case of a gravitational field.
See below.

> These ordering is equivalent to the existence of empirical temperature
> scales.
> None of these conditions seem to be violated in GR, so that there must
> be an empirical temperature \theta also for a neutron star. I think it is
> perfectly ok to take the temperature of the radiation that is received
> from every system in the star as measured at a common reference point.
> As it is a common quantity of all the systems which are in equilibrium.
> If the energy density of the radiation at the point where it is
> generated is E_i, the energy density as the reference point would be
> \theta_i= E_i \exp((\Phi_i)/c^2), or, more generally \theta=E/\sqrt(g_{44}).
>
>
> The second law says in one of its famous formulations by Clausius that
> it is impossible to transfer heat from a colder to a hotter body
> without doing work. Your temperature scale would violate this principle.

That formulation too must be revised in the case of general relativity
being taken into account.
And not only in general relativity: any force that works on all forms of
energy equally would have such an effect.
It is just that no others are known.
Adding E = mc^2 as an ad-hoc postulate to Newtonian physics
has the same consequences.

The opposite may even be the case: energy may be gained by transferring
heat from a colder to a hotter body through a gravitational potential
gradient, when the gravitational energy gain wins.
And a forteriori: heat will flow spontaneously in an isothermal body
to establish a temperature gradient matching the gravitational potential
by the formula derived above.

And no: it is not 'my' temperature scale. It is -the- absolute
temperature. There is nothing to choose.

To see the thruth of these assertions consider the following
gedankenexperiment.
Take a long collumn, in thermal equilibrium in a gravitational field,
and a small test body.
(to make everything simple you can take both to be blackbody cavities,
with the traditional kohlenstoffpartikel inside to equibrilate the
modes)
Put the small body at half height, connect it through small holes and a
waveguide, and wait for equilibrium.
Now lift the testbody adiabatically to the top, and connect there.
Heat will then flow from the testbody into the collumn.
Conversely, lower the testbody to the bottom.
Then heat will flow the oppposite way.
It is therfore impossible to establish a temperature ordering
based on the direction of heat flow alone.
And one can easily go on to construct carnot machinery on the same
principles.

> You are right when you say that absolute temperature may be defined by
> the use of Carnot cycles, but note that it also defines the
> transformation law of temperature.

There is no absolute temperature beyond that defined by using carnot
cycles.
And every other 'temperature' must either be related to it,
or be no more than a formal concept without physical relevance.

As I explained in my earlier posting
there is no problem at all with thermodynamics,
if properly generalised to take account of the weight of heat and work,
and of energy in general.
There are just some too narrow attemps at axiomatization
which need to be reformulated.

Best,

Jan

alistair

May25-04, 03:56 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>\nFlorian Dufey <dufey@jupiter.t30.physik.tu-muenchen.de> wrote:\n\n> J. J. Lodder schrieb:\n> [snip]\n> >\n> > You can easily do it for yourself.\n> > (natural units, newtonian approximation, skipping some steps)\n> > Energy = frequency = temperature = gravitational mass, so\n>\n>\n> I don\'t see any reason why temperature should be proportional to energy\n> in GR.\n\n\nIt certainly is here:\n\nT = hc^3/8pi^2 kGM becomes kT = hc^3/8pi^2GM = Energy\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>Florian Dufey <dufey@jupiter.t30.physik.tu-muenchen.de> wrote:

> J. J. Lodder schrieb:
> [snip]
> >
> > You can easily do it for yourself.
> > (natural units, newtonian approximation, skipping some steps)
> > Energy = frequency = temperature = gravitational mass, so
>
>
> I don't see any reason why temperature should be proportional to energy
> in GR.

It certainly is here:

T = hc^3/8pi^2 kGM becomes kT = hc^3/8pi^2GM = Energy

Peter Tobias

May25-04, 03:56 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>\nJ. J. Lodder:\n> Temperature is nothing but energy per degree of freedom,\n> and best measured in the same unit.\n\nThis is wrong - if the energy k*T is not large compared to the energy\nspacing of a degree of freedom, the energy in this degree of freedom\nis smaller than in other degrees of freedom where k*T is large\ncompared to the energy spacing.\n\nRegards,\n\nPeter\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>J. J. Lodder:
> Temperature is nothing but energy per degree of freedom,
> and best measured in the same unit.

This is wrong - if the energy k*T is not large compared to the energy
spacing of a degree of freedom, the energy in this degree of freedom
is smaller than in other degrees of freedom where k*T is large
compared to the energy spacing.

Regards,

Peter