View Full Version : hawking radiation - conservation of energy
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>I\'ve just read the excellent FAQ on conservation of energy in GR,\nwhich certainly changed my world view, and set me thinking. As I\nunderstand it, in GR the most naive formulation of conservation of\nenergy is true only locally in curved space, in differential form. In\nintegral form (i.e. conservation of energy in a finite volume), either\n"energy" is not conserved, or we have to come up with a different\ninterpretation of energy, which has no single good definition.\n\nA black hole is just such a region of curved space. Therefore,\nasymptotically far from the BH, we do not necessarily observe the\nconservation of energy over the volume inside + flux going outside.\nThe "amount" of non-conservation I think is proportional to the\nintegral of the curvature over the enclosed volume (although I don\'t\nhave the mathematical tools to calculate this properly), but probably\ncalculable by working out the commutator of the curvature of space\nwith energy in some Noether\'s theorem formulation (???)\n\nNow if the answer we get turns out to be the same as the amount of\nHawking radiation (hand-waving argument suggests it at least scales\nthe same way with BH radius), and both necessarily scale with Planck\'s\nconstant correctly. Then my question is this: just because a BH emits\nHawking radiation from the horizon, does it necessarily follow that\nthe BH is losing mass i.e. evaporating? I always assumed it did from\nconservation of energy, but it looks like I was wrong. Maybe if I\ncould calculate the "other side" of the equation for mass-energy of\nthe BH on its own, I would find that it was actually constant? That\nis, the total energy of BH + Hawking radiation simply is not conserved\nand increases with time. Has anyone done this calculation?\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>I've just read the excellent FAQ on conservation of energy in GR,
which certainly changed my world view, and set me thinking. As I
understand it, in GR the most naive formulation of conservation of
energy is true only locally in curved space, in differential form. In
integral form (i.e. conservation of energy in a finite volume), either
"energy" is not conserved, or we have to come up with a different
interpretation of energy, which has no single good definition.
A black hole is just such a region of curved space. Therefore,
asymptotically far from the BH, we do not necessarily observe the
conservation of energy over the volume inside + flux going outside.
The "amount" of non-conservation I think is proportional to the
integral of the curvature over the enclosed volume (although I don't
have the mathematical tools to calculate this properly), but probably
calculable by working out the commutator of the curvature of space
with energy in some Noether's theorem formulation (???)
Now if the answer we get turns out to be the same as the amount of
Hawking radiation (hand-waving argument suggests it at least scales
the same way with BH radius), and both necessarily scale with Planck's
constant correctly. Then my question is this: just because a BH emits
Hawking radiation from the horizon, does it necessarily follow that
the BH is losing mass i.e. evaporating? I always assumed it did from
conservation of energy, but it looks like I was wrong. Maybe if I
could calculate the "other side" of the equation for mass-energy of
the BH on its own, I would find that it was actually constant? That
is, the total energy of BH + Hawking radiation simply is not conserved
and increases with time. Has anyone done this calculation?
Florian Dufey
Jun18-04, 01:25 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>I would like to ask a maybe related question.\nI started some weeks ago a thread about the definition of temperature\nwhen gravitizing objects are present.\nI argued that observers at different gravitational potential will have\ndifferent absolute temperature scales.\nIf one defines a temperature by the 0. law, that is, two systems in\nthermal equilibrium have to have the same temperature, an observer could\ndefine the temperature of the body he observes as the temperature of\nthe thermal radiation of the body which he will measure at his\nobservation point. As the radiation will be red or blue shifted with\nrespect to the location where it was produced, the temperature will be\ndifferent from that measured by an observer located at the point where\nthe radiation was emitted.\nAs far as I understand it, the Hawking temperature is the temperature\nthat an observer located at the event horizon would observe.\nWhat temperature would an observer measure who is far away from the\nblack hole?\n\nI would have argued that the redshift of radiation emitted at the event\nhorizon will be infinite so that the temperature of this radiation far\naway from the black hole will tend to 0. Hence, temperature will also\ntend to 0.\n\nThere seems to me to be another problem with the Hawking radiation.\nAccording to the 0. law, temperature is a property of two systems in\nequilibrium with each other. That means, it should not change with time.\nHowever, even if the black hole is stationary, I see no possibility to\nfix a thermometer at the event horizon\nas the force on it will be infinite.\nAfter a short time, it will have fallen inside the event horizon.\nHow can I be sure that it has reached thermal equilibrium?\nTherefore, I doubt that "Hawking temperature" is really a temperature\nin the thermodynamical sense.\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>I would like to ask a maybe related question.
I started some weeks ago a thread about the definition of temperature
when gravitizing objects are present.
I argued that observers at different gravitational potential will have
different absolute temperature scales.
If one defines a temperature by the . law, that is, two systems in
thermal equilibrium have to have the same temperature, an observer could
define the temperature of the body he observes as the temperature of
the thermal radiation of the body which he will measure at his
observation point. As the radiation will be red or blue shifted with
respect to the location where it was produced, the temperature will be
different from that measured by an observer located at the point where
the radiation was emitted.
As far as I understand it, the Hawking temperature is the temperature
that an observer located at the event horizon would observe.
What temperature would an observer measure who is far away from the
black hole?
I would have argued that the redshift of radiation emitted at the event
horizon will be infinite so that the temperature of this radiation far
away from the black hole will tend to . Hence, temperature will also
tend to .
There seems to me to be another problem with the Hawking radiation.
According to the . law, temperature is a property of two systems in
equilibrium with each other. That means, it should not change with time.
However, even if the black hole is stationary, I see no possibility to
fix a thermometer at the event horizon
as the force on it will be infinite.
After a short time, it will have fallen inside the event horizon.
How can I be sure that it has reached thermal equilibrium?
Therefore, I doubt that "Hawking temperature" is really a temperature
in the thermodynamical sense.
carlip@no-physics-spam.ucdavis.edu
Jun21-04, 02:51 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[...]\n> I started some weeks ago a thread about the definition of temperature\n> when gravitizing objects are present.\n> I argued that observers at different gravitational potential will have\n> different absolute temperature scales.\n> If one defines a temperature by the 0. law, that is, two systems in\n> thermal equilibrium have to have the same temperature, an observer could\n> define the temperature of the body he observes as the temperature of\n> the thermal radiation of the body which he will measure at his\n> observation point. As the radiation will be red or blue shifted with\n> respect to the location where it was produced, the temperature will be\n> different from that measured by an observer located at the point where\n> the radiation was emitted.\n\nThat\'s right. There\'s a nice paper by Martinez, Phys.Rev.D54 (1996) 6302,\ngr-qc/9609048, that goes carefully through the modifications needed to\nmake sense of thermodynamics a a gravitational field. Position-dependent\ntemperatures are one such modification.\n\n> As far as I understand it, the Hawking temperature is the temperature\n> that an observer located at the event horizon would observe.\n\nNo, it\'s the temperature as measured at infinity. As you approach the\nhorizon, the temperature measured by a stationary observer is blue-shifted\nto infinity.\n\nSteve Carlip\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:
[...]
> I started some weeks ago a thread about the definition of temperature
> when gravitizing objects are present.
> I argued that observers at different gravitational potential will have
> different absolute temperature scales.
> If one defines a temperature by the . law, that is, two systems in
> thermal equilibrium have to have the same temperature, an observer could
> define the temperature of the body he observes as the temperature of
> the thermal radiation of the body which he will measure at his
> observation point. As the radiation will be red or blue shifted with
> respect to the location where it was produced, the temperature will be
> different from that measured by an observer located at the point where
> the radiation was emitted.
That's right. There's a nice paper by Martinez, Phys.Rev.D54 (1996) 6302,
http://www.arxiv.org/abs/gr-qc/9609048, that goes carefully through the modifications needed to
make sense of thermodynamics a a gravitational field. Position-dependent
temperatures are one such modification.
> As far as I understand it, the Hawking temperature is the temperature
> that an observer located at the event horizon would observe.
No, it's the temperature as measured at infinity. As you approach the
horizon, the temperature measured by a stationary observer is blue-shifted
to infinity.
Steve Carlip
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