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View Full Version : How do black holes change in scalar-tensor gravity?


Christoph Schiller
Aug12-04, 08:30 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n\nFor the horizon of a non-rotating black hole in general relativity,\nthere is a famous relation between the energy E, the area A, and the\nsurface gravity a:\n\nE = a A (c^2/8 pi G)\n\nIs this relation also valid in the scalar-tensor theory of gravity?\nAn acquaintance of mine is adamant on this point, and maintains\nthat Schwarzschild black holes - or at least their\nhorizons - are the same for both theories.\n\nOn the other hand, the scalar tensor theory has an additional\nfundamental constant, which does not appear in the equation.\nHow is this possible?\n\nChristoph Schiller\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>For the horizon of a non-rotating black hole in general relativity,
there is a famous relation between the energy E, the area A, and the
surface gravity a:

E = a A (c^2/8 \pi G)

Is this relation also valid in the scalar-tensor theory of gravity?
An acquaintance of mine is adamant on this point, and maintains
that Schwarzschild black holes - or at least their
horizons - are the same for both theories.

On the other hand, the scalar tensor theory has an additional
fundamental constant, which does not appear in the equation.
How is this possible?

Christoph Schiller