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alistair
Aug7-04, 05:06 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>By writing the length contraction equation\n\nL2 = L1 (1 - v^2/c^2)^1/2 as:\n\nL2 = L1 ( 1 - v^2 / c^2 + small constant)^1/2\n\nT ab still equals T ba in the stress energy tensor\nand the magnitude of the four momentum of a photon is still zero.\nOne can imagine a proton falling into a black hole and\nreaching the speed of light at the expected position of the\nsingularity -the proton keeping a finite diameter under\nthis transformation.\nWould this mean the singularity in a black hole does not exist?\nAlso using the classical angular momentum mvr for a spinning\nblack hole and (1 - v^2/c^2 + small constant)^1/2, I\ncalculated a maximum angular momentum of 10^42 for\na black hole of five solar masses.This compares well\nwith the value from general relativity for the maximum\nwhich is 10^41 (GM^2/c).\nLet me know if you\'d like to see the details of this\ncalculation.\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">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>By writing the length contraction equation

L2 = L1 (1 - v^2/c^2)^1/2[/itex] as:

L2 = L1 ( 1 - v^2 / c^2 + small [itex]constant)^1/2

T ab still equals T ba in the stress energy tensor
and the magnitude of the four momentum of a photon is still zero.
One can imagine a proton falling into a black hole and
reaching the speed of light at the expected position of the
singularity -the proton keeping a finite diameter under
this transformation.
Would this mean the singularity in a black hole does not exist?
Also using the classical angular momentum mvr for a spinning
black hole and (1 - v^2/c^2 + small constant)^1/2, I
calculated a maximum angular momentum of 10^42 for
a black hole of five solar masses.This compares well
with the value from general relativity for the maximum
which is 10^41 (GM^2/c).
Let me know if you'd like to see the details of this
calculation.

tessel@tum.bot
Aug12-04, 08:29 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>On Sat, 7 Aug 2004, alistair wrote:\n\n&gt; T ab still equals T ba in the stress energy tensor and the magnitude of\n&gt; the four momentum of a photon is still zero.\n\nNo, in conventional str kinematics, the momentum of a photon is -not-\nzero. See:\n\nauthor = {Edwin F. Taylor and John Archibald Wheeler},\ntitle = {Spacetime Physics: Introduction to Special Relativity},\nedition = {Second},\npublisher = {W. H. Freeman},\nyear = 1992}\n\nThen you can read this:\n\nauthor = {Geroch, Robert},\ntitle = {General relativity from {A} to {B}},\npublisher = {University of Chicago Press},\nyear = 1978}\n\nHTH,\n\n"T. Essel" (hiding somewhere in cyberspace)\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Sat, 7 Aug 2004, alistair wrote:

> T ab still equals T ba in the stress energy tensor and the magnitude of
> the four momentum of a photon is still zero.

No, in conventional str kinematics, the momentum of a photon is -not-
zero. See:

author = {Edwin F. Taylor and John Archibald Wheeler},
title = {Spacetime Physics: Introduction to Special Relativity},
edition = {Second},
publisher = {W. H. Freeman},
year = 1992}

Then you can read this:

author = {Geroch, Robert},
title = {General relativity from {A} to {B}},
publisher = {University of Chicago Press},
year = 1978}

HTH,

"T. Essel" (hiding somewhere in cyberspace)