View Full Version : cosmological constant
alistair
Jun22-04, 05:01 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>A uniformly accelerated observer (Rindler acceleration) can conclude\nthat there is energy in a volume of space that is empty of energy to\nan inertial observer.\nThe temperature measured by the accelerated observer would be given\nby\nT = acceleration / 2 pi k\n\nIn reality, can an observer move with an acceleration that yields a\ntemperature which corresponds to the observed energy density of the\nvacuum, which is about 10^-9 Joules per cubic metre?\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>A uniformly accelerated observer (Rindler acceleration) can conclude
that there is energy in a volume of space that is empty of energy to
an inertial observer.
The temperature measured by the accelerated observer would be given
by
T = acceleration / 2 \pi k
In reality, can an observer move with an acceleration that yields a
temperature which corresponds to the observed energy density of the
vacuum, which is about 10^-9 Joules per cubic metre?
Rob Woodside
Jun25-04, 03:49 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>Note to moderator: why do the postings in this thread have no "follow\nup" buttons?\n\n[Note from moderator: I don\'t know. This sounds to me like something\nspecific to the way your news server and/or your news reading\nsoftware are set up. -TB]\n\nHere\'s an abbreviated motivation for Einstein\'s equations and the\nCosmological constant L. Once you think that there is a connection\nbetween Geometry and Mechanics through Gravity and want a local\ntheory, you ask what is the most mechanical information and the most\ngeometric information that can be carried by a point.\n\nWith mechanics you come up with the second rank stress-energy tensor\nfor a continuous fluid. The divergence of a stess-energy tensor is the\nexternal force density on the fluid. The total stress-energy tensor T\nmust have a vanishing divergence as the are no other external agents\nto push on it. To locally conserve angular momentum you want T\nsymmetric.\n\nWith geometry you come up with the metric g or the full Riemann\ncurvature R. Specifying either one at every point fixes the geometry.\nMechanics requires a second rank symmetric and divergence free tensor.\nThe metric has these properties, but the curvature is a fourth rank\ntensor. A little index gynastics and Bianchi\'s second identity shows\nthat the trace of the double dual of the Riemann curvature (The\n"Einstein\'s tensor" G)is a symmetric, divergence free, second rank\ntensor representation of Curvature.\n\nThe most general symmetric, divergence free, second rank geometric\ntensor is then G + Lg. This should then couple to T with the constant\nk to give Einstein\'s equations: G + Lg = kT\n\nNotice that the Cosmological constant L is giving the relative\nweighting of metric to curvature that is required by T. Curvature and\nmetric are apples and oranges, but this Blunder with positive L\nallowed Einstein a non collapsing universe. Hubble\'s dynamic universe\nhad no need for the constraining L\n\nNow we are forced to accept a universe with accelerating expansion. A\nquick fix is to postulate a new type of matter called dark energy with\na stress-energy tensor Td, satisfying kTd = -Lg. The total stress\nenergy tensor is now T\'=T+Td and Einstein\'s equations become G=kT\'.\nDark energy is very bizarre with a negative energy density that is\nminus the isotropic pressure. It is divergence free so nothing can\npush on it. I doubt that this will be the ultimate explanation of the\naccelerating expansion, but it is a place to start and may stimulate\nsome experiments that will get us thinking on the right track.\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>Note to moderator: why do the postings in this thread have no "follow
up" buttons?
[Note from moderator: I don't know. This sounds to me like something
specific to the way your news server and/or your news reading
software are set up. -TB]
Here's an abbreviated motivation for Einstein's equations and the
Cosmological constant L. Once you think that there is a connection
between Geometry and Mechanics through Gravity and want a local
theory, you ask what is the most mechanical information and the most
geometric information that can be carried by a point.
With mechanics you come up with the second rank stress-energy tensor
for a continuous fluid. The divergence of a stess-energy tensor is the
external force density on the fluid. The total stress-energy tensor T
must have a vanishing divergence as the are no other external agents
to push on it. To locally conserve angular momentum you want T
symmetric.
With geometry you come up with the metric g or the full Riemann
curvature R. Specifying either one at every point fixes the geometry.
Mechanics requires a second rank symmetric and divergence free tensor.
The metric has these properties, but the curvature is a fourth rank
tensor. A little index gynastics and Bianchi's second identity shows
that the trace of the double dual of the Riemann curvature (The
"Einstein's tensor" G)is a symmetric, divergence free, second rank
tensor representation of Curvature.
The most general symmetric, divergence free, second rank geometric
tensor is then G + Lg. This should then couple to T with the constant
k to give Einstein's equations: G + Lg = kT
Notice that the Cosmological constant L is giving the relative
weighting of metric to curvature that is required by T. Curvature and
metric are apples and oranges, but this Blunder with positive L
allowed Einstein a non collapsing universe. Hubble's dynamic universe
had no need for the constraining L
Now we are forced to accept a universe with accelerating expansion. A
quick fix is to postulate a new type of matter called dark energy with
a stress-energy tensor Td, satisfying kTd = -Lg. The total stress
energy tensor is now T'=T+Td and Einstein's equations become G=kT'.
Dark energy is very bizarre with a negative energy density that is
minus the isotropic pressure. It is divergence free so nothing can
push on it. I doubt that this will be the ultimate explanation of the
accelerating expansion, but it is a place to start and may stimulate
some experiments that will get us thinking on the right track.
alistair
Jun27-04, 06:02 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:<4d06b9d7.0406231715.dd9750b@posting.google.c om>...\n\n> The most general symmetric, divergence free, second rank geometric\n> tensor is then G + Lg. This should then couple to T with the constant\n> k to give Einstein\'s equations: G + Lg = kT\n>\n> Notice that the Cosmological constant L is giving the relative\n> weighting of metric to curvature that is required by T. Curvature and\n> metric are apples and oranges, but this Blunder with positive L\n> allowed Einstein a non collapsing universe. Hubble\'s dynamic universe\n> had no need for the constraining L\n>\n> Now we are forced to accept a universe with accelerating expansion. A\n> quick fix is to postulate a new type of matter called dark energy with\n> a stress-energy tensor Td, satisfying kTd = -Lg. The total stress\n> energy tensor is now T\'=T+Td and Einstein\'s equations become G=kT\'.\n> Dark energy is very bizarre with a negative energy density that is\n> minus the isotropic pressure. It is divergence free so nothing can\n> push on it. I doubt that this will be the ultimate explanation of the\n> accelerating expansion, but it is a place to start and may stimulate\n> some experiments that will get us thinking on the right track.\n\nALISTAIR writes:\n\nT changes slowly with time as the space dimensions of the universe\nexpand (for example the energy density of a large mass reduces).\nTd does not change with time because more dark energy is created as\nthe space dimensions expand.Over a short period of time it is\ncorrect to write T\'= T + Td but is it okay to do this in the long run?\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:<4d06b9d7.0406231715.dd9750b@posting.google.com>...
> The most general symmetric, divergence free, second rank geometric
> tensor is then G + Lg. This should then couple to T with the constant
> k to give Einstein's equations: G + Lg = kT
>
> Notice that the Cosmological constant L is giving the relative
> weighting of metric to curvature that is required by T. Curvature and
> metric are apples and oranges, but this Blunder with positive L
> allowed Einstein a non collapsing universe. Hubble's dynamic universe
> had no need for the constraining L
>
> Now we are forced to accept a universe with accelerating expansion. A
> quick fix is to postulate a new type of matter called dark energy with
> a stress-energy tensor Td, satisfying kTd = -Lg. The total stress
> energy tensor is now T'=T+Td and Einstein's equations become G=kT'.
> Dark energy is very bizarre with a negative energy density that is
> minus the isotropic pressure. It is divergence free so nothing can
> push on it. I doubt that this will be the ultimate explanation of the
> accelerating expansion, but it is a place to start and may stimulate
> some experiments that will get us thinking on the right track.
ALISTAIR writes:
T changes slowly with time as the space dimensions of the universe
expand (for example the energy density of a large mass reduces).
Td does not change with time because more dark energy is created as
the space dimensions expand.Over a short period of time it is
correct to write T'= T + Td but is it okay to do this in the long run?
Rob Woodside
Jun29-04, 05:44 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>alistair@goforit64.fsnet.co.uk (alistair) wrote in message news:<861c1b21.0406260750.7098b797@posting.google. com>...\n> ALISTAIR writes:\n>\n> T changes slowly with time as the space dimensions of the universe\n> expand (for example the energy density of a large mass reduces).\n> Td does not change with time because more dark energy is created as\n> the space dimensions expand.Over a short period of time it is\n> correct to write T\'= T + Td but is it okay to do this in the long run?\n\nSuggest an experiment. Why should dark energy be continously created\nwhen Td is divergence free? What else other than T\'= T + Td would you\nwrite?\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>alistair@goforit64.fsnet.co.uk (alistair) wrote in message news:<861c1b21.0406260750.7098b797@posting.google.com>...
> ALISTAIR writes:
>
> T changes slowly with time as the space dimensions of the universe
> expand (for example the energy density of a large mass reduces).
> Td does not change with time because more dark energy is created as
> the space dimensions expand.Over a short period of time it is
> correct to write T'= T + Td but is it okay to do this in the long run?
Suggest an experiment. Why should dark energy be continously created
when Td is divergence free? What else other than T'= T + Td would you
write?
vBulletin® v3.8.7, Copyright ©2000-2012, vBulletin Solutions, Inc.