View Full Version : cosmological constant in Planck units
John Baez
Oct27-04, 10:56 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>\n\nWhat\'s the cosmological constant in Planck units?\n\nA figure of 10^{-120} is often bandied about, but\nI\'ve recently seen 10^{-123}. If we use the WMAP\ndata and all the current conventional wisdom on\ncosmology, what do we actually get? (I don\'t\nreally want to hear all the caveats.)\n\nIf you give it to me in MKS units, I\'ll do the rest.\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>What's the cosmological constant in Planck units?
A figure of 10^{-120} is often bandied about, but
I've recently seen 10^{-123}. If we use the WMAP
data and all the current conventional wisdom on
cosmology, what do we actually get? (I don't
really want to hear all the caveats.)
If you give it to me in MKS units, I'll do the rest.
ebunn@lfa221051.richmond.edu
Oct28-04, 01:28 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>>What\'s the cosmological constant in Planck units?\n>\n>A figure of 10^{-120} is often bandied about, but\n>I\'ve recently seen 10^{-123}. If we use the WMAP\n>data and all the current conventional wisdom on\n>cosmology, what do we actually get? (I don\'t\n>really want to hear all the caveats.)\n>\n>If you give it to me in MKS units, I\'ll do the rest.\n\nThe consensus value from WMAP and other cosmological data is\nOmega_Lambda = 0.7. By definition, Omega_anything is the energy\ndensity of "anything" over the critical density:\n\nOmega_Lambda = rho_Lambda / rho_crit.\n\nThe energy density rho_Lambda = Lambda / 8 pi G, and the critical\ndensity rho_crit = 3 H^2 / 8 pi G, so\n\nOmega_Lambda = Lambda / 3 H^2.\n\nSo I guess the cosmological constant is\n\nLambda = 3 Omega_Lambda H^2 = 3 (0.7) (9.25 x 10^27 cm / h)^(-2).\n\nThat last number is the Hubble distance. (The speed of light is 1, as\neveryone knows, so the Hubble constant is the reciprocal of the Hubble\ndistance.) h isn\'t Planck\'s constant; it\'s the Hubble constant in\nunits of 100 km/(s Mpc). Best available data says h = 0.7 or so, so\n\nLambda = 1.2 x 10^(-56) cm^(-2).\n\nThe Planck length is 1.62 x 10^(-33) cm, so\n\nLambda = 3.2 x 10^(-122)\n\nin Planck units.\n\nOf course, if you wanted the energy density rho_Lambda instead, you\'d\ndivide by 8 pi G. G is one in Planck units, so you just have to\ndivide by 8 pi: rho_Lambda = 1.3 x 10^(-123) in Planck units.\n\nBy the way, Appendix A of Kolb & Turner\'s book "The Early Universe"\nis a great thing to have by your side in this sort of calculation.\n\n-Ted\n\n--\n[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]\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>>What's the cosmological constant in Planck units?
>
>A figure of 10^{-120} is often bandied about, but
>I've recently seen 10^{-123}. If we use the WMAP
>data and all the current conventional wisdom on
>cosmology, what do we actually get? (I don't
>really want to hear all the caveats.)
>
>If you give it to me in MKS units, I'll do the rest.
The consensus value from WMAP and other cosmological data is
\Omega_Lambda = .7. By definition, \Omega_anything is the energy
density of "anything" over the critical density:
\Omega_Lambda = \rho_Lambda / \rho_crit[/itex].
The energy density \rho_Lambda = \Lambda / 8 \pi G, and the critical
density \rho_crit = 3 H^2 / 8 \pi G, so
\Omega_Lambda = \Lambda / 3 H^2.
So I guess the cosmological constant is
\Lambda = 3 \Omega_Lambda H^2 = 3 (0.7) (9.25 x 10^27 cm / h)^(-2).
That last number is the Hubble distance. (The speed of light is 1, as
everyone knows, so the Hubble constant is the reciprocal of the Hubble
distance.) h isn't Planck's constant; it's the Hubble constant in
units of 100 km/(s Mpc). Best available data says h = .7 or so, so
\Lambda = 1.2 x 10^(-56) cm^(-2).
The Planck length is 1.62 x 10^(-33) cm, so
\Lambda = 3.[itex]2 x 10^(-122)
in Planck units.
Of course, if you wanted the energy density \rho_Lambda instead, you'd
divide by 8 \pi G. G is one in Planck units, so you just have to
divide by 8 \pi: \rho_Lambda = 1.3 x 10^(-123) in Planck units.
By the way, Appendix A of Kolb & Turner's book "The Early Universe"
is a great thing to have by your side in this sort of calculation.
-Ted
--
[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]
<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"John Baez" <baez@galaxy.ucr.edu> wrote in message\nnews:clkbil\\$gl5\\$1@glue.ucr.edu...\n>\ n>\n> What\'s the cosmological constant in Planck units?\n>\n> A figure of 10^{-120} is often bandied about, but\n> I\'ve recently seen 10^{-123}. If we use the WMAP\n> data and all the current conventional wisdom on\n> cosmology, what do we actually get? (I don\'t\n> really want to hear all the caveats.)\n\nActually, in my previous post I forgot the factor 8 Pi (This sometimes\nhappens, when switching between units G=1 and 8 Pi G = 1). Therefore:\n\nWith Omega_Lambda = 0.73, h = 0.71 we have:\n\nLambda = 3.37 * 10^-122\n\nWith h = 0.63\n\nLambda = 2.61 * 10^-122\n\nBest MP\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>"John Baez" <baez@galaxy.ucr.edu> wrote in message
news:clkbil$gl5$1@glue.ucr.edu...
>
>
> What's the cosmological constant in Planck units?
>
> A figure of 10^{-120} is often bandied about, but
> I've recently seen 10^{-123}. If we use the WMAP
> data and all the current conventional wisdom on
> cosmology, what do we actually get? (I don't
> really want to hear all the caveats.)
Actually, in my previous post I forgot the factor 8 \Pi (This sometimes
happens, when switching between units G=1 and 8 \Pi G = 1). Therefore:
With \Omega_Lambda = .73, h = .71 we have:
\Lambda = 3[/itex].37 * 10^-122
With h = .63
\Lambda = 2.[itex]61 * 10^-122
Best MP
ebunn@lfa221051.richmond.edu
Nov3-04, 09:45 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>[Moderator\'s note: There was a typo in (a) previous version(s) of this\npost. -P.H.]\n\nIn article <clkbil\\$gl5\\$1@glue.ucr.edu>, John Baez <baez@galaxy.ucr.edu> wrote:\n\n>What\'s the cosmological constant in Planck units?\n>\n>A figure of 10^{-120} is often bandied about, but\n>I\'ve recently seen 10^{-123}. If we use the WMAP\n>data and all the current conventional wisdom on\n>cosmology, what do we actually get? (I don\'t\n>really want to hear all the caveats.)\n>\n>If you give it to me in MKS units, I\'ll do the rest.\n\nThe consensus value from WMAP and other cosmological data is\nOmega_Lambda = 0.7. By definition, Omega_anything is the energy\ndensity of "anything" over the critical density:\n\nOmega_Lambda = rho_Lambda / rho_crit.\n\nThe energy density rho_Lambda = Lambda / 8 pi G, and the critical\ndensity rho_crit = 3 H^2 / 8 pi G, so\n\nOmega_Lambda = Lambda / 3 H^2.\n\nSo I guess the cosmological constant is\n\nLambda = 3 Omega_Lambda H^2 = 3 (0.7) (9.25 x 10^27 cm / h)^(-2).\n\nThat last number is the Hubble distance. (The speed of light is 1, as\neveryone knows, so the Hubble constant is the reciprocal of the Hubble\ndistance.) h isn\'t Planck\'s constant; it\'s the Hubble constant in\nunits of 100 km/(s Mpc). Best available data says h = 0.7 or so, so\n\nLambda = 1.2 x 10^(-56) cm^(-2).\n\nThe Planck length is 1.62 x 10^(-33) cm, so\n\nLambda = 3.2 x 10^(-122)\n\nin Planck units.\n\nOf course, if you wanted the energy density rho_Lambda instead, you\'d\ndivide by 8 pi G. G is one in Planck units, so you just have to\ndivide by 8 pi: rho_Lambda = 1.3 x 10^(-123) in Planck units.\n\nBy the way, Appendix A of Kolb & Turner\'s book "The Early Universe"\nis a great thing to have by your side in this sort of calculation.\n\n-Ted\n\n--\n[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]\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>[Moderator's note: There was a typo in (a) previous version(s) of this
post. -P.H.]
In article <clkbil$gl5$1@glue.ucr.edu>, John Baez <baez@galaxy.ucr.edu> wrote:
>What's the cosmological constant in Planck units?
>
>A figure of 10^{-120} is often bandied about, but
>I've recently seen 10^{-123}. If we use the WMAP
>data and all the current conventional wisdom on
>cosmology, what do we actually get? (I don't
>really want to hear all the caveats.)
>
>If you give it to me in MKS units, I'll do the rest.
The consensus value from WMAP and other cosmological data is
\Omega_Lambda = .7. By definition, \Omega_anything is the energy
density of "anything" over the critical density:
\Omega_Lambda = \rho_Lambda / \rho_crit[/itex].
The energy density \rho_Lambda = \Lambda / 8 \pi G, and the critical
density \rho_crit = 3 H^2 / 8 \pi G, so
\Omega_Lambda = \Lambda / 3 H^2.
So I guess the cosmological constant is
\Lambda = 3 \Omega_Lambda H^2 = 3 (0.7) (9.25 x 10^27 cm / h)^(-2).
That last number is the Hubble distance. (The speed of light is 1, as
everyone knows, so the Hubble constant is the reciprocal of the Hubble
distance.) h isn't Planck's constant; it's the Hubble constant in
units of 100 km/(s Mpc). Best available data says h = .7 or so, so
\Lambda = 1.2 x 10^(-56) cm^(-2).
The Planck length is 1.62 x 10^(-33) cm, so
\Lambda = 3.[itex]2 x 10^(-122)
in Planck units.
Of course, if you wanted the energy density \rho_Lambda instead, you'd
divide by 8 \pi G. G is one in Planck units, so you just have to
divide by 8 \pi: \rho_Lambda = 1.3 x 10^(-123) in Planck units.
By the way, Appendix A of Kolb & Turner's book "The Early Universe"
is a great thing to have by your side in this sort of calculation.
-Ted
--
[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]
<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>"John Baez" <baez@galaxy.ucr.edu> wrote in message\nnews:clkbil\\$gl5\\$1@glue.ucr.edu...\n>\ n>\n> What\'s the cosmological constant in Planck units?\n>\n> A figure of 10^{-120} is often bandied about, but\n> I\'ve recently seen 10^{-123}. If we use the WMAP\n> data and all the current conventional wisdom on\n> cosmology, what do we actually get? (I don\'t\n> really want to hear all the caveats.)\n\n\nCricital Density = 3 / (8 Pi) H^2\n\nWith H = 71 km s^-1 MPc^-1 (assumed by WMAP) this is (in Planck units)\n\nrho_critical = 1.8358 * 10^-123\n\nWith Omega_Lambda = 0.73:\n\nLambda = 1.34 * 10^-123\n\n\nCaveats: (Although you don\'t want to hear them)\n\nHubble-Constant is probably is lower than WMAP-value of H = 71 (in obvious\nunits). Absolute measurements (SZ-effect, lensing, etc) favor H =60-65. Fit\nto recent supernova-luminosity distances consistently give H = 60-63 if one\ncalibrates to the LMC-cepheids and the well known SN1997A supernova\n(geometrical distance to SN 1997A can be measured by determining the\nextension of the expanding gas envelope)\n\nA realistic value might rather be H = 62.5. Therefore multiply rho_crit with\n(62.5/71)^2, which gives:\n\nLambda = 1.04 * 10^-123\n\nAlso Lambda depends on Planck-units, whose error is dominated by the\nmeasurement error of Newtons constant G. Don\'t expect any result depending\non Planck units to be better than the the error of G, i.e. 0.1 % (unless G\ncancels out in the equations, as is the case for some thermodynamic\nrelations in GR)\n\nFinally keep in mind, that the determination of Lambda is model-dependent.\nPositive Lambda is required (at roughly 3 sigma level) in the isotropic\nFRW-models of the universe. There are other solutions of the field\nequations, which are capable of reproducing many features of the observable\nuniverse, but that require other values of Lambda. One therefore has to\ncarefully differentiate between the notion of the "cosmological constant"\n(which value is model-dependent, and therefore fails if the model fails)\nwith the notion of "dark energy", i.e. negative pressure (=tension). It is\nthe negative pressure, that is observed. Negative pressure quite certainly\nis real (3 sigma). But it is not necessarily comes in form of a cosmological\nconstant. Other models parametrize "dark energy" in different ways. In my\nopinion, the theoretically most promising models assume that dark energy is\nnothing else than string tension. These models explain observations best,\nwhen Lambda = 0.\n\nBest, MP\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>"John Baez" <baez@galaxy.ucr.edu> wrote in message
news:clkbil$gl5$1@glue.ucr.edu...
>
>
> What's the cosmological constant in Planck units?
>
> A figure of 10^{-120} is often bandied about, but
> I've recently seen 10^{-123}. If we use the WMAP
> data and all the current conventional wisdom on
> cosmology, what do we actually get? (I don't
> really want to hear all the caveats.)
Cricital Density = 3 / (8 \Pi) H^2
With H = 71 km s^-1 MPc^-1 (assumed by WMAP) this is (in Planck units)
\rho_critical = 1.[/itex]8358 * 10^-123
With \Omega_Lambda = .73:
\Lambda = 1.34 * 10^-123
Caveats: (Although you don't want to hear them)
Hubble-Constant is probably is lower than WMAP-value of H = 71 (in obvious
units). Absolute measurements (SZ-effect, lensing, etc) favor H =60-65. Fit
to recent supernova-luminosity distances consistently give H = 60-63 if one
calibrates to the LMC-cepheids and the well known SN1997A supernova
(geometrical distance to SN 1997A can be measured by determining the
extension of the expanding gas envelope)
A realistic value might rather be H = 62.5. Therefore multiply \rho_crit with
(62.5/71)^2, which gives:
\Lambda = 1.[itex]04 * 10^-123
Also \Lambda depends on Planck-units, whose error is dominated by the
measurement error of Newtons constant G. Don't expect any result depending
on Planck units to be better than the the error of G, i.e. .1 % (unless G
cancels out in the equations, as is the case for some thermodynamic
relations in GR)
Finally keep in mind, that the determination of \Lambda is model-dependent.
Positive \Lambda is required (at roughly 3 \sigma level) in the isotropic
FRW-models of the universe. There are other solutions of the field
equations, which are capable of reproducing many features of the observable
universe, but that require other values of \Lambda. One therefore has to
carefully differentiate between the notion of the "cosmological constant"
(which value is model-dependent, and therefore fails if the model fails)
with the notion of "dark energy", i.e. negative pressure (=tension). It is
the negative pressure, that is observed. Negative pressure quite certainly
is real (3 \sigma). But it is not necessarily comes in form of a cosmological
constant. Other models parametrize "dark energy" in different ways. In my
opinion, the theoretically most promising models assume that dark energy is
nothing else than string tension. These models explain observations best,
when \Lambda = .
Best, MP
<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>John Baez Wrote:\n> What\'s the cosmological constant in Planck units?\n>\n> A figure of 10^{-120} is often bandied about, but\n> I\'ve recently seen 10^{-123}. If we use the WMAP\n> data and all the current conventional wisdom on\n> cosmology, what do we actually get? (I don\'t\n> really want to hear all the caveats.)\n>\n> If you give it to me in MKS units, I\'ll do the rest.\n\nI think the current estimate of Lambda in Planck units is\n4.0 E-123\n\nThe estimated dark energy density would be one third of that, expressed\nin Planck units, namely\n1.3 E-123\n\nAnyone who wishes to can calculate this from the value of the Hubble\nparameter 71 km/s per Megaparsec.\nThis is an inverse time, and in MKS units it would be 2.3 E-18 per\nsecond,\nor 2.30 E-18, if we postpone rounding off till later.\nThis MKS value is equivalent to 1.24 E-61 reciprocal Planck time\nunits.\n\nPutting G=c=1,\nthe critical density, rho_crit, is (3/8pi)H^2\nThe current estimate of the dark energy density is 73 percent of that.\nSo in Planck terms the dark energy density is\n0.73 (3/8pi)H^2 = 0.73 (3/8pi)(1.24 E-61)^2 = 1.3 E-123\n\nBut according to the conventions about Lamda, the dark energy density\nis given not by Lambda but by Lambda/3---so you still have to multiply\nby 3.\n\nLambda = 0.73 (9/8pi)H^2 = 0.73 (9/8pi)(1.24 E-61)^2 = 4.0 E-123\n\n------------------------------------------------------------------------\nThis post submitted through the LaTeX-enabled physicsforums.com\nTo view this post with LaTeX images:\nhttp://www.physicsforums.com/showthread.php?t=49948#post355861\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>John Baez Wrote:
> What's the cosmological constant in Planck units?
>
> A figure of 10^{-120} is often bandied about, but
> I've recently seen 10^{-123}. If we use the WMAP
> data and all the current conventional wisdom on
> cosmology, what do we actually get? (I don't
> really want to hear all the caveats.)
>
> If you give it to me in MKS units, I'll do the rest.
I think the current estimate of \Lambda in Planck units is
4. E-123
The estimated dark energy density would be one third of that, expressed
in Planck units, namely
1.3 E-123
Anyone who wishes to can calculate this from the value of the Hubble
parameter 71 km/s per Megaparsec.
This is an inverse time, and in MKS units it would be 2.3 E-18 per
second,
or 2.30 E-18, if we postpone rounding off till later.
This MKS value is equivalent to 1.24 E-61 reciprocal Planck time
units.
Putting G=c=1,
the critical density, \rho_crit, is (3/8pi)H^2
The current estimate of the dark energy density is 73 percent of that.
So in Planck terms the dark energy density is
.73 (3/8pi)H^2 = .73 (3/8pi)(1.24 E-61)^2 = 1.3 E-123
But according to the conventions about Lamda, the dark energy density
is given not by \Lambda but by \Lambda/3---so you still have to multiply
by 3.
\Lambda =[/itex] .73 (9/8pi)H^2 = .[itex]73 (9/8pi)(1.24 E-61)^2 = 4. E-123
------------------------------------------------------------------------
This post submitted through the LaTeX-enabled physicsforums.com
To view this post with LaTeX images:
http://www.physicsforums.com/showthread.php?t=49948#post355861
vBulletin® v3.8.7, Copyright ©2000-2012, vBulletin Solutions, Inc.