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Cosmological ConstantVacuum Energy Density, or How Can Nothing Weigh Something?Recently two different groups have measured the apparent brightness ofsupernovaewith redshifts near z = 1. Based on this datathe old idea of a cosmological constant is making a comeback.Einstein Static CosmologyEinstein's original cosmological model was a static, homogeneous modelwith spherical geometry. The gravitational effect of matter caused anacceleration in this model which Einstein did not want, since at the timethe Universe was not known to be expanding. Thus Einstein introduced acosmological constant into his equations for General Relativity.This term acts to counteract the gravitational pull of matter, and soit has been described as an anti-gravity effect.Why does the cosmological constant behave this way?This term acts like a vacuum energy density, an idea which hasbecome quite fashionable in high energy particle physics models sincea vacuum energy density of a specific kind is used in the Higgs mechanism forspontaneous symmetry breaking. Indeed, the inflationary scenario for the firstpicosecond after the Big Bang proposes that a fairly large vacuum energydensity existed during the inflationary epoch.The vacuum energy density must be associated with a negative pressurebecause:The vacuum energy density must be constant because there is nothing for itto depend on.If a piston capping a cylinder of vacuum is pulled out, producing morevacuum, the vacuum within the cylinder then has more energy which must havebeen supplied by a force pulling on the piston.If the vacuum is trying to pull the piston back into the cylinder, it must have a negative pressure, since a positive pressure would tend topush the piston out. The animation above shows the piston moving in the cylinder filled witha "vacuum" containing quantum fluctuations, while the region outside thecylinder has "nothing" with zero density and pressure. Of course thepolitically correct terms are "false vacuum" in the cylinder and "truevacuum" outside, but the physics is the same.The magnitude of the negative pressure needed for energy conservation iseasily found to be P = -u = -rho*c2where P is the pressure, u is the vacuum energy density,and rho is the equivalent mass density using E = m*c2. An alternate derivation uses theargument that the stress-energy tensor of the vacuum must be Lorentzinvariant and thus must be a multiple of the metric tensor. Here are the technical details ofthis argument.But in General Relativity, pressure has weight, which means thatthe gravitational acceleration at the edge of a uniform density sphereis not given byg = GM/R2 = (4*pi/3)*G*rho*Rbut is rather given byg = (4*pi/3)*G*(rho+3P/c2)*RNow Einstein wanted a static model, which means that g = 0, buthe also wanted to have some matter, so rho > 0, and thushe needed P < 0. In fact, by settingrho(vacuum) = 0.5*rho(matter)he had a total density of 1.5*rho(matter) and a total pressureof -0.5*rho(matter)*c2 since the pressure from ordinary matter is essentially zero (compared to rho*c2).Thus rho+3P/c2 = 0 and the gravitational accelerationwas zero, g = (4*pi/3)*G*(rho(matter)-2*rho(vacuum))*R = 0allowing a static Universe.Einstein's Greatest BlunderHowever, there is a basic flaw in this Einstein static model: it is unstable- like a pencil balanced on its point. For imagine that the Universegrew slightly: say by 1 part per million in size. Then the vacuum energydensity stays the same, but the matter energy density goes down by 3 parts permillion. This gives a net negative gravitational acceleration, which makes theUniverse grow even more! If instead the Universe shrank slightly, one gets anet positive gravitational acceleration, which makes it shrink more! Any smalldeviation gets magnified, and the model is fundamentally flawed.In addition to this flaw of instability, the static model's premise of a staticUniverse was shown by Hubble to be incorrect.This led Einstein to refer to the cosmological constant as his greatestblunder, and to drop it from his equations.But it still exists as a possibility -- a coefficient that should be determinedfrom observations or fundamental theory.The Quantum Expectation The equations of quantum field theory describing interacting particlesand anti-particles of mass M are very hard to solve exactly.With a large amount of mathematical work it is possible to prove that theground state of this system has an energy that is less than infinity.But there is no obvious reason why the energy of this ground state should bezero. One expects roughly one particle in every volume equal to theCompton wavelength of the particle cubed, which gives a vacuum density ofrho(vacuum) = M4c3/h3 = 1013 [M/proton mass]4 gm/ccFor the highest reasonable elementary particle mass, the Planck mass of20 micrograms, this density is more than 1091 gm/cc.So there must be a suppression mechanism at work now that reduces the vacuumenergy density by at least 120 orders of magnitude.A Bayesian ArgumentWe don't know what this mechanism is, but it seems reasonable that suppression by 122 orders of magnitude, which would make the effect of thevacuum energy density on the Universe negligible, is just as probable assuppression by 120 orders of magnitude. And 124, 126, 128 etc.orders of magnitude should all be just as probable as well, and all givea negligible effect on the Universe. On the other hand suppressions by118, 116, 114, etc. orders of magnitude are ruled out by the data.Unless there are data to rule out suppression factors of 122, 124, etc.orders of magnitude then the most probable value of the vacuum energy densityis zero.The Dicke Coincidence ArgumentIf the supernova data and the CMB data are correct, then the vacuum density is about 73% of the total density now. But at redshift z=2, which occurred 10 Gyr ago for this model if Ho = 71,the vacuum energy density was only 9% of the total density. And 10 Gyr in the future the vacuum density will be 96% of the total density. Whyare we alive coincidentally at the time when the vacuum density is in themiddle of its fairly rapid transition from a negligible fraction to thedominant fraction of the total density? If, on the other hand, thevacuum energy density is zero, then it is always 0% of the total density andthe current epoch is not special.What about Inflation?During the inflationary epoch, the vacuum energy density was large:around 1071 gm/cc. So in the inflationary scenario thevacuum energy density was once large, and then was suppressed by a largefactor. So non-zero vacuum energy densities are certainly possible.Observational LimitsSolar SystemOne way to look for a vacuum energy density is to study the orbits ofparticles moving in the gravitational field of known masses.Since we are looking for a constant density,its effect will be greater in a large volume system.The Solar System is the largest system where we really know what the massesare, and we can check for the presence of a vacuum energy density by a carefultest of Kepler's Third Law: that the period squared is proportional to thedistance from the Sun cubed. The centripetal acceleration of a particle movingaround a circle of radius R with period P isa = R*(2*pi/P)2which has to be equal to the gravitational acceleration worked out above:a = R*(2*pi/P)2 = g = GM(Sun)/R2 - (8*pi/3)*G*rho(vacuum))*RIf rho(vacuum) = 0 then we get(4*pi2/GM)*R3 = P2which is Kepler's Third Law. But if the vacuum density is not zero, thenone gets a fractional change in period ofdP/P = (4*pi/3)*R3*rho(vacuum)/M(sun) = rho(vacuum)/rho(bar)where the average density inside radius R is rho(bar) = M/((4*pi/3)*R3).This can only be checked for planets where we have an independent measurementof the distance from the Sun. The Voyager spacecraft allowed very precisedistances to Uranus and Neptune to be determined, and Anderson et al.(1995, ApJ, 448, 885) found that dP/P = (1+/-1) parts per millionat Neptune's distance from the Sun. This gives us a Solar System limitofrho(vacuum) = (5+/-5)*10-18 < 2*10-17 gm/ccThe cosmological constant will also cause a precession of the perihelionof a planet.Cardona and Tejeiro (1998, ApJ, 493, 52) claimed that thiseffect could set limits on the vacuum density only ten or so timeshigher than the critical density, but their calculation appears tobe off by afactor of 3 trillion. The correct advance of the perihelionis 3*rho(vacuum)/rho(bar) cycles per orbit. Because theranging data to the Viking landers on Mars is so precise, a very goodlimit on the vacuum density is obtained:rho(vacuum) < 2*10-19 gm/ccMilky Way GalaxyIn larger systems we cannot make part per million verifications of the standardmodel. In the case of the Sun's orbit around the Milky Way, we only say thatthe vacuum energy density is less than half of the average matter densityin a sphere centered at the Galactic Center that extends out to the Sun'sdistance from the center. If the vacuum energy density were more than this,there would be no centripetal acceleration of the Sun toward the GalacticCenter. But we compute the average matter density assuming that the vacuumenergy density is zero, so to be conservative I will drop the "half" andjust sayrho(vacuum) < (3/(4*pi*G))(v/R)2 = 3*10-24 gm/ccfor a circular velocity v = 220 km/sec and a distanceR = 8.5 kpc.Large Scale Geometry of the UniverseThe best limit on the vacuum energy density comes from the largest possiblesystem: the Universe as a whole. The vacuum energy density leads to anaccelerating expansion of the Universe. If the vacuum energy densityis greater than the critical density, then the Universe will not have gonethrough a very hot dense phase when the scale factor was zero (the Big Bang).We know the Universe went through a hot dense phase because of the lightelement abundances and the properties of the cosmic microwave background.These require that the Universe was at least a billion times smaller in thepast than it is now, and this limits the vacuum energy density torho(vacuum) < rho(critical) = 8*10-30 gm/ccThe recent supernova results suggest thatthe vacuum energy density is close to this limit:rho(vacuum) = 0.75*rho(critical) = 6*10-30 gm/cc.The ratio of rho(vacuum) to rho(critical) is calledlambda. This expresses the vacuum energy density on the samescale used by the density parameter Omega. Thus the supernova data suggest that lambda = 0.75. If we useOmegaM to denote the ratio of ordinary matter density tocritical density, then the Universe is openif OmegaM + lambda is less than one,closed if it is greater than one, and flat if it is exactly one.If lambda is greater than zero, then the Universe will expand forever unless the matter density OmegaM is muchlarger than current observations suggest. For lambda greater thanzero, even a closed Universe can expand forever. The figure above shows the regions in the(OmegaM, lambda) plane that are suggested by thedata in 1998.The green region in the upper left is ruled out because there would notbe a Big Bang in this region, leaving the CMB spectrum unexplained.The red and green ellipses with yellow overlap region show the LBLteam's allowed parameters (red) and the Hi-Z SN Team's allowedparameters (green). The blue wedge shows the parameter space regionthat gives the observed Doppler peak position in the angular powerspectrum of the CMB. The purple region is consistent with the CMBDoppler peak position and the supernova data. The big pink ellipseshows the possible systematic errors in the supernova data. The figure above shows the scale factor as a function of time forseveral different models. The colors of the curves are keyed to thecolors of the circular dots in the (OmegaM, lambda) plane Figure.The purple curve is for the favoredOmegaM = 0.25,lambda = 0.75 model.The blue curve is the Steady State model, which has lambda = 1 but no Big Bang.Because the time to reach a given redshift is larger in theOmegaM = 0.25,lambda = 0.75 model than in the OmegaM = 1model, the angular size distance and luminosity distance are larger inthe lambda model, as shown in the space-time diagram below: The OmegaM = 1 model is on the left, theOmegaM = 0.25,lambda = 0.75 model is on the right.The green line across each space-time diagram shows the time when theredshift was z = 1, which corresponds to approximately to themost distant of the supernovae observed to date.Using a ruler you can see that the angular size distance to z =1 is 1.36 times larger in the right hand diagram, which makes theobserved supernovae 1.84 times fainter (0.66 magnitudes fainter).Since 1998 both the CMB and the supernova data have improved. The figurebelow repeats the diagram above with new error ellipses for the supernova data and a new CMB allowed region shown. The 3 year WMAP"open"-CDM Monte Carlo Markov chain gives the dots, and this chain wascut off a priori at lambda=0. The allowed region consistent with both the CMB and the supernova datahas shrunk dramatically toward a flat butvacuum energy dominated model. The CMB models also give a Hubble constant,which is shown by the color coding of the dots.The flat vacuum dominated model is also consistent with theHST key project value ofHo = 72 +/- 8 km/sec/Mpc.ConclusionIn the past, we have had only upper limits on the vacuum density andphilosophical arguments based on the Dicke coincidence problem andBayesian statistics that suggested that the most likely valueof the vacuum density was zero. Now we have the supernova data thatsuggests that the vacuum energy density is greater than zero. This result isvery important if true. We need to confirm it using other techniques,such as theWMAP satellite which has observedthe anisotropy of the cosmic microwave background with angularresolution and sensitivity that are sufficient to measure the vacuum energydensity. CMB data combined with the measured Hubble constant do confirmthe supernova data: there is a positive but small vacuum energy density.Ned Wright's Home Page FAQ | Tutorial : Part 1 | Part 2 | Part 3 | Part 4 | Age | Distances | Bibliography | Relativity © 1998-2006 EdwardL. Wright. Last modified 10 Sep 2006 |
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