Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes...

45
LIVING WITH THE COSMOLOGICAL CONSTANT Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here

Transcript of Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes...

Page 1: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

LIVING WITH THE COSMOLOGICAL CONSTANT

Marek NowakowskiDepartamento de Fisica, Universidad de los Andes

Representant of all Dark Energy aspects notdiscussed here

Page 2: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

1 The accelerated Universe explained by a positivecosmological constant Λ

a reminder of the state of art

2 The cosmological aspectUniverses with Λ: critical value set by Ωm0

3 Scales of Λ

connection to Newtonian limit and hydrostaticequilibrium (Buchdahl inequalities); coincidenceproblems; the following scales wil be of importance:density: ρΛ, ρvac

distance: rΛ = 1/√

Λ

small mass: mΛ =√

Λ

big mass: MΛ = 1/GN

√Λ

4 The astrophysical aspect: ρvac

equilibrium concepts necessary since Λ is asource of an external repulsive force

5 Aspects of the Schwarzshild-de Sitter metric: rΛ

combination of Schwarzschild radius and thecosmological constant in a central infall

6 The semi-classical aspect: mΛ

Generalized uncertainty relation and Λ

Page 3: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

The work is based on: M.N., The consistent Newtonian limit ofEinstein’s gravity with a cosmological constant, Int. J. Mod. Phys. D10 (2000)649

M.N., J.-C. Sanabria and A. Garcia, Gravitational equilibrium in the presence of apositive cosmological constant, Phys. Rev. D16 (2001) 023003

A. Balaguera-Antolinez, C. G. Bohmer and M.N., On astrophysical bounds of thecosmological constant, Int. J. Mod. Phys. D14 (2005) 1507

ibid, Scales set by the cosmological constant, Class. Quant. Grav. 23 (2006) 485

A. Balaguera-Antolinez and M.N., Equilibrium of large astrophysical structures inthe Newton-Hooke spacetime, Astron. and Astrophys. 441 (2005) 23

A. Balaguera-Antolinez, D. Mota and M.N., Ellipsoidal configurations in spacetimewith a cosmological constant, Class. Quant. Grav. 23 (2006) 4497

M. N. and A. Balaguear-Antolinez, Dark Energy in an astrophysical context, AIPConf. Proc. 61 (2006) 1001

M. N. and A. Balaguear-Antolinez, From global to local dynamics: effects of theexpansion on astrophysical structures, Class. Quant. Grav. 24 (2007) 2677

A. Balaguera-Antolinez, D. F. Mota and M.N., Astrophysical configurations with

background cosmology:

Probing Dark Energy at astrophysical scales

Monthly Notices of Royal Astronomical Society,in print

Page 4: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

1. The accelerated Universe explained by Λ

Observation of standard candles like type ISupernova led us to conclude that the expansionof the Universe as compared to the standardFriedman model is accelerated (the Supernovaeare dimer than expected in a standard Friedmanmodel, hence they have to be further away thanexpected, hence the Universe must have expandedfaster than expected). Of course there couldbe other explanations also, like dust and dimingdue to photon-axion conversion in the presence ofmagnetic fields....

B Field

Photon Axion

a

γ

γ

However, the plot seems to thicken now. Forexample, the age of globular clusters is too big as

Page 5: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

compared to the age of the Universe (this is actuallyan old problem) unless we assume e.g. a positivecosmological constant Λ. ΛCDM (Cold Dark Matter)models have been also sucessful...

General Evidence based on:Supernovae surveyWMAP (CMB)Cluster formation (LambdaCDM)Baryon acoustic oscillations (BAO)Age of the UniverseWeak LensingAll evidence in agreement with a positivecomsological constant: equation of state

p = p(ρvac) = wρvac, w = −1

Page 6: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

All what we need to understand the accelerationis Einstein’s equations leading to the equationsgoverning the Universe:

Gµν = κTµν, κ = 8πGN

ds2 = −dt2 + a2(t)R20

[dr2

1− kr2+ r2dΩ

]

H2 ≡

( ·a

a

)2

=8πGN

3ρ +

Λ3− k

a2R0, k = ±1, 0

a

a= −4πGN

3(ρ + 3p(ρ)) +

Λ3

·ρ= −3(ρ + p(ρ))

·a

a

Page 7: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

Example of a flat Universe with k = 0

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

-1 -0.5 0 0.5 1 1.5 2 2.5 3

Sca

le F

acto

r R

(τ)

τ = H0 t

Flat Universe

ΩΛ0 = 0.3 , ΩM0 = 0.7

ΩΛ0 = 0 , ΩM0 = 1

Figure 1: Two fates of matter-filled Universes. Thepresent epoch is at τ = 0.

Friedmann equation at the present epoch

Ωm0 + ΩΛ0 − Ωk0 = 1 → k = sgn(Ωm0 + ΩΛ0 − 1)

Page 8: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

Ωm0 =ρ0

ρcrit, ρcrit =

3H20

8πGN

Ωk =k

R20H

20

ΩΛ0 =ρvac

ρcrit, Λ = 8πGNρvac

An intuitive understanding of the acceleratedexpansion is given by the Newtonian limit. For aspherically symmetric object with mass M we canuse the Schwarzschild metric

ds2 = −eν(r)dt2 + e−ν(r)dr2 + r2dθ2 + r2 sin2 θdφ2

g00 = eν(r) = 1−2rs

r− r2

3(rΛ)2, rs = GNM, rΛ =

1√Λ

and its connection to the gravitational potential

g00 ' −(1 + 2Φ)

to get

Φ(r) = −rs

r− 1

6r2

(rΛ)2

Page 9: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

The last term plays the role of a repulsive externalforce! The Galilean spacetime gets replaced byNewton-Hooke spacetime where each two spacepoints go apart due to the cosmological constant(this is the part of the cosmological expansion whichsurvives the Newtonian limit).

Taking account:

Figure 2: The balance of the Universe.

Conclusions: Solid State Physics immaterial; weare just a dust in the wind.

Page 10: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

2. The cosmological aspect:Universes with Λ: critical value set by Ωm0

J. E. Felten and R.Isaacsman, Rev. Mod. Phys. 58 (1986) 689 We knowalready that Λ > 0 leads to acceleration. Are thereother new aspect of Λ-Universes?

0

0.5

1

1.5

2

2.5

-1 0 1 2 3 4 5

Sca

le F

acto

r R

(τ)

τ = H0 t

Matter-Dominated Universe with Λ = 0

0

0.81.5

ΩM0 = 2

3

1030100

Figure 3: Λ = 0 Universes.

Page 11: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

0

0.5

1

1.5

2

2.5

-5 -4 -3 -2 -1 0 1 2 3 4 5

Sca

le F

acto

r R

(τ)

τ = H0 t

Matter-Dominated Universe with Λ

ΩMO = 0.1

ΩΛ0 = -3

-0.3

0

1.34

1.35

1.36

2

3.5

7

20

0

0.5

1

1.5

2

2.5

-5 -4 -3 -2 -1 0 1 2 3 4 5

Sca

le F

acto

r R

(τ)

τ = H0 t

Matter-Dominated Universe with Λ

ΩMO = 0.3

ΩΛ0 = - 3

- 0.3

0

1.712

1.713486

1.714

2

3.57

20

Page 12: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

New features appear:(i) Λ > Λcrit : NO Big Bang(ii) Λ ∼ Λcrit : ’semi-static’ coasting Universes

Λcrit/H20 = 12Ωm0x

3

x solution of

x3 − 34x + (Ωm0 − 1)/Ωm0 = 0

(Remark: The first of the three times we willencounter a third order equation in problems withΛ .) Solutions:

0 < Ωm0 ≤12

: x = cosh[13

cosh−1 1− Ωm0

Ωm0

]12≤ Ωm0 ≤ 1 : x = cos

[13

cos−1 1− Ωm0

Ωm0

]Ωm0 > 1 : x = cos

[13

cos−1 1− Ωm0

Ωm0+

34π

]in addition to the root given in the second case(’semi-static’ Universe surfing beyond the time 0)

Page 13: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

4. Scales of ΛCosmological coincidences and other curiosities

Λ = 3(

ρvac

ρcrit

)H2

0

From this we can get different scales:

• Density:

ρobsvac ' 0.7ρcrit ∼ ρcrit ∼ 2h2

0 × 10−29gcm−3

ρPl = G−2N ∼ 5× 1093gcm−3

ρvac/ρPl ∼ 10122

• Length:

rΛ =1√Λ

=1√3

(ρvac

ρcrit

)−1/2

H−10 ∼ H−1

0 ∼ 1028h−10 cm

Page 14: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

rPl = G1/2N ∼ 1.5× 10−33cm

rΛ/rPl ∼ 1061

• large Mass:

MΛ =1

GN

√Λ

= 3.6× 1022h−10

(ρvac

ρcrit

)−1/2

M

with H0 = 70h70kms−1Mpc−1 = 100h0kms−1Mpc−1

and h70 = 1.0± 0.15

mPl = G−1/2N ∼ 1019GeV ∼ 2× 10−5g

MΛ/mPl ∼ 1060

• small mass:

mΛ =√

Λ ∼ 3× 10−42GeV,mp

mΛ∼ 1041 ∼ Dirac

Page 15: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

Dirac’s large number1

GNmpme∼ 1041

mPl/mΛ ∼ 1060

MΛ mPl mΛ

• Time:TΛ =

1√Λ∼ TH0

Concidences:

• Dirac’s large number also with the cosmologicalconstant.

Page 16: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

• Cosmological coincidence no.1: ρvac ∼ ρcrit

• Cosmological coincidence no.2: rΛ ca. extensionof the visible Universe !

• Cosmological coincidence no.3: MΛ ca. mass ofthe visible Universe !

The coincidence is that we are living right now ina Universe whose mass, length and density scalesare dominated by a constant Λ. After all we couldhave been living in a different Universe or at aearlier/later epoch in the same Universe. ΩΛ whichis now of order one, was different in the past and willgrow away from one in the future. The transitions inboth directions are quite steep...

The coindcidence problem appears even strangerrealizing that the Λ-scales appear as limiting valuesin two completely different (non-cosmological)settings, namely in the Newtonian Limit andHydrostatic equilibrium.

Page 17: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

Newtonian Limit: condition for the approximation tobe valid

|Φ(r)| 1 → rs d(r) ≡ r − 16

r3

(rΛ)2

d(r) has a local maximum at r+ =√

2rΛ. Hence:

Mmax =2√

23

MΛ M

Solvingrs = d(r)

(Remark: This is the 2nd time we have to solve acubic equation.), one gets

Rmax =√

6rΛ r rs = Rmin

The Newtonian Limit for non-spherically symmetricobjects is given by

∇2Φ = 4πGNρ− Λ

Page 18: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

We have a problem to put the Dirichlet boundarycondition at infinity...

Hydrostatic equilibrium: existence of a globalsolution (general relativistic spherically symmetricobject in hydrostatic equilibrium) requires (Buchdahlinequalities)

3rs ≤23R + R

√49− 1

3R2

(rΛ)2

We have to satisfy

R ≤√

43rΛ ∼ Rmax

Hence we get also

Mmax ∼23

√49MΛ ≥ M

Page 19: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

Combination of scales:very often the relevant physical scale emerge as acombination of two different scales:

ρvac =18π

√ρΛρPl. ∼

√small× large

ρΛ = mΛ/r3Λ, ρPl. = G−2

M = m4Pl.

and for the Scwarzschild radius

rs = 2(

rPl.

rM

)rPl., rM =

1M

We have seen already

MUniv. ∼1

GN

√Λ

=(

mPl.

)mPl.

Does the combination

√mPl.mΛ ' 3× 10−3 eV ', K

appear anywhere in physics?

Page 20: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

Λ on the MOND: the acceleration connectionThe acceleration crisis:

• inflation: R + αR2 gravity etc.

• recent Universe: Λ

• Pioneer10 anomlous acceleration:

a0 ∼ 10−8cm s−2 ∼ (Λ/3)1/2

which is another COINCIDENCE!

• MOND (Modified Newtonian Dynamics) ofMilgrom versus DARK MATTER:

a2/a0 = MGNr−2, a0 a

a = aN = MGNr−2, a a0

interpolating function µ

x = a/a0, µ(x)a = aN

Page 21: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

for exampleµ(x) = x/(1 + x)

from which it follows

a = GNM/2r2 +√

G2NM2/4r4 + a0GNM/r2

or due to the coincidence

a = GNM/2r2√

r2s/4r4 + (rs/rΛ)1/r2

which forr

√rsrΛ

becomes

a '√

rs/rΛ

r= v2/r → v(r) ' const

Page 22: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

4. The astrophysical aspect: densityIf the cosmological constant alone sets onlycosmological scales, how does it happen that itcan also set relevant astrophysical scales? Incase of the density this happens for non-sphericalobjects with at least two length scales l1 and l2.Then in equilibria concepts we often get (l1/l2)n.Gravitational equilibrium via virial equationsThe standard non-relativistic virial theorem reads

d2Ijk

dt2= 4Kjk + 2Wjk

where Ijk is the inertial tensor, Kjk the kinetic andWjk the gravitational potential tensor. If an externalforce is exerted on the object, we have to add to theright hand side of the equation the term

Vjk = −12

∫ρ

(xk

∂Φext

∂xj+ xj

∂Φext

∂xk

)d3x

where Φext is the external potential and the case of

Page 23: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

a cosmological constant corresponds to

Φext = −16Λr2

Therefore the new virial theorem which accountsfor the cosmological constant takes the form

d2Ijk

dt2= 4Kjk + 2Wjk +

23IjkΛ

It is very often more convenient to consider a lessdemanding task by simply noting that the trace Wof Wjk is negative whereas the trace K of Kjk ispositive definite. Then the gravitational equilibriumi.e. d2Ijk/dt2 = 0 yields the inequality

−13ΛI + |W | ≥ 0

where I denotes the trace of the inertial tensorIjk. To appreciate the meaning of this inequality wespecialize to the case of constant density. It is then

Page 24: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

easy to show that

8πGNρ ≥ AΛ, ρ ≥ Aρvac

where the quantity A depends only on the geometryof the object under consideration.

A =16π

3

∫r2d3x∫ |ΦN|ρ d3x

where ΦN is the Newtonian part of the non-relativistic gravitational potential. For sphericallysymmetric objects one easily calculates A = 2 andtherefore the virial inequality is simply

4πGNρ ≥ Λ, ρ ≥ 2ρvac

Where is the ratio of two length scales to a powern? In principle it is in A if we consider non-sphericalobjects. Instead of the inequality we can alsocalculate from the virial theorem the mean velocity

Page 25: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

of the objects

〈v2〉 =|W |M

− 8π

3ρvac

MI

To appreciate the effect of Λ let us assume aconstant density and the shape of the astrophysicalobject to be an ellipsoid. The mean velocity can benow written as

〈v2〉ellipsoid =32π

45Mρρvaca1a2a3

(a21 + a2

2 + a23

(34

ρ

ρvacΓellipsoid − 1

)The prolate case (a1 = a2 < a3, e =

√1− a2

1/a23)

gives

Γprolate =

(a1a3

)3

1 + 2(

a1a3

)2

ln(

1+e1−e

)e

Note now that for a flattened prolate ellipsoid we can

Page 26: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

approximate

Γprolate '(

a1

a3

)3

ln(

1 + e

1− e

)

Since the nowadays preferred value of ρvac is (0.7−0.8)ρcrit we can say that if the constant ρ/ρcrit is,say, 103, it suffices for the ellipsoid to have theratio a1/a3 ∼ 10−1 in order that the mean velocityof its components approaches zero. This is validalways under the assumption that the object is ingravitational equilibrium. This effect is due to therelatively large cosmological constant. In generalwe can say that in flattened astrophysical systemsin gravitational equilibrium, the mean velocity getsaffected by the cosmological constant. The denserthe system, the bigger should be the deviation fromspherical symmetry to have a sizeable effect. Itis interesting if such an effect can be observedin reality which would confirm the existence of Λ.One can paraphrase this also by looking at theresults from a different perspective. If we are certainthat a given astrophysical object is in gravitational

Page 27: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

equilibrium, then the above equations would put astringent bound on Λ in the case of strong deviationfrom spherical symmetry. Hydrostatic equilibriumfor spherically symmetric objects:Is the hydrostatic equilibrium related to thegravitational one? Note that in the virial equationsused above no pressure appeared (indeed, withpressure the virial equations change), but thehydrostatic equilibrium is

∇P = −ρ∇Φ, ∇2Φ = 4πGNρ− Λ

This condition leads to

P ′(r) = −rρ(r)(

GNm(r)r3

− Λ3

)which is sometimes called the “fundamentalequation of Newtonian astrophysics”. The massfunction is as usual defined by

m(r) =∫ r

0

4πρ(s)s2ds

Page 28: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

Furthermore let the mean density be defined by

ρ(r) =34π

m(r)r3

ThenP ′(r) = −r

ρ(r)3

(4πρ(r)− Λ

)For any physically reasonable astrophysical object,the pressure and density must be monotonicallydecreasing functions of the object’s radius. Hencenegativity of the derivative of the pressure implies(1)

Λ < 4πGN ρb

This equation we also obtained via the virialcondition (2)! Indeed, one can arrive at it also viathe general relativistic Tolman-Oppenheimer-Volkoffequation (3)

P ′(r) = −rρ(r)3

(1+

P (r)ρ(r)

)(12πP (r) + 4πGN ρ(r)− Λ1− 8π

3 ρ(r)r2 − Λ3r2

)

(via a postive definite denominator and theboundary condition Pboundary = 0 ) and by

Page 29: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

demanding stability of circular orbits (4). Thiscalls for two things: include pressure in thevirial equations and generalize the hydrostaticequilibrium concept to non spherically symmetricobjects.

Let me quote at the end of this section from abook “The measure of the Universe” by North whichdescribes the history of cosmology in general andthe work of some authors on the cosmologicalconstant: “The essential difficulty with a relativistictheory in which λ [the Cosmological Constant] ispositive is that of accounting for the formation andcondensation in terms of gravitational instability; for,to use the ‘force’ metaphor, the present expansionindicates that the force of cosmic repulsion exceedsthose of gravitational attraction. This is not likely todisturb the stability of systems (such as the galaxy)of high average density, but it is likely to prevent newcondensation in regions of low density.”

Page 30: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

5. Apects the Schwarzschild-de Sitter metric:lengthTo obtain astrophysically relevant length scales fromΛ we have to combine rΛ with another smallerlength scale which is rs.

Consider the motion of test particles in sphericallysymmetric and static space-time with a cosmologicalconstant. The Schwarzschild metric takes the form

ds2 = −eν(r)dt2 + e−ν(r)dr2 + r2dθ2 + r2 sin2 θdφ2

eν(r) = 1− 2rs

r− r2

3(rΛ)2One would suspect that the inclusion of Λ isirrelevant in this setting. Note, however, that thereare now indeed two scales involved, rs and rΛ. Thecombination of the two can lead to new results.Indeed, the equation of motion for a massiveparticle with proper time τ in the Schwarzschildmetric is given by

12

(dr

)2

+Ueff =12

(E2 +

L2Λ3− 1)≡ C = constant

Page 31: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

where E (which should not be confused with theenergy ) and L are conserved quantities defined by

E = eν(r) dt

dτ, L = r2dΦ

dτwhere Φ is the azimuthal angle and Ueff is definedby

Ueff(r) = −rs

r− 1

6r2

(rΛ)2+

L2

2r2− rsL

2

2r3

is the analog of effective potential potential inclassical mechanics.

We now consider radial motion (L = 0). From thedefinition of C we obtain the inequality

C > −12

which will play a crucial role later in the derivation.For the limiting value C = −1

2, we have E = 0 whichsignals an artifact of the Schwarzschild coordinates.This means that there exist some r = r? whichsatisfies the cubic equation

y3 − 3y + 6x = 0

Page 32: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

y =r?

x =rs

rΛ= 1.94× 10−23

(M

M

)h70

(ρvac

ρcrit

)1/2

1

(Remark: this is the third time we arrive at a cubicequation.) The two positive roots are

r(1)? =

√3rΛ − rs, r(2)

? = 2rs

(1− 1

6

(rs

)2)

In other words, the condition C = −1/2 is satisfiedat the Schwarzschild radius and at the edge of theuniverse. With the same limiting value, we have|Ueff(r?)| = 1

2. Motion with |Ueff(r)| ≥ 12 becomes

unphysical since it corresponds to allowing themotion of test particles inside the Schwarzschildradius and beyond the observed universe. Thelatter is a result of the coincidence in the sensethat rΛ sets the scale of the horizon of the universe.Hence, the particles are allowed to be at some rsuch that

Rmin ∼ rs < r <√

3rΛ ∼ Rmax

Page 33: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

with|C| < |Ueff(r)| < 1

2(for negative C and Ueff). It is clear that at certaindistance, the terms −rs/r and r2/(rΛ)2 will becomecomparable leading to a local maximum located at

rmax =(3rsr

)1/3 ' 10−4

(M

M

)1/3(ρcrit

ρvac

)1/3

h−2/370 Mpc

Ueff(rmax) = −7.51×10−16

(M

M

)2/3(ρvac

ρcrit

)1/3

h2/370

Beyond rmax, Ueff is a continuously decreasingfunction. This implies that rmax is the maximumvalue within which we can find bound solutions forthe orbit of a test body.

Consider now the following chain of matterconglomeration of astrophysical objects: thesmallest are star clusters (globular and open) withstars as members (M = M) and a mass of106M. We proceed to galaxies and galacticclusters. Within this chain, we find for rmax the

Page 34: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

following values as a function of mass:

M/M rmax/α (pc)1 75

106 7.5× 103

1011 3.5× 105

1013 1.6× 106

with α = h−2/30 (ρvac/ρcrit)

−1/3.

• The value at the first line is of the order ofmagnitude of the tidal radius of globular clusters.

• The second line agrees with the extension of anaverage galaxy. Unlike the other three ones,where the argument M of rmax has been takento be the mass of the average members of theastrophysical object, it might appear unjustified totake the mass of the globular cluster to obtain theextension of the galaxy. However, in view of thefact that globular clusters are very old objects andare thought to be of importance in the formation

Page 35: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

of the galaxy, this choice seems justified. Indeed,with Λ > 0, our result strengthens the belief thatglobular clusters are relics of the formation of thegalaxy. For instance, rmax for open star clusterswith a mass M = 250M is only 0.5 kpc.

• The next two values are about the size of agalaxy cluster. The value 1013M correspondsto a giant elliptic galaxy encountered often at thecenter of the clusters.

Conclusions:Hence, rΛ in combination with rs gives ussurprisingly accurate and natural astrophysicalscales. The combination rmax = (3rs(rΛ)2)1/3

from which these scales where calculated is notan arbitrary combination with length dimension, butit is the distance beyond which we cannot findbound orbits. Therefore we would expect that rmax

sets a relevant astrophysical scale. Of course, weare talking here about scales neglecting dynamicalaspects of many body interactions, but no doubtrmax is roughly the scale to be set for bound

Page 36: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

systems. Indeed, the agreement of the result in thetable with values encountered in nature is striking.

What happens in the case of rl ≡ L 6= 0 ?. Look fora saddle point i.e.

dUeff

dr=

d2Ueff

dt2= 0

The two conditions lead to the position of the saddlepoint and a condition on one parameter, say xl ≡

r2l

(rΛ)2

x4l −

(3rs

4rΛ

)4

xl − 12(

3rs

4rΛ

)6

= 0

To solve this fourth order we have to solve theassociated third order (and this is the fourth time wedeal with third order polynomial equations). Afterhandling hyperbolic functions and their inverses,going through complex numbers and their roots,one gets a simple expression

rmaxl = rcrit

l = 0.9(r2srΛ)1/3

Page 37: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

provided rs/rΛ 1. For rl ≥ rcritl the minimum and

maximum fall together and there are no more boundorbits! Taking now from non-relativistic mechanicsthe expression for the order of magnitude of a boundorbit we get

Rorbit ∼r2l

rs→ Rmax

orbit ∼ 0.55rmax

which is a very satisfying result as it does notchange the order of magnitude of the estimate withzero angular momentum!

Note: to ensure the existence of the first maximumand minimum of Ueff one has to require

rminl = 2

√3rs

Page 38: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

-15

-12.5

-10

-7.5

-5

-2.5

0

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

Uef

f(r)

r [arbitrary units]

Effective potential From Schwarszchild’s Metric with Cosmological Constant

Λ not 0Λ = 0

Page 39: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

6. The semi-classical aspect: mΛ

Generalized Uncertianty Relation (GUP) and BlackBody radiation used to see whether mΛ or any otherscale like

√mPl.mΛ enter physics. In particular,

is mΛ the mass of the Black Hole (BH) remnant?Recall, mPl. mΛ.

GUP FOR Λ = 0 case: Let E = p be the photon’senergy, then

aG =GNE

r2

As an order of magnitude estimate, we can write

∆xG 'GNE

r2L2 ' GNE = GNp

where we used r ∼ L. Using ∆p ∼ p. we arrive atGUP

∆x ≥ 12∆p

+ GN∆p

Identifying now

∆x ∼ 2rs = 2GNM

Page 40: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

∆p ∼ E ∼ T

and using GUP we obtain

2GNM = 2M

mPl.=

12T

+T

mPl.

Solving this equation for T = T (M) and introducinga calibration factor (2π)−1 gives

T =12π

(M −

√M2 −m2

Pl./2)

Conclusions:

• This reduces to Hawking’s formula for large M

• BH remnant

M > Mmin =mPl.√

2

Page 41: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

Black body radiation as a check:

0 < g00 = 1− 2GNM

R= 1− (8π/3)GNρR2

Henceρ <

38π

1GNR2

Using Stefan-Boltzmann law ρ = σT 4 one gets

T 4 <38π

1σGNR2

Using the quantum mechanical result R > 1/T onegets

T < Tmax =

√458π2

mPl.

Inserted into Hawking’s formula yields

M ′min =

(25

18π

)mPl.

Page 42: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

Conclusions:

• A BH remnannt is confirmed in a completedifferent way.

• In 1966 Andrei Sakharov derived the result T <Tmax ∼ mPl. ∼ 1032 K. Obviously, only nowin the context of Hawking’s radiation and semi-classical quantum gravity such a results makessense.

Repeating the above steps with Λ 6= 0:

0 < g00 = 1− 2rs

R− R2

3rΛ2

0 < ρ <38π

m2Pl.

R2− 1

338π

m2Pl.mΛ

2

Finally we obtain

1√3mΛ = Tmin < T < Tmax

Page 43: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

Back to Hawking’s formula (recall: Tmax ↔ Mmin

and Tmin ↔Mmax ) gives

Mmax ∼ MΛ =(

mPl.

)mPl.

Repeating the procedure of GUP, now including Λ ,we conclude :

• Mmax gets confirmed as an upper bound.This is the maximal Black Hole due to Λ

• mPl. remains the BH remnant.This is the minimal Black Hole.

Page 44: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

Instead of a Conclusion:The fearful sphere or how centuries influence eachother

• 500 B.C.Xenophanes of Colophon:GOD IS AN ETERNAL SPHERE

• 40 years later Parmenides:THE DIVINE BEING IS LIKE THE MASS OFA WELL-ROUNDED SPHERE WHOSE FORCEIS CONSTANT FROM THE CENTRE IN ANYDIRECTION

• 300 B.C. Hermes Trismegistus in CorpusHermeticum, rediscovered by Alain de Lille in1300 :GOD IS AN INTELLIGIBLE SPHERE, WHOSECENTRE IS EVERYWHERE AND WHOSECIRCUMFERENCE IS NOWHERE.

• 1580 Giordano Bruno:WE CAN ASSERT WITH CERTAINTY THATTHE UNIVERSE IS ALL CENTRE, OR THAT

Page 45: Marek Nowakowski - if.ufrgs.br · Marek Nowakowski Departamento de Fisica, Universidad de los Andes Representant of all Dark Energy aspects not discussed here. 1 The accelerated Universe

THE CENTRE OF THE UNIVERSE IS EVERYWHEREAND ITS CIRCUMFERENCE NOWHERE.

• 1650 Blaise Pascal:NATURE IS AN INFINITE SPHERE WHOSECENTRE IS EVERYWHERE, WHOSE CIRCUMFERENCEIS NOWHERE.

• today :THE UNIVERSE HAS A SPHERICAL SYMMETRY.THERE IS NO CENTRE.