B.Sadoulet

Phys 250 -4 GR 1

The Dark Side

Gravitational RedshiftEquivalence PrincipleClocks in accelerated framesTidal forces cannot be suppressedEinstein equationsApplications to Robertson-WalkerThermal history of the universe

cf Peebles section 10, Misner/Thorne/Wheeler,Kolb&Turner and Strauman(we have the “same”notations/signs)Weinberg (we have different sign for metric and R)

General Relativity:A Quick Introduction

B.Sadoulet

Phys 250 -4 GR 2

The Dark Side

Photons Are Sensitive to Gravity

Compatibility of gravity with relativityLet us consider an uniform gravitational field and:

a source 1 of photon s of frequency ν1

a receiver 2 at an height l above the emitterBoth are stationary in the laboratory.

Let us call ϕ the gravitational potential per unit massE=mc2 => ϕ has to enter in definition of mass

Compatibility with equivalence of mass and energyBefore emission

After emission

If energy is conserved

Photons are sensitive to gravityget redder as they go out of a potential well

Emitter

Receiver

1

2

l

ET = m1c2 + m1 1 + m2c2 + m2 2

2

1=

1+ 1

c2

1 + 2

c2

ET = m1 −h 1

c2

c2 + m1 −

h 1

c2

1 + m2 +

h 2

c2

c2 + m2 +

h 2

c2

2

B.Sadoulet

Phys 250 -4 GR 3

The Dark Side

The Principle of Equivalence

PrincipleThe inertial mass and the gravitational mass (“charge”) are proportional.

=> can be taken as equal (definition of G)all objects fall equally in a gravitational field

Experimentally true to a high degree of precisionNewton 10-3

Eötvös: (torsion balance checking the direction of geff)Dicke 10-12

Free Falling Systems=> it is possible to zero out a constant gravitational field by going to a free

falling system

Postulate of General RelativityIn free fall, all physical laws reduce to those in an inertial frame.In particular, free particles (including photons) travel on straight lines

= geodesics of flat space (=auto-parallel)

Effect of gravity: distortion of space => curved geodesics=> automatically explains equality of inertial and gravitational mass.

z

g

z → ′ z = z − 1

2gt2

m˙ z = mg → ′ ˙ z = 0

B.Sadoulet

Phys 250 -4 GR 4

The Dark Side

Gravitational Redshift (2)

Derivation from Equivalence PrincipleLet us consider an uniform gravitational field and:

a source 1 of photon s of frequency ν1

a receiver 2 at an height h above the emitterBoth are stationary in the laboratory.

If we go to a free falling system which is a rest withrespect to the lab at the time of emission of aphoton.

The photon reaches the receiver after a time

At that time the receiver is moving in the freefalling system with a velocity

This leads to a Doppler shift

The frequency shift is then

Same result as before

Emitter

Receiver

1

2

h

t =h

c

v = g t = gh

c

2

1=

1 −v

c

1 + vc

≈ 1−gh

c2

= −gh

c2 = − 2 − 1( )c2

B.Sadoulet

Phys 250 -4 GR 5

The Dark Side

Clocks in Accelerated Frames

Time ticks differently!Consider a free falling frame and an accelerated frame which is instantaneously at rest with

respect to this frame.

Gravitational redshift (subtle!)Coming back to emission and absorption of photons in a gravitational field, it is easy to show

that the time of travel of each wave crest from the point of emission to the point of absorptiondoes not vary with “time” (the gravitational field is constant and geodesics are parallel!).

In other words the period of the photons dt1=dt2 does not change!What is going on?

It has to be that the clocks in the accelerated frame are not ticking at the same rhythm as inthe free falling frame which is instantaneously at rest with respect to them.

Looking at the clock tick intervals ∆τ and ∆t in the free falling and lab framesIn free falling frames clocks have to have same tick intervals

We get the same results as before if

d i = 0 i = 1,2,3 ⇒ dxi = 0 i = 1,2,3

d 2 invariant ⇒ dtaccel. =1

good =

1

goodtfree

goo ≈ 1+ 2c2

⇒ as measured in lab frame 2

1

=∆t2

∆t1=

goo 1( )goo(2)

t

x1 2

∆ 2 → ∆t2 = ∆ 2

goo(2) 2 = rest frame clock freqency

number of clock ticks/photon period= 1

dt2 / ∆t2

= ∆t2

dt2

=↑ b Periods in lab frame dt1 = dt2 (time of propagation is constant)

∆ 1 → ∆t1 =∆ 1

goo(1) 1 =

rest frame clock freqency

number of clock ticks/photon period=

1

dt1 / ∆t1

=∆t1

dt1

2

1

=goo 1( )goo(2)

≈1 + 2 1

c2

1+ 2 2

c2

≈1 − 2 − 1( )c2 if goo ≈1 + 2

c2

B.Sadoulet

Phys 250 -4 GR 6

The Dark Side

Consequences

We have derived a first order effect of gravity!

Time delayPhotons are slowed down in an gravitational field

Apparent velocity ≠ velocity of lightIndex

Can be deduced e.g. from the most general stationary and spherically symmetric solution ofthe Einstein’s equations: Schwarzschild metric! See also slide 12.

Twin paradoxThe accelerated twin ages less rapidly!

Moral of the story: when in doubt, go to free falling frameThe physics is well defined

goo ≈ 1+ 2c2

d 2 = 1 − 2GM

c2r

dt2 −

1

c2

dr2

1− 2GM

c2r

+ r2d 2 + r 2 sin2 d 2

n ≈ 1− 2c2

B.Sadoulet

Phys 250 -4 GR 7

The Dark Side

Tidal Forces in Newtonian GravityGravitational gradients cannot be zeroed out:

free falling frame varies with position!e.g. in Einstein’s free falling elevator Responsible for tidestest objects will get closer

Deviation of trajectoriesConsider neighboring trajectories

we get

Poisson equationStarting with Gauss’ theorem

but flux of a vector through a surface is equal to the integral over theenclosed volume of its divergence (Stokes)

Free

Fal

ling

r r

r d

r R

ME

M

xi t , n( )

d2xi t , n( )dt2 = − ∂

∂xi

∂∂n

d2xi t , n( )dt 2

= −

∂∂n

∂∂xi

= −

∂2

∂xi∂x j∂x j

∂n

Calling the deviation j =∂x j

∂nd2 i

dt2 = −∂2

∂xi∂x jj

dr S

−

r ∇ ( )

S∫ ⋅ d

r S = −4 GM ⇐

1

r 2 law

r u

S∫ ⋅ d

r S =

r ∇ ⋅

r u

V∫ dV ⇒ −

r ∇ ( )

S∫ ⋅ d

r S = −∇2( )

V∫ dV = −4 G

V∫ dV

⇒ ∇2 = 4 G Poisson equation

t

r x t , n( )

n

j =∂x j

∂n

B.Sadoulet

Phys 250 -4 GR 8

The Dark Side

Momentum Tensorcf. Weinberg Chap 1, Peebles section 10

DefinitionLocal 4-current of 4-momentum. The sum is on the particles passing by xi at time t=xo .

Conservation of energy and momentum

Proof

T x( ) = pn

dxn

dt3 r

x −r x n t( )( ) =

n∑ pn∫

dxn

dt4 x − xn t( )( )dt =

n∑ pn∫ un

v 4 x − xn n( )( )d nn

∑

with u = dxd

, p = mu for massive particles

∂ T ≡ T ,v = 0

∂iTi x( ) = pn

dxni

dt

∂∂x i

3 r x − r

x n t( )( ) =n∑ pn

∂∂t

3 r x − r

x n t( )( ) =n

∑ − ∂∂t

T 0 + dpn

dt3 r

x − r x n t( )( )

n∑

⇒ ∂ T x( ) = 0 if no long range force.

Collisions strictly localized in space are OK ⇐ 3 r x −

r x c t( )( ) d

dtpn

n∈c∑ = 0

B.Sadoulet

Phys 250 -4 GR 9

The Dark Side

Energy Momentum Tensor(2)

Ideal fluid at restReplace sum by integral over densityIdeal= point like particles and isotropic distribution in fluid frame

(mean free path << scales used by observer)

Proof: has to be diagonal

General expression for ideal fluid

T x( ) =

0 0 0

0 p 0 0

0 0 p 0

0 0 0 p

where is the mass density and p the pressure

T 00 = n < mu0 >= (n = number density < u0 >= 1)

T ii = n < muiu0ui >=v2

3= p (v = mass weighted r.m.s. 3 velocity)

T = +p

c2

u u − g

p

c2

Contravariant!

B.Sadoulet

Phys 250 -4 GR 10

The Dark Side

Heuristic Derivation of Einstein Equations

Deviation of geodesics (3-d)Consider again a family of geodesics

is not perpendicular to u !

We also have

Proof:

Choose normal coordinate system aroundGeodesic=> t=τOrthogonal spatial coordinates n1 n2 n3

of the same form as the Newtonian mechanics

n

j =∂x j

∂n

,n( ) u =∂ ,n( )

∂=

∂ ,n( )∂n

But = − < ,u > u is < u,u >= 1( )

∇u∇u = R u,( )u

∇u = ∇u ⇐ ∇u < ,u > = < ∇u ,u > + < ,∇uu > = < ∇u,u >= 0 ∇uu = 0 + Torsion = 0( ) R u,( )u = R u,( )u− < ,u > R u,u( )u = R u,( )u

,0( )

d2 i

dt 2 = R00 ji j

d2 i

dt2 = −∂2

∂xi∂x jj

B.Sadoulet

Phys 250 -4 GR 11

The Dark Side

Heuristic Derivation of Einstein Equations 21st attempt

Make association

Wrong because

2nd attempt

Most general 2nd order

R0 j 0i ↔ ∂ i∂ j permutation of indices to change sign( )

Introducing the Ricci tensor R = R , R00 ↔ ∇2

∇2 = 4 G would correspond to R00 = 4 GT00 ⇒ R = 4 GT

T ; =1

4 GR ; ≠ 0. This theory would not conserve energy!

Introducing the curvature R = R and the Einstein tensor G = R − 1

2g R, G ; = 0

Correcting coefficient to coincide with Newton for weak fields G = R −1

2g R = 8 GT

or taking trace on both sides R = 8 G T −1

2g T

R − 1

2g R − Λg = 8 GT where Λ = cosmological constant

or R −1

2g R = 8 G T +

Λ8 G

g

⇒ Λ = substance with p = − c2

B.Sadoulet

Phys 250 -4 GR 12

The Dark Side

Limitscf Peebles sect 10.

Small scale limitConsider again deviation of geodesics in free falling frame

In GR, not only the acceleration is due not only to the mass density but also tothe pressure!

For large enough negative pressure gravity can become repulsive!Note that shows that our coefficient 8πG is the right one!

Weak field, zero pressure limit

Acceleration in free falling frame =d2 i

dt2= R00 j

i j = Figravity m = 1( )

r ∇ .

r F = R00 i

i = −R00 = −8 G T00 −1

2T

= −4πG + 3

p

c2

Einstein equations + gauge transformation (see Peebles)⇒ ∇2 hoo = 8 G ∇2hij = 8 G ij

∇2h0 j = 0

⇒ d 2 = 1+2

c2

dt 2 −

1

c2 1 −2

c2

d

r x 2

goo ≈ 1+ 2c2Our old friend !

= change of coordinates (4 functions)

B.Sadoulet

Phys 250 -4 GR 13

The Dark Side

Homogenous and Isotropic Space

Robertson-Walker

Einstein equationsCompute Γ’sCompute theStraightforward but long (especially if one does not use symmetry)

d 2 = dt2 − a2 (t)c2

11− kr2 / c2

dr 2 + r2 d 2 + r2 sin2 d 2

ord 2 = dt2 −a2 (t)

c2 dr x 2 +

k

c2

r x .d

r x ( )2

1 − kr2 / c2

with k =

c2

R2 (dimension [T-2])

R , R = R and R

R00 = −3˙ a

a

Rij = −˙ a a

+ 2˙ a 2

a2+ 2

ka2

g ij

R = −6˙ a

a+ 2

˙ a 2

a2 + 2k

a2

g ij

Einstein 00 ⇒˙ a 2

a2+

k

a2−

Λ3

=8πG

3

Einstein ii ⇒ 2˙ a a

+˙ a 2

a2+ k

a2− Λ = −8πG

pc2

B.Sadoulet

Phys 250 -4 GR 14

The Dark Side

Friedmann EquationFrom previous slide

Note that the second equation is integral of the first one

Newtonian Derivation (without and pressure)but correct (Birkhoff's theorem)Consider a spherical shell of fixed comoving radius r

(e.g. at rest with respect to expansion)

GR gives the value of k as comoving curvature

Sum of kinetic and potential energies is constant

⇒˙ a

a= − 4 G

3+ 3

p

c2

+

Λ3

˙ a 2

a2=

8πG

3+

Λ3

−k

a2 Friedmann equation

⇒˙ a

a

2=

8π3

G −k

a2

acceleration ˙ a r = − GM

a2r2 with M = mass inside

⇒ ˙ a = − GM

a2r3 multiply by ˙ a : ˙ a a = − ˙ a

a2GM

r3

by integration 12

˙ a 2 = 1a

GM

r3 − k (M constant)

if = density : M = 4π3

ar( )3

a(t )r

B.Sadoulet

Phys 250 -4 GR 15

The Dark Side

Critical Density

Incorporate into the mass density

k = a2 H 2 8πG3H 2

−1

=

c2

R2= constant

Introducing the critical density c =3H 2

8πG and Ω t =

c

k = a2 H 2 Ωt −1( )

Ωt > 1 closed universe

Ω t = 1 spatially flat

Ωt <1 open universe

B.Sadoulet

Phys 250 -4 GR 16

The Dark Side

Conservation of Energy

Covariant derivative of energy-momentum tensorFrom principle of general relativity

CalculationFrom the general transformation of covariant tensor (similar to g chap 3 slide 17)

Interpretation

i.e. the energy per comoving volume is changed by the work of pressure+ the entropy per comoving volume is constant (for constant N see slide 22 )

T , = 0 should be replaced by T ; = 0 (covariant 4- divergence)

Let us compute the 0 componentT 0

; = ∂0T00 + Γij

0T ij + Γ0ii T 00 where we have used T 0i = 0 Γ0

0 = 0It is easy to show that for the Robertson metric

Γij0 = −

˙ a

agij Γ0 j

i =˙ a

a ji

Using gijTij = −3

pc2

we finally get

T 0; = ∂

∂t+3

˙ a a

pc2

+ 3˙ a a

= 0

or d c2 a3( )

dt= −p

d a3( )dt

T ; = ∂ T + Γ T + Γ T

The energy in volume a3 varies as dU = −pdV

B.Sadoulet

Phys 250 -4 GR 17

The Dark Side

Equations of State

Assume an equation of state of the form

Three main cases (dominant at various phases)Matter: pressureless

Radiation : relativistic gas

Vacuum energy :

p = 0 => ˙ m = -3 m˙ a

a⇔ m

mo=

a

ao

−3

p = rc2

3 => ˙ r = -4 r

˙ a

a⇔ r

r o=

a

ao

−4

( =energy density u

c2 )

v = constant ⇒pv

c2 = − v < 0

Introducing m = m

c, r = r

c, = v

c≡

3H2

⇒ k = a2H2m + r + −1( )

p = w c2 w = constantThe expression of previous slide gives

a3d + da 3 = −w da3 ⇒d

= −3 1+ w( ) da

aor = a−3 1+ w( )

B.Sadoulet

Phys 250 -4 GR 18

The Dark Side

Evolution of the ’sThe quantities evolve very rapidly with z

Starting from

and using the Friedmann equation

we obtain in the three casesfor a matter dominated universe:

for a radiation dominated universe:

for a vacuum energy dominated universe:

e.g. for matter dominated H2 = H 2Ωm − ka2 = H 2Ωm − ao

2 Ho2

a2 Ωmo − 1( )

dividing by H 2Ωm we obtain Ωm−1 − 1 = Ωmo

−1 −1( ) ao2Ho

2Ωmo

a2 H 2Ωm

= Ωmo−1 −1( ) a

ao

mo, ro, o

ΩmH 2 =8πG

3=

8πG o

3

ao

a

3

= H02Ωmo

a

ao

−3

Ω r H2 =

8πG

3=

8πG o

3

ao

a

4

= H02Ω ro

a

ao

−4

H2ΩΛ =1

3= Ho

2ΩΛo

m−1 −1 = mo

−1 −1( ) 1

1 + z

r−1 − 1 = ro

−1 −1( ) 1

1 + z( )2

−1 −1 = o−1 −1( ) 1 + z( )2

B.Sadoulet

Phys 250 -4 GR 19

The Dark Side

Curved spaces curve even moreFlat without Open

Flat with Λ starts todominate very recentlySpecial epoch!

0

0.2

0.4

0.6

0.8

1

0.01 0.1 1 10 100 1000 1e+04 1e+05

m =1- r

Ωm, Ωr, ΩΛ

1+z

Ωm

Ωr

ΩΛ

Time

Now

0

0.2

0.4

0.6

0.8

1

0.01 0.1 1 10 100 1000 1e+04 1e+05

mo = 0.3

Ωm, Ωr, ΩΛ

1+z

Ωm

Ωr

ΩΛ

Now

0

0.2

0.4

0.6

0.8

1

0.01 0.1 1 10 100 1000 1e+04 1e+05

mo =0.3o=1- mo- ro

Ωm, Ωr, ΩΛ

1+z

Ωm

Ωr

ΩΛ

B.Sadoulet

Phys 250 -4 GR 20

The Dark Side

Calculation of t(z), r(z)

where index o indicates present values, the Friedmann equation canbe written as

Finally a redshift z corresponds to a time distance from us

Starting from k = ao2Ho

2mo + ro + o −1( )

˙ a 2

a2 = Ho2 m

co+ r

co+ o +

ao2

a2 1 − mo + ro + o( )[ ]

Taking into account that m

mo=

a

ao

−3

, r

ro=

a

ao

−4

˙ a 2

ao2 = Ho

2mo

aao

−1

+ roa

ao

−2

+ oaao

2

+ 1− mo + ro + o( )[ ]

Using the fact that ˙ a = dadt

⇒ dt = da˙ a

dt =

da

ao

Ho moa

ao

−1

+ roa

ao

−2

+ oaao

2

+ 1− mo + ro + o( )[ ]

to - t1 =1

Ho

dx

mo

x+ ro

x2 + ox 2 + 1− mo + ro + o( )[ ]

1

1+z

1

∫

B.Sadoulet

Phys 250 -4 GR 21

The Dark Side

Distances

Distance of an event at redshift zSimilarly taking into account the fact that

the comoving distance from us of a point at redshift z is

where the sin or sinh has to be used depending whether ε>0 or <0

The proper distance is

aor = aoRsin

sinh

c

aoRHo

dx

mox + ro + ox4 + 1 − mo + ro + o( )[ ]x2( )1

1+z

1

∫

dr

1 −r2

R2

= cdt

a t( )

aoR =c

Ho 1− mo + ro + o( )

d z( ) = aodr

1 −r2

R2

0

r z( )∫

=c

Ho

dx

mox + ro + ox4 + 1 − mo + ro + o( )[ ]x2( )1

1+ z

1

∫

B.Sadoulet

Phys 250 -4 GR 22

The Dark Side

Evolution of entropyLet us come back to our statement of slide 17 that the entropy per

comoving volume is constant!We started from the first law of thermodynamics. In thermal equilibrium

This proof should be sufficientHowever can be proven explicitly (cf Weinberg p 532 or Kolb and Turner p 66)

TdS = d uV( ) + pdV Watch out : identify independent variables!

⇒ dS V,T( ) = 1

Td u T( )V[ ] + p T( )dV = 1

TudV + V

du

dTdT + pdV

⇒∂S V,T( )

∂V=

1

Tu + p( ), ∂S V,T( )

∂T=

1

TV

du

dT

∂ 2S V,T( )∂T∂V

= − 1T 2

u + p( ) + 1T

dudT

+ 1T

dpdT

∂ 2S V,T( )∂V∂T

= 1

T

du

dT

Integrability condition ∂ 2S V,T( )

∂T∂V=

∂ 2S V,T( )∂V∂T

⇒dp

dT=

u + p

T

d ua3( ) = − pd a3( ) ⇔ d u + p( )a3[ ] = a3dp = a3 u + p

TdT

⇒ du + p( )a3

T

=

1

Td u + p( )a3[ ] − u + p( )a3

T 2dT = 0 ⇒ u + p( )a3

T= constant

dS =d ua3( ) + pda3

T=

d u + p( )a3[ ] − a3dp

T=

d u + p( )a3[ ]T

− a3 u + p( )T 2

dT = du + p( )a3

T

dU = d ua3( ) = TdS − pda3 (u = c2 no chemical potential)

If dU = − pda3 ⇒ dS = 0 (in comoving volume a3 )

Can be generalizedto case with chemical

potentialsKolb and Turner p66

Maxwell identity

B.Sadoulet

Phys 250 -4 GR 23

The Dark Side

Relativistic Bosons and Fermions(cf.Kolb,Turner p.67-82)

Important for behavior of early universe (energy density =>expansion)Suppose that particles are non degenerate (µ<<τ)

Occupation number

Density of energy with for ultra- relativistic

where g is the spin multiplicity

Bosons Fermions

Effective number of degrees of freedom for a relativistic fermion = g x 7/8!

Similarly the number densities are

for Bosons Fermions

f ( ) ≈1

expkT

±1

u = c2 = D( ) f ( )d0

∞

∫ D( )d =

g

2

2d2c3h3

u =

g

2

kT( )4

2c3h3x3dx

ex ±10

∞

∫x3dx

ex −10

∞∫ =

4

15

x3dx

ex +10

∞∫ =

7

8

4

15

g

2

π2

15h3c3 kT( )4 =g

2aBT 4

7

8

g

2

π2

15h3c3 kT( )4 =7

8

g

2aBT 4

u = gbosons +7

8gfermions

∑ aB

2T4 = g *

aB

2T4 = 3p

n = g

3( )2h3c3 kT( )3

n =

3

4g

3( )2h3c3 kT( )3

The entropy density is s = gbosons +

7

8gfermions

∑ 2π2

45h3c3 k 4T 3

B.Sadoulet

Phys 250 -4 GR 24

The Dark Side

Evolution of Temperature

Very early universeNo ambiguity: high temperature => all particles are relativistic and in

thermal equilibrium => a single temperature

Therefore for relativistic particles

Simple interpretation: comoving number density is constant, energy is redshifted. True forall relativistic species even if they have dropped out of thermal equilibrium.

Change of number of degrees of freedomWhat happens if g* changes?e.g.,

We use the fact that comoving entropy density is constant

In above example the effective number of degrees of freedom in equilibrium beforeannihilation was 11/2 (2 for the photons, 7/8x4 for e+ e-). After only 2: ratio of temperature of = ratio of temperature of photons and neutrinos

u = c2 ∝ a−4 ⇐ w =1 /3

u = g *aB

2T 4 ⇐ Stefan - Boltzmann

T ∝ a t( )−1

e+ + e- ↔ + : e± disappear when T < 300MeV

scomov ∝ a t( )3g *s T 3 = constant ⇒ T ∝ g *s

− 1

3 a t( )−1

11

4

1 / 3

T

a

v

γ11

4

1 / 3

B.Sadoulet

Phys 250 -4 GR 25

The Dark Side

Standard Model of Particle PhysicsSU(3)xSU(2)xU(1) T<100GeVAbove quark hadron transition (deconfinement)

Bosons γ (2 states)8 gluons (2 states)

Fermionse 2 spin statesνe 1 spin state3x u 2 spin states + charge conjugate3x d 2 spin states

µ 2 spin statesνµ 1 spin state3x s 2 spin states + charge conjugate3x c 2 spin states

τ 2 spin statesντ 1 spin state3x t 2 spin states + charge conjugate3x b 2 spin states

Below deconfinement temperature (≈200 MeV)only γ, ν’s, e±, µ± and hadrons in form of π+,π-,πo, p, n

B.Sadoulet

Phys 250 -4 GR 26

The Dark Side

Number of Degrees of Freedom

T = Temperature of photons. If all species not at the same temperature

g* = gbosons

Ti

T

4

+7

8gfermions

Ti

T

4

i∑ ⇒ u = g *

aB

2T 4

g *S = gbosons

Ti

T

3

+7

8gfermions

Ti

T

3

i∑ ⇒ S = g *s

2π2

45h3c3 T 3

B.Sadoulet

Phys 250 -4 GR 27

The Dark Side

Thermal History