Einstein's Theory of Gravitykokkotas/...bIf Qkl 6= 0 the potential will contain a term proportional...

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Einstein’s Theory of Gravity June 10, 2009 Einstein’s Theory of Gravity

Transcript of Einstein's Theory of Gravitykokkotas/...bIf Qkl 6= 0 the potential will contain a term proportional...

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Einstein’s Theory of Gravity

June 10, 2009

Einstein’s Theory of Gravity

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Newtonian Gravity

Poisson equation

∇2U(~x) = 4πGρ(~x) → U(~x) = −G

∫d3~x ′

ρ(~x)

|~x − ~x ′|For a spherically symmetric mass distribution of radius R

U(r) = −1

r

∫ R

0

r ′2ρ(r ′)dr ′ for r > R

U(r) = −1

r

∫ r

0

r ′2ρ(r ′)dr ′ −∫ R

r

r ′ρ(r ′)dr ′ for r < R

Einstein’s Theory of Gravity

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For a non-spherical distribution the term 1/|~x − ~x ′| can be expanded as

1

|~x − ~x ′|=

1

r+∑

k

xkx ′k

r3+

1

2

∑k

∑l

(3x ′kx ′l −~r ′2δl

k

) xkx l

r5+ . . .

U(~x) = −GM

r− G

r3

∑k

xkDk − G

2

∑kl

Qkl xkx l

r5+ . . .

Gravitational Multipoles

M =

∫ρ(~x ′)d3x ′ Mass

Dk =

∫x ′kρ(~x ′)d3x ′ Mass Dipole momenta

Qkl =

∫ (3x ′kx ′l − ~r ′

2δl

k

)ρ(~x ′)d3x ′ Mass Quadrupole tensorb

aIf the center of mas is chosen to coincide with the origin of the coordinatesthen Dk = 0 (no mass dipole).

bIf Qkl 6= 0 the potential will contain a term proportional to ∼ 1/r 3 and thegravitational force will deviate from the inverse square law by a term ∼ 1/r 4.

Einstein’s Theory of Gravity

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The Earth’s polar and equatorial diameters differ by 3/1000. Thisdeviation produces a quadrupole term in the gravitational potential,which causes perturbations in the elliptical Kepler orbits of satellites.Usually, we define the dimensionless parameter

J2 = − Q33

2MR2(1)

as a convenient measure of the oblateness of a nearly spherical body. Forthe Sun the oblateness due to rotation gives J2 ≈ 10−7.The main perturbation is the precession of Kepler’s ellipse and this canbe used for precise determinations of the multipole moments and themass distribution in the Earth.

What about the Sun?Einstein’s Theory of Gravity

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Equivalence Principle

In GR gravitational phenomena arise not from forces and fields, but fromthe curvature of the 4-dim spacetime. The starting point for thisconsideration is the Equivalence Principle which states that theGravitational and the Inertial masses are equal. The equality mG = mI isone of the most accurately tested principles in physics!Now it is known experimentally that: γ = mG−mI

mG< 10−13

Experimental verification

Galileo (1610) , Newton (1680) γ < 10−3

Bessel (19th century) γ < 2 × 10−5

Eotvos (1890) & (1908) γ < 3 × 10−9

Dicke et al. (1964) γ < 3 × 10−11

Braginsky et al. (1971) γ < 9 × 10−13,

Kuroda and Mio (1989) γ < 8 × 10−10,

Adelberger et al. (1990) γ < 1 × 10−11

Su et al. (1994) γ < 1 × 10−12 (torsional)

Williams et al. (‘96), Anderson & Williams (‘01) γ < 1 × 10−13

Einstein’s Theory of Gravity

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Equivalence Principle II

Weak Equivalence Principle : The motion of a neutral testbody released at a given point in space-time is independent ofits composition

Strong Equivalence Principle :

The results of all local experiments in a frame in free fall areindependent of the motionThe results are the same for all such frames at all places andall timesThe results of local experiments in free fall are consistent withSTR

Einstein’s Theory of Gravity

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Equivalence Principle : Dicke’s Experiment

The experiment is based on measuring the effect of the gravitational fieldon two masses of different material in a torsional pendulum.

GMm(E)G

R2=

m(E)I v2

R⇒ v2 =

GM

R

(mG

mI

)(E)

The forces acting on both masses are:

F (j) +GMm

(j)G

R2=

m(j)I v2

R⇒ F (j) =

GMm(j)I

R2

[(mG

mI

)(E)

−(

mG

mI

)(j)]

and the total torque applied is: L =(F (1) − F (2)

)` = GM

R2

[m

(1)G −m

(2)G

]`

here we assumed that : m(1)I = m

(2)I = m.

Einstein’s Theory of Gravity

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Einstein’s equivalence principle:

Gravitational and Inertial forces are equivalent and they cannot bedistinguished by any physical experiment.

This statement has the 3 implications:1. Gravitational accelerations are described in the same way asthe inertial ones.• This means that the motion of a freely moving particle, observed

from an inertial frame, will be described by d2xµ

dt2 = 0 while from anon-inertial frame its movement will be described by the geodesicequation

d2xµ

ds2+ Γµρσ

dxρ

ds

dxσ

ds= 0 .

• The 2nd term appeared due to the use of a non-inertial frame, i.e.the inertial accelerations will be described by the Christoffel symbols.• But according to Einstein the gravitational accelerations as well willbe described by the Christoffel symbols.• This leads to the following conclusion: The metric tensor should playthe role of the gravitational potential since Γµρσ is a function of the metrictensor and its derivatives.

Einstein’s Theory of Gravity

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• When gravitational accelerations are present the space cannotbe flat since the Christoffel symbols are non-zero and the Riemanntensor is not zero as well.In other words, the presence of the gravitational field forces the space tobe curved. I.e. there is a direct link between the presence of agravitational field and the geometry of the space.

• Consequence: if gravity is present there cannot exist inertialframes.If it was possible then one would have been able to discriminate amonginertial and gravitational accelerations which against the ”generalizedequivalence principle” of Einstein.The absence of ”special” coordinate frames (like the inertial) and theirsubstitution from ”general” (non-inertial) coordinate systems lead innaming Einstein’s theory for gravity ”General Theory of Relativity”.

Einstein’s Theory of Gravity

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Einstein’s Equations

Since the source of the gravitational field is a tensor (Tµν) the fieldshould be also described by a 2nd order tensor e.g. Fµν . Since the role ofthe gravitational potential is played by the metric tensor then Fµν shouldbe a function of the metric tensor gµν and its 1st and 2nd orderderivatives. Moreover, the law of energy-momenum conservation impliesthat Tµν

;µ = 0 which suggests that Fµν ;µ = 0 Then since Fµν should bea linear function of the 2nd derivative of gµν we come to the followingform of the field equations (how & why?):

Fµν = Rµν + agµνR + bgµν = κTµν (2)

where κ = 8πGc4 . Then since Fµν ;µ = 0 there should be

(Rµν + agµνR + bgµν);µ = 0 (3)

which is possible only for a = −1/2. Thus the final form of Einstein’sequations is:

Rµν − 1

2gµνR + Λgµν = κTµν . (4)

where Λ = 8πGc2 ρv is the so called cosmological constant.

Einstein’s Theory of Gravity

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Newtonian Limit

In the absence of strong gravitational fields and for small velocities bothEinstein & geodesic equations reduce to the Newtonian ones.

Geodesic equations:

d2xµ

dt2+ Γµαβ

dxα

dt

dxβ

dt= − d2t/ds2

(dt/ds)2

dxµ

dt⇒ d2x j

dt2≈ g00,j (5)

If g00 ≈ η00 + h00 = 1 + h00 = 1 + 2 Uc2 then d2xk

dt2 ≈ ∂U∂xk

Einstein equations

Rµν = −κ(

Tµν −1

2gµνT

)⇒ ∇2U =

1

2κρ (6)

where κ = 8πG .

Here we have used the following approximations:

Γj00 ≈

1

2g00,j and R00 ≈ Γj

00,j ≈ ∇2U

Einstein’s Theory of Gravity

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Solutions of Einstein’s Equations

• A static spacetime is one for which a timelike coordinate x0 has thefollowing properties:

(I) all metric components gµν are independent of x0

(II) the line element ds2 is invariant under the transformationx0 → −x0.

Note that the 1st property does not imply the 2nd (e.g. the time reversalon a rotating star changes the sense of rotation, but the metriccomponents are constant in time).• A spacetime that satisfies (I) but not (II) is called stationary.

• The line element ds2 of a static metric depends only on rotational

invariants of the spacelike coordinates x i and their differentials, i.e. the

metric is isotropic.

Einstein’s Theory of Gravity

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The only rotational invariants of the spacelike coordinates x i and theirdifferentials are

~x · ~x ≡ r2, ~x · d~x ≡ rdr , d~x · d~x ≡ dr2 + r2dθ2 + r2 sin2 θdφ2

Thus the more general form of a spatially isotropic metric is:

ds2 = A(t, r)dt2 − B(t, r)dt (~x · d~x)− C (t, r) (~x · d~x)2 − D(t, r)d~x2

= A(t, r)dt2 − B(t, r)r dt dr − C (t, r)r2dr2

−D(t, r)(dr2 + r2dθ2 + r2 sin2 θdφ2

)(7)

= A(t, r)dt2 − B(t, r)dt dr − C (t, r)dr2 − D(t, r)(dθ2 + sin2 θdφ2

)= A′(t, r)dt2 − B ′(t, r)dt dr − C ′(t, r)dr2 − r2

(dθ2 + sin2 θdφ2

)(8)

where we have set r2 = D(t, r) and we redefined the A, B and C . Thenext step will be to introduce a new timelike coordinate t as

dt = Φ(t, r)

[A′(t, r)dt − 1

2B ′(t, r)dr

]where Φ(t, r) is an integrating factor that makes the right-hand side an

exact differential.Einstein’s Theory of Gravity

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By squaring we obtain

dt2 = Φ2

(A′2dt2 − A′B ′dtdr +

1

4B ′2dr2

)from which we find

A′dt2 − B ′dtdr =1

A′Φ2dt2 − B ′

4A′dr2

Thus by defining the new functions A = 1/(A′Φ)2 and B = C + B ′/(4A′)the metric 8 becomes diagonal

ds2 = A(t, r)dt2 − B(t, r)dr2 − r2(dθ2 + sin2 θdφ2

)(9)

or by dropping the ‘hats’ and ‘tildes’

ds2 = A(t, r)dt2 − B(t, r)dr2 − r2(dθ2 + sin2 θdφ2

)(10)

• Thus the general isotropic metric is specified by two functions of t andr , namely A(t, r) and B(t, r).• Also, surfaces for t and r constant are 2-spheres (isotropy of themetric).

• Since B(t, r) is not unity we cannot assume that r is the radial

distance.Einstein’s Theory of Gravity

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Schwarzschild Solution

A typical solution of Einstein’s equations describing spherically symmetricspacetimes has the form:

ds2 = eν(t,r)dt2 − eλ(t,r)dr2 − r2(dθ2 + sin2 θdφ2

)(11)

Then the components of the Ricci tensor will be:

R11 = −ν′′

2− ν′2

4+ν′λ′

4+λ′

r+ eλ−ν

2+λ2

4− λν

4

](12)

R00 = eν−λ[ν′′

2+ν′2

4− ν′λ′

4+ν′

r

]− λ

2− λ2

4+λν

4(13)

R10 = λ/r (14)

R22 = = −e−λ[1 +

r

2(ν′ − λ′)

]+ 1 (15)

R33 = sin2 θR22 (16)

in the absence of matter (outside the source) Rµν = 0 we can prove thatλ(r) = −ν(r), i.e. the solution is independent of time (how and why?).

Einstein’s Theory of Gravity

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ds2 =

(1− 2GM

rc2

)c2dt2 −

(1− 2GM

rc2

)−1

dr2 − r2(dθ2 + sin2 θdφ2

)

Sun: M ≈ 2× 1033gr and R = 696.000km

2GM

rc2≈ 4× 10−6

Neutron star : M ≈ 1.4M and R ≈ 10− 15km

2GM

rc2≈ 0.3− 0.5

Neutonian limit:

g00 ≈ η00 + h00 = 1 +2U

c2⇒ U =

GM

r

Einstein’s Theory of Gravity

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Schwarzschild Solution: Geodesics

r − 1

2ν′r2 − reν θ2 − reν sin2 θφ2 +

1

2e2νν′t2 = 0 (17)

θ +2

rr θ − sin θ cos θφ2 = 0 (18)

φ+2

rr φ+ 2

cos θ

sin θθφ = 0 (19)

t + ν′r t = 0 (20)

and from gλµxλxµ = 1 (here we assume a massive particle) we get :

−eν t2 + eλr2 + r2θ + r2 sin2 θφ2 = 1. (21)

If a geodesic is passing through a point P in the equatorial plane(θ = π/2) and has a tangent at P situated also in this plane (θ = 0 atP) then from (18) we get θ = 0 at P and all higher derivatives are alsovanishing at P. That is, the geodesic lies entirely in the plane defined byP, the tangent at P and the center of symmetry of the space. Since thesymmetry planes are equivalent to each other, it will be sufficient todiscuss the geodesics lying on one of these planes e.g. the equatorialplane θ = π/2.

Einstein’s Theory of Gravity

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The geodesics on the equatorial plane are:

r − 1

2ν′r2 − reν φ2 +

1

2e2νν′t2 = 0 (22)

φ+2

rr φ = 0 (23)

t + ν′r t = 0 (24)

−eν t2 + eλr2 + r2φ2 = 1 (25)

Then from equations (23) and (24) we can easily prove (how?) that:

d

(r2φ)

= 0 ⇒ r2φ = L = const : (Angular Momentum)(26)

d

(eν t)

= 0 ⇒ eν t = E = const : (Energy) (27)

• For a massive particle with unit rest mass, by assuming that τ is theaffine parameter for its motion we get pµ = xµ. Thus

p0 = g00t = eν t = E and pφ = gφφφ = −r2φ = −L (28)

Einstein’s Theory of Gravity

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• An observer with 4-velocity Uµ will find that the energy of a particlewith 4-momentum pµ is:

E = pµUµ

An observer at infinity, Uµ = (1, 0, 0, 0) will find E = p0 = E .Actually, for a particle with rest mass m0 we get E = E/(m0c

2) i.e. it isthe energy per unit rest-mass.

————Finally, by substituting eqns (27) and (26) into (25) we get the “energyequation” for the r -coordinate

r2 +eν

r2L2 + eν = E 2 (29)

which suggests that at r →∞ we get that E = 1.

Einstein’s Theory of Gravity

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By combining the 2 integrals of motion and eqn (25) we can eliminate theproper time τ to derive an equation for a 3-d path of the particle (how?)(

d

dφu

)2

+ u2 =E 2 − 1

L2+

2Mu

L2+ 2Mu3 (30)

where we have used u = 1/r and

r =dr

dτ=

dr

dτ=

L

r2

dr

The previous equation can be written in a form similar to Kepler’sequation of Newtonian mechanics (how?) i.e.

d2

dφ2u + u =

M

L2+ 3Mu2 (31)

The term 3Mu2 is the relativistic correction to the Newtonian equation.This term for the trajectories of planets in the solar system, whereM = M = 1.47664 km, and for orbital radius r ≈ 4.6× 107km(Mercury) and r ≈ 1.5× 108km (Earth) gets very small values≈ 3× 10−8 and ≈ 10−8 correspondingly.

Einstein’s Theory of Gravity

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Radial motion of massive particles

For the radial motion φ is constant, which implies that L = 0 and eqn(29) reduces to

r2 = E 2 − eν (32)

and by differentiation we get an equation which reminds the equivalentone of Newtonian gravity i.e.

r =M

r2. (33)

• If a particle is dropped from the rest at r = R we get thatE 2 = eν(R) = 1− 2M/R and (32) will be written

r2 = 2M

(1

r− 1

R

)(34)

which is again similar to the Newtonian formula for the gain of kinetic

energy due to the loss in gravitational potential energy for a particle (of

unit mass) falling from rest at r = R.

Einstein’s Theory of Gravity

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• For a particle dropped from the rest at infinity E = 1 and thegeodesic equations are simplified

dt

dτ= e−ν and

dr

dτ= −

√2M

r(35)

The component of the 4-velocity will be

uµ =dxµ

dτ=

(e−ν , −

√2M

r, 0 , 0

)(36)

Then by integrating the second of (35) and by assuming that at τ = τ0

that r = r0 we get

τ =2

3

√r30

2M− 2

3

√r3

2M(37)

which suggests that for r = 0 we get τ → 23

√r 3

0

2M ie the particle takesfinite proper time to reach r = 0.

Einstein’s Theory of Gravity

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If we want to map the trajectory of the particle in the (r , t) coordinateswe need to solve the equation

dr

dt=

dr

dt= −e−ν

√2M

r(38)

The integration leads to the relation

t =2

3

(√r30

2M−√

r3

2M

)+ 4M

(√r0

2M−√

r

2M

)

+ 2M ln

∣∣∣∣∣(√

r/2M + 1√r/2M − 1

)(√r0/2M − 1√r0/2M + 1

)∣∣∣∣∣ (39)

where we have chosen that for t = 0 to set r = r0.Notice that when r → 2M then t →∞. In other words, for an observerat infinity, it takes infinite time for a particle to reach r = 2M.———-

QUESTION : Can you find what will be the velocity of the radially

falling particle for a stationary observer at coordinate radius r?

Einstein’s Theory of Gravity

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Figure: Radial fall from rest towards a Schwarschild BH as described by acomoving observer (proper time τ) and by a distant observer(Schwarschild coordinate time t)

τ =2

3

√r30

2M− 2

3

√r3

2M

t =2

3

0@s r30

2M−

sr3

2M

1A + 4M

„rr0

2M−r

r

2M

«+ 2M ln

˛˛ p

r/2M + 1pr/2M − 1

! pr0/2M − 1pr0/2M + 1

!˛˛

Einstein’s Theory of Gravity

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Circular motion of massive particles

The motion of massive particles in the equatorial plane is described byeqn (31)

d2

dφ2u + u =

M

L2+ 3Mu2 (40)

For circular motions r = 1/u=const and r = r = 0. Thus we get

L2 =r2M

r − 3M(41)

if we also put r = 0 in eqn (29) we get :

E =1− 2M/r√

1− 3M/r(42)

Circular orbits will be bound for E/(m0c2) = E < 1, so the limit on r for

an orbit to be bound is given by E = 1 which leads to

(1− 2M/r)2 = 1− 3M/r true when r = 4M or r =∞ (43)

Thus over the range 4M < r <∞ circular orbits are bound.

Einstein’s Theory of Gravity

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Figure: The variation of E = E/(m0c2) as a function of r/M for a

circular orbit of a massive particle in the Schwarzchild geometry

Einstein’s Theory of Gravity

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Figure: The shape of a bound orbit outside a spherical star or ablack-hole

Einstein’s Theory of Gravity

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From the integral of motion r2φ = L and eqn (41) we get(dφ

)2

=M

r2(r − 3M)(44)

NOTICE : This equation cannot be satisfied for circular orbits withr < 3M. Such orbits cannot be geodesics and cannot be followed byfreely falling particles.We can also calculate an expression for Ω = dφ/dt

Ω2 =

(dφ

dt

)2

=

(dφ

dt

)2

=e2ν

E 2

(dφ

)2

=M

r3(45)

which is equivalent to Kepler’s law in Newtonian gravity.

Einstein’s Theory of Gravity

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Stability of massive particle orbits

According to the previous discussion theclosest bound orbit around a massivebody is at r = 4M, however we cannotyet determine whether this orbit isstable.In Newtonian theory the particle motionin a central potential is described by:

1

2

(dr

dt

)2

+ Veff(r) = E 2 (46)

Veff(r) = −M

r+

L2

2r2(47)

The bound orbits have two turning points while the circular orbit

corresponds to the special case where the particle sits in the minimum of

the of the effective potential.

Einstein’s Theory of Gravity

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In GR the ‘energy’ equation (29) is

r2 +eν

r2L2 + eν = E 2 (48)

which leads to an effective potential of the form

Veff(r) =eν

r2L2 + eν = −M

r+

L2

2r2− ML2

r3(49)

Circular orbits occur where dVeff/dr = 0 that is:

dVeff

dr=

M

r2− L2

r3+

3ML2

r4(50)

so the extrema are located at the solutions of the eqnMr2 − Lr + 3ML2 = 0 which occur at

r =L

2M

(L±

√L2 − 12M2

)

Einstein’s Theory of Gravity

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Note that if L =√

12M = 2√

3Mthen there is only one extremum andno turning points in the orbit forlower values of L. Thus theinnermost stable circular orbit(ISCO) has

rISCO = 6M and L = 2√

3M

and it is unique in satisfying bothdVeff/dr = 0 and d2Veff/dr2 = 0,the latter is the condition formarginal stability of the orbit.

Figure: The dots indicate thelocations of stable circular orbitswhich occur at the local minimum ofthe potential. The local maxima inthe potential are the locations of theunstable circular orbits

Einstein’s Theory of Gravity

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Figure: Orbits for L = 4.3 and different values of E .

The upper shows circular orbits. A stable (outer) and an unstable(inner) one.

The lower shows bound orbits , the particle moves between twoturning points marked by dotted circles.

Einstein’s Theory of Gravity

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Figure: Orbits for L = 4.3 and different values of E .

The upper shows a scattering orbit the particle comes in frominfinity passes around the center of attraction and moves out toinfinity again.

The lower shows plunge orbit , in which the particle comes in frominfinity.

Einstein’s Theory of Gravity

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Trajectories of photons

Photons as any zero rest mass particle move on null geodesics. In thiscase we cannot use the proper time τ as the parameter to characterizethe motion and thus we will use some affine parameter σ.We will study photon orbits on the equatorial plane and the equation ofmotion will be in this case

eν t = E (51)

eν t2 − e−ν r2 − r2φ2 = 0 (52)

r2φ = L (53)

The equivalent to eqn (29) for photons is

r2 +eν

r2L2 = E 2 (54)

while the equivalent of eqn (31) is

d2

dφ2u + u = 3Mu2 (55)

Einstein’s Theory of Gravity

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Radial motion of photons

For radial motion φ = 0 and we get

eν t2 − e−ν r2 = 0

from which we obtain

dr

dt= ±

(1− 2M

r

)(56)

integration leads to:Outgoing Photon

t = r + 2M ln∣∣∣ r

2M− 1∣∣∣+ const

Incoming Photon

t = −r − 2M ln∣∣∣ r

2M− 1∣∣∣+ const Figure: Radially infalling particle

emitting a radially outgoing photon.

Einstein’s Theory of Gravity

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Circular motion of photons

For circular orbits we have r=constant and thus from eqn (55) we seethat the only possible radius for a circular photon orbit is:

r = 3M

(or r =

3GM

c2

)(57)

There are no such orbits around typical stars because their radius is much

larger than 3M (in geometrical units). But outside the black hole there

can be such an orbit.

Einstein’s Theory of Gravity

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Stability of photon orbits

We can rewrite the “energy” equation (54) as

r2

L2+

r2=

1

D2(58)

where D = L/E and Veff = eν/r

Figure: The effective potential for photon orbits.

We can see that Veff has a single maximum at r = 3M where the valueof the potential is 1/(27M2). Thus the circular orbit r = 3M is unstable.

There are no stable circular photon orbits in the Schwarzschild

geometry.Einstein’s Theory of Gravity

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The slow-rotation limit : Dragging of inertial frames

ds2 = ds2Schw +

4Ma

rsin2 θdtdφ (59)

where J = Ma is the angular momentum.The contravariant components of the particles 4-momentum will be

pφ = gφµpµ = gφtpt + gφφpφ and pt = g tµpµ = g ttpt + g tφpφ

If we assume a particle with zero angular momentum, i.e. pφ = 0 alongthe geodesic then the particle’s trajectory is such that

dt=

pt=

g tφ

g tt=

2Ma

r3= ω(r)

which is the coordinate angular velocity of a zero-angular-momentumparticle.A particle dropped “straight in” from infinity ( pφ = 0) is dragged justby the influence of gravity so that acquires an angular velocity in thesame sense as that of the source of the metric.The effect weakens with the distance and makes the angularmomentum of the source measurable in practice.

Einstein’s Theory of Gravity

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Useful constants

Useful Constants in geometrical units

Speed of light c =299,792.458 km/s = 1Planck’s constant ~ = 1.05× 10−27 erg ·s = 2.612× 10−66cm2

Gravitation constant G = 6.67× 10−8 cm3/g ·s2 =1Energy eV=1.602×10−12erg = 1.16× 104K

=1.7823×10−33g =1.324×10−56 kmDistance 1 pc=3.09×1013km=3.26 lyTime 1 yr = 3.156×107 secLight year 1 ly = 9.46×1012kmAstronomical unit (AU) 1AU = 1.5× 108 kmEarth’s mass M⊕ = 5.97× 1027 gEarth’s radius (equator) R⊕ = 6378 kmSolar Mass M = 1.99× 1033 g =1.47664 kmSolar Radius R = 6.96× 105 km

Einstein’s Theory of Gravity

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The Classical Tests: Perihelion Advance

For a given value of angular momentum Kepler’s equation (31)

d2

dφ2u + u =

M

L2+ 3Mu2 (31)

(without the GR term 3Mu2) admits a solution given by

u =M

L2[1 + e cos(φ+ φ0)] (60)

where e and φ0 are integration constants.• We can set φ0 = 0 by rotating the coordinate system by φ0.• The other constant e is the eccentricity of the orbit which is anellipse if e < 1.Now since the term 3Mu2 is small we can use perturbation theory to geta solution of equation (31).

d2

dφ2u + u ≈ M

L2+

3M3

L4[1 + e cos(φ)]2 ≈ M

L2+

3M3

L4+

6eM3

L4cos(φ)

Because, 3M3/L4 M/L2 and can be omitted and its corrections will besmall periodic elongations of the semiaxis of the ellipse.

Einstein’s Theory of Gravity

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The term 6eM3/L4 cos(φ) is also small but has an accumulative effectwhich can be measured. Thus the solution of the relativistic form ofKepler’s equation (61) becomes (k = 3M2/L2)

u =M

L2

[1 + e cosφ+

3eM2

L2φ sinφ

]≈ M

L21 + e cos[φ(1− k)] (61)

The perihelion of the orbit can be found be maximizing u that is whencos[φ(1− k)] = 1 or better if φ(1− k) = 2nπ. This mean that after eachrotation around the Sun the angle of the perihelion will increase by

φn =2nπ

1− k⇒ φn+1 − φn =

1− k≈ 2π(1 + k) = 2π +

6πM2

L2(62)

Einstein’s Theory of Gravity

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i.e. in the relativistic orbit the perihelion is no longer a fixed point, as itwas in the Newtonian elliptic orbit but it moves in the direction of themotion of the planet, advancing by the angle

δφ ≈ 6πM2

L2(63)

By using the well know relation from Newtonian Celestial MechanicsL2 = Mr0(1− e2) connecting the perihelion distance r0 with L we derivea more convenient form for the perihelion advance

δφ ≈ 6πM

r0(1− e2)(64)

Solar system measurements

Mercury (43.11± 0.45)′′ 43.03′′

Venus (8.4± 4.8)′′ 8.6′′

Earth (5.0± 1.2)′′ 3.8′′

In binary pulsars separated by 106km the perihelion advance is extremely

important and it can be up to ∼ 2o/year (about 1000 orbital rotations)

Einstein’s Theory of Gravity

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Sources of the precession of perihelion for Mercury

Amount (arcsec/century) Gause5025.6 Coordinate ( precession of the equinoxes)a

531.4 Gravitational tugs of the other planets0.0254 Oblateness of the Sun42± 0.04 General relativity5600.0 Total5599.7 Observed

aPrecession refers to a change in the direction of the axis of a rotatingobject. There are two types of precession, torque-free and torque-induced

Einstein’s Theory of Gravity

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The Classical Tests : Deflection of Light Rays

Deflection of light rays due to presence of a gravitational field is aprediction of Einstein dated even before the GR and has been verified in1919.Photons follow null geodesics, this means that ds = 0 and the integralsof motion (E and L) are divergent but not their ratio L/E . Thus thephoton’s equation of motion on the equatorial plane of Schwarszchildspacetime will be:

d2u

dφ2+ u = 3Mu2 . (65)

with an approximate solution (we omit at the moment the term 3Mu2)

u =1

bcos(φ+ φ0) (66)

which describes a straight line, where b and φ0 are integrationconstants. Actually, with an appropriate rotation φ0 = 0 and the length bis the distance of the line from the origin.

Einstein’s Theory of Gravity

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Since the term 3Mu2 is very small we can substitute u with theNewtonian solution (66) and we need to solve the non-homogeneous ODE

d2u

dφ2+ u =

3M

b2cos2 φ . (67)

admitting a solution of the form

u =cosφ

b+

M

b2

(1 + sin2 φ

)(68)

which for distant observers (r →∞) i.e. u → 0, and we get a relationbetween φ, M and the parameter b

cosφ

b+

M

b2

(1 + sin2 φ

)= 0 (69)

and since r →∞, this means that φ→ π/2 + ε and cosφ→ 0 + ε andsinφ→ 1− ε we get:

ε ≈ −2M

b(70)

Since, φ→ π/2 + 2M/b for r →∞ on the one side & φ→ 3π/2− 2M/bon the other side the total deviation will be the sum of the two i.e.

δφ =4M

b. (71)

Einstein’s Theory of Gravity

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For a light ray tracing the surface of the Sun gives a deflection of∼ 1.75′′.The deflection of light rays is a quite common phenomenon in Astronomyand has many applications. We typically observe “crosses” or “rings”

Figure: Einstein Cross (G2237+030) is the most characteristic case of gravitational lens where a galaxy at a

distance 5 × 108 lys focuses the light from a quasar who is behind it in a distance of 8 × 109 lys. The focusingcreates 4 symmetric images of the same quasar. The system has been discovered by John Huchra.

Figure: “Einstein rings” are observed when the source, the focusing body and Earth are on the same line ofsight. This ring has been discovered by Hubble space telescope.

Einstein’s Theory of Gravity

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The Classical Tests : Gravitational Redshift

Figure: Let’s assume 3 static observers on a Schwarszchild spacetime, one very close to the source of the fieldthe other in a medium distance from the source and the third at infinity.

The clocks of the 3 observers ticking with different rates. The clocks ofthe two closer to the source are ticking slower than the clock of theobserver at infinity who measures the so called ‘ coordinate time” i.e.

dτ1 =(

1− 2Mr1

)1/2

dt and dτ2 =(

1− 2Mr2

)1/2

dt this means that

dτ2

dτ1=

(1− 2M

r2

)1/2

(1− 2M

r1

)1/2≈ 1 +

M

r1− M

r2. (72)

Einstein’s Theory of Gravity

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If the 1st observer sends light signals on a specific wavelegth λ1 fromc = λ/τ we get a relation between the wavelength of the emitted andreceived signals

λ2

λ1≈ 1 +

M

r1− M

r2⇒ λ2 − λ1

λ1=

∆λ

λ1=

M

r1− M

r2. (73)

A similar relation can be found for the frequency of the emitted signal:

ν1

ν2=

(1− 2M

r2

)1/2

(1− 2M

r1

)1/2. (74)

While the photon redshift z is defined by

1 + z =ν1

ν2(75)

QUESTION : What will be the redshift for signals emitted from thesurface of the Sun, a neutron star and a black hole?

Einstein’s Theory of Gravity

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The Classical Tests : Radar Delay

A more recent test (late 60s) where the delay of the radar signals causedby the gravitational field of Sun was measured. This experimentsuggested and performed by I.I. Shapiro and his collaborators.

The line element ds2 = gµνdxµdxν for the light rays i.e. for ds = 0 onthe equatorial plane has the form

0 =

(1− 2M

r

)−(

1− 2M

r

)−1(dr

dt

)2

− r2

(dφ

dt

)2

(76)

For the study of the radial motion we should substitute the term dφ/dtfrom the integrals of motion (26) and (27) i.e. we can create thequantity:

D =L

E= r2

(1− 2M

r

)−1dφ

dt(77)

Einstein’s Theory of Gravity

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Then eqn (76) becomes(1− 2M

r

)−(

1− 2M

r

)−1(dr

dt

)2

− D2

r2

(1− 2M

r

)2

= 0 . (78)

At the point of the closest approach to the Sun, r0, there should be

dr/dt = 0 and thus we get the value of D2 =r 2

0

1−2M/r0. Leading to an

equation for the radial motion:

dr

dt=

(1− 2M

r

)[1−

( r0

r

)2 1− 2M/r

1− 2M/r0

]1/2

(79)

leading to

t1 =

∫ r1

r0

dr(1− 2M

r

)√1−

(r0

r

)2 1−2M/r1−2M/r0

=√

r21 − r2

0 + 2M ln

[r1

√r21 − r2

0

r0

]+ M

√r1 − r0

r1 + r0

≈ r1 + 2M ln

(2r1

r0

)+ M

Einstein’s Theory of Gravity

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For flat space we have t1 = r1 i.e. the term t1 = 2M ln(2r1/r0) + M isthe relativistic correction for the first part of the orbit in same way we geta similar contribution as the signal returns to Earth. Thus the total“extra time” is:

∆T = 2 (t1 + t2) = 4M

[1 + ln

(4r1r2

r20

)]. (80)

Einstein’s Theory of Gravity

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Figure: Comparison of the experimental results with the prediction of thetheory. The results are from I.I. Shapiro’s experiment (1970) using Venusas reflector.

Einstein’s Theory of Gravity