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Einstein’s Equations

July 1, 2008


Newtonian Gravity I

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


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 ′

2δlk

)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.


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 Equations

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


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


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 Equations I

Einstein’s g equivalence principle:

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

This statement has the 3 implications:

Gravitational accelerations are described in the same way as theinertial ones. This means that the motion of a freely movingparticle, observed from an inertial frame, will be described byd2xµ

dt2 = 0 while from a non-inertial frame its movement will be

described by the geodesic equation d2xµ

ds2 + Γµρσdxρ

dsdxσ

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 Christoffelsymbols. But according to Einstein the gravitational accelerationsas well will be described by the Christoffel symbols. This leads tothe following conclusion: The metric tensor should play the role ofthe gravitational potential since Γµρσ is a function of the metrictensor and its derivatives.


When gravitational accelerations are present the space cannot beflat since the Christoffel symbols are non-zero and the Riemanntensor is not zero as well. In other words, in the presence of thegravitational field forces the space to be curved. I.e. there is adirect link between the presence of a gravitational field and thegeometry of the space.

Consequence: if gravity is present there cannot exist inertial frames.If it was possible then one would have been able to discriminateamong inertial and gravitational accelerations which against the”generalized equivalence 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”.


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.


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


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.


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 Equations

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 Equations

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

[λ

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?).


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


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.


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

dτ

(r2φ)

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

d

dτ

(eν t)

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

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 (28)

where u = 1/r .


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 (29)

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 radii r ≈ 4.6× 107km (Mercury) andr ≈ 1.5× 108km (Earth) gets verys small values ≈ 3× 10−8 and ≈ 10−8.Finally, by substituting eqns (27) and (26) into (25) we get the “energyequation” for the r -coordinate

r2 +eν

r2L2 + eν = E 2 (30)

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


Radial motion of massive particles

For the radial motion φ is constant, which implies that L = 0 and eqn(47) reduces to

r2 = E 2 − eν (31)

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

r =M

r2. (32)

If a particle is dropped from the rest at r = R we get thatE 2 = eν(R) = 1− 2M/R and (31) will be written

r2 = 2M

(1

r− 1

R

)(33)

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.


For a particle dropped from the rest at infinity E = 1 and the geodesicequations are simplified

dt

dτ= e−ν and

dr

dτ= −

√2M

r(34)

The component of the 4-velocity will be

uµ =dxµ

dτ=(e−ν , −

√2M/r , 0 , 0

)(35)

Then by integrating the second of (34) and by assuming that at τ = τ0

that r = r0 we get

τ =2

3

√r30

2M− 2

3

√r3

2M(36)

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

√r3

0

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


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

dr

dt=

dr

dτ

dτ

dt= −e−ν

√2M

r(37)

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

)∣∣∣∣∣ (38)

where we have chosen that when r = r0 to have t = 0. Notice that whenr → 2M then t →∞. In other words it takes infinite time for a particleto reach r = 2M for an observer at infinity.


Circular motion of massive particles

The motion of massive particles in the equatorial plane is described beeqn (29)

d2

dφ2u + u =

M

L2+ 3Mu2 (39)

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

L2 =r2M

r − 3M(40)

if we also put r = 0 in eqn (47) we get :

E =1− 2M/r√

1− 3M/r(41)

Circular orbits will be bound for E < 1, so the limit on r for an orbit tobe bound is given by E = 1 which leads to

(1− 2M/r)2 = 1− 3M/r → r = 4M or r =∞ (42)

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


From the integral of motion r2φ = L and eqn (40) we get(dφ

dτ

)2

=M

r2(r − 3M)(43)

THis equation cannot be satisfied for circular orbits with R < 3M. Suchorbits cannot be geodesics and cannot be followed by freely fallingparticles.We can also calculate an expression for dφ/dt(

dφ

dt

)2

=

(dφ

dτ

dτ

dt

)2

=e2ν

E 2

(dφ

dτ

)2

=M

r3(44)

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


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 particlemotion in a central potential isdescribed by:

1

2

(dr

dt

)2

+ Veff(r) = E 2 (45)

Veff(r) = −M

r+

L2

2r2(46)

The bound orbits have two turningpoints while the circular orbitcorresponds to the special case wherethe particle sits in the minimum of theof the effective potential.


In GR the ‘energy’ equation is

r2 +eν

r2L2 + eν = E 2 (47)

which leads to an effective potential of the form

Veff(r) =eν

r2L2 + eν = −M

r+

L2

2r2− ML2

r3(48)

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

dVeff

dr=

M

r2− L2

r3+

3ML2

r4(49)

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

r =L

2M

(L±

√L2 − 12M2

)


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 orbit has

rmin = 6M and L = 2√

3M

and it is unique 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


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 studyphoton orbits on the equatorial plane and the equation of motion will bein this case

eν t = E (50)

eν t2 − e−ν r2 − r2φ2 = 0 (51)

r2φ = L (52)

The equivalent to eqn (47) for photons is

r2 +eν

r2L2 = E 2 (53)

while the equivalent of eqn (29) is

d2

dφ2u + u = 3Mu2 (54)


• Radial motion of photonsFor radial motion φ = 0 and we get

eν t2 − e−ν r2 = 0

from which we obtaindr

dt= 1− 2M

r(55)

integration leads to:

t = r + 2M ln∣∣∣ r

2M− 1∣∣∣+ const (outgoing photon) (56)

t = −r − 2M ln∣∣∣ r

2M− 1∣∣∣+ const (incoming photon) (57)

• Circular motion of photons For circular orbits we have r=constantand thus from eqn (54) we see that the only possible radius for a circularphoton orbit is:

r = 3M (58)

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

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

there can be such an orbit.Einstein’s Equations

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


The Classical Tests: Perihelion Advance

For a given value of angular momentum Kepler’s equation (29) (withoutthe term 3Mu2) admits a solution given by

u =M

L2[1 + e cos(φ+ φ0)] (59)

where e and φ0 are integration constants. We can set φ0 = 0 by rotatingthe coordinate system by φ0. The other constant e is the eccentricity ofthe orbit which is an ellipse if e < 1. Now since the term 3Mu2 is smallwe can use perturbation theory to get a solution of equation (29).

d2

dφ2u + u ≈ M

L2+

3M3

L4[1 + e cos(φ)]2 ≈ M

L2+

3M3

L4+

6eM3

L4cos(φ)

3M3/L4 << M/L2 and can be omitted and its corrections will be smallperiodic elongations of the semiaxis of the ellipse. The term6eM3/L4 cos(φ) is also small but has an accumulative effect which canbe measured. Thus the solution of the relativistic form of Kepler’sequation (60) becomes (k = 3M2/L2)

u =M

L2

[1 + e cosφ+

3eM2

L2φ sinφ

]≈ M

L2{1 + e cos[φ(1− k)]} (60)


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 =

2π

1− k≈ 2π(1 + k) = 2π +

6πM2

L2(61)

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(62)


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)(63)

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)


The Classical Tests : Deflection of Light Rays

Deflection of light rays due to presence of a gravit. field is a prediction ofEinstein dated even before the GR and has been verified in 1919.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 . (64)

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

u =1

bcos(φ+ φ0) (65)

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.


Since the term 3Mu2 is very small we can substitute u with theNewtonian solution (65) and we need to solve the non-homogeneous ODE

d2u

dφ2+ u =

3M

b2cos2 φ . (66)

admitting a solution of the form

u =cosφ

b+

M

b2

(1 + sin2 φ

)(67)

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 (68)

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

ε ≈ −2M

b(69)

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. (70)


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.


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 closed 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. (71)


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⇒ λ2 − λ1

λ1=

∆λ

λ1=

M

r1. (72)

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. (73)

While the photon redshift z is defined by

1 + z =ν1

ν2(74)

What will be the redshift for signals emitted from the surface of the sun,a neutron star and a black hole?


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

(75)

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(76)


Then eqn (75) becomes(1− 2M

r

)−(

1− 2M

r

)−1(dr

dt

)2

− D2

r2

(1− 2M

r

)2

= 0 . (77)

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 =r2

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

(78)

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


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 totalduration of the “trip” is:

∆T = 2 (t1 + t2) = 4M

[1 + ln

(4r1r2

r20

)]. (79)


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


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