General Relativity Fall 2019
Lecture 17: three classic tests of linearized gravity
Yacine Ali-Haı̈moud
October 29th 2019
In the previous lecture, we found that, far from a quasi-Newtonian source, and to leading order in rsrc /r, the metric
can be cast in the form
ds2 = −(1 − 2M/r)dt2 + (1 + 2M/r)d~x2 = −(1 + 2Φ)dt2 + (1 − 2Φ)d~x2 . (1)
where M is the mass-energy of the source, which is constant at linear order in perturbations, r ≡ ~x and Φ =
−M/r + O(M rsrc /r3 ) is the Newtonian potential. In this lecture we will assume that Φ is independent of time.
The gravitational wave part hTT TT 2
ij also scales as 1/r far from the source, but is of order hij ∼ Q̈ij /r ∼ v M/r M/r
for sources with small characteristic velocities v 1.
To derive the above metric, we have assumed that the metric is nearly-Minkowski everywhere, including inside the
source, which allowed us to define M in terms of an integral of the stress-energy tensor on a flat background. In
fact, one can show that far away from any source (even a black hole), in the asymptotically flat region, the metric
coefficients take the same form, where M can now be thought of an integration constant. For a relativistic source,
M cannot (and should not!) be related to any kind of integrals over the curved spacetime inside the source, as such
integrals are not well defined. The discussion below therefore applies to the asymptotic flat spacetime of any source.
GRAVITATIONAL REDSHIFT
We already derived the gravitational redshift directly from the equivalence principle. Here we will do it again using
the tools and notions we have learned.
Consider an observer with 4-velocity U µ . In the observer’s instantaneous rest frame (i.e. using Fermi normal
coordinates centered around the observer’s worldline), the energy of a particle is the zero-th contravariant component
of its 4-momentum, Eobs = p0̂ . In that frame, Uµ̂ = (−1, 0, 0, 0), so
Eobs = −Uµ pµ . (2)
This expression is covariant, and holds in any frame. It is an important expression to remember (or at least, the
reasoning that led to it).
Suppose now the observer is at rest in the coordinates (t, ~x) in which the metric components are given by Eq. (1).
Its 4-velocity is then U µ = (U 0 , 0, 0, 0). It is normalized so gµν U µ U ν = −1 = −(1 + 2Φ)(U 0 )2 , implying, to linear
order, U 0 = 1 − Φ. The energy this observer measures is
Eobs = −U µ pµ = −U 0 p0 = −(1 − Φ)p0 . (3)
The fact that ∂0 gµν = 0 implies that ∂0 is a Killing vector field and that the covariant component p0 is conserved
along geodesics. Therefore, the fractional change in energy between two observers sitting at different values of Φ is
Eobs (2) − Eobs (1) −(1 − Φ(2)) + (1 − Φ(1))
= = Φ(1) − Φ(2) + O(Φ2 ). (4)
Eobs (1) −(1 − Φ(1))
DEFLECTION OF LIGHT
Consider a null geodesic with 4-momentum p = d/dλ, where λ is an affine parameter. The spatial part of the
geodesic equation is
dpi
= −pµ pν Γiµν . (5)
dλ
2
The relevant Christoffel symbols are, for this metric,
Γi00 = ∂i Φ, Γi0j = 0, Γijk = −δij ∂k Φ − δik ∂j Φ + δjk ∂i Φ. (6)
So we get
dpi
= −(p0 )2 ∂i Φ − p~2 δij − 2pi pj ∂j Φ, p~2 ≡ δjk pj pk .
(7)
dλ
Now, for null geodesics we have gµν pµ pν = 0, so
(1 + 2Φ)(p0 )2 = (1 − 2Φ)~
p2 ⇒ p~2 = (1 + 4Φ)(p0 )2 . (8)
To lowest order in Φ, we may use p~2 = (p0 )2 , implying
dpi
p2 (δ ij − p̂i p̂j )∂j Φ = −2~
= −2~ p2 ∇i⊥ Φ, (9)
dλ
~ ⊥ is the gradient perpendicular to p~, and p̂ ≡ p~/|~
where ∇ p2 |1/2 . We then find
p2
d~ dpi
= 2δik pk p2 pk δik (δ ij − p̂i p̂j )∂j Φ = 0.
= −4~ (10)
dλ dλ
Therefore, p~2 is constant along null geodesics, to lowest order in Φ. So the direction p̂ changes with rate
dp̂ ~ ⊥ Φ = −2 dt ∇
~ ⊥ Φ = −2p0 ∇ ~ ⊥ Φ ⇒ dp̂ = −2∇
~ ⊥Φ .
= −2|~
p|∇ (11)
dλ dλ dt
Hence, we find the deflection angle for a general potential
Z
∆p̂ = −2 ~ ⊥Φ .
dt ∇ (12)
traj
Let us apply this result to a spherically symmetric source, for which Φ(r) depends on r only. Consider a photon
trajectory as depicted in Fig. 1. Set the z axis along the lens-source direction, and set the lens at coordinate z = 0.
The trajectory is planar (check this explictly ). Assume it is in the x − z plane (i.e. y = 0). The perpendicular
gradient is
~ ⊥ Φ = dΦ (r̂ − ẑ(ẑ · r̂)) = x dΦ .
∇ (13)
dr r dr
To linear order in Φ, we may integrate along the unperturbed trajectory (this is in fact assuming that the lengthscale
of variation of Φ is longer than the characteristic deviation of the trajectory, or, far from the source, that ∆b/b 1).
We then approximate x ≈ b, and obtain the following deflection angle:
Z ∞ Z ∞
b p du p
∆p̂ ≈ −2 dz √ Φ0 z 2 + b2 = −2b √ Φ0 (b 1 + u2 ), [u ≡ z/b]. (14)
2
b +z 2 1+u 2
−∞ −∞
Let’s first apply this to a point mass: Φ = −M/r. We then get
M ∞
Z
du M
|∆p̂| ≈ 2 =4 . (15)
b −∞ (1 + u2 )3/2 b
Let us now apply this to a light ray grazing the surface of the Sun, so that b = R . Outside the Sun, we have
Φ(r) = Φ(R )R /r, so the integral is identical, with the substitution M → R Φ(R ):
M
|∆p̂| ≈ 4Φ(R ) ∼ 4 ∼ 10−5 rad. (16)
R
The exact evaluation of the deflection of light rays grazing the Sun’s surface gives |∆p̂| ≈ 1.75 arcseconds. Measuring
this deflection was one of the first succesful experimental tests of GR.
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FIG. 1. Geometry of the problem of light deflection by a mass
Let us now check under which condition it is self-consistent to integrate along the unperturbed trajectory. Define
the distances DLS , DS and DL , as shown in Fig. 1, and the angles α and β. In the small-angle limit (which is the
limit self-consistent with Φ 1), we have
|∆p̂|DLS M DLS
α≈ =4 , (17)
DS DS b
therefore, the fractional change in impact parameter is of order
∆b αDL M DLS DL
≈ ≈4 . (18)
b b DS b2
If we apply this to light rays from distant stars grazing the Suns’s surface, we have b = R ≈ 2 second, M = M ≈
5 × 10−5 second, DLS ≈ DS DL ≈ 1 AU ≈ 8 minutes ≈ 500 seconds. We thus have
∆b 5 × 10−5 × 500
∼4× = 2.5 × 10−2 1. (19)
b 22
Hence we were justified in computing the integral along the unperturbed trajectory.
SHAPIRO TIME DELAY
p| is constant along null geodescis, to first order in Φ. We also saw that p0 = (1 − 2Φ)|~
We just saw that |~ p| to first
order in Φ. Consider two points with coordinate separation |∆~x| = DS . The coordinate time for a light signal to
travel between them is
dx0
Z Z Z Z Z
∆t = dλ = dλ p0 = dλ(1 − 2Φ)|~
p| = d` (1 − 2Φ) = `traj − 2 d` Φ, (20)
traj dλ traj traj traj traj
p
p| = d`/dλ, where d` ≡ δij dxi dxj . The distance along the trajectory, `traj , is slightly larger than
where we used |~
the unperturbed separation DS . This difference is a geometric time delay. We can estimate this by approximating
photon trajectories as straight lines, as shown in the figure. We saw that change in impact parameter is of order
4
p p
∆b ≈ 4M DLS DL /(DS b). The distance travelled is then `traj ≈ 2 + ∆b2 +
DLS DL2 + ∆b2 . Taylor expanding, we
find that the geometric time delay is approximately
1 ∆b2 1 ∆b2 M 2 DLS DL
∆tgeom = `traj − DS ≈ + =8 . (21)
2 DLS 2 DL DS b2
R
In addition, the second term −2 d` Φ is the Shapiro time delay – it is positive (hence delay) because Φ < 0.
Let us estimate it by computing the integral along the unperturbed trajectory. Place the origin of coordinates at the
lens, and. Assume the source is at coordinate z = −DLS and the observer at coordinate z = DL , we then have
Z Z DS
dz DLS DL
∆tShapiro = −2 dλ Φ ≈ 2M √ = 2M [arcsinh(DS /b) − arcsinh(−DLS /b)] ≈ 2M log 4 ,
−DLS b2 + z 2 b2
(22)
where we assumed b DLS , DL , and again integrated along the unperturbed trajectory. Therefore we have
∆tgeom M DLS DL ∆b
∼ ∼ . (23)
∆tShapiro DS b2 b
The Shapiro time delay was measured by bouncing off radio signals from the surface of solar system planets (Mercury,
Mars, Venus). In this case, M = M ∼ 10−5 sec ∼ 105 cm, and characteristic distances are DL ∼ DS ∼ DLS ∼ AU
∼ 1013 cm, and, at closest, b ∼ R ∼ 1011 cm, so that the Shapiro delay (which dominates over the geometric delay)
is of order ∼ 10−4 seconds.
