763695S GENERAL RELATIVITY
Exercise 1
Autumn 2016
1. Newton’s law of gravity as a field
According to Newton’s law of gravity, the force on a body 1 (mass m1 ) at r1 caused
by a second body (mass m2 ) at r2 is
F = −G
m1 m2
r̂.
r2
(1)
where r = r1 − r2 and G is the constant of gravity. In this exercise we try to show
that this implies that the acceleration a of a test particle at r in the presence of an
arbitrary mass distribution with density ρ(r) is given by
a(r) = −∇V (r)
(2)
where the potential V satisfies the Poisson equation
∇2 V (r) = 4πGρ(r).
(3)
a) Based on (1) and (2) show that the potential caused by one point particle of
mass M at the origin corresponds to the potential
V (r) = −G
M
.
r
(4)
b) Show that the potential (4) satisfies the Poisson equation for r 6= 0.
c) By integrating the Poisson equation over a small sphere of radius around the
origin, and transforming the left hand side to a surface integral, show that the
potential (4) satisfies the Poisson equation also at r = 0.
d) Based on the above, try to justify the claim in the introduction for an arbitrary
mass distribution (not just one point mass).
You can use the following formulas appropriate for spherical coordinates (r, θ, φ):
∇Φ = r̂
∂Φ
1 ∂Φ
1 ∂Φ
+ θ̂
+ φ̂
∂r
r ∂θ
r sin θ ∂φ
1 ∂
∂Φ
1
∂
∂Φ
1
∂ 2Φ
r2
+ 2
sin θ
+ 2 2
∇ Φ= 2
r ∂r
∂r
r sin θ ∂θ
∂θ
r sin θ ∂φ2
!
2
(5)
!
(6)
The motivation for this exercise is that equation (3) has some similarity with the
corresponding equation in general relativity, that will be discussed later in this
course.
Solution
a) The force on a test particle (mass m) at r caused by the particle (mass M ) at
the origin is
F = −G
mM
r̂.
r2
(7)
Because F = ma, the acceleration a of the test particle is given by
a=−
GM
r̂.
r2
(8)
The corresponding potential V is obtained from (2):
∇V (r) =
1 ∂V
1 ∂V
GM
∂V
GM
+ θ̂
+ φ̂
= 2 r̂,
r̂ ⇒ r̂
2
r
∂r
r ∂θ
r sin θ ∂φ
r
where the angular derivatives are equal to 0 due to the spherical symmetry of the
potential. V (r) is then obtained as follows:
Z V (r)
Z r
GM
GM 0
dV
M
= 2 ⇒
.
dV =
dr
⇒
V
(r)
=
−G
dr
r
r
0
∞ (r 0 )2
b) Substituting (4) to the left-hand side of the Poisson equation (3) and noting that
the angular derivatives are equal to 0 again, one obtains
"
#
1 ∂
M
∂
∇ V (r) = 2
r2
−G
r ∂r
∂r
r
2
=
1 ∂
(GM ) = 0.
r2 ∂r
At r 6= 0, the density ρ(r) = 0 and the right-hand side of (3) also becomes 0. The
potential (4) thus satisfies the Poisson equation at r 6= 0.
c) Gauss’s theorem relates the volume integral of the divergence of a vector field f
to the surface integral of f :
Z
V
(∇ · f )dV =
I
S
f · dS.
(9)
Applying (9), the left-hand side of (3) can be integrated over a small sphere of radius
and volume V :
Z
V
∇2 V (r)dV =
=
Z
Ω
Z
V
∇ · ∇V (r)dV =
I
S
∇V (r) · dS
Z
GM
2
r̂ · r dΩr̂ = GM dΩ = 4πGM.
r2
Ω
The integration over the right-hand side of (3) becomes
4πG
Z
ρ(r)dV = 4πGM,
V
as the volume integral gives the mass of the point particle. Because both the leftand right-hand side of the Poisson equation are equal to 4πGM , the equation is
satisfied for a small sphere of radius . Thus, (4) also satisfies the Poisson equation
when → 0.
d) A simple and sufficient answer is as follows. An arbitrary mass distribution can
be expressed as a sum of mass points with masses mk located at points rk . Each of
them separately produces the potential
Vk (r) = −G
mk
.
|r − rk |
(10)
Because (3) is linear, the total potential that solves this equation is the sum of the
potentials coming from the mass points,
V (r) =
X
k
Vk (r) = −G
X
k
mk
.
|r − rk |
(11)
A more formal version of the same answer uses the delta function. An arbitrary
mass distribution can be expressed as a sum of mass points with masses mk located
at points rk . This means writing the density as
ρ(r) =
X
k
mk δ(r − rk )
(12)
where δ(r) is the delta function that has the properties
δ(r) = 0 everywhere except at r = 0,
Z
d3 r δ(r) = 1.
(13)
Here d3 r means integration over the whole three-dimensional space. The delta
function can be depicted as a function that is zero everywhere else except in a small
region around the origin (r < ), where it is so large (if constant for r < , its
value is 3/4π3 ) that when integrated over that region, the integral becomes unity.
In items b) and c) we in fact showed that (4) is the solution of the equation
R
∇2 V (r) = 4πGM δ(r).
(14)
Using this one can then see that (10) is the solution of
∇2 Vk (r) = 4πGmk δ(r − rk ).
(15)
Substituting (11) to the left hand sides (3), and using (15) and (12) shows that (11)
indeed is the solution of the equation (3).
Equation (3) is the field equation of Newtonian gravity: it determines the potential
V (r) for a given mass distribution ρ(r). Later in this course we will derive Einstein
field equations that generalize the field equation (3) to general relativity. We will
derive that under certain conditions the more general equations reduce to (3).
2. Four vector manipulation in special relativity
Consider the four-vectors
λµ = (2, 1, 1, 0) and σ α = (1, 3, 0, 0).
(16)
a) Calculate λµ , σν , λµ λµ , σ α σα , σ ν λν .
b) Draw a sketch of the four vector λµ (16) in coordinate axes (λ1 , λ2 , λ0 ), where
the λ0 axis is drawn vertical. Can λµ represent a difference of events for the same
particle? Do the same for σ µ . Sketch also the surface λµ λµ = 0 (known as the light
cone).
0
0
c) Consider the Lorentz-transformation λµ = Λµν λν , where

0



[Λµν ] = 
γ
− vc γ
v
−cγ
γ
0
0
0
0
0
0
1
0
0
0
0
1





(17)
q
and γ = 1/ 1 − v 2 /c2 . Show that when this is applied to event four vector xµ =
(ct, x, y, z), it is equivalent to the familiar form of the Lorentz transformation
x0 = q
x − vt
1 − v 2 /c2
y0 = y
z0 = z
t − (v/c2 )x
t0 = q
,
1 − v 2 /c2
(18)
0
0
0
d) Using (16) and Lorentz transformation with v = c/2, calculate λµ , σ µ and λµ σµ0 .
Compare the last one with σ ν λν .
Solution
a)




λµ = ηµν λν = 
1 0
0
0
0 −1 0
0
0 0 −1 0
0 0
0 −1





2
1
1
0





= (2, −1, −1, 0),
σν = ηνα σ α = (1, −3, 0, 0),
λµ λµ = ηµν λν λµ = 22 − 12 − 12 − 02 = 2,
σ α σα = σα σ α = ηαν σ ν σ α = 12 − 32 − 02 − 02 = −8,
σ ν λν = σν λν = ηνµ σ µ λν = 1 · 2 − 3 · 1 − 0 · 1 − 0 · 0 = −1.
b) Figure 1 shows the vectors λµ and σ µ plotted with the lightcone xµ xµ = 0. The
four-vector λµ can represent a difference of events for the same particle because it is
inside the light cone. σ µ is outside the light cone and cannot represent events of the
3
λµ
2
x0
σµ
1
0 −3
−2
−1
x
1
0
1
2
0
−1
−2
3 −3
1
2
3
x2
Kuva 1: Vectors λµ = (2, 1, 1, 0) and σ µ = (1, 3, 0, 0) plotted with the lightcone xµ xµ = 0.
same particle. This was also shown in a), where it was calculated that λµ λµ = 2 > 0,
meaning that the vector is timelike, and σ α σα = −8 < 0, which indicates that the
vector is spacelike.
c) The familiar form of the Lorentz transformation follows directly from
ct
γ
− vc γ 0 0
v
 x 

0
0
γ
γ
0
0
−





xµ = Λµν xν =  c
 0
0
1 0  y 
z
0
0
0 1
0
0 0 0
= (γct − γ(v/c)x, −γvt + γx, y, z) = (ct , x , y , z ).



d) In a system
√ of units where c = 1, the velocity and Lorentz factor are v = 1/2
and γ = 2/ 3.

√
√

2/ √3 −1/√ 3 0 0
2


 −1/ 3 2/ 3 0 0  
0
0
1 



λµ = Λµν λν = 

 1 

0
0
1
0


0
0
0
0 1
√
√
√
√
√
= (2 · 2/ 3 − 1 · 1/ 3, −2 · 1/ 3 + 1 · 2/ 3, 1 · 1, 0 · 1) = ( 3, 0, 1, 0),
√
√
0
0
σ µ = Λµν σ ν = (−1/ 3, 5/ 3, 0, 0),
0
0
0
0
λµ σµ0 = λµ0 σ µ = ηµ0 ν 0 λν σ µ = −1.
The last result is equal to σ ν λν , which was expected due to the invariance of the
inner product.
