The geodesic equation as Newton’s Law of Universal Gravitation
February 22, 2015
1
The geodesic equation as Newton’s second law
We have seen that curves of extremal proper length or time satisfy the geodesic equation,
d2 xα
= −Γαµβ uµ uβ
dτ 2
where uα =
dxα
dτ
(1)
is the 4-velocity. Since the acceleration may be written using the chain rule as
d2 xα
dτ 2
duα
dτ
dxβ ∂uα
=
dτ ∂xβ
∂uα
= uβ β
∂x
=
we may write the geodesic equation as
0
∂uα
+ Γαµβ uµ uβ
β
∂x
α
∂u
µ α
+
u
Γ
= uβ
µβ
∂xβ
= uβ
= uβ Dβ uα
This describes an autoparallel : the parallel transport of the vector uα along its own direction.
In the form of eq.(1), the geodesic equation gives the acceleration of a particle in terms of geometric
quantities. This suggests that it might be possible to explain some force using the connection. If we multiply
the geodesic equation by the mass of a particle, the left side becomes the proper time rate of change of the
momentum,
dpα
= −mΓαµβ uµ uβ
dτ
where pα = muα . The question is, can we make the connection term look like a force?
2
Nearly flat space
Suppose the metric is nearly that of a flat space, differing

φ
 0
gµν = ηµν + 
 0
0
1
only in the time component,

0 0 0
0 0 0 

0 0 0 
0 0 0

−c2 + φ 0 0 0

0
1 0 0 

= 

0
0 1 0 
0
0 0 1

p
and suppose the function φ depends only on r = x2 + y 2 + z 2 . Then the interval of proper time is
1
1
2
dτ = 1 − 2 φ dt2 − 2 dx2 + dy 2 + dz 2
c
c
We would like to find the connection, which is built
derivatives of the metric are

φ,i 0
 0 0
g00,i = 
 0 0
0 0
from derivatives of gµν . The only nonvanishing

0 0
0 0 

0 0 
0 0
where i = 1, 2, 3 and
φ,i
dφ ∂r
dr ∂xi
dφ xi
dr r
=
=
This means the only nonvanishing connection components must have two time indices, 0, and one spatial
one, i:
1
(g00,i + g0i,0 − g0i,0 )
2
1 dφ xi
=
2 dr r
1
(gi0,0 + gi0,0 − g00,i )
=
2
1 dφ xi
= −
2 dr r
Γ00i = Γ0i0
=
Γi00
Raising the first index on each, we have
Γ00i = Γ0i0
= g 0α Γαi0
= g 00 Γ0i0
1
dφ xi
= −
2 (1 − φ) dr r
and
Γi00
= g iα Γα00
= g ij Γj00
1 dφ xi
2 dr r
=
Now return to the geodesic equation. Writing the 4-velocity as uα = γ c, v i and computing each
component of the equation separately, we have
dp0
dτ
= −mΓ0µβ uµ uβ
2
= −mΓ000 u0 u0 − mΓ00i u0 ui − mΓ0i0 ui u0 − mΓ0ij ui uj
= −2mΓ00i u0 ui
1
dφ xi
= −2m −
γ 2 cv i
2 (1 − φ) dr r
mc
dφ
=
γ 2 v i xi
r (1 − φ) dr
for the time component and
dpi
dτ
= −mΓiµβ uµ uβ
= −mΓi00 u0 u0 − mΓi0j u0 uj − mΓij0 uj u0 − mΓijk uj uk
= −mΓi00 u0 u0
1
dφ xi
= − mγ 2 c2
2
dr r
2
We consider the case of non-relativistic velocities, so that we may set γ ≈ 1. Then, noting that xr is just
a unit vector in the radial direction, we can choose φ so that this last expression looks like the Newton’s
gravitational force,
1
dφ
− mc2 r̂
2
dr
dφ
dr
GM m
r̂
r2
2GM
r2 c2
= −
=
Integrating,
φ=−
Notice that
2GM
r
2GM
rc2
is just the square of the classical escape velocity,
so the magnitude of φ is very small,
2
vescape
c2
1
mv 2
2
=
v2
=
GM m
r
2GM
r
1 for planets and ordinary stars. With this choice for φ, the
2
vescape
,
c2
metric differs from flat space by only a term of order
dp
dt
−
=
and we reproduce Newton’s law of gravity,
GM m
r̂
r2
at normal velocities v c.
Now consider the time component of the geodesic equation. Substituting this result for φ into the 0
component of the geodesic equation,
dp0
dτ
Since p0 =
E
c
=
1
1+
2GM
rc2
2GM m
v · r̂
r2 c
and τ ≈ t,
dE
dt
=
1
1+
2GM
rc2
3
2GM m
v · r̂
r2
=
1
2
vescape
c2
2GM m
v · r̂
r2
1+
2GM m
≈
r̂ · v
r2
= 2FN ewton · v
This differs from the actual rate of change of energy of a falling particle by a factor of 2. This factor is
eliminated when we use a more realistic metric.
Exercise: Repeat these arguments using the metric

−c2 + φ
0

0
1
+
λφ
gµν = 

0
0
0
0

0
0

0
0


1 + λφ
0
0
1 + λφ
where λ is constant. Can you choose λ to get the correct expression for the change in energy?
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