Derivation of the Christoffel symbols directly from the geodesic equation
We start by considering the action for a point particle:
S[xσ ]
1
= m
2
Z
dxµ dxν
gµν (xσ )dλ.
dλ dλ
where the orbit is parameterized by λ, and we have stressed that the metric depends on
position. Then:
Z dxν
d
dxµ
dxµ dxν ∂gµν (xσ ) ρ
d
1
σ
µ
σ
ν
gµν (x )
δx +
gµν (x )
δx −
δx dλ
=− m
2
dλ
dλ
dλ
dλ
dλ dλ
∂xρ
Z ν
µ
1
d
dxµ dxν ∂gµν (xσ ) ρ
d
σ dx
ρ
σ dx
ρ
=− m
gρν (x )
δx +
gµρ (x )
δx −
δx dλ
2
dλ
dλ
dλ
dλ
dλ dλ
∂xρ
Z 2 ν
∂gρν (xσ ) ∂gµρ (xσ ) ∂gµν (xσ ) dxµ dxν
d 2 xµ
1
d x
σ
σ
gρν (x ) +
gµρ (x ) +
δxρ dλ.
=− m
+
−
2
∂xµ
∂xν
∂xρ
dλ dλ
dλ2
dλ2
δS[xσ ]
Thus, the equation of motion is:
2 ν
∂gρν (xσ ) ∂gµρ (xσ ) ∂gµν (xσ ) dxµ dxν
d 2 xµ
d x
1
σ
σ
gρν (x ) +
gµρ (x ) +
= 0.
m
+
−
2
∂xµ
∂xν
∂xρ
dλ dλ
dλ2
dλ2
Multiplying by g τ ρ (xσ ) and, in the second term, using the fact that gµρ (xσ ) = gρµ (xσ ), we
find:
2 τ
1 τ ρ σ ∂gρν (xσ ) ∂gµρ (xσ ) ∂gµν (xσ ) dxµ dxν
d x
+ g (x )
m
= 0.
+
−
2
∂xµ
∂xν
∂xρ
dλ dλ
dλ2
Now, inside the large square brackets, the second term can be identified as:
dx
Γτµν (xσ )
µ dxν
dλ dλ
,
since
Γτµν (xσ )
1 τ ρ σ ∂gρν (xσ ) ∂gµρ (xσ ) ∂gµν (xσ )
= g (x )
+
−
,
2
∂xµ
∂xν
∂xρ
while first term inside the large square brackets can be rewritten as:
dxµ ∂
dλ ∂xµ
dxτ
dλ
.
Together these can be arranged as:
τ
τ
dxµ ∂
dxν
dxµ
dx
dx
τ
σ
+ Γµν (x )
=
,
∇µ
µ
dλ ∂x
dλ
dλ
dλ
dλ
since the two terms in the large square brackets represent the covariant derivative of the vector
dxτ/dλ. This variational problem gives the easiest way to find all the Γτµν (xσ ) symbols.