5.5 Geodesics as Extremum Curves
71
Since F is a function of only T the derivative dF/dT is a function of only T , and
since T is a constant along the curve C the function dF/dT is also a constant, so we
may factor it out of (5.34). This leaves
d
ds
∂ T
∂ ˙
x α
−
∂ T
∂ x α = 0.
(5.36)
That is T must obey the Euler-Lagrange equations on the extremum curve. We have
thus shown algebraically that T and any monotonic function of it lead to the same
extremum curve (one might expect this intuitively). This means we may study the
extremum problem using not the square root of T but T itself, which is often easier.
Now we need to find the extremum of the quantity S and show that the extremum
curve is the same as the geodesic curve previously defined as self-parallel. As above
we choose the curve parameter to be the arc length, so the quantity to be extremized
is
S =
f
i
T ds =
f
i
g αβ ˙
x
α
˙
x
β ds.
(5.37)
The Euler-Lagrange equations are obtained as follows
∂ T
∂ ˙
x λ = 2g λα ˙
x
α
,
d
ds
∂ T
∂ ˙
x λ = 2g αλ ¨
x
α
+ 2g αλ,ρ ˙
x
α
˙
x
ρ
,
∂ T
∂ x λ = g αβ,λ ˙
x
α
˙
x
β
,
g αλ ¨
x
α
+ g αλ,β ˙
x
α
˙
x
β
−
1
2
g αβ,λ ˙
x
α
˙
x
β
= 0.
(5.38)
Next, we multiply through by g
μλ and juggle indices using the symmetry of the
metric, to obtain
¨
x
μ
+
1
2
g
μλ
(g λβ,α + g αλ,β − g αβ,λ ) ˙
x
α
˙
x
β
= 0, ¨
x
μ
+
μ
αβ ˙
x
α
˙
x
β
= 0.
(5.39)
Thus the extremum curve is a geodesic since it satisfies the same differential equation
(5.28) as we previously obtained for the geodesic.
Several features of this result are worth noting. We specialized to the arc length
as the curve parameter, but it is evident from the geodesic equation that any constant
multiple of the arc length will give the same equation, that is d p proportional to ds.
Also it is obvious that our approach cannot be used if the geodesic is a null curve,
one along which the line element is zero, ds = 0. We will consider this special case
later. Finally note that in Euclidean space the interesting geodesics are the shortest
curves between points, while in the Minkowski space of special relativity they are
the longest curves between points, as we discussed in Sect. 3.4.
Example 5.2 It is illustrative to study a simple example of the extremum
approach—to get the geodesics in Euclidean 2-space. Using polar coordinates
71
Since F is a function of only T the derivative dF/dT is a function of only T , and
since T is a constant along the curve C the function dF/dT is also a constant, so we
may factor it out of (5.34). This leaves
d
ds
∂ T
∂ ˙
x α
−
∂ T
∂ x α = 0.
(5.36)
That is T must obey the Euler-Lagrange equations on the extremum curve. We have
thus shown algebraically that T and any monotonic function of it lead to the same
extremum curve (one might expect this intuitively). This means we may study the
extremum problem using not the square root of T but T itself, which is often easier.
Now we need to find the extremum of the quantity S and show that the extremum
curve is the same as the geodesic curve previously defined as self-parallel. As above
we choose the curve parameter to be the arc length, so the quantity to be extremized
is
S =
f
i
T ds =
f
i
g αβ ˙
x
α
˙
x
β ds.
(5.37)
The Euler-Lagrange equations are obtained as follows
∂ T
∂ ˙
x λ = 2g λα ˙
x
α
,
d
ds
∂ T
∂ ˙
x λ = 2g αλ ¨
x
α
+ 2g αλ,ρ ˙
x
α
˙
x
ρ
,
∂ T
∂ x λ = g αβ,λ ˙
x
α
˙
x
β
,
g αλ ¨
x
α
+ g αλ,β ˙
x
α
˙
x
β
−
1
2
g αβ,λ ˙
x
α
˙
x
β
= 0.
(5.38)
Next, we multiply through by g
μλ and juggle indices using the symmetry of the
metric, to obtain
¨
x
μ
+
1
2
g
μλ
(g λβ,α + g αλ,β − g αβ,λ ) ˙
x
α
˙
x
β
= 0, ¨
x
μ
+
μ
αβ ˙
x
α
˙
x
β
= 0.
(5.39)
Thus the extremum curve is a geodesic since it satisfies the same differential equation
(5.28) as we previously obtained for the geodesic.
Several features of this result are worth noting. We specialized to the arc length
as the curve parameter, but it is evident from the geodesic equation that any constant
multiple of the arc length will give the same equation, that is d p proportional to ds.
Also it is obvious that our approach cannot be used if the geodesic is a null curve,
one along which the line element is zero, ds = 0. We will consider this special case
later. Finally note that in Euclidean space the interesting geodesics are the shortest
curves between points, while in the Minkowski space of special relativity they are
the longest curves between points, as we discussed in Sect. 3.4.
Example 5.2 It is illustrative to study a simple example of the extremum
approach—to get the geodesics in Euclidean 2-space. Using polar coordinates