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Geodesics and Length
Adrian Down
October 17, 2006
1 Review
Today we will finish our discussion of geodesics. We will prove an important
theorem which states that any curve that minimizes the distance between
two points on a surface is a geodesic.
1.1 Tangent vectors
Recall the previous formulae,
gij = 〈xi, xj 〉 Lij = 〈xij , n〉 Γ k ij =^
l
g kl 〈xij , xl〉
We derived an intrinsic formula for the Christoffel symbols Γ (^) ijk ,
k ij =
l
g kl
∂gil
∂uj^
∂gjl
∂ui^
∂gij
∂ul
Note. By the way in which it is defined, the Christoffel symbols are invariant
under permutation of the lower indices,
k ij = Γ^
k ji
This symmetry can be used as an aid to remember the intrinsic formula for
Γ (^) ijk above.
In general, the derivatives of the tangent vectors, xij , are neither normal
nor tangent to the surface. Gauss’s theorem shows that these vectors can be
decomposed into a component in the tangent plane and a component normal
to the surface,
xij = Lij n +
k
k ij xk
1.2 Curves on a surface
Previously, we discussed unit speed curves of the form
γ = x(γ 1 (s), γ 2 (s))
We can construct the Frenet-Serret apparatus of a curve on a surface as with
any other curve studied previously. From this Frenet-Serret apparatus, we
defined the particular vector S = n × T. With this definition, the second
derivative of the curve γ can be written in terms of a normal component and
a component in the direction of S. The coefficients of these components are
defined to be the normal and geodesic curvatures,
γ′′^ =
Lij
γi
γj^
κn
n +
γk
Γ (^) ijk (γi)′(γj^ )′
xk
︸ ︷︷ ︸ κg S
1.3 Geodesics
Geodesics are the analog of straight lines on an arbitrary surface. A curve γ
is a geodesic if κg ≡ 0 everywhere on the curve. Using Gauss’s formula above,
we saw that a curve γ is a geodesic if and only if γ satisfies the differential
equation,
( γ k)′′^
ijk
k ij (γ
i )
γ j )′^ = 0 k = 1, 2
The summation in these differential equations can be expanded,
( γ 1
1 11
γ 1
γ 1
1 12
γ 1
γ 2
1 22
γ 2
γ 2
γ 2
2 11
γ 1
γ 1
2 12
γ 1
γ 2
2 22
γ 2
γ 2
A solution to this differential equation is guaranteed by a basic existence
theorem for ordinary differential equations, which we will not pursue here.
The theorem states that we can uniquely solve this differential equation for
− < 0 < for some > 0.
Note. The previous theorem that we used required the Lipshitz condition
to be satisfied. In this case, there is no Lipshitz condition because we are
finding a solution only for a small time interval.
2.4 γ remains unit speed
Although we have claimed that we can solve the differential equation describ-
ing the geodesic, we must further ensure that the resulting curve remains unit
speed at all times. We will see that this fact is built in to the apparatus of
the curve.
Note. There is also a possibility that the quadratic term in the differential
equation could be problematic. Considering the simple differential equation
x˙ = x^2 with x(0) = x 0 > 0 shows that the solution to a quadratic differential
equation can diverge in finite time. Our solution will show that quadratic
terms are not problematic in this case.
To show that γ remains unit speed at all time, we need to prove that
|γ ′ (s) = 1| ∀ − < s <
If this can be shown, then γ(s) remains unit speed at all times.
We assume that the curve γ is unit speed at t = 0,
|γ ′ (0)| = |X| = 1
As per usual, to show that |γ′| = 1 at subsequent times, differentiate the
length of γ and show that it is equal to 0 to prove that the length of γ is
constant. To this end, define f (s) = |γ(s)|^2.
Note. When differentiating lengths, it is almost always preferable to differ-
entiate the length squared rather than the length itself.
f (s) can be written in component terms,
γ(s) = x
γ 1 (s), γ 2 (s)
γ ′ (s) =
γ i
xi
|γ ′ (s)| 2 = 〈γ ′ (s), γ ′ (s)〉 =
γ i
γ j
〈xi, xj 〉 ︸ ︷︷ ︸ gij
f (s) = |γ′(s)|^2 =
∑^2
ij
gij
γ^1 , γ^2
γi
γj^
We would like to show that f ′(s) = 0 for − < s < . Differentiating f (s)
componentwise and using the chain rule repeatedly,
f ′ (s) =
ijk
∂gij
∂uk
γ k
γ i
γ j
ij
gij
γ i
γ j
ij
gij
γ i
γ j
Although this equation appears formidable, it can be simplified using the
following identity,
∂gij
∂uk^
l
gilΓ l jk +^ gjkΓ^
l ik
We omit the proof of this identity. With this substitution, f ′(s) becomes,
f ′ (s) =
ijkl
gilΓ l jk
γ k
γ i
γ j
ijkl
gjkΓ l ik
γ k
γ i
γ j
ij
gij
γ i
γ j
ij
gij
γ i
γ j
Since summation indices are arbitrary, the indices in the last two summations
can be redefined to allow for combination with the first two summations
3.3 Create a family of variations
We will now create a family of curves with the same endpoints c and d as
γ in the neighborhood of s 0. Each of these variations has a different length
than initial curve, which we will see will be useful, since we are trying to
study the length of γ. To this end, define the quantity λ, which measure
the extent to which a variation deviates from the initial curve at a particular
value of s,
λ(s) ≡ (s − c)(d − s)κg(s)
Note. By definition, λ(c) = λ(d) = 0
We can write S = n × T in terms of components along a variation,
λ(s)S =
v i (s)xi
Note. As in the case of λ, v^1 (c) = v^2 (c) = v^1 (c) = v^2 (d) = 0.
For small t, define the curve,
α(s, t) = x
γ 1 (s) + tv 1 (s), γ 2 (s) + tv 2 (s)
3.4 Define the length of the variations
We can define the length from C = γ(c) and D = γ(d) along a particular
variation. The variation is parameterized by the variable t,
L(t) ≡
∫ (^) d
c
∂α
∂s
∂α
∂s
〉^1
2 ds
Note. Think of t as measuring the size of the perturbation of the variation
from the original curve γ. t = 0 corresponds to the original curve.
The key point is that L(0) ≤ L(t) for all small t, since the initial curve
γ(s) was assumed to be a curve of shortest length between C and D,
L(0) = length of α(s, 0) = γ(s) between C and D
Thus L(t) has a minimum at t = 0, meaning that L′(t) = 0 at t = 0. This
fact is very important, and contains all of the geometric information that we
would like to extract. We will use that fact that L ′ (0) = 0 to obtain the
desired result that κg(s) ≡ 0 for c < s < d.
3.5 Differentiate the length
To extract this geometric information, we must compute the value of the
derivative of L at t = 0,
d
dt
L(t) =
d
dt
∫ (^) d
c
∂α
∂s
∂α
∂s
〉^1
2 ds
∫ (^) d
c
d
dt
∂α
∂s
∂α
∂s
〉^1
2
ds
∫ (^) d
c
∂α ∂s ,^
∂α ∂s
2
∂^2 α
∂s∂t
∂α
∂s
∫ (^) d
c
∂^2 α ∂s∂t ,^
∂α ∂s
∂α ∂s ,^
∂α ∂s
〉^1
2
Now let t = 0. By definition, α(t = 0) = γ. γ is a unit speed curve, and so
the denominator of the above expression for L ′ (t) is equal to the length of
the tangent vector T, which is equal to 1. Thus,
dL(t)
dt
t=
∫ (^) d
c
∂^2 α
∂s∂t
∂α
∂s
t=
ds
In the next lecture, we will integrate this integrand by parts and relate the
resulting expression to the curvature.
