3Geodesics

III Riemannian Geometry 3.5 Variations of arc length and energy
This section is mostly a huge computation. As we previously saw, geodesics are
locally length-minimizing, and we shall see that another quantity, namely the
energy is also a useful thing to consider, as minimizing the energy also forces
the parametrization to be constant speed.
To make good use of these properties of geodesics, it is helpful to compute
explicitly expressions for how length and energy change along variations. The
computations are largely uninteresting, but it will pay off.
Definition (Energy). The energy function E : Ω(p, q) R is given by
E(γ) =
1
2
Z
T
0
|˙γ|
2
dt,
where γ : [0, T ] M.
Recall that Ω(
p, q
) is defined as the space of piecewise
C
1
curves. Often, we
will make the simplifying assumption that all curves are in fact
C
1
. It doesn’t
really matter.
Note that the length of a curve is independent of parametrization. Thus,
if we are interested in critical points, then the critical points cannot possibly
be isolated, as we can just re-parametrize to get a nearby path with the same
length. On the other hand, the energy
E
does depend on parametrization. This
does have isolated critical points, which is technically very convenient.
Proposition.
Let
γ
0
: [0
, T
]
M
be a path from
p
to
q
such that for all
γ
Ω(
p, q
) with
γ
: [0
, T
]
M
, we have
E
(
γ
)
E
(
γ
0
). Then
γ
0
must be a
geodesic.
proof is just the application of Cauchy-Schwartz.
Proof. By the Cauchy-Schwartz inequality, we have
Z
T
0
|˙γ|
2
dt
Z
T
0
|˙γ(t)| dt
!
2
with equality iff |˙γ| is constant. In other words,
E(γ)
`(γ)
2
2T
.
So we know that if
γ
0
minimizes energy, then it must be constant speed. Now
given any
γ
, if we just care about its length, then we may wlog it is constant
speed, and then
`(γ) =
p
2E(γ)T
p
2E(γ
0
)T = `(γ
0
).
So γ
0
minimizes length, and thus γ
0
is a geodesic.
We shall consider smooth variations
H
(
t, s
) of
γ
0
(
t
) =
H
(
t,
0). We require
that
H
: [0
, T
]
×
(
ε, ε
)
M
is smooth. Since we are mostly just interested
in what happens “near”
s
= 0, it is often convenient to just consider the
corresponding vector field along γ:
Y (t) =
H
s
s=0
= (dH)
(t,0)
s
,
Conversely, given any such vector field
Y
, we can generate a variation
H
that
gives rise to Y . For example, we can put
H(t, s) = exp
γ
0
(t)
(sY (t)),
which is valid on some neighbourhood of [0
, T
]
× {
0
}
. If
Y
(0) = 0 =
Y
(
T
), then
we can choose H fixing end-points of γ
0
.
Theorem (First variation formula).
(i) For any variation H of γ, we have
d
ds
E(γ
s
)
s=0
= g(Y (t), ˙γ(t))|
T
0
Z
T
0
g
Y (t),
dt
˙γ(t)
dt. ()
(ii) The critical points, i.e. the γ such that
d
ds
E(γ
s
)
s=0
for all (end-point fixing) variation H of γ, are geodesics.
(iii) If |˙γ
s
(t)| is constant for each fixed s (ε, ε), and |˙γ(t)| 1, then
d
ds
E(γ
s
)
s=0
=
d
ds
`(γ
s
)
s=0
(iv)
If
γ
is a critical point of the length, then it must be a reparametrization of
a geodesic.
This is just some calculations.
Proof.
We will assume that we can treat
s
and
t
as vector fields on an
embedded submanifold, even though H is not necessarily a local embedding.
The result can be proved without this assumption, but will require more
technical work.
(i) We have
1
2
s
g( ˙γ
s
(t), ˙γ
s
(t)) = g
ds
˙γ
s
(t), ˙γ
s
(t)
= g
dt
H
s
(t, s),
H
t
(t, s)
=
t
g
H
s
,
H
t
g
H
s
,
dt
H
t
.
Comparing with what we want to prove, we see that we get what we want
by integrating
R
T
0
dt, and then putting s = 0, and then noting that
H
s
s=0
= Y,
H
t
s=0
= ˙γ.
(ii) If γ is a geodesic, then
dt
˙γ(t) = 0.
So the integral on the right hand side of (
) vanishes. Also, we have
Y (0) = 0 = Y (T ). So the RHS vanishes.
Conversely, suppose γ is a critical point for E. Then choose H with
Y (t) = f(t)
dt
˙γ(t)
for some f C
[0, T ] such that f(0) = f(T ) = 0. Then we know
Z
T
0
f(t)
dt
˙γ(t)
2
dt = 0,
and this is true for all f. So we know
dt
˙γ = 0.
(iii)
This is evident from the previous proposition. Indeed, we fix [0
, T
], then
for all H, we have
E(γ
s
) =
`(γ
s
)
2
2T
,
and so
d
ds
E(γ
s
)
s=0
=
1
T
`(γ
s
)
d
ds
`(γ
s
)
s=0
,
and when s = 0, the curve is parametrized by arc-length, so `(γ
s
) = T .
(iv)
By reparametrization, we may wlog
|˙γ|
1. Then
γ
is a critical point for
`, hence for E, hence a geodesic.
Often, we are interested in more than just whether the curve is a critical
point. We want to know if it maximizes or minimizes energy. Then we need
more than the “first derivative”. We need the “second derivative” as well.
Theorem
(Second variation formula)
.
Let
γ
(
t
) : [0
, T
]
M
be a geodesic with
|˙γ| = 1. Let H(t, s) be a variation of γ. Let
Y (t, s) =
H
s
(t, s) = (dH)
(t,s)
s
.
Then
(i) We have
d
2
ds
2
E(γ
s
)
s=0
= g
Y
ds
(t, 0), ˙γ
T
0
+
Z
T
0
(|Y
0
|
2
R(Y, ˙γ, Y, ˙γ)) dt.
(ii) Also
d
2
ds
2
`(γ
s
)
s=0
= g
Y
ds
(t, 0), ˙γ(t)
T
0
+
Z
T
0
|Y
0
|
2
R(Y, ˙γ, Y, ˙γ) g( ˙γ, Y
0
)
2
dt,
where R is the (4, 0) curvature tensor, and
Y
0
(t) =
Y
dt
(t, 0).
Putting
Y
n
= Y g(Y, ˙γ) ˙γ
for the normal component of Y , we can write this as
d
2
ds
2
`(γ
s
)
s=0
= g
Y
n
ds
(t, 0), ˙γ(t)
T
0
+
Z
T
0
|Y
0
n
|
2
R(Y
n
, ˙γ, Y
n
, ˙γ)
dt.
Note that if we have fixed end points, then the first terms in the variation
formulae vanish.
Proof. We use
d
ds
E(γ
s
) = g(Y (t, s), ˙γ
s
(t))|
t=T
t=0
Z
T
0
g
Y (t, s),
dt
˙γ
s
(t)
dt.
Taking the derivative with respect to s again gives
d
2
ds
2
E(γ
s
) = g
Y
ds
, ˙γ
T
t=0
+ g
Y,
ds
˙γ
s
T
t=0
Z
T
0
g
Y
ds
,
dt
˙γ
s
+ g
Y,
ds
dt
˙γ

dt.
We now use that
ds
dt
˙γ
s
(t) =
dt
ds
˙γ
s
(t) + R
H
s
,
H
t
˙γ
s
=
dt
2
Y (t, s) + R
H
s
,
H
t
˙γ
s
.
We now set s = 0, and then the above gives
d
2
ds
2
E(γ
s
)
s=0
= g
Y
ds
, ˙γ
T
0
+ g
Y,
˙γ
ds
T
0
Z
T
0
"
g
Y,
dt
2
Y
!
+ R( ˙γ, Y, ˙γ, Y )
#
dt.
Finally, applying integration by parts, we can write
Z
T
0
g
Y,
dt
2
Y
!
dt = g
Y,
dt
Y
T
0
+
Z
T
0
Y
dt
2
dt.
Finally, noting that
ds
˙γ(s) =
dt
Y (t, s),
we find that
d
2
ds
2
E(γ
s
)
s=0
= g
Y
ds
, ˙γ
T
0
+
Z
T
0
|Y
0
|
2
R(Y, ˙γ, Y, ˙γ)
dt.
It remains to prove the second variation of length. We first differentiate
d
ds
`(γ
s
) =
Z
T
0
1
2
p
g( ˙γ
s
, ˙γ
s
)
s
g( ˙γ
s
, ˙γ
s
) dt.
Then the second derivative gives
d
2
ds
2
`(γ
s
)
s=0
=
Z
T
0
"
1
2
2
s
2
g( ˙γ
s
, ˙γ
s
)
s=0
1
4
s
g( ˙γ
s
, ˙γ
s
)
2
s=0
#
dt,
where we used the fact that g( ˙γ, ˙γ) = 1.
We notice that the first term can be identified with the derivative of the
energy function. So we have
d
2
ds
2
`(γ
s
)
s=0
=
d
2
ds
2
E(γ
s
)
s=0
Z
T
0
g
˙γ
s
,
ds
˙γ
s
s=0
2
dt.
So the second part follows from the first.