These notes are set so that you get to prove the main results
by solving smaller problems that when put together give the big result.
The answers to the problems are in the videos.
You will get the most out of these notes if you do (or try) the problems
before looking at the videos.
Here is the problem we wish to understand. Let \( M\) be a surface in
\(\mathbb R^3\) and let \(p\) and \(q\) be points of \(M\). Then we
want a description of the curve in \(M\) of shortest length.
That is for all curves \(\mathbf c\colon [a,b]\to M\) with
\(\mathbf c(a) = p\) and \(\mathbf c(b) = q\) we want the one
which is the shortest. (The more rigorously included among will
point out that we should first prove that such a shortest curve
exists. I am not going to worry about that, but am going to assume
it exists and describe its properties.)
Before starting on this we review some vector calculus that will
be used.
Problem 1:
Let \( \pmb \beta \colon [a,b] \to \mathbb R^3\) be a \( C^1\) curve.
Than at points where \(\pmb \beta(t)\ne \mathbf 0\) show
\[
\frac d{dt} \| \pmb\beta (t) \| =\left( \frac{\pmb\beta(t)}{\|\pmb\beta(t)\|}\right)
\cdot \pmb \beta'(t)=\widehat{\pmb\beta}(t)\cdot \pmb\beta'(t).
\]
where \( \widehat{\pmb\beta}(t)\) is the unit vector in the direction
of \(\pmb \beta(t)\).
Solution:
Let us note for future reference that if we have a function
\( \mathbf f \colon (a,b)\times (c,d) \to \mathbb R^3\) of two variables
with values in \(\mathbb R^3\) that holding \(s\) fixed and letting
and we let
\[
\pmb \beta (t) = \frac{\partial \mathbf f}{\partial s}(s,t) = \mathbf f_s(s,t)
\]
\( \pmb \beta(t) = \mathbf f(s,t)\) then Problem 1 gives us that
\[
\frac{\partial}{\partial t} \| \mathbf f_s\| = \left(\frac{\mathbf f_s}{\| \mathbf f_s\|}\right)
\cdot \mathbf f_{st}
\]
as in this case \(\pmb \beta'(s)= \mathbf f_{st}\).
Now assume that \(\mathbf c \colon [a,b] \to M \) is a \(C^2\) curve in our surface
\(M\) such that it has the shortest length of all the curves that connect \(
\mathbf c(a)\) and \(\mathbf c(b)\). We can parameterize \(\mathbf c\) by
arclength and thus assume it is unit speed. That is \(\| \mathbf c'(s)\|=1 \)
for all \(s\). We now look at a bunch of curves that are close to \(\mathbf
c\) and have the same endpoints. An explicit method of doing this is to let \(
\delta \gt 0\) be a small positive number and let \( \mathbf f\colon [a,b]
\times (-\delta,\delta) \to M \) be a function such that
\[
\mathbf f(s,0) = \mathbf c(s) \qquad \text{(that is \(t=0\) gives our minimizing curve.)}
\]
\[
\mathbf f(a,t) = \mathbf c(a) \qquad \text{ for all \(t\in (-\delta,\delta)\)}
\]
\[
\mathbf f(b,t) = \mathbf c(b) \qquad \text{ for all \(t\in (-\delta,\delta)\)}
\]
Such a function is called a variation of \(\mathbf c\) with fixed endpoints
. One way to think of these variations is that
for each fixed \(t\) the curve \(\mathbf c^t(s) = \mathbf f(s,t)\) (we use
superscripts rather than subscripts to avoid confusion with partial
derivatives) is a curve with the same endpoints as \(\mathbf c\)
and with \(\mathbf c^0=\mathbf c\). Thus as \(t\) varies over \((-\delta,
\delta)\) the family of curves \( \{ \mathbf c^t\}_{t\in (-\delta,\delta)}\)
is a bunch of curves with the same endpoints as \(\mathbf c\) and
with \(\mathbf c^0=\mathbf c\).
Problem 2:
Draw some picture of some variations of a curve with fixed endpoints.
For what we are doing probably the best way to do this is to start with
a curve \(\mathbf c\) and then draw the curves \( \mathbf c^t(s) = \mathbf f(s,t)\)
for some values of \(t\).
Solution:
We are assuming that \(\mathbf c\) is the shortest curve between the
points \(\mathbf c(a)\) and \( \mathbf c(b)\). As all the curves
\(\mathbf c^t\) have the same endpoints as \(\mathbf c\) the
function \( L \colon (-\delta,\delta) \to \mathbb R\) given by
\[
L(t) = \text{Length of the curve \(\mathbf c^t\)}
\]
has a minimum at $t=0$.
Problem 3: Show that \(L(t)\) is given by the formula
\[
L(t) = \int_a^b \| \mathbf f_s(s,t)\|\,ds = \int_a^b \|\mathbf f_s\|\,ds.
\]
Solution:
You have likely guessed the next step: since \(L(t)\) has a minimum at
\(t=0\) so, by the first derivative test \(L'(0)=0\).
Problem 4:
Use that \(\mathbf c\) is unit speed, so that \(\|\mathbf c'(s)\|
= \| \mathbf f_s(s,0)\|\) to show
\[
L'(0)= \int_a^b \mathbf c'(s)\cdot \mathbf f_{st}(s,0) \,ds.
\]
We wish simplify this formula for \(L'(0)\). This will involve
do an integration by parts. Recall that our variation \(\mathbf f\)
of \(c\) has its endpoints fixed. That is
\[
\mathbf f(a,t)= \mathbf c(a), \qquad \mathbf f(b,t) = \mathbf c(b).
\]
Taking the derivatives of these equation with respect to \(t\) and using
that they are constant with respect to \(t\) gives
\[
\mathbf 0 = \frac{\partial \mathbf f}{\partial t}(a,t) = \mathbf f_t(a,0),
\qquad
\mathbf 0 = \frac{\partial \mathbf f}{\partial t}(b,t) = \mathbf f_t(b,0)
\]
Problem 5:
Using these equations and our formula and integration by parts to
show that our formula for \(L'(0)\) can be rewritten as
\[
L'(0) = -\int_a^b \mathbf c''(s) \cdot \mathbf f_t(s,0)\,ds.
\]
Solution:
So far we have not used any of the geometry of surfaces (first and second
fundamental forms etc.) We now do that. As \( \mathbf c\) is a curve
in our surface if \( \text{II} \) is the second fundamental form
of \(M\) we have
\[
\mathbf c''(s) = \nabla_{\mathbf c'(s)} \mathbf c'(s) + \text{II}(\mathbf c'(s),
\mathbf c'(s))\mathbf n.
\]
where \(\mathbf n\) is the unit normal to the surface and
\( \nabla_{\mathbf c'(s)} \mathbf c'(s)\) is the projection of \(\mathbf c''(s)\)
onto the tangent plane to \(M\) at \(\mathbf c(s)\).
Problem 6: Using the formula above for \(\mathbf c''(s)\) show that
the formula for \(L'(0)\) can be rewritten as
\[
L'(0) = - \int_a^b \left(\nabla_{\mathbf c'(s)} \mathbf c'(s)\right) \cdot \mathbf f_t(s,0)\,ds.
\]
Solution:
We and summarize what we have done so far as
Theorem. Let \(\mathbf c [a,b] \to M\) be a curve
that minimizes length for all curves in \(M\) connecting \(\mathbf
c(a)\) and \(\mathbf c(b)\). Then for all variations, \(\mathbf f\)
with endpoints fixed we have
\[
L'(0)= -\int_a^b \left( \nabla_{\mathbf c'(s)} \mathbf c'(s)\right)\cdot \mathbf f_t(s,0)\,ds=0.
\]
Note if \( \nabla_{\mathbf c'(s)} \mathbf c'(s)=\mathbf 0\) for all \(s\in [a,b]\)
then we will have \(L'(0)=0\). There is a converse to this.
Theorem. Let \(\mathbf c [a,b] \to M\) be a curve
such that for all variations, \(\mathbf f\)
with endpoints fixed we have
\[
-\int_a^b \left( \nabla_{\mathbf c'(s)} \mathbf c'(s)\right)\cdot \mathbf f_t(s,0)\,ds=0.
\]
Then \( \nabla_{\mathbf c'(s)} \mathbf c'(s)=\mathbf 0\).
This is not outrageously hard to prove, the idea is to constrict
very many variations of \(\mathbb c\) with fixed endpoints
so that the only way for the integral above to vanish for all of them
is that if \( \nabla_{\mathbf c'(s)} \mathbf c'(s)=\mathbf 0\).
But the methods are more in line with Math 554 than our class so we will
not give a proof. But this does suggest giving a name to the curves
that make this true.
Definition. A curve \( \mathbf c \colon [a,b] \to M\)
such that
\[
\nabla_{\mathbf c'(s)} \mathbf c'(s)=\mathbf 0
\]
for all \(s\) is a geodesic of \(M\).
In light of the definition of \( \nabla_{\mathbf c'(s)} \mathbf c'(s)\)
being the component of \(\mathbf c''(s)\) that is tangent to \(M\)
anther way to say that \(\mathbf c\) is geodesic is to say that
\(i\mathbf c''(s)\) is perpendicular to \(M\) at all its points.
Problem 7:
Show that if \(\mathbf c\) is a geodesic of \(M\), then it has
constant speed. That is \( \| \mathbf c'(s)\| \) is a constant.
Solution: