Part IB Variational Principles
Based on lectures by P. K. Townsend
Notes taken by Dexter Chua
Easter 2015
These notes are not endorsed by the lecturers, and I have modified them (often
significantly) after lectures. They are nowhere near accurate representations of what
was actually lectured, and in particular, all errors are almost surely mine.
Stationary points for functions on
R
n
. Necessary and sufficient conditions for minima
and maxima. Importance of convexity. Variational problems with constraints; method
of Lagrange multipliers. The Legendre Transform; need for convexity to ensure
invertibility; illustrations from thermodynamics. [4]
The idea of a functional and a functional derivative. First variation for functionals,
Euler-Lagrange equations, for both ordinary and partial differential equations. Use of
Lagrange multipliers and multiplier functions. [3]
Fermat’s principle; geodesics; least action principles, Lagrange’s and Hamilton’s equa-
tions for particles and fields. Noether theorems and first integrals, including two forms
of Noether’s theorem for ordinary differential equations (energy and momentum, for
example). Interpretation in terms of conservation laws. [3]
Second variation for functionals; associated eigenvalue problem. [2]
Contents
0 Introduction
1 Multivariate calculus
1.1 Stationary points
1.2 Convex functions
1.2.1 Convexity
1.2.2 First-order convexity condition
1.2.3 Second-order convexity condition
1.3 Legendre transform
1.4 Lagrange multipliers
2 Euler-Lagrange equation
2.1 Functional derivatives
2.2 First integrals
2.3 Constrained variation of functionals
3 Hamilton’s principle
3.1 The Lagrangian
3.2 The Hamiltonian
3.3 Symmetries and Noether’s theorem
4 Multivariate calculus of variations
5 The second variation
5.1 The second variation
5.2 Jacobi condition for local minima of F [x]
0 Introduction
Consider a light ray travelling towards a mirror and being reflected.
z
We see that the light ray travels towards the mirror, gets reflected at
z
, and hits
the (invisible) eye. What determines the path taken? The usual answer would
be that the reflected angle shall be the same as the incident angle. However,
ancient Greek mathematician Hero of Alexandria provided a different answer:
the path of the light minimizes the total distance travelled.
We can assume that light travels in a straight line except when reflected.
Then we can characterize the path by a single variable
z
, the point where the
light ray hits the mirror. Then we let
L
(
z
) to be the length of the path, and we
can solve for z by setting L
0
(z) = 0.
This principle sounds reasonable - in the absence of mirrors, light travels in
a straight line - which is the shortest path between two points. But is it always
true that the shortest path is taken? No! We only considered a plane mirror, and
this doesn’t hold if we have, say, a spherical mirror. However, it turns out that
in all cases, the path is a stationary point of the length function, i.e. L
0
(z) = 0.
Fermat put this principle further. Assuming that light travels at a finite
speed, the shortest path is the path that takes the minimum time. Fermat’s
principle thus states that
Light travels on the path that takes the shortest time.
This alternative formulation has the advantage that it applies to refraction as
well. Light travels at different speeds in different mediums. Hence when they
travel between mediums, they will change direction such that the total time
taken is minimized.
We usually define the refractive index
n
of a medium to be
n
= 1
/v
, where
v
is the velocity of light in the medium. Then we can write the variational
principle as
minimize
Z
path
n ds,
where d
s
is the path length element. This is easy to solve if we have two distinct
mediums. Since light travels in a straight line in each medium, we can simply
characterize the path by the point where the light crosses the boundary. However,
in the general case, we should be considering any possible path between two
points. In this case, we could no longer use ordinary calculus, and need new
tools - calculus of variations.
In calculus of variations, the main objective is to find a function
x
(
t
) that
minimizes an integral
R
f
(
x
) d
t
for some function
f
. For example, we might
want to minimize
R
(
x
2
+
x
) d
t
. This differs greatly from ordinary minimization
problems. In ordinary calculus, we minimize a function
f
(
x
) for all possible
values of
x R
. However, in calculus of variations, we will be minimizing the
integral
R
f(x) dt over all possible functions x(t).
1 Multivariate calculus
Before we start calculus of variations, we will first have a brief overview of
minimization problems in ordinary calculus. Unless otherwise specified,
f
will be a function
R
n
R
. For convenience, we write the argument of
f
as
x
= (
x
1
, ··· , x
n
) and
x
=
|x|
. We will also assume that
f
is sufficiently smooth
for our purposes.
1.1 Stationary points
The quest of minimization starts with finding stationary points.
Definition
(Stationary points)
.
Stationary points are points in
R
n
for which
f = 0, i.e.
f
x
1
=
f
x
2
= ··· =
f
x
n
= 0
All minima and maxima are stationary points, but knowing that a point is
stationary is not sufficient to determine which type it is. To know more about
the nature of a stationary point, we Taylor expand
f
about such a point, which
we assume is 0 for notational convenience.
f(x) = f(0) + x · f +
1
2
X
i,j
x
i
x
j
2
f
x
i
x
j
+ O(x
3
).
= f(0) +
1
2
X
i,j
x
i
x
j
2
f
x
i
x
j
+ O(x
3
).
The second term is so important that we have a name for it:
Definition (Hessian matrix). The Hessian matrix is
H
ij
(x) =
2
f
x
i
x
j
Using summation notation, we can write our result as
f(x) f(0) =
1
2
x
i
H
ij
x
j
+ O(x
3
).
Since
H
is symmetric, it is diagonalizable. Thus after rotating our axes to a
suitable coordinate system, we have
H
0
ij
=
λ
1
0 ··· 0
0 λ
2
··· 0
.
.
.
.
.
.
.
.
.
.
.
.
0 0 ··· λ
n
,
where
λ
i
are the eigenvalues of
H
. Since
H
is real symmetric, these are all real.
In our new coordinate system, we have
f(x) f(0) =
1
2
n
X
i=1
λ
i
(x
0
i
)
2
This is useful information. If all eigenvalues
λ
i
are positive, then
f
(
x
)
f
(
0
)
must be positive (for small
x
). Hence our stationary point is a local minimum.
Similarly, if all eigenvalues are negative, then it is a local maximum.
If there are mixed signs, say
λ
1
>
0 and
λ
2
<
0, then
f
increases in the
x
1
direction and decreases in the
x
2
direction. In this case we say we have a saddle
point.
If some
λ
= 0, then we have a degenerate stationary point. To identify the
nature of this point, we must look at even higher derivatives.
In the special case where
n
= 2, we do not need to explicitly find the
eigenvalues. We know that
det H
is the product of the two eigenvalues. Hence if
det H
is negative, the eigenvalues have different signs, and we have a saddle. If
det H is positive, then the eigenvalues are of the same sign.
To determine if it is a maximum or minimum, we can look at the trace of
H
, which is the sum of eigenvalues. If
tr H
is positive, then we have a local
minimum. Otherwise, it is a local maximum.
Example. Let f(x, y) = x
3
+ y
3
3xy. Then
f = 3(x
2
y, y
2
x).
This is zero iff
x
2
=
y
and
y
2
=
x
. This is satisfied iff
y
4
=
y
. So either
y
= 0,
or y = 1. So there are two stationary points: (0, 0) and (1, 1).
The Hessian matrix is
H =
6x 3
3 6y
,
and we have
det H = 9(4xy 1)
tr H = 6(x + y).
At (0
,
0),
det H <
0. So this is a saddle point. At (1
,
1),
det H >
0,
tr H >
0.
So this is a local minimum.
1.2 Convex functions
1.2.1 Convexity
Convex functions is an important class of functions that has a lot of nice
properties. For example, stationary points of convex functions are all minima,
and a convex function can have at most one minimum value. To define convex
functions, we need to first define a convex set.
Definition
(Convex set)
.
A set
S R
n
is convex if for any distinct
x, y
S, t
(0
,
1), we have (1
t
)
x
+
ty S
. Alternatively, any line joining two points
in S lies completely within S.
non-convex convex
Definition (Convex function). A function f : R
n
R is convex if
(i) The domain D(f) is convex
(ii) The function f lies below (or on) all its chords, i.e.
f((1 t)x + ty) (1 t)f(x) + tf(y) ()
for all x, y D(f), t (0, 1).
A function is strictly convex if the inequality is strict, i.e.
f((1 t)x + ty) < (1 t)f(x) + tf(y).
x y
(1 t)x + ty
(1 t)f(x) + tf (y)
A function f is (strictly) concave iff f is (strictly) convex.
Example.
(i) f(x) = x
2
is strictly convex.
(ii) f(x) = |x| is convex, but not strictly.
(iii) f(x) =
1
x
defined on x > 0 is strictly convex.
(iv) f
(
x
) =
1
x
defined on
R
=
R \{
0
}
is not convex. Apart from the fact that
R
is not a convex domain. But even if we defined, like
f
(0) = 0, it is
not convex by considering the line joining (
1
,
1) and (1
,
1) (and in fact
f(x) =
1
x
defined on x < 0 is concave).
1.2.2 First-order convexity condition
While the definition of a convex function seems a bit difficult to work with, if
our function is differentiable, it is easy to check if it is convex.
First assume that our function is once differentiable, and we attempt to find
a first-order condition for convexity. Suppose that
f
is convex. For fixed
x, y
,
we define the function
h(t) = (1 t)f(x) + tf(y) f((1 t)x + ty).
By the definition of convexity of
f
, we must have
h
(
t
)
0. Also, trivially
h(0) = 0. So
h(t) h(0)
t
0
for any t (0, 1). So
h
0
(0) 0.
On the other hand, we can also differentiate h directly and evaluate at 0:
h
0
(0) = f(y) f(x) (y x) · f(x).
Combining our two results, we know that
f(y) f(x) + (y x) · f(x) ()
It is also true that this condition implies convexity, which is an easy result.
How can we interpret this result? The equation
f
(
x
) + (
y x
)
· f
(
x
) = 0
defines the tangent plane of
f
at
x
. Hence this condition is saying that a convex
differentiable function lies above all its tangent planes.
We immediately get the corollary
Corollary.
A stationary point of a convex function is a global minimum. There
can be more than one global minimum (e.g. a constant function), but there is at
most one if the function is strictly convex.
Proof. Given x
0
such that f(x
0
) = 0, () implies that for any y,
f(y) f(x
0
) + (y x
0
) · f(x
0
) = f(x
0
).
We can write our first-order convexity condition in a different way. We can
rewrite () into the form
(y x) · [f(y) f(x)] f(x) f(y) (x y) · f(y).
By applying (
) to the right hand side (with
x
and
y
swapped), we know that
the right hand side is 0. So we have another first-order condition:
(y x) · [f(y) f(x)] 0,
It can be shown that this is equivalent to the other conditions.
This condition might seem a bit weird to comprehend, but all it says is that
f
(
x
) is a non-decreasing function. For example, when
n
= 1, the equation
states that (
y x
)(
f
0
(
y
)
f
0
(
x
))
0, which is the same as saying
f
0
(
y
)
f
0
(
x
)
whenever y > x.
1.2.3 Second-order convexity condition
We have an even nicer condition when the function is twice differentiable. We
start with the equation we just obtained:
(y x) · [f(y) f(x)] 0,
Write y = x + h. Then
h · (f(x + h) f(x)) 0.
Expand the left in Taylor series. Using suffix notation, this becomes
h
i
[h
j
j
i
f + O(h
2
)] 0.
But
j
i
f = H
ij
. So we have
h
i
H
ij
h
j
+ O(h
3
) 0
This is true for all
h
if the Hessian
H
is positive semi-definite (or simply positive),
i.e. the eigenvalues are non-negative. If they are in fact all positive, then we say
H is positive definite.
Hence convexity implies that the Hessian matrix is positive for all
x D
(
f
).
Strict convexity implies that it is positive definite.
The converse is also true if the Hessian is positive, then it is convex.
Example. Let f(x, y) =
1
xy
for x, y > 0. Then the Hessian is
H =
1
xy
2
x
2
1
xy
1
xy
2
y
2
The determinant is
det H =
3
x
4
y
4
> 0
and the trace is
tr H =
2
xy
1
x
2
+
1
y
2
> 0.
So f is convex.
To conclude that
f
is convex, we only used the fact that
xy
is positive,
instead of
x
and
y
being individually positive. Then could we relax the domain
condition to be
xy >
0 instead? The answer is no, because in this case, the
function will no longer be convex!
1.3 Legendre transform
The Legendre transform is an important tool in classical dynamics and thermo-
dynamics. In classical dynamics, it is used to transform between the Lagrangian
and the Hamiltonian. In thermodynamics, it is used to transform between
the energy, Helmholtz free energy and enthalpy. Despite its importance, the
definition is slightly awkward.
Suppose that we have a function
f
(
x
), which we’ll assume is differentiable.
For some reason, we want to transform it into a function of the conjugate variable
p
=
df
dx
instead. In most applications to physics, this quantity has a particular
physical significance. For example, in classical dynamics, if
L
is the Lagrangian,
then
p
=
L
˙x
is the (conjugate) momentum.
p
also has a context-independent
geometric interpretation, which we will explore later. For now, we will assume
that p is more interesting than x.
Unfortunately, the obvious option
f
(
p
) =
f
(
x
(
p
)) is not the transform we
want. There are various reasons for this, but the major reason is that it is ugly.
It lacks any mathematical elegance, and has almost no nice properties at all.
In particular, we want our f
(p) to satisfy the property
df
dp
= x.
This says that if
p
is the conjugate of
x
, then
x
is the conjugate of
p
. We will
soon see how this is useful in the context of thermodynamics.
The symmetry is better revealed if we write in terms of differentials. The
differential of the function f is
df =
df
dx
dx = p dx.
So we want our f
to satisfy
df
= x dp.
How can we obtain this? From the product rule, we know that
d(xp) = x dp + p dx.
So if we define
f
=
xp f
(more explicitly written as
f
(
p
) =
x
(
p
)
p f
(
x
(
p
))),
then we obtain the desired relation d
f
=
x
d
p
. Alternatively, we can say
df
dp
= x.
The actual definition we give will not be exactly this. Instead, we define it in
a way that does not assume differentiability. We’ll also assume that the function
takes the more general form R
n
R.
Definition
(Legendre transform)
.
Given a function
f
:
R
n
R
, its Legendre
transform f
(the “conjugate” function) is defined by
f
(p) = sup
x
(p · x f(x)),
The domain of
f
is the set of
p R
n
such that the supremum is finite.
p
is
known as the conjugate variable.
This relation can also be written as
f
(
p
) +
f
(
x
) =
px
, where
x
(
p
) is the
value of x that maximizes the function.
To show that this is the same as what we were just talking about, note
that the supremum of
p · x f
(
x
) is obtained when its derivative is zero, i.e.
p
=
f
(
x
). In particular, in the 1D case,
f
(
p
) =
px f
(
x
), where
x
satisfies
f
0
(x) = p. So p is indeed the derivative of f with respect to x.
From the definition, we can immediately conclude that
Lemma. f
is always convex.
Proof.
f
((1 t)p + tq) = sup
x
((1 t)p · x + tq · x f(x)
.
= sup
x
(1 t)(p · x f(x)) + t(q · x f(x))
(1 t) sup
x
[p · x f(x)] + t sup
x
[q · x f(x)]
= (1 t)f
(p) + tf
(q)
Note that we cannot immediately say that
f
is convex, since we have to show
that the domain is convex. But by the above bounds,
f
((1
t
)
p
+
tq
) is bounded
by the sum of two finite terms, which is finite. So (1
t
)
p
+
tq
is also in the
domain of f
.
This transformation can be given a geometric interpretation. We will only
consider the 1D case, because drawing higher-dimensional graphs is hard. For
any fixed
x
, we draw the tangent line of
f
at the point
x
. Then
f
(
p
) is the
intersection between the tangent line and the y axis:
x
y
slope = p
f
(p)
px
f
(p) = px f(x)
f(x)
Example.
(i)
Let
f
(
x
) =
1
2
ax
2
for
a >
0. Then
p
=
ax
at the maximum of
px f
(
x
). So
f
(p) = px f(x) = p ·
p
a
1
2
a
p
a
2
=
1
2a
p
2
.
So the Legendre transform maps a parabola to a parabola.
(ii) f(v) =
1 v
2
for |v| < 1 is a lower semi-circle. We have
p = f
0
(v) =
v
1 v
2
So
v =
p
p
1 + p
2
and exists for all p R. So
f
(p) = pv f(v) =
p
2
p
1 + p
2
+
1
p
1 + p
2
=
p
1 + p
2
.
A circle gets mapped to a hyperbola.
(iii)
Let
f
=
cx
for
c >
0. This is convex but not strictly convex. Then
px f
(
x
) = (
p c
)
x
. This has no maximum unless
p
=
c
. So the domain
of f
is simply {c}. One point. So f
(p) = 0. So a line goes to a point.
Finally, we prove that applying the Legendre transform twice gives the
original function.
Theorem.
If
f
is convex, differentiable with Legendre transform
f
, then
f
∗∗
= f.
Proof.
We have
f
(
p
) = (
p · x
(
p
)
f
(
x
(
p
)) where
x
(
p
) satisfies
p
=
f
(
x
(
p
)).
Differentiating with respect to p, we have
i
f
(p) = x
i
+ p
j
i
x
j
(p)
i
x
j
(p)
j
f(x)
= x
i
+ p
j
i
x
j
(p)
i
x
j
(p)p
j
= x
i
.
So
f
(p) = x.
This means that the conjugate variable of p is our original x. So
f
∗∗
(x) = (x · p f
(p))|
p=p(x)
= x · p (p · x f(x))
= f(x).
Note that strict convexity is not required. For example, in our last example
above with the straight line,
f
(
p
) = 0 for
p
=
c
. So
f
∗∗
(
x
) = (
xp f
(
p
))
|
p=c
=
cx = f(x).
However, convexity is required. If
f
∗∗
=
f
is true, then
f
must be convex,
since it is a Legendre transform. Hence
f
∗∗
=
f
cannot be true for non-convex
functions.
Application to thermodynamics
Given a system of a fixed number of particles, the energy of a system is usually
given as a function of entropy and volume:
E = E(S, V ).
We can think of this as a gas inside a piston with variable volume.
There are two things that can affect the energy: we can push in the piston
and modify the volume. This corresponds to a work done of
p
d
V
, where
p
is
the pressure. Alternatively, we can simply heat it up and create a heat change
of T dS, where T is the temperature. Then we have
dE = T dS p dV.
Comparing with the chain rule, we have
E
S
= T,
E
V
= p
However, the entropy is a mysterious quantity no one understands. Instead, we
like temperature, defined as
T
=
E
S
. Hence we use the (negative) Legendre
transform to obtain the conjugate function Helmholtz free energy.
F (T, V ) = inf
S
[E(S, V ) T S] = E(S, V ) S
E
S
= E ST,
Note that the Helmholtz free energy satisfies
dF = S dT p dV.
Just as we could recover
T
and
p
from
E
via taking partial derivatives with
respect to
S
and
V
, we are able to recover
S
and
p
from
F
by taking partial
derivatives with respect to
T
and
V
. This would not be the case if we simply
defined F (T, V ) = E(S(T, V ), V ).
If we take the Legendre transform with respect to
V
, we get the enthalpy
instead, and if we take the Legendre transform with respect to both, we get the
Gibbs free energy.
1.4 Lagrange multipliers
At the beginning, we considered the problem of unconstrained maximization.
We wanted to maximize
f
(
x, y
) where
x, y
can be any real value. However,
sometimes we want to restrict to certain values of (
x, y
). For example, we might
want x and y to satisfy x + y = 10.
We take a simple example of a hill. We model it using the function
f
(
x, y
)
given by the height above the ground. The hilltop would be given by the
maximum of f, which satisfies
0 = df = f · dx
for any (infinitesimal) displacement dx. So we need
f = 0.
This would be a case of unconstrained maximization, since we are considering all
possible values of x and y.
A problem of constrained maximization would be as follows: we have a path
p defined by p(x, y) = 0. What is the highest point along the path p?
We still need
f ·
d
x
= 0, but now d
x
is not arbitrary. We only consider
the d
x
parallel to the path. Alternatively,
f
has to be entirely perpendicular
to the path. Since we know that the normal to the path is
p
, our condition
becomes
f = λp
for some lambda
λ
. Of course, we still have the constraint
p
(
x, y
) = 0. So what
we have to solve is
f = λp
p = 0
for the three variables x, y, λ.
Alternatively, we can change this into a single problem of unconstrained
extremization. We ask for the stationary points of the function
φ
(
x, y, λ
) given
by
φ(x, y, λ) = f(x, y) λp(x, y)
When we maximize against the variables
x
and
y
, we obtain the
f
=
λp
condition, and maximizing against λ gives the condition p = 0.
Example.
Find the radius of the smallest circle centered on origin that intersects
y = x
2
1.
(i)
First do it the easy way: for a circle of radius
R
to work,
x
2
+
y
2
=
R
2
and y = x
2
1 must have a solution. So
(x
2
)
2
x
2
+ 1 R
2
= 0
and
x
2
=
1
2
±
r
R
2
3
4
So R
min
=
3/2.
(ii)
We can also view this as a variational problem. We want to minimize
f
(
x, y
) =
x
2
+
y
2
subject to the constraint
p
(
x, y
) = 0 for
p
(
x, y
) =
yx
2
+1.
We can solve this directly. We can solve the constraint to obtain
y
=
x
2
1.
Then
R
2
(x) = f(x, y(x)) = (x
2
)
2
x
2
+ 1
We look for stationary points of R
2
:
(R
2
(x))
0
= 0 x
x
2
1
2
= 0
So
x
= 0 and
R
= 1; or
x
=
±
1
2
and
R
=
3
2
. Since
3
2
is smaller, this is
our minimum.
(iii)
Finally, we can use Lagrange multipliers. We find stationary points of the
function
φ(x, y, λ) = f(x, y) λp(x, y) = x
2
+ y
2
λ(y x
2
+ 1)
The partial derivatives give
φ
x
= 0 2x(1 + λ) = 0
φ
y
= 0 2y λ = 0
φ
λ
= 0 y x
2
+ 1 = 0
The first equation gives us two choices
x
= 0. Then the third equation gives
y
=
1. So
R
=
p
x
2
+ y
2
= 1.
λ
=
1. So the second equation gives
y
=
1
2
and the third gives
x = ±
1
2
. Hence R =
3
2
is the minimum.
This can be generalized to problems with functions
R
n
R
using the same
logic.
Example. For x R
n
, find the minimum of the quadratic form
f(x) = x
i
A
ij
x
j
on the surface |x|
2
= 1.
(i)
The constraint imposes a normalization condition on
x
. But if we scale up
x, f (x) scales accordingly. So if we define
Λ(x) =
f(x)
g(x)
, g(x) = |x|
2
,
the problem is equivalent to minimization of Λ(
x
) without constraint. Then
i
Λ(x) =
2
g
A
ij
x
j
f
g
x
i
So we need
Ax = Λx
So the extremal values of Λ(
x
) are the eigenvalues of
A
. So Λ
min
is the
lowest eigenvalue.
This answer is intuitively obvious if we diagonalize A.
(ii)
We can also do it with Lagrange multipliers. We want to find stationary
values of
φ(x, λ) = f(x) λ(|x|
2
1).
So
0 = φ A
ij
x
j
= λx
i
Differentiating with respect to λ gives
φ
λ
= 0 |x|
2
= 1.
So we get the same set of equations.
Example.
Find the probability distribution
{p
1
, ··· , p
n
}
satisfying
P
i
p
i
= 1
that maximizes the information entropy
S =
n
X
i=1
p
i
log p
i
.
We look for stationary points of
φ(p, λ) =
n
X
i=1
p
i
ln p
i
λ
n
X
i=1
p
i
+ λ.
We have
φ
p
i
= ln p
i
(1 + λ) = 0.
So
p
i
= e
(1+λ)
.
It is the same for all i. So we must have p
i
=
1
n
.
2 Euler-Lagrange equation
2.1 Functional derivatives
Definition
(Functional)
.
A functional is a function that takes in another real-
valued function as an argument. We usually write them as
F
[
x
] (square brackets),
where
x
=
x
(
t
) :
R R
is a real function. We say that
F
[
x
] is a functional of
the function x(t).
Of course, we can also have functionals of many functions, e.g.
F
[
x, y
]
R
for x, y : R R. We can also have functionals of a function of many variables.
Example.
Given a medium with refractive index
n
(
x
), the time taken by a
path x(t) from x
0
to x
1
is given by the functional
T [x] =
Z
x
1
x
0
n(x) dt.
While this is a very general definition, in reality, there is just one particular
class of functionals we care about. Given a function
x
(
t
) defined for
α t β
,
we study functional of the form
F [x] =
Z
β
α
f(x, ˙x, t) dt
for some function f .
Our objective is to find a stationary point of the functional
F
[
x
]. To do so,
suppose we vary
x
(
t
) by a small amount
δx
(
t
). Then the corresponding change
δF [x] of F [x] is
δF [x] = F [x + δx] F [x]
=
Z
β
α
f(x + δx, ˙x + δ ˙x, t) f(x, ˙x, t)
dt
Taylor expand to obtain
=
Z
β
α
δx
f
x
+ δ ˙x
f
˙x
dt + O(δx
2
)
Integrate the second term by parts to obtain
δF [x] =
Z
β
α
δx
f
x
d
dt
f
˙x

dt +
δx
f
˙x
β
α
.
This doesn’t seem like a very helpful equation. Life would be much easier if
the last term (known as the boundary term)
h
δx
f
˙x
i
β
α
vanishes. Fortunately, for
most of the cases we care about, the boundary conditions mandate that the
boundary term does indeed vanish. Most of the time, we are told that
x
is fixed
at
t
=
α, β
. So
δx
(
α
) =
δx
(
β
) = 0. But regardless of what we do, we always
choose boundary conditions such that the boundary term is 0. Then
δF [x] =
Z
β
α
δx
δF [x]
δx(t)
dt
where
Definition (Functional derivative).
δF [x]
δx
=
f
x
d
dt
f
˙x
is the functional derivative of F [x].
If we want to find a stationary point of F , then we need
δF [x]
δx
= 0. So
Definition (Euler-Lagrange equation). The Euler-Lagrange equation is
f
x
d
dt
f
˙x
= 0
for α t β.
There is an obvious generalization to functionals F [x] for x(t) R
n
:
f
x
i
d
dt
f
˙x
i
= 0 for all i.
Example
(Geodesics of a plane)
.
What is the curve
C
of minimal length between
two points A, B in the Euclidean plane? The length is
L =
Z
C
d`
where d` =
p
dx
2
+ dy
2
.
There are two ways we can do this:
(i)
We restrict to curves for which
x
(or
y
) is a good parameter, i.e.
y
can be
made a function of x. Then
d` =
p
1 + (y
0
)
2
dx.
Then
L[y] =
Z
β
α
p
1 + (y
0
)
2
dx.
Since there is no explicit dependence on y, we know that
f
y
= 0
So the Euler-Lagrange equation says that
d
dx
f
y
0
= 0
We can integrate once to obtain
f
y
0
= constant
This is known as a first integral, which will be studied more in detail later.
Plugging in our value of f, we obtain
y
0
p
1 + (y
0
)
2
= constant
This shows that y
0
must be constant. So y must be a straight line.
(ii)
We can get around the restriction to “good” curves by choosing an arbitrary
parameterization
r
= (
x
(
t
)
, y
(
t
)) for
t
[0
,
1] such that
r
(0) =
A
,
r
(1) =
B
.
So
d` =
p
˙x
2
+ ˙y
2
dt.
Then
L[x, y] =
Z
1
0
p
˙x
2
+ ˙y
2
dt.
We have, again
f
x
=
f
y
= 0.
So we are left to solve
d
dt
f
˙x
=
d
dt
f
˙y
= 0.
So we obtain
˙x
p
˙x
2
+ ˙y
2
= c,
˙y
p
˙x
2
+ ˙y
2
= s
where
c
and
s
are constants. While we have two constants, they are not
independent. We must have
c
2
+
s
2
= 1. So we let
c
=
cos θ
,
s
=
sin θ
.
Then the two conditions are both equivalent to
( ˙x sin θ)
2
= ( ˙y cos θ)
2
.
Hence
˙x sin θ = ±˙y cos θ.
We can choose a θ such that we have a positive sign. So
y cos θ = x sin θ + A
for a constant A. This is a straight line with slope tan θ.
2.2 First integrals
In our example above,
f
did not depend on
x
, and hence
f
x
= 0. Then the
Euler-Lagrange equations entail
d
dt
f
˙x
= 0.
We can integrate this to obtain
f
˙x
= constant.
We call this the first integral. First integrals are important in several ways. The
most immediate advantage is that it simplifies the problem a lot. We only have
to solve a first-order differential equation, instead of a second-order one. Not
needing to differentiate
f
˙x
also prevents a lot of mess arising from the product
and quotient rules.
This has an additional significance when applied to problems in physics.
If we have a first integral, then we get
f
˙x
= constant. This corresponds to
a conserved quantity of the system. When formulating physics problems as
variational problems (as we will do in Chapter 3), the conservation of energy
and momentum will arise as constants of integration from first integrals.
There is also a more complicated first integral appearing when
f
does not
(explicitly) depend on
t
. To find this out, we have to first consider the total
derivative
df
dt
. By the chain rule, we have
df
dt
=
f
t
+
dx
dt
f
x
+
d ˙x
dt
f
˙x
=
f
t
+ ˙x
f
x
+ ¨x
f
˙x
.
On the other hand, the Euler-Lagrange equation says that
f
x
=
d
dt
f
˙x
.
Substituting this into our equation for the total derivative gives
df
dt
=
f
t
+ ˙x
d
dt
f
˙x
+ ¨x
f
˙x
=
f
t
+
d
dt
˙x
f
˙x
.
Then
d
dt
f ˙x
f
˙x
=
f
t
.
So if
f
t
= 0, then we have the first integral
f ˙x
f
˙x
= constant.
Example.
Consider a light ray travelling in the vertical
xz
plane inside a
medium with refractive index
n
(
z
) =
a bz
for positive constants
a, b
. The
phase velocity of light is v =
c
n
.
According the Fermat’s principle, the path minimizes
T =
Z
B
A
d`
v
.
This is equivalent to minimizing the optical path length
cT = P =
Z
B
A
n d`.
We specify our path by the function z(x). Then the path element is given by
d` =
p
dx
2
+ dz
2
=
p
1 + z
0
(x)
2
dx,
Then
P [z] =
Z
x
B
x
A
n(z)
p
1 + (z
0
)
2
dx.
Since this does not depend on x, we have the first integral
k = f z
0
f
z
0
=
n(z)
p
1 + (z
0
)
2
.
for an integration constant k. Squaring and putting in the value of n gives
(z
0
)
2
=
b
k
2
(z
0
z),
where z
0
= (a k
2
)/b. This is integrable and we obtain
dz
z
0
z
= ±
b
k
dx.
So
z z
0
= ±
b
2k
(x x
0
),
where x
0
is our second integration constant. Square it to obtain
z = z
0
b
4k
2
(x x
0
)
2
,
which is a parabola.
Example
(Principle of least action)
.
Mechanics as laid out by Newton was
expressed in terms of forces and acceleration. While this is able to describe a lot
of phenomena, it is rather unsatisfactory. For one, it is messy and difficult to
scale to large systems involving many particles. It is also ugly.
As a result, mechanics is later reformulated in terms of a variational principle.
A quantity known as the action is defined for each possible path taken by
the particle, and the actual path taken is the one that minimizes the action
(technically, it is the path that is a stationary point of the action functional).
The version we will present here is an old version proposed by Maupertuis
and Euler. While it sort-of works, it is still cumbersome to work with. The
modern version is from Hamilton who again reformulated the action principle
to something that is more powerful and general. This modern version will be
discussed more in detail in Chapter 3, and for now we will work with the old
version first.
The original definition for the action, as proposed by Maupertuis, was mass
×
velocity
×
distance. This was given a more precise mathematical definition
by Euler. For a particle with constant energy,
E =
1
2
mv
2
+ U(x),
where v = |
˙
x|. So we have
mv =
p
2m(E U(x)).
Hence we can define the action to be
A =
Z
B
A
p
2m(E U(x)) d`,
where d` is the path length element. We minimize this to find the trajectory.
For a particle near the surface of the Earth, under the influence of gravity,
U = mgz. So we have
A[z] =
Z
B
A
p
2mE 2m
2
gz
p
1 + (z
0
)
2
dx,
which is of exactly the same form as the optics problem we just solved. So the
result is again a parabola, as expected.
Example
(Brachistochrone)
.
The Brachistochrone problem was one of the
earliest problems in the calculus of variations. The name comes from the Greek
words br´akhistos (“shortest”) and khr´onos (“time”).
The question is as follows: suppose we have a bead sliding along a frictionless
wire, starting from rest at the origin
A
. What shape of wire minimizes the time
for the bead to travel to B?
x
y
A
B
The conservation of energy implies that
1
2
mv
2
= mgy.
So
v =
p
2gy
We want to minimize
T =
Z
d`
v
.
So
T =
1
2g
Z
p
dx
2
+ dy
2
y
=
1
2g
Z
s
1 + (y
0
)
2
y
dx
Since there is no explicit dependence on x, we have the first integral
f y
0
f
y
0
=
1
p
y(1 + (y
0
)
2
)
= constant
So the solution is
y(1 + (y
0
)
2
) = c
for some positive constant c.
The solution of this ODE is, in parametric form,
x = c(θ sin θ)
y = c(1 cos θ).
Note that this has x = y = 0 at θ = 0. This describes a cycloid.
2.3 Constrained variation of functionals
So far, we’ve considered the problem of finding stationary values of
F
[
x
] without
any restraint on what
x
could be. However, sometimes there might be some
restrictions on the possible values of
x
. For example, we might have a surface in
R
3
defined by
g
(
x
) = 0. If we want to find the path of shortest length on the
surface (i.e. geodesics), then we want to minimize
F
[
x
] subject to the constraint
g(x(t)) = 0.
We can again use Lagrange multipliers. The problem we have to solve is
equivalent to finding stationary values (without constraints) of
Φ
λ
[x] = F [x] λ(P [x] c).
with respect to the function x(t) and the variable λ.
Example (Isoperimetric problem). If we have a string of fixed length, what is
the maximum area we can enclose with it?
We first argue that the region enclosed by the curve is convex. If it is not,
we can “push out” the curve to increase the area enclosed without changing the
length. Assuming this, we can split the curve into two parts:
y
2
y
1
α
β
x
y
We have dA = [y
2
(x) y
1
(x)] dx. So
A =
Z
β
α
[y
2
(x) y
1
(x)] dx.
Alternatively,
A[y] =
I
y(x) dx.
and the length is
L[y] =
I
d` =
I
p
1 + (y
0
)
2
dx.
So we look for stationary points of
Φ
λ
[y] =
I
[y(x) λ
p
1 + (y
0
)
2
] dx + λL.
In this case, we can be sure that our boundary terms vanish since there is no
boundary.
Since there is no explicit dependence on x, we obtain the first integral
f y
0
f
y
0
= constant = y
0
.
So
y
0
= y λ
p
1 + (y
0
)
2
+
λ(y
0
)
2
p
1 + (y
0
)
2
= y
λ
p
1 + (y
0
)
2
.
So
(y y
0
)
2
=
λ
2
1 + (y
0
)
2
(y
0
)
2
=
λ
2
(y y
0
)
2
1
(y y
0
)y
0
p
λ
2
(y y
0
)
2
= ±1.
d
h
p
λ
2
(y y
0
)
2
± x
i
= 0.
So we have
λ
2
(y y
0
)
2
= (x x
0
)
2
,
or
(x x
0
)
2
+ (y y
0
)
2
= λ
2
.
This is a circle of radius
λ
. Since the perimeter of this circle will be 2
πλ
, we
must have λ = L/(2π). So the maximum area is πλ
2
= L
2
/(4π).
Example
(Sturm-Liouville problem)
.
The Sturm-Liouville problem is a very
general class of problems. We will develop some very general theory about these
problems without going into specific examples. It can be formulated as follows:
let
ρ
(
x
),
σ
(
x
) and
w
(
x
) be real functions of
x
defined on
α x β
. We will
consider the special case where
ρ
and
w
are positive on
α < x < β
. Our objective
is to find stationary points of the functional
F [y] =
Z
β
α
(ρ(x)(y
0
)
2
+ σ(x)y
2
) dx
subject to the condition
G[y] =
Z
β
α
w(x)y
2
dx = 1.
Using the Euler-Lagrange equation, the functional derivatives of F and G are
δF [y]
δy
= 2
(ρy
0
)
0
+ σy
δG[y]
δy
= 2(wy).
So the Euler-Lagrange equation of Φ
λ
[y] = F [y] λ(G[y] 1) is
(ρy
0
)
0
+ σy λwy = 0.
We can write this as the eigenvalue problem
Ly = λwy.
where
L =
d
dx
ρ
d
dx
+ σ
is the Sturm-Liouville operator. We call this a Sturm-Liouville eigenvalue
problem. w is called the weight function.
We can view this problem in a different way. Notice that
Ly
=
λwy
is linear
in
y
. Hence if
y
is a solution, then so is
Ay
. But if
G
[
y
] = 1, then
G
[
Ay
] =
A
2
.
Hence the condition
G
[
y
] = 1 is simply a normalization condition. We can get
around this problem by asking for the minimum of the functional
Λ[y] =
F [y]
G[y]
instead. It turns out that this Λ has some significance. To minimize Λ, we
cannot apply the Euler-Lagrange equations, since Λ is not of the form of an
integral. However, we can try to vary it directly:
δΛ =
1
G
δF
F
G
2
δG =
1
G
(δF ΛδG).
When Λ is minimized, we have
δΛ = 0
δF
δy
= Λ
δG
δy
Ly = Λwy.
So at stationary values of Λ[y], Λ is the associated Sturm-Liouville eigenvalue.
Example
(Geodesics)
.
Suppose that we have a surface in
R
3
defined by
g
(
x
) = 0,
and we want to find the path of shortest distance between two points on the
surface. These paths are known as geodesics.
One possible approach is to solve
g
(
x
) = 0 directly. For example, if we have
a unit sphere, a possible solution is
x
=
cos θ cos φ
,
y
=
cos θ sin φ
,
z
=
sin θ
.
Then the total length of a path would be given by
D[θ, φ] =
Z
B
A
q
dθ
2
+ sin
2
θdφ
2
.
We then vary θ and φ to minimize D and obtain a geodesic.
Alternatively, we can impose the condition
g
(
x
(
t
)) = 0 with a Lagrange
multiplier. However, since we want the constraint to be satisfied for all
t
, we
need a Lagrange multiplier function
λ
(
t
). Then our problem would be to find
stationary values of
Φ[x, λ] =
Z
1
0
|
˙
x| λ(t)g(x(t))
dt
3 Hamilton’s principle
As mentioned before, Lagrange and Hamilton reformulated Newtonian dynamics
into a much more robust system based on an action principle.
The first important concept is the idea of a configuration space. This
configuration space is a vector space containing generalized coordinates
ξ
(
t
) that
specify the configuration of the system. The idea is to capture all information
about the system in one single vector.
In the simplest case of a single free particle, these generalized coordinates
would simply be the coordinates of the position of the particle. If we have
two particles given by positions
x
(
t
) = (
x
1
, x
2
, x
3
) and
y
(
t
) = (
y
1
, y
2
, y
3
), our
generalized coordinates might be
ξ
(
t
) = (
x
1
, x
2
, x
3
, y
1
, y
2
, y
3
). In general, if we
have N different free particles, the configuration space has 3N dimensions.
The important thing is that the generalized coordinates need not be just the
usual Cartesian coordinates. If we are describing a pendulum in a plane, we do
not need to specify the
x
and
y
coordinates of the mass. Instead, the system
can be described by just the angle
θ
between the mass and the vertical. So the
generalized coordinates is just
ξ
(
t
) =
θ
(
t
). This is much more natural to work
with and avoids the hassle of imposing constraints on x and y.
3.1 The Lagrangian
The concept of generalized coordinates was first introduced by Lagrange in 1788.
He then showed that
ξ
(
t
) obeys certain complicated ODEs which are determined
by the kinetic energy and the potential energy.
In the 1830s, Hamilton made Lagrange’s mechanics much more pleasant. He
showed that the solutions of these ODEs are extremal points of a new “action”,
S[ξ] =
Z
L dt
where
L = T V
is the Lagrangian, with T the kinetic energy and V the potential energy.
Law
(Hamilton’s principle)
.
The actual path
ξ
(
t
) taken by a particle is the path
that makes the action S stationary.
Note that
S
has dimensions
ML
2
T
1
, which is the same as the 18th century
action (and Plank’s constant).
Example.
Suppose we have 1 particle in Euclidean 3-space. The configuration
space is simply the coordinates of the particle in space. We can choose Cartesian
coordinates x. Then
T =
1
2
m|
˙
x|
2
, V = V (x, t)
and
S[x] =
Z
t
B
t
A
1
2
m|
˙
x|
2
V (x, t)
dt.
Then the Lagrangian is
L(x,
˙
x, t) =
1
2
m|
˙
x|
2
V (x, t)
We apply the Euler-Lagrange equations to obtain
0 =
d
dt
L
˙
x
L
x
= m
¨
x + V.
So
m
¨
x = −∇V
This is Newton’s law
F
=
ma
with
F
=
−∇V
. This shows that Lagrangian
mechanics is “the same” as Newton’s law. However, Lagrangian mechanics has
the advantage that it does not care what coordinates you use, while Newton’s
law requires an inertial frame of reference.
Lagrangian mechanics applies even when
V
is time-dependent. However, if
V is independent of time, then so is L. Then we can obtain a first integral.
As before, the chain rule gives
dL
dt
=
L
t
+
˙
x ·
L
x
+
¨
x ·
˙
x
=
L
t
+
˙
x ·
L
x
d
dt
L
˙
x

| {z }
δS
δx
= 0
+
˙
x ·
d
dt
L
˙
x
+
¨
x ·
L
˙
x
| {z }
d
dt
˙
x ·
L
˙
x
So we have
d
dt
L
˙
x ·
L
˙
x
=
L
t
.
If
L
t
= 0, then
˙
x ·
L
˙
x
L = E
for some constant E. For example, for one particle,
E = m|
˙
x|
2
1
2
m|
˙
x|
2
+ V = T + V = total energy.
Example.
Consider a central force field
F
=
−∇V
, where
V
=
V
(
r
) is inde-
pendent of time. We use spherical polar coordinates (r, θ, φ), where
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ.
So
T =
1
2
m|
˙
x|
2
=
1
2
m
˙r
2
+ r
2
(
˙
θ
2
+ sin
2
θ
˙
φ
2
)
So
L =
1
2
m ˙r
2
+
1
2
mr
2
˙
θ
2
+ sin
2
θ
˙
φ
2
V (r).
We’ll use the fact that motion is planar (a consequence of angular momentum
conservation). So wlog θ =
π
2
. Then
L =
1
2
m ˙r
2
+
1
2
mr
2
˙
φ
2
V (r).
Then the Euler Lagrange equations give
m¨r mr
˙
φ
2
+ V
0
(r) = 0
d
dt
mr
2
˙
φ
= 0.
From the second equation, we see that
r
2
˙
φ
=
h
is a constant (angular momentum
per unit mass). Then
˙
φ = h/r
2
. So
m¨r
mh
2
r
3
+ V
0
(r) = 0.
If we let
V
eff
= V (r) +
mh
2
2r
2
be the effective potential, then we have
m¨r = V
0
eff
(r).
For example, in a gravitational field, V (r) =
GM
r
. Then
V
eff
= m
GM
r
+
h
2
2r
2
.
3.2 The Hamiltonian
In 1833, Hamilton took Lagrangian mechanics further and formulated Hamilto-
nian mechanics. The idea is to abandon
˙
x
and use the conjugate momentum
p
=
L
˙
x
instead. Of course, this involves taking the Legendre transform of the
Lagrangian to obtain the Hamiltonian.
Definition
(Hamiltonian)
.
The Hamiltonian of a system is the Legendre trans-
form of the Lagrangian:
H(x, p) = p ·
˙
x L(x,
˙
x),
where
˙
x is a function of p that is the solution to p =
L
˙
x
.
p
is the conjugate momentum of
x
. The space containing the variables
x, p
is known as the phase space.
Since the Legendre transform is its self-inverse, the Lagrangian is the Legendre
transform of the Hamiltonian with respect to p. So
L = p ·
˙
x H(x, p)
with
˙
x =
H
p
.
Hence we can write the action using the Hamiltonian as
S[x, p] =
Z
(p ·
˙
x H(x, p)) dt.
This is the phase-space form of the action. The Euler-Lagrange equations for
these are
˙
x =
H
p
,
˙
p =
H
x
Using the Hamiltonian, the Euler-Lagrange equations put
x
and
p
on a much
more equal footing, and the equations are more symmetric. There are also many
useful concepts arising from the Hamiltonian, which are explored much in-depth
in the II Classical Dynamics course.
So what does the Hamiltonian look like? Consider the case of a single particle.
The Lagrangian is given by
L =
1
2
m|
˙
x|
2
V (x, t).
Then the conjugate momentum is
p =
L
˙
x
= m
˙
x,
which happens to coincide with the usual definition of the momentum. How-
ever, the conjugate momentum is often something more interesting when we
use generalized coordinates. For example, in polar coordinates, the conjugate
momentum of the angle is the angular momentum.
Substituting this into the Hamiltonian, we obtain
H(x, p) = p ·
p
m
1
2
m
p
m
2
+ V (x, t)
=
1
2m
|p|
2
+ V.
So H is the total energy, but expressed in terms of x, p, not x,
˙
x.
3.3 Symmetries and Noether’s theorem
Given
F [x] =
Z
β
α
f(x, ˙x, t) dt,
suppose we change variables by the transformation
t 7→ t
(
t
) and
x 7→ x
(
t
).
Then we have a new independent variable and a new function. This gives
F [x] 7→ F
[x
] =
Z
β
α
f(x
, ˙x
, t
) dt
with α
= t
(α) and β
= t
(β).
There are some transformations that are particularly interesting:
Definition
(Symmetry)
.
If
F
[
x
] =
F
[
x
] for all
x
,
α
and
β
, then the transfor-
mation is a symmetry.
This transformation could be a translation of time, space, or a rotation, or
even more fancy stuff. The exact symmetries
F
has depends on the form of
f
.
For example, if
f
only depends on the magnitudes of
x
,
˙x
and
t
, then rotation
of space will be a symmetry.
Example.
(i)
Consider the transformation
t 7→ t
and
x 7→ x
+
ε
for some small
ε
. Then
F
[x
] =
Z
β
α
f(x + ε, ˙x, t) dx =
Z
β
α
f(x, ˙x, t) + ε
f
x
dx
by the chain rule. Hence this transformation is a symmetry if
f
x
= 0.
However, we also know that if
f
x
= 0, then we have the first integral
d
dt
f
˙x
= 0.
So
f
˙x
is a conserved quantity.
(ii)
Consider the transformation
t 7→ t ε
. For the sake of sanity, we will also
transform x 7→ x
such that x
(t
) = x(t). Then
F
[x
] =
Z
β
α
f(x, ˙x, t ε) dt =
Z
β
α
f(x, ˙x, t) ε
f
t
dt.
Hence this is a symmetry if
f
t
= 0.
We also know that if
f
t
= 0 is true, then we obtain the first integral
d
dt
f ˙x
f
˙x
= 0
So we have a conserved quantity f ˙x
f
˙x
.
We see that for each simple symmetry we have above, we can obtain a first
integral, which then gives a constant of motion. Noether’s theorem is a powerful
generalization of this.
Theorem
(Noether’s theorem)
.
For every continuous symmetry of
F
[
x
], the
solutions (i.e. the stationary points of
F
[
x
]) will have a corresponding conserved
quantity.
What does “continuous symmetry” mean? Intuitively, it is a symmetry we
can do “a bit of”. For example, rotation is a continuous symmetry, since we can
do a bit of rotation. However, reflection is not, since we cannot reflect by “a
bit”. We either reflect or we do not.
Note that continuity is essential. For example, if
f
is quadratic in
x
and
˙x
,
then
x 7→ x
will be a symmetry. But since it is not continuous, there won’t be
a conserved quantity.
Since the theorem requires a continuous symmetry, we can just consider
infinitesimally small symmetries and ignore second-order terms. Almost every
equation will have some O(ε
2
) that we will not write out.
We will consider symmetries that involve only the
x
variable. Up to first
order, we can write the symmetry as
t 7→ t, x(t) 7→ x(t) + εh(t),
for some
h
(
t
) representing the symmetry transformation (and
ε
a small number).
By saying that this transformation is a symmetry, it means that when we
pick
ε
to be any (small) constant number, the functional
F
[
x
] does not change,
i.e. δF = 0.
On the other hand, since
x
(
t
) is a stationary point of
F
[
x
], we know that if
ε
is non-constant but vanishes at the end-points, then
δF
= 0 as well. We will
combine these two information to find a conserved quantity of the system.
For the moment, we do not assume anything about
ε
and see what happens
to F [x]. Under the transformation, the change in F [x] is given by
δF =
Z
f(x + εh, ˙x + ε
˙
h + ˙εh, t) f(x, ˙x, t)
dt
=
Z
f
x
εh +
f
˙x
ε
˙
h +
f
˙x
˙εh
dt
=
Z
ε
f
x
h +
f
˙x
˙
h
dt +
Z
˙ε
f
˙x
h
dt.
First consider the case where
ε
is a constant. Then the second integral vanishes.
So we obtain
ε
Z
f
x
h +
f
˙x
˙
h
dt = 0
This requires that
f
x
h +
f
˙x
˙
h = 0
Hence we know that
δF =
Z
˙ε
f
˙x
h
dt.
Then consider a variable
ε
that is non-constant but vanishes at end-points. Then
we know that since x is a solution, we must have δF = 0. So we get
Z
˙ε
f
˙x
h
dt = 0.
We can integrate by parts to obtain
Z
ε
d
dt
f
˙x
h
dt = 0.
for any ε that vanishes at end-points. Hence we must have
d
dt
f
˙x
h
= 0.
So
f
˙x
h is a conserved quantity.
Obviously, not all symmetries just involve the
x
variable. For example, we
might have a time translation
t 7→ t ε
. However, we can encode this as a
transformation of the x variable only, as x(t) 7→ x(t ε).
In general, to find the conserved quantity associated with the symmetry
x
(
t
)
7→ x
(
t
) +
εh
(
t
), we find the change
δF
assuming that
ε
is a function of time
as opposed to a constant. Then the coefficient of ˙ε is the conserved quantity.
Example.
We can apply this to Hamiltonian mechanics. The motion of the
particle is the stationary point of
S[x, p] =
Z
(p ·
˙
x H(x, p)) dt,
where
H =
1
2m
|p|
2
+ V (x).
(i)
First consider the case where there is no potential. Since the action depends
only on
˙
x (or p) and not x itself, it is invariant under the translation
x 7→ x + ε, p 7→ p.
For general ε that can vary with time, we have
δS =
Z
h
p · (
˙
x +
˙
ε) H(p)
p ·
˙
x H(p)
i
dt
=
Z
p ·
˙
ε dt.
Hence p (the momentum) is a constant of motion.
(ii)
If the potential has no time-dependence, then the system is invariant under
time translation. We’ll skip the tedious computations and just state that
time translation invariance implies conservation of
H
itself, which is the
energy.
(iii)
The above two results can also be obtained directly from first integrals
of the Euler-Lagrange equation. However, we can do something cooler.
Suppose that we have a potential
V
(
|x|
) that only depends on radius. Then
this has a rotational symmetry.
Choose any favorite axis of rotational symmetry
ω
, and make the rotation
x 7→ x + εω × x
p 7→ p + εω × p,
Then our rotation does not affect the radius
|x|
and momentum
|p|
. So
the Hamiltonian
H
(
x, p
) is unaffected. Noting that
ω ×p ·
˙
x
= 0, we have
δS =
Z
p ·
d
dt
(x + εω × x) p ·
˙
x
dt
=
Z
p ·
d
dt
(εω × x)
dt
=
Z
p ·
ω ×
d
dt
(εx)

dt
=
Z
(p · [ω × ( ˙εx + ε
˙
x)]) dt
=
Z
( ˙εp ·(ω × x) + εp · (ω ×
˙
x)) dt
Since p is parallel to
˙
x, we are left with
=
Z
( ˙εp ·(ω × x)) dt
=
Z
˙εω · (x × p) dt.
So
ω ·
(
x ×p
) is a constant of motion. Since this is true for all
ω
,
L
=
x ×p
must be a constant of motion, and this is the angular momentum.
4 Multivariate calculus of variations
So far, the function
x
(
t
) we are varying is just a function of a single variable
t
.
What if we have a more complicated function to consider?
We will consider the most general case
y
(
x
1
, ··· , x
m
)
R
n
that maps
R
m
R
n
(we can also view this as
n
different functions that map
R
m
R
).
The functional will be a multiple integral of the form
F [y] =
Z
···
Z
f(y, y, x
1
, ··· , x
m
) dx
1
···dx
m
,
where y is the second-rank tensor defined as
y =
y
x
1
, ··· ,
y
x
m
.
In this case, instead of attempting to come up with some complicated generalized
Euler-Lagrange equation, it is often a better idea to directly consider variations
δy of y. This is best illustrated by example.
Example
(Minimal surfaces in
E
3
)
.
This is a natural generalization of geodesics.
A minimal surface is a surface of least area subject to some boundary conditions.
Suppose that (
x, y
) are good coordinates for a surface
S
, where (
x, y
) takes
values in the domain
D R
2
. Then the surface is defined by
z
=
h
(
x, y
), where
h is the height function.
When possible, we will denote partial differentiation by suffices, i.e.
h
x
=
h
x
.
Then the area is given by
A[h] =
Z
D
q
1 + h
2
x
+ h
2
y
dA.
Consider a variation of h(x, y): h 7→ h + δh(x, y). Then
A[h + δh] =
Z
D
q
1 + (h
x
+ (δh)
x
)
2
+ (h
y
+ (δh)
y
)
2
dA
= A[h] +
Z
D
h
x
(δh)
x
+ h
y
(δh)
y
q
1 + h
2
x
+ h
2
y
+ O(δh
2
)
dA
We integrate by parts to obtain
δA =
Z
D
δh
x
h
x
q
1 + h
2
x
+ h
2
y
+
y
h
y
q
1 + h
2
x
+ h
2
y
dA + O(δh
2
)
plus some boundary terms. So our minimal surface will satisfy
x
h
x
q
1 + h
2
x
+ h
2
y
+
y
h
y
q
1 + h
2
x
+ h
2
y
= 0
Simplifying, we have
(1 + h
2
y
)h
xx
+ (1 + h
2
x
)h
yy
2h
x
h
y
h
xy
= 0.
This is a non-linear 2nd-order PDE, the minimal-surface equation. While it is
difficult to come up with a fully general solution to this PDE, we can consider
some special cases.
There is an obvious solution
h(x, y) = Ax + By + C,
since the equation involves second-derivatives and this function is linear.
This represents a plane.
If |∇h|
2
1, then h
2
x
and h
2
y
are small. So we have
h
yy
+ h
yy
= 0,
or
2
h = 0.
So we end up with the Laplace equation. Hence harmonic functions are
(approximately) minimal-area.
We might want a cylindrically-symmetric solution, i.e.
h
(
x, y
) =
z
(
r
), where
r =
p
x
2
+ y
2
. Then we are left with an ordinary differential equation
rz
00
+ z
0
+ z
03
= 0.
The general solution is
z = A
1
cosh(Ar) + B,
a catenoid.
Alternatively, to obtain this,this we can substitute
h
(
x, y
) =
z
(
r
) into
A
[
h
]
to get
A[z] = 2π
Z
r
p
1 + (h
0
(r))
2
dr,
and we can apply the Euler-Lagrange equation.
Example
(Small amplitude oscillations of uniform string)
.
Suppose we have a
string with uniform constant mass density ρ with uniform tension T .
y
Suppose we pull the line between
x
= 0 and
x
=
a
with some tension
T
. Then we
set it into motion such that the amplitude is given by y(x; t). Then the kinetic
energy is
T =
1
2
Z
a
0
ρv
2
dx =
ρ
2
Z
a
0
˙y
2
dx.
The potential energy is the tension times the length. So
V = T
Z
d` = T
Z
a
0
p
1 + (y
0
)
2
dx = (T a) +
Z
a
0
1
2
T (y
02
) dx.
Note that y
0
is the derivative wrt x while ˙y is the derivative wrt time.
The
T a
term can be seen as the ground-state energy. It is the energy initially
stored if there is no oscillation. Since this constant term doesn’t affect where
the stationary points lie, we will ignore it. Then the action is given by
S[y] =
ZZ
a
0
1
2
ρ ˙y
2
1
2
T (y
0
)
2
dx dt
We apply Hamilton’s principle which says that we need
δS[y] = 0.
We have
δS[y] =
ZZ
a
0
ρ ˙y
t
δy T y
0
x
δy
dx dt.
Integrate by parts to obtain
δS[y] =
ZZ
a
0
δy(ρ¨y T y
00
) dx dt + boundary term.
Assuming that the boundary term vanishes, we will need
¨y v
2
y
00
= 0,
where
v
2
=
T
. This is the wave equation in two dimensions. Note that this is a
linear PDE, which is a simplification resulting from our assuming the oscillation
is small.
The general solution to the wave equation is
y(x, t) = f
+
(x vt) + f
(x + vt),
which is a superposition of a wave travelling rightwards and a wave travelling
leftwards.
Example
(Maxwell’s equations)
.
It is possible to obtain Maxwell’s equations
from an action principle, where we define a Lagrangian for the electromagnetic
field. Note that this is the Lagrangian for the field itself, and there is a separate
Lagrangian for particles moving in a field.
We have to first define several quantities. First we have the charges:
ρ
represents the electric charge density and
J
represents the electric current
density.
Then we have the potentials:
φ
is the electric scalar potential and
A
is the
magnetic vector potential.
Finally the fields:
E
=
−∇φ
˙
A
is the electric field, and
B
=
× A
is the
magnetic field.
We pick convenient units where
c
=
ε
0
=
µ
0
= 1. With these concepts in
mind, the Lagrangian is given by
S[A, φ] =
Z
1
2
(|E|
2
|B|
2
) + A · J φρ
dV dt
Varying A and φ by δA and δφ respectively, we have
δS =
Z
E ·
δφ +
t
δA
B · × δA + δA · J ρδφ
dV dt.
Integrate by parts to obtain
δS =
Z
δA · (
˙
E × B + J) + δφ( · E ρ)
dV dt.
Since the coefficients have to be 0, we must have
× B = J +
˙
E, · E = ρ.
Also, the definitions of E and B immediately give
· B = 0, × E =
˙
B.
These four equations are Maxwell’s equations.
5 The second variation
5.1 The second variation
So far, we have only looked at the “first derivatives” of functionals. We can
identify stationary points, but we don’t know if it is a maximum, minimum or a
saddle. To distinguish between these, we have to look at the “second derivatives”,
or the second variation.
Suppose x(t) = x
0
(t) is a solution of
δF [x]
δy(x)
= 0,
i.e. F [x] is stationary at y = y
0
.
To determine what type of stationary point it is, we need to expand
F
[
x
+
δx
]
to second order in
δx
. For convenience, let
δx
(
t
) =
εξ
(
t
) with constant
ε
1.
We will also only consider functionals of the form
F [x] =
Z
β
α
f(x, ˙x, t) dt
with fixed-end boundary conditions, i.e.
ξ
(
α
) =
ξ
(
β
) = 0. We will use both dots
( ˙x) and dashes (x
0
) to denote derivatives.
We consider a variation x 7→ x + δx and expand the integrand to obtain
f(x + εξ, ˙x + ε
˙
ξ, t) f(x, ˙x, t)
= ε
ξ
f
x
+
˙
ξ
f
˙x
+
ε
2
2
ξ
2
2
f
x
2
+ 2ξ
˙
ξ
2
f
x∂ ˙x
+
˙
ξ
2
2
f
˙x
2
+ O(ε
3
)
Noting that 2ξ
˙
ξ = (ξ
2
)
0
and integrating by parts, we obtain
= εξ
f
x
d
dt
f
˙x

+
ε
2
2
ξ
2
2
f
x
2
d
dt
2
f
x∂ ˙x

+
˙
ξ
2
f
˙x
2
.
plus some boundary terms which vanish. So
F [x + εξ] F [x] =
Z
β
α
εξ
f
x
d
dt
f
˙x

dt +
ε
2
2
δ
2
F [x, ξ] + O(ε
3
),
where
δ
2
F [x, ξ] =
Z
β
α
ξ
2
2
f
x
2
d
dt
2
f
x∂ ˙x

+
˙
ξ
2
2
f
˙x
2
dt
is a functional of both x(t) and ξ(t). This is analogous to the term
δx
T
H(x)δx
appearing in the expansion of a regular function
f
(
x
). In the case of normal
functions, if
H
(
x
) is positive,
f
(
x
) is convex for all
x
, and the stationary point
is hence a global minimum. A similar result holds for functionals.
In this case, if
δ
2
F
[
x, ξ
]
>
0 for all non-zero
ξ
and all allowed
x
, then a
solution x
0
(t) of
δF
δx
= 0 is an absolute minimum.
Example
(Geodesics in the plane)
.
We previously shown that a straight line is
a stationary point for the curve-length functional, but we didn’t show it is in
fact the shortest distance! Maybe it is a maximum, and we can get the shortest
distance by routing to the moon and back.
Recall that f =
p
1 + (y
0
)
2
. Then
f
y
= 0,
f
y
0
=
y
0
p
1 + (y
0
)
2
,
2
f
y
02
=
1
p
1 + (y
0
)
2
3
,
with the other second derivatives zero. So we have
δ
2
F [y, ξ] =
Z
β
α
˙
ξ
2
(1 + (y
0
)
2
)
3/2
dx > 0
So if we have a stationary function satisfying the boundary conditions, it is an
absolute minimum. Since the straight line is a stationary function, it is indeed
the minimum.
However, not all functions are convex
[citation needed]
. We can still ask whether
a solution
x
0
(
t
) of the Euler-Lagrange equation is a local minimum. For these,
we need to consider
δ
2
F [x
0
, ξ] =
Z
β
α
(ρ(t)
˙
ξ
2
+ σ(t)ξ
2
) dt,
where
ρ(t) =
2
f
˙x
2
x=x
0
, σ(t) =
2
f
x
2
d
dt
2
f
x∂ ˙x

x=x
0
.
This is of the same form as the Sturm-Liouville problem. For
x
0
to minimize
F [x] locally, we need δ
2
F [x
0
, ξ] > 0. A necessary condition for this is
ρ(t) 0,
which is the Legendre condition.
The intuition behind this necessary condition is as follows: suppose that
ρ
(
t
) is negative in some interval
I
[
α, β
]. Then we can find a
ξ
(
t
) that makes
δ
2
F
[
x
0
, ξ
] negative. We simply have to make
ξ
zero outside
I
, and small but
crazily oscillating inside
I
. Then inside
I
,
˙x
2
wiill be very large while
ξ
2
is kept
tiny. So we can make δ
2
F [y, ξ] arbitrarily negative.
Turning the intuition into a formal proof is not difficult but is tedious and
will be omitted.
However, this is not a sufficient condition. Even if we had a strict inequality
ρ(t) > 0 for all α < t < β, it is still not sufficient.
Of course, a sufficient (but not necessary) condition is
ρ
(
t
)
>
0
, σ
(
t
)
0, but
this is not too interesting.
Example. In the Branchistochrone problem, we have
T [x]
Z
β
α
r
1 + ˙x
2
x
dt.
Then
ρ(t) =
2
f
˙x
2
x
0
> 0
σ(t) =
1
2x
2
p
x(1 + ˙x
2
)
> 0.
So the cycloid does minimize the time T .
5.2 Jacobi condition for local minima of F [x]
Legendre tried to prove that
ρ >
0 is a sufficient condition for
δ
2
F >
0. This is
known as the strong Legendre condition. However, he obviously failed, since it is
indeed not a sufficient condition. Yet, it turns out that he was close.
Before we get to the actual sufficient condition, we first try to understand
why thinking ρ > 0 is sufficient isn’t as crazy as it first sounds.
If
ρ >
0 and
σ <
0, we would want to create a negative
δ
2
F
[
x
0
, ξ
] by choosing
ξ
to be large but slowly varying. Then we will have a very negative
σ
(
t
)
ξ
2
while
a small positive ρ(t)
˙
ξ
2
.
The problem is that
ξ
has to be 0 at the end points
α
and
β
. For
ξ
to take a
large value, it must reach the value from 0, and this requires some variation of
ξ
,
thereby inducing some
˙
ξ
. This is not a problem if
α
and
β
are far apart - we
simply slowly climb up to a large value of
ξ
and then slowly rappel back down,
maintaining a low
˙
ξ
throughout the process. However, it is not unreasonable to
assume that as we make the distance
β α
smaller and smaller, eventually all
ξ
will lead to a positive
δ
2
F
[
x
0
, ξ
], since we cannot reach large values of
ξ
without
having large
˙
ξ.
It turns out that the intuition is correct. As long as
α
and
β
are sufficiently
close,
δ
2
F
[
x
0
, ξ
] will be positive. The derivation of this result is, however, rather
roundabout, involving a number of algebraic tricks.
For a solution x
0
to the Euler Lagrange equation, we have
δ
2
F [x
0
, ξ] =
Z
β
α
ρ(t)
˙
ξ
2
+ σ(t)ξ
2
dt,
where
ρ(t) =
2
f
˙x
2
x=x
0
, σ(t) =
2
f
x
2
d
dt
2
f
x∂ ˙x

x=x
0
.
Assume
ρ
(
t
)
>
0 for
α < t < β
(the strong Legendre condition) and assume
boundary conditions ξ(α) = ξ(β) = 0. When is this sufficient for δ
2
F > 0?
First of all, notice that for any smooth function w(x), we have
0 =
Z
β
α
(wξ
2
)
0
dt
since this is a total derivative and evaluates to
wξ
(
α
)
wξ
(
β
) = 0. So we have
0 =
Z
β
α
(2wξ
˙
ξ + ˙
2
) dt.
This allows us to rewrite δ
2
F as
δ
2
F =
Z
β
α
ρ
˙
ξ
2
+ 2wξ
˙
ξ + (σ + ˙w)ξ
2
dt.
Now complete the square in ξ and
˙
ξ. So
δ
2
F =
Z
β
α
"
ρ
˙
ξ +
w
ρ
ξ
2
+
σ + ˙w
w
2
ρ
ξ
2
#
dt
This is non-negative if
w
2
= ρ(σ + ˙w). ()
So as long as we can find a solution to this equation, we know that
δ
2
F
is
non-negative. Could it be that
δ
2
F
= 0? Turns out not. If it were, then
˙
ξ =
w
ρ
ξ. We can solve this to obtain
ξ(x) = C exp
Z
x
α
w(s)
ρ(s)
ds
.
We know that ξ(α) = 0. But ξ(α) = Ce
0
. So C = 0. Hence equality holds only
for ξ = 0.
So all we need to do is to find a solution to (
), and we are sure that
δ
2
F >
0.
Note that this is non-linear in
w
. We can convert this into a linear equation
by defining w in terms of a new function u by w = ρ ˙u/u. Then () becomes
ρ
˙u
u
2
= σ
ρ ˙u
u
0
= σ
(ρ ˙u)
0
u
+ ρ
˙u
u
2
.
We see that the left and right terms cancel. So we have
(ρ ˙u)
0
+ σu = 0.
This is the Jacobi accessory equation, a second-order linear ODE.
There is a caveat here. Not every solution
u
will do. Recall that
u
is used
to produce
w
via
w
=
ρ ˙u/u
. Hence within [
α, β
], we cannot have
u
= 0 since
we cannot divide by zero. If we can find a non-zero
u
(
x
) satisfying the Jacobi
accessory equation, then
δ
2
F >
0 for
ξ 6
= 0, and hence
y
0
is a local minimum of
F .
A suitable solution will always exists for sufficiently small
β α
, but may
not exist if β α is too large, as stated at the beginning of the chapter.
Example
(Geodesics on unit sphere)
.
For any curve
C
on the sphere, we have
L =
Z
C
q
dθ
2
+ sin
2
θ dφ
2
.
If θ is a good parameter of the curve, then
L[φ] =
Z
θ
2
θ
1
q
1 + sin
2
θ(φ
0
)
2
dθ.
Alternatively, if φ is a good parameter, we have
L[θ] =
Z
φ
2
φ
1
q
(θ
0
)
2
+ sin
2
θ dφ.
We will look at the second case.
We have
f(θ, θ
0
) =
q
(θ
0
)
2
+ sin
2
θ.
So
f
θ
=
sin θ cos θ
q
(θ
0
)
2
+ sin
2
θ
,
f
θ
0
=
θ
0
q
(θ
0
)
2
+ sin
2
θ
.
Since
f
φ
= 0, we have the first integral
const = f θ
0
f
θ
0
=
sin
2
θ
q
(θ
0
)
2
+ sin
2
θ
So a solution is
c sin
2
θ =
q
(θ
0
)
2
+ sin
2
θ.
Here we need c 1 for the equation to make sense.
We will consider the case where
c
= 1 (in fact, we can show that we can
always orient our axes such that
c
= 1). This occurs when
θ
0
= 0, i.e.
θ
is a
constant. Then our first integral gives
sin
2
θ
=
sin θ
. So
sin θ
= 1 and
θ
=
π/
2.
This corresponds to a curve on the equator. (we ignore the case
sin θ
= 0 that
gives θ = 0, which is a rather silly solution)
There are two equatorial solutions to the Euler-Lagrange equations. Which,
if any, minimizes L[θ]?
We have
2
f
(θ
0
)
2
θ=π/2
= 1
and
2
f
θ∂θ
0
= 1,
2
θ∂θ
0
= 0.
So ρ(x) = 1 and σ(x) = 1. So
δ
2
F =
Z
φ
2
φ
1
((ξ
0
)
2
ξ
2
) dφ.
The Jacobi accessory equation is
u
00
+
u
= 0. So the general solution is
u
sin φ γ cos φ. This is equal to zero if tan φ = γ.
Looking at the graph of
tan φ
, we see that
tan
has a zero every
π
radians.
Hence if the domain
φ
2
φ
1
is greater than
π
(i.e. we go the long way from
the first point to the second), it will always contain some values for which
tan φ
is zero. So we cannot conclude that the longer path is a local minimum (it
is obviously not a global minimum, by definition of longer) (we also cannot
conclude that it is not a local minimum, since we tested with a sufficient and
not necessary condition). On the other hand, if
φ
2
φ
1
is less than
π
, then we
will be able to pick a γ such that u is non-zero in the domain.