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PX242 Hamiltonian Mechanics
Compact Lecture Notes
2000
These are working notes, and not intended to be a full and complete account.
Lecture 1.
Traditional and very logical approach to this subject:
Newton 1660s
Lagrange 1780s
Hamilton 1830s
really focussing around the the crowning achievement of the nineteenth Century.
New Millenium edition: start from the approach of
Feynman 1950s
which is a rather general approach to Quantum Mechanics. This is the real world, so
that is where we should get classical mechanics from!
There is a (fair) price to pay, and that is that we will have to do the most difficult bit
of physics in lecture 1, and the most difficult maths in lecture 2. The latter is called
the 'Calculus of Variations': we will take it gently, and you will know you are over
the worst!
Themes from Physics in Year 1:
Mechanics (Relativistic, Newtonian)
Waves and interference
Quantum Phenomena - matter <-> waves
How do these all fit together?
Out of this question I hope you will obtain:
a greater overview of 'classical' mechanics
the general significance of 'momentum'
where conservation laws come from
considerable economies in calculation
ideas about reversibility vs chaos
Clues to the answer come from various 'Variational Priniciples' in (conservative)
dynamics:
Optics: Rays trace paths of minimal (sometimes maximal)
Optical Path
 n(r)ds
With a bit of work you can check that this corresponds to Snell's law of refraction
n1
h2
n1 sin 1  n2 sin 2
when you minimise the path
subject to fixed
h1
1
2
n2
h1n1 cos(1 )  h2 n2 cos(2 )
h1n1 tan(1 )  h2 n2 tan(2 )
Waves:
High intensity corresponds to constructive interference, that is different paths
contributing amplitude with the same phase
Relativity:
Trajectories in free space (and -GR- in a gravitational field) correspond to maximum
elapsed proper time:
x
   d t    dt 1  u 2 c 2
where it is at least obvious that straight paths, giving lower u give higher 
Newtonian Mechanics:
y
For a simple particle trajectories extremise an 'Action'
A   T (v)  V ( x) dt
/
kinetic energy
\
potential energy
where L  T  V (note the minus sign!!) is the simplest example of a Lagrangian.
This result will be central to this course - I am not supposing you would understand it
already.
In all of these cases:
there is something you can integrate along a path
and this quantity is extremised by the physically important paths
and it all comes from Quantum Mechanics.
Feynman's Path Integral:
A rather general formulation of Quantum Mechanics, arguably the 'Theory of
Everything' (though he never called it that..). I will describe it terms of a simple
particle characterised by its position x .
The starting point is:
The state of our system at time t is described by a wavefunction  ( x, t ) . This
being Quantum Mechanics, there is some amplitude (=value of ) for every position
x.
The Physics is how amplitude from position x 0 at time t 0 leads to amplitude at
position x1 later time t1 . Feyman proposed:
 ( x1 , t1 ) 
 i t1 
dx  

exp
L x(t ), dt  ( x0 , t 0 )


t 
dt  
all possiblepaths
 0
from x to x
0
1
This means that for every conceivable trajectory from x 0 at time t 0 to x1 at later time
t1 , we have to compute the exponential and then 'sum' this result over all possible
paths. The mathematics of making sense of that summation as a functional integral
is outside the scope of this course - but if you want to follow it up those are the key
words.
If you think of the integral in the exponent as (the limit of) a sum, then the
exponential can in turn be interpreted as a product of separate factors
i 
dx  
exp  L x(t ), dt  compounding through time along the path. As we would wish
dt  
 
for 'Causality', these factors do not mix different times.
dx 

The nature of our system is of course hidden inside the 'Lagrangian' L x (t ),  , but
dt 

I have already let out that for a simple classical particle
2
dx 

 dx 
L x(t ),   T  V  12 m   V ( x) .
dt 

 dt 
In principle you can take x to be 'the configuration of the Universe', and then all the
laws of Nature are contained in the form of L.
Feynman showed that  ( x, t ) calculated his way does obey Schrödinger's Equation of
Quantum Mechanics.
Bizarre though the above may sound (and there remains of course the usual difficulty
of where outside the possible states of the Universe the values of  reside!), this is
our current most powerful picture of how microscopic physics works. The Big Bang,
Quantum Gravity, String Theory .... have all been looked at this (but not only this)
way.
Classical Physics follows as just constructive interference. If a lot of nearby
t1
dx  , then through constructive
trajectories all give the same phase 1 
L x(t ), dt

 t0 
dt 
interference this will tend to dominate how amplitude propagates. This happens most
where the phase is stationary with respect to variations in the path, and those are the
trajectories we are used to seeing:
 t1 
dx  
   L x(t ), dt   0  Newton' s Laws
t 
dt  
0
as we shall confirm later.
Lecture 2.
Examples of Lagrangian:
2
Newtonian Mechanics L  12 mv  V (x)
Special Relativity
L 
Electromagnetic Field
L
m0 c 2

2
2
2
  m0 c 1  v c
 2
1 2 
dV  0 E 
B 
 2
2  0 
Simple check of stationary phase <-> Newtonian Mechanics (skipped 99)
2
Particle in gravitational field. L  12 mv  mgx
t 0  0 , x0  0 ;
Trial trajectory
t1  T , x1  X
x(t) = t/T X + b t(T-t)
x =height
/
gets end points right
\
adds curvature
v  dx dt  X / T  b(T  2t)
Both of these can be substituted into L and we integrate with respect to time to
obtain the phase:
m
2 3
3
Ldt  13 b T  mg16 bT 
2
Minimising this with respect to b gives b=g/2 leading to the expected result
x(t )  v0t  12 gt 2
where v0  (X / T  gT / 2)

General d=1 Case: Calculus of Variations
We consider the variation in value of
A   Ldt
when we vary the trajectory from x(t)  x(t)  a(t)
where we will consider the variation a(t) to be small and examine its consequences
only to first order in expansion of A.
The velocity will be correspondingly changed v(t )  v(t )  a (t )
and hence the Lagrangian changes
L
L
L  L( x, v)  a(t )
 a (t )
x
v
giving us the change in the phase as


 A   dt a(t)
L da L 

 x dt  v 
The second term we integrate by parts:
da L   L 
d  L 
dt
 a
 dt a
dt  v  v 
dt v 
where given that the value of x(t) was constrained at the end points of the trajectory,
the variation a(t) must be zero at the endpoints and so the integrated part vanishes.
This then leaves us with
 L d L 
 A  dt a(t)

  x dt  v 
Therefore the condition for A to be stationary with respect to small variations a(t) in
the trajectory reduces to
 L d  L

0
"Euler-Lagrange Equation"
x dt v 



First let us recap what we have done. We asked, purely as a mathematical exercise,
for A to be stationary and got the equation above.
Now let us put in some physics such as
L  T  V  12 mv2  V(x)
from which:
L
V

F
x
x
L
 mv  p
v
and
where F is the force and p the momentum in standard Newtonian Mechanics
interpretation. Putting these expressions back into the E-L equation then gives us
F  dp / dt  0 ,
or Newton's familiar second law of motion.
Thus we have shown that:
stationary A using L  T  V
==>
F  ma
Simple Applications
You did not of course need a new way to deduce F  ma , but actual examples
exhibit the power of the method. It is often relatively easy to write down T, V and
hence L, and the equations of motion are the Euler-Lagrange equations. It is crucial
that the statement  A  0 does not presuppose any particular coordinate system: we
can use any convenient set of coordinates (so long as we can express L in terms of
them).
Simple Pendulum - use angle of swinging about the pivot.
2
V  mg cos
T  12 I 
where I = moment of inertia about the pivot
and
 distance from pivot to centre of mass of pendulum.

Now we make A   Ldt stationary with respect to variations in  (t) , giving us
L
d L d 
    I  mg sin  
dt    dt

 
Roller Coaster - use arc length s as coordinate to obtain
1
L  ms 2  mg hs 
2
leading to the equation of motion
d L
dh  L

 ms  mg 
dt   s 
ds  s
Lecture 3.
Time Dependence (and lack of it)
So far our discussion has been quite general in respect of time dependence. For
example if you consider a rotating object with time dependent moment of inertia,
2
L  T  V  12 I (t )  ,
you will get from the Euler-Lagrange equation the expected equation of motion
d
I (t )  0 , corresponding to conservation of angular momentum I (t ) .
dt



A special result obtains when (unlike the above) L does not depend explicitly on
time, but only varies with time through its dependence on the coordinates and their
velocities. This means L(x(t),v(t), not t).
The result in this case is that the 'Hamiltonian'
L
Hv
 L is a constant of the motion.
v
This means that H does not change with time when the Euler-Lagrange equations are
obeyed.
The proof is straightforward:
dH d   L L dv L dx
d L   L

v


 v

v0
dt dt  v   v dt  x dt
dt  v  x
where the last step uses the Euler-Lagrange equation.
The physical interpretation is quite familiar as conservation of energy. Given
L  T  V  12 mv2  V(x) , it is easy to evaluate that H  12 mv2  V(x)  T  V .
A note on notation: why bother with the notation H when we could just have used E
(for energy)? It is generally helpful to reserve H for the expression in terms of
2
coordinates, e.g. H  12 mv  V(x) , and E for the constant value which this keeps.
Then we can meaningfully write H  E as the statement of conservation of energy.
For completeness, if L does depend explicitly upon time you can easily verify that
dH
L

dt
 t x ,v
Application of H  E
Most of the practical importance of the result H  E (when it applies) is that it is an
dH
integral of the motion (from
 0 ), containing only first order time derivatives of
dt
the coordinates whereas the Euler-Lagrange equations are second order.
dh
Thus for the Roller-Coaster example, instead of ms  mg
we have
ds
1
2
2 ms  mgh( s )  E : one time derivative has been 'integrated out'.
To 'solve' the motion in these one dimensional cases we have to integrate once more,
leading in the Roller-Coaster example (by the method of separation of variables) to
ds m 2

 dt .
E  mgh( s) 
Notice there is no trick to get this last integral in closed form, unless we exploit
special cases of the 'potential' mgh(s) . Numerically it is in general fairly
straightforward to compute, with care at the 'turning points' E  mgh(s)  0 .
Example of Special Relativity
We have already noted that the Lagrangian here is (with a potential added now)
L  m0c 2 /  V (x)  m0c 2 1  v 2 / c 2  V (x)
which will serve to remind us of the formal structure of things because it cannot be
interpreted as 'T-V'.
The corresponding Hamiltonian, which will be a constant of the motion, is then
L
2
2
2
2
Hv
 L  m0 c
1 v / c  V(x)   m0 c  V(x)
v
where you should certainly check the intermediate working for yourself, but the
answer is reassuringly familiar.
Multiple Coordinates
Hitherto we have worked in terms of just one coordinate for simplicity, be it
cartesian position x(t), an angle (t) or an arc length s(t). The extension to more
coordinates is straightforward. We begin by discussing examples with just two (or
three) coordinates rather informally, as a prelude to addressing the general case more
formally.
For a system with two coordinates, say x(t) and y(t ) , we have correspondingly two
velocities x (t ) and y (t ) and in general the phase will be
t1
A   L( x, y, x, y )dt .
t0
The trajectory should now be thought of as the pair x(t),y(t) as a function of time
and constraining the end points means fixing values of x0 , y0  and x1 ,y1  . Making A
stationary with respect to an arbitrary variation in x(t),y(t) is the same as making it
stationary with respect to both variation in x(t) and variation in y(t ) , as the general
case is just a linear combination of these. Thus
L d L
L d L
0
A  0     0 and
 
 x dt   x 
 y dt   y 
where we have to be clear that, for example,
L
L
means
x
x
.
x , y , y
Two coordinates can correspond to one particle moving in two dimensions, for
example
L  T  V  12 mx 2  y 2   V ( x, y )
 V ( x, y )
d
d
my     V ( x, y) .
leading as expected to dt mx     x
and
dt
y
However it could also correspond to two particles each moving in one dimension,
where x is the coordinate of particle 1 and y is the coordiante of particle 2.
Then
L  T  V  12 m1 x 2  12 m2 y 2  V ( x, y )
 V ( x, y )
d
leading to the equations of motion dt m1 x     x
and
d
m2 y     V ( x, y) . Notice that in the case of a simple interaction between the
dt
y
 V(x,y)
 V(x,y)
two particles, V(x, y)  U(x  y) , we have
and
 U (x  y)  
y
x
the particles duly experience equal and opposite forces in accordance with Newton's
Third Law.
Examples.
Consider the pair of three spoke magnetic gears shown in the lecture. In terms of
angles of rotation the kinetic energy is obviously T  12 I 12  2 2 . A plausible


model (no more than that) of the potential energy is V  C1  cos31 1 cos3 2 ,
designed to give a peak whenever both wheels are correctly aligned, modulo three for
three spokes.
You can find the Euler-Lagrange equatiosn without difficulty, e.g.
d 
I 1  3C sin 3 1 1  cos 3 2 
dt
and since L=T-V does not depend explicitly on time we have one integral of the
motion,
2
2
E  12 I 1  2 C1  cos 31 1  cos 3 2 .
 


Notice how conservation of energy is no longer enough to eliminate all second order
time derivatives: for example you can use it to give 2 but you are still left with the
second order Euler-Lagrange equation for 1 . We will discuss this and related
scenarios in more detail in terms of Hamilton's equations later in the course; the
message in the meantime is that:
one coordinate (+one velocity) - one integral of motion (Energy) -> simple bahaviour
two coordinates (+two velocities) - one integral of motion -> no easy formulae and
rich dynamics.
The flexible Pendulum..... question 9 on sheet.